a \ centerlineRow 1 6 Row{a 2} i-dieresis-u . sub e y-d to the power of m-e E-a n s-O-dieresis-t sub n a-T e-l h-a n-n i-d T-a sub o-l U-s u i-n-n T-e sub r-e s-r z-i t-i sub o-y g-breve lu \ [MDepartment−S\ begin { ofarray Mathematics}{ c} 6 \\ i−d i e r e s i s −u \end{ array } l−d { e y−d ˆ{ m−e E−a } n } s−O−d i e r e s i s −t { n a−T } e−l h−a n−n i−d T−a { o−l Ua−s u i−n−n } T−e { r−e s−r z−i } t−i { o−y \breve{g}} lu \ ] 6 M−S l−dey−dm−eE−ans − O − dieresis − tna−Te − lh − an − ni − dT − ao−lU−sui−n−nT − er−es−rz−it − io−y˘glu \noindent iDepartment− dieresis − u of Mathematics Department of Mathematics CONTENTS \ centerlineIntroduction{CONTENTS .... period .... period ....} period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period .... period\noindent .... periodIntroduction .... period ....\ periodh f i l l ..... period\ h f i ....l l period. \ h f .... i l l period. \ h .... f i lperiod l . \ ....h f period i l l . ....\ h 5 f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l . \ h f i l l 5 1 period .. Almost bounded sets and operators ..CONTENTS period .. period .. period .. period .. period .. period .. period .. period .. period .. period ..\noindent periodIntroduction .. period1 . ..\ .quad period . .Almost .. . period . . . bounded .. . period . . .. . sets period . . and ... period . operators . . .. . period . .\ ..quad . period ...... \quad period ...... \ periodquad ...... period .\quad . . .. 56. \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 6 22 . period1\quad . .. Almost EidelheitEidelheit bounded quoteright ’ sets s and theorem s theorem operators\ ..quad period .. . ..\quad period .. ...\ periodquad . .. . period\quad . .. .. period\quad . .. . period. \ .quad .. period.. \ .quad .. period . . \ ..quad period. ..\quad period. .. \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 1 3 period. .. period . . .. period . . .. period . . .. period . 6 .. 2period . Eidelheit .. period ’ s.. theorem period .. period . . .. period . . .. period . . .. period . . .. period . .. period .. period .. period\noindent...... 13 .. period3 .. . period\quad ..Nuclear period .. 1 K 3 $ \ddot{o} $ the quotients \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 20 43 . period3\quad . .. Nuclear NuclearNuclear K Ko¨ o-dieresisKthe quotients $ \ddot the quotients{o .} $ . the.. period. subspaces . .. period . . ..and period . completing . .. period . . .. sequences period . . .. period .\quad ... period .\quad .. . period. \ ..quad period. ..\ periodquad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 22 .. period. .. period . . .. period . . .. period . .. . period . .. . period . .. period . 20 .. period 4 . .. Nuclear period K ..o¨ periodthe subspaces .. period and .. period completing .. period .. period .. period .. period\noindentsequences .. period5 .. . 20\ .quad .Applications . . . .\quad . .. \quad . .. \ .quad .. .\quad .. .\quad 22 . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 25 64 . period5\quad . .. Applications NuclearSpaces K of o-dieresis . continuous . the . subspaces . functions . and . completing\ .quad . sequences. .\quad . ... period\quad . .. . period. \quad . .. . period. \ .quad .. .period. .\quad .. period. . ..\quad period. ..\ periodquad ... \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 30 periodReferences. .. period . ...\quad period .. ...\ periodquad . .. .. period\quad . .. .. period\ .quad .. . period. .\quad .. . period. 25\quad ..6 . 22 Spaces. \quad of continuous. \quad functions. \quad .. \ .quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad 35 5 period...... 30 .. Applications .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period\ hspaceReferences .. period∗{\ f i l.. l period} .1 9 . 9 .. period1 . Mathematics . .. period . . .. periodSubject . . .. period . Classification . .. period . . .. period . : Primary. .. period . . 46 .. period .A 3 . ,.. 46period . A . 4 .. period, . 46 A .. 45period ; Secondary .. period .. 46 A 1 1 , period. .. period . ... period . ... period . .. . period . .. . 25 . . . . . 35 \noindent6 period ..46A13 Spaces1 of9 9 continuous 1 ,46E10Mathematics functions Subject . .. Classification period .. period: Primary .. period 46 A .. 3 period , 46 A 4.. , period 46 A 45 .. ;period Secondary .. period 46 A 1 .. 1 ,period .. period .. period .. period46 .. periodA 1 3 , ..46 period E 1 0 . .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. 30 \ centerlineReferences ..{ Received period .. period 1 0Received .. . period6 . 1 1 0 .. . 99 6period . 1 1 99 ; .. 1 revised ; period revised .. version versionperiod 1 ..2 . period 1 1 2 . 1 . .. 99 1 period 1 1 . . .. 1 period 99 1 .. . period} .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. period .. 35 1 9 9 1 Mathematics Subject Classification : Primary 46 A 3 comma 46 A 4 comma 46 A 45 semicolon Secondary 46 A 1 1 comma 46 A 1 3 comma 46 E 1 0 period Received 1 0 period 6 period 1 99 1 semicolon revised version 1 2 period 1 1 period 1 99 1 period Introduction \ centerlineIn this article{ Introduction our purpose will} be t o examine the conditions under which a locally convex space has a subspace or a quotient space which i s isomorphic to Ina this Fr acute-e article chet space our period purpose .. The will concrete be t Fr o e-acute examine chet spacesthe conditions we have in mind under are the which nuclear a locallyK o-dieresis convex the spaces space or the has space a subspace omega simeq orIntroduction K a to quotient the power of space N period which .. By i anuclear s isomorphic K o-dieresis to the space we mean a nuclear aFr Fr acute-e $In\ thisacute chet article space{e} ourwhich$ purpose chet has a space basis will be and . t ao\quadcontinuous examineThe the norm concrete conditions period Fr under $ which\acute a locally{e} $ convex chet space spaces has a we have in mind are the nuclear In opensubspace square or bracket a quotient 8 closing space squarewhich i bracket s isomorphic Eidelheit to a has Fr provede ´ chet spacethat every . The proper concrete Fr acute-e Fre ´ chet space spaces has we omega as a quotient \noindentspacehave period inK mind .. $ For are\ddot another the{ nuclearo} proof$ theof this spaces and results or about the space kernels of $ surjections\omega onto\simeq omega commaK ˆ{ N } . $ \quad By a nuclear K N $ \weddotK refero ¨{othe t} o$ spaces Vogt the .. or openspace the space square weω bracket mean' K . a 19 nuclearBy closing a nuclear square K bracketo ¨ the space period we .. mean The asearch nuclear for Fr nucleare ´ chet K space o-dieresis which the quotients of Fr acute-e chetFr spaceshas $ \ aacute basis{ ande} a$ continuous chet space norm . which has a basis and a continuous norm . was ..In initiated [ 8 ] Eidelheit by .. Bellenot has proved .. and that .. Dubinsky every proper .. open Fre ´ squarechet space bracket has 1ω closingas a quotient square space bracket . period For another .. The present .. authors .. in .. openIn [ squareproof 8 ] ofEidelheitbracket this and 1 3 results closing has about proved square kernels bracket that of surjections.. every open parenthesis proper onto ω, Frwe cf referperiod $ \ tacute o Vogt{e} [ 19$ ] . chet The space search for has nuclear $ \omega $ as a quotient spacealsoK openo ¨ . the\ squarequad quotients bracketFor of another Fr 1 2e ´ closingchet proofspaces square was of bracket this initiated closing and by parenthesisresults Bellenot about showed and kernels that Dubinsky the assumption of [surjections 1 ] . of Theseparability present onto in the $ \ theoremomega of Bellenot, $ weand referauthors Dubinsky t o in i Vogts redundant [ 1 3\quad ] ( period cf[ . 19 .. ]More . \ preciselyquad The comma search they have for proved nuclear that K either $ \ everyddot{o} $ the quotients of Fr $ \acute{e} $ chetcontinuousalso spaces [ 1 2 operator ] ) showed from that a given the assumption Fr acute-e of chet separability space E into in the any theorem nuclear of Fr Bellenot e-acute andchet Dubinsky space comma i s redundant waswhich.\quad admits Moreinitiated precisely a continuous , they by norm\ havequad comma provedB e il l s that e boundedn o t either\quad or every Eand has continuous a\ nuclearquad Dubinsky operator K dieresis-o from\ thequad aquotient given[ Fr1 periode ]´ chet . \quad space EThe present \quad authors \quad in \quad [ 1 3 ] \quad ( c f . Sinceinto we any will nuclear encounter Fre ´ similarchet space dichotomies , which also admits in this a continuous work comma norm let , us i s make bounded a short or E has a nuclear Ko ¨ the \noindentdetourquotient periodalso . Since [ we1 2 will ] encounter ) showed similar that dichotomies the assumption also in this of work separability , let us make a in short the detour theorem . of Bellenot andIf every DubinskyIf continuous every continuous i s operator redundant operator T : E right.T :\quad arrowE →MoreF Fis bounded precisely we will will , write write they as as have usual usual L proved(E,F ) that = LB either(E,F ). everyIf continuousL openthis holds parenthesis operator and E neither comma fromE Fnor a closingF givenis parenthesis a Fr normed $ \ space =acute LB , thenopen{e} parenthesis$ of chetcourse space EE commacannot $ F E closing b$ e i into somorphic parenthesis any to nuclear period a .. Fr If this $ \ holdsacute ..{ ande} $ neitherchetsubspace space E nor F , ofis ..F a normedand F spacecannot comma b e i somorphic to a quotient space of E. Pairs of Fre ´ chet spaces E and whichthenF ..for admits of which courseL a ..(E,F continuous E cannot) = LB ..(E,F b enorm i) somorphic have , been i s to completely bounded .. a subspace characterized or of $ F E .. $ and by has F Vogt .. cannota [ nuclear 1 8 ] .. . b e We K shall $ \ddot{o} $ the quotient . Sincei somorphicalso we refer will to t o a the quotient encounter articles space by similar Bonet of E period [ 3 ] dichotomies and .. Pairs Bonet of – Fr Galbis acute-e also [ 4 in chet ] in this spacesthe general work E and case , F let for . which us make a short detourL openIn parenthesis . the case of E comma subspaces F closing , Bessaga parenthesis and Pe = LBlczy openn´ ski parenthesis [ 2 ] proved E comma that F a closing Fre ´ chet parenthesis space i have s either been completely characterized by Vogtisomorphic open square to ω bracket× Banach 1 8 closingor it has square a nuclear bracket Ko ¨periodthe subspace .. We shall . The existence of common nuclear Ko ¨ the Ifalso everysubspaces refer continuous t o was the examined articles operator by in Bonet [ 1 5 ] open and $ T [ square 1 6 ]: . bracket ETo deal\ 3rightarrow with closing this square question bracketF in $ the and is case bounded Bonet of general endash we locally Galbiswill convex write open square as usual bracket 4 closing square$L(E,F)=LB(E,F)spaces bracket we in introduce the general the case concept period of almost boundedness in Section 1 , .$ and examine\quad theIf properties this holds of almost\quad and neither $ E $ norIn thebounded $ F case $ of set is subspaces s\ andquad operators commaa normed . .. Bessaga Although space and , by Pe means l-suppress of these sub results czy to we the can power derive of some acute-n interesting ski .. open theorems square bracket 2 closing square bracketthenfor ..\quad example provedo that f about course a Fr the acute-e structure\quad chet of$ spaces E $ which cannot have\ωquadas a subspaceb e i ,somorphic most of them to will\quad b e generalizeda subspace further of $ F $ \quad and $ Fspace $in thei\ squad either subsequentcannot isomorphic sections\quad to omega. b e times Banach or it has a nuclear K dieresis-o the subspace period .. The iexistence somorphicSection of common to2 is a on quotient nuclear Eidelheit K ’ o-dieresis s space theorem of the . subspaces $ Our E main . wastool $ examined\ therequad isPairs the in openconcept o squaref Fr of completing bracket $ \acute 1 sequences 5{ closinge} $ square of chet De bracket spaces and $E$ open square and bracket$ F $Wilde 1 for6 closing [ 5 which ] . square The bracketexistence period of an .. unbounded To completing $Ldeal with ( this E question , in F the ) case = of general LB locally ( Econvex , spaces F we )$ introduce havebeencompletelycharacterizedbyVogt[18] . \quad We s h a l l the concept of almost boundedness in Section 1 comma and examine the properties of \noindentalmost boundedalso set refer s and t operators o the period articles .. Although by Bonet by means [ 3 of ] these and results Bonet we−− canGalbis [ 4 ] in the general case . derive some interesting theorems for example about the structure of spaces which Inhave the omega case as of a subspace subspaces comma , \ mostquad ofBessaga them will band e generalized Pe $ \ l further{ czy in the ˆ{\ subsequentacute{n}}}$ s k i \quad [ 2 ] \quad proved that a Fr $ \sectionsacute{ periode} $ chet spaceSection i 2 s is eitheron Eidelheit isomorphic quoteright s to theorem $ \omega period .. Our\times main tool$ Banach there is the or concept it has of a nuclear K $ \ddot{o} $ the subspace . \quad The existencecompleting sequences of common of De nuclear Wilde open K square$ \ddot bracket{o} 5$ closing the square subspaces bracket was period examined .. The existence in [ 1 of 5 an ] unbounded and [ 1 completing 6 ] . \quad To deal with this question in the case of general locally convex spaces we introduce the concept of almost boundedness in Section 1 , and examine the properties of almost bounded set s and operators . \quad Although by means of these results we can derive some interesting theorems for example about the structure of spaces which have $ \omega $ as a subspace , most of them will b e generalized further in the subsequent s e c t i o n s .
Section 2 is on Eidelheit ’ s theorem . \quad Our main tool there is the concept of completing sequences of De Wilde [ 5 ] . \quad The existence of an unbounded completing 6 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentsequence in6 a\ loquad callyS convex $ . space\ddot gives{ usO} a means$ nal of constructingandT . Terzio a surj ection $ \check{g} $ lu onto omega which lift s bounded set s .. open parenthesis Theorem 2 period 3 closing parenthesis period .. The spirit of the construction \noindent sequence in a lo cally convex space gives us a means of constructing a surj ection i s similar t¨ o the proof of Eidelheit quoteright s theorem period .. The class of webbed spaces i s the ontonatural6 $ contextS\omega.O nal of and$ completing T which . Terzio sequences liftgˇ lu s boundedperiod .. For set certain s \quad webbed( spaces Theorem E we 2 again . 3 ) . \quad The spirit of the construction ihave ssequence similar the dichotomy in t a olocally the that convex proof either Lspace of open Eidelheit gives parenthesis us a means ’ E s comma of theorem constructing omega . \ closingquad a surj parenthesis ectionThe class onto =ω ofwhich LB webbed open lift parenthesiss bounded spaces E i comma s the omega closing parenthesisnaturalset s or context ( omega Theorem i s of i 2 somorphic .completing 3 ) . Theto what spirit sequences of the construction . \quad For i s similar certain t o the webbed proof of spaces Eidelheit ’$ s E theorem $ we . again haveweThe call the a class faithful dichotomy of webbed quotient that spaces of E eitheropen i s the parenthesis natural $ L context Corollary ( of E completing 2 period , \omega 5 sequencesclosing parenthesis) . =For certain period LB webbed.. ( In Section E spaces , 3 commaE\omega .. we examine) $ or the $ \omega $ i sexistence iwe somorphic again of nuclear have the to K dichotomy whatdieresis-o the that quotients either L( periodE, ω) = LB(E, ω) or ω i s i somorphic to what In Sectionwe call a4 faithfulcomma quotientwe look at of theE question( Corollary of Fr 2 . acute-e 5 ) . chet In Section subspaces 3 , by we using examine the concepts the existence of nuclear K \noindentof almosto¨ the quotients boundednesswe call . a and faithful completing quotient sequences of period $E .. Here ( we $ also Corollary generalize 2some . 5 ) . \quad In Section 3 , \quad we examine the existenceof the resultsIn Section of of nuclear Section 4 , we look1by K atusing $ the\ddot an question idea{o} of of$ Valdivia Fr thee ´ chet open quotients subspaces square bracket by . using 1 the 7 closing concepts square of almost bracket boundedness period .. Again we have the dichotomyand completing that either sequences L open .parenthesis Here we E also comma generalize F closing some parenthesis of the results = LB of open Section parenthesis 1 by using E comma an idea F closingof parenthesis for every FrIn acute-e SectionValdivia chet 4 [space 1 , 7 we ] E . or look Again F has at we a the have questionthe dichotomy of that Fr either $ \acuteL(E,F{)e =} LB$(E,F chet) for subspaces every Fre ´ chet by using space E the concepts ofproper almostor F Frhas acute-e boundedness a chet subspace and open completing parenthesis sequences Theorem 4 period . \quad 5 closingHere parenthesis we also period generalize some ofIn the Sectionproper results Fr 5 commae ´ chet of subspace we Section apply ( our Theorem 1 results by using 4 t . o 5 inductive ) . an idea limits of of FrValdivia acute-e chet [ 1 spaces 7 ] comma . \quad proAgain hyphen we have the dichotomythateitherj ectiveIn limits Section of DF 5 , hyphen we apply spaces $L our results and ( the t E projo inductive ,ective F limits limits ) of of DFS = Fre ´ hyphen LBchet spaces( spectra ,E pro of -sequence , j ective F limits )$ of foreveryFrDF - $ \acute{e} $ chetspacesspaces space period and $E$ .. the In proj the or final ective $F$section limits we of hasDFS consider a - spectra the space of sequence C open parenthesis spaces . X In closing the final parenthesis section we of consider continuous the real hyphen valued functionsspace C on(X a) completely of continuous regular real t - opological valued functions space X on period a completely .. Whenever regular possible t opological we space X. Whenever \noindentrelatepossible the existenceproper we relate Frof the concrete existence $ \acute Fr acute-eof{ concretee} $ chet Frchet subspacese ´ chet subspace subspaces and quotient ( and Theorem quotient spaces of 4spaces C . open 5 of )C parenthesis .(X) t o the X t closing opological parenthesis t oproperties the t opological of X. properties of X period InOur Section notationOur notation 5 and , we t erminology and apply t erminology our i s quite results i s standard quite t standard o period inductive . .. We We refer refer limits to to the the of books books Fr by by $ K\oacute ¨ the [{ 1e 1} ] and$ chetJarchow spaces , pro − jK ective o-dieresis[ 1 0 ] for limits the the open general of square DF theory− bracketspaces of lo cally 1 1 and closing convex the square spaces proj bracket . ective We and consider limits Jarchow locallyopen of convex DFS square− spacesspectra bracket over 1 the 0 of closing field sequenceK squareof bracket for the general theoryspacesreal of lo or . cally\ complexquad convexIn numbers spaces the final, period which .. aresection We assumed we t o consider be Hausdorff the unless space stated $ otherwise C ( . X ) $ of continuous real − valued functionsconsiderFinally locally on ,we a convex completelywould spaces like t over o thank regular the thefield Scientific K t of opological real and or complex Technical space numbers Research $ X comma Council . which $ of\ Turkeyquad forWhenever partial support possible we relateare. assumed the t existence o be Hausdorff of unless concrete stated Fr otherwise $ \acute period{e} $ chet subspaces and quotient spaces of $ C ( X ) $Finally comma we would like1 . t o thank Almost the Scientific bounded and Technical sets Research and operators Council tof o Turkey theWe t for denote opological partial by U support(E) aproperties base period of neighborhoods of $X of a . lo $ cally convex space ( lcs ) of E consisting of barrels and 1 periodby F( ..E) Almost the collection .. bounded of finite sets -and dimensional operators subspaces . Our notationA subset A andof E twill erminology be called i s quitealmost standard bounded . if\quad for eachWeU refer∈ U(E to) the books by there i s We denote by0 U open parenthesis E⊥ closing parenthesis a base of neighborhoods of a lo cally convex space open parenthesis lcs closing parenthesisK $L\∈ddot F of(E E{)o and} $ρ > the0 with [ 1A 1∩ L ] and⊂ ρU. Jarchow [ 1 0 ] for theA general bounded subset theory i s almost of lo bounded cally . convex spaces . \quad We considerTrivially locally , any subset convex of E[σ spaces(E,E0)] i over s almost the bounded field . $ K $ of real or complex Therefore numbers if P : E → ,E which consisting of barrels and by F open parenthesis E closingI parenthesis the collection of finite hyphen dimensional subspaces period areA subseti assumed s a proj A of ection Et will o be whose be Hausdorffcalled range .... is almost isomorphic unless bounded statedto K if for, the each otherwise product U in U of openI . copies parenthesis of the field E closingK, then parenthesis any subset .... of there i s L inP F(E open) i s alsoparenthesis almost Ebounded to the power . This of prime i s the closing reason parenthesis why we consider and rho almost greater bounded 0 with subsets A cap L . to theWe powerfirst of bottom subset rho U periodFinallyneed .... A , two bounded we rather would subset technical like i s almost lemmas t o thank bounded the period Scientific and Technical Research Council ofTrivially Turkeyin order comma tfor o dualize any partial subset this concept supportof E open . square . bracket sigma open parenthesis E comma E to the power of prime closing parenthesis closing square bracket1 . 1 i . sLemma almost bounded . Let periodA be an .... absolutely Therefore convex if P : cE los right ed arrow subset E of a lcs E . Let L ∈ F(E). Then \ centerlinei s aA proj+ L ectionis{ c1 los . whose ed\quad . rangeAlmost is isomorphic\quad tobounded K to the power sets of and I comma operators the product} of I copies of the field KP comma r o o f then . any If B subseti s of compact P open parenthesis then B E+ closingL i parenthesis s already i closed s also .almost We bounded shall period exploit .. Thisthis i s the reason why Wedenotesimple by fact $U . Let (R E: )E0∗ $ a→ basesp (A of◦)∗ neighborhoodsbe the restriction of a lo map cally , where convexA◦ space ( lcs ) of $E$ we consider almost bounded0 subsets period .. We first need two rather technical lemmas0∗ 0 consistingin orderi s the t polaro dualize of in barrelsE this. concept and period byWe $ F equip ( these E spaces ) $ with the the collectiont opologies σ(E of,E finite)− dimensional and subspaces . 1 period 1 period Lemma period Let A be an absolutely convex c los ed subset of a lcs E period .. Let L in \noindentF open parenthesisAsubset E closing $A$ parenthesis of $E$ period .. will Then be A plus called L is c\ losh f ed i l l periodalmost bounded if for each $ U \ in U( EP )r o $ o f\ periodh f i l l .. Ifther B .. e i si .. s compact .. then .. B plus L .. i s .. already closed period .. We .. shall .. exploit this .. simple .. fact period .. Let .. R : E to the power of prime * right arrow .. sp open parenthesis A to the power of circ closing parenthesis to\noindent the power of$ * L.. be ..\ in the ..F restriction ( E .. map ˆ{\ commaprime ..} where) $ .. A and to the $ power\rho of circ> 0 $ with $ A \cap L ˆ{\bot } \subseti s the polar\rho in E toU the power. $ \ ofh prime f i l l A period bounded .... We subset equip these i s spaces almost with bounded the t opologies . sigma open parenthesis E to the power of prime * comma E to the power of prime closing parenthesis .... and \noindent Trivially , any subset of $ E [ \sigma ( E , E ˆ{\prime } ) ] $ i s almost bounded . \ h f i l l Therefore if $ P : E \rightarrow E $
\noindent i s a proj ection whose range is isomorphic to $ K ˆ{ I } , $ the product of $ I $ copies of the field $K , $ then any subset of $P ( E ) $ i s also almost bounded . \quad This i s the reason why we consider almost bounded subsets . \quad We first need two rather technical lemmas
\noindent in order t o dualize this concept .
1 . 1 . Lemma . Let $A$ be an absolutely convex c los ed subset of a lcs E . \quad Let $ L \ in $ $ F ( E ) . $ \quad Then $A + L$ isclosed.
P r o o f . \quad I f $ B $ \quad i s \quad compact \quad then \quad $ B + L $ \quad i s \quad already closed . \quad We \quad s h a l l \quad e x p l o i t t h i s \quad simple \quad f a c t . \quad Let \quad $ R : E ˆ{\prime ∗ } \rightarrow $ \quad sp $ ( A ˆ{\ circ } ) ˆ{ ∗ }$ \quad be \quad the \quad restriction \quad map , \quad where \quad $ A ˆ{\ circ }$
\noindent i s the polar in $Eˆ{\prime } . $ \ h f i l l We equip these spaces with the t opologies $ \sigma ( E ˆ{\prime ∗ } , E ˆ{\prime } ) $ \ h f i l l and Subspaces and quotient spaces of locally convex spaces .. 7 \ hspacesigma∗{\ openf parenthesis i l l } Subspaces sp open and parenthesis quotient A to spacesthe power of of circ locally closing convex parenthesis spaces to the power\quad of7 * comma sp open parenthesis A to the power of circ closing parenthesis closing parenthesis period .. Since RA i s bounded comma RA i s compact period .. Hence RA plus RL i s \noindentclosed period$ ..\sigma By the bipolar( $ theorem sp $ comma ( A we ˆ{\ havecirc RE cap} RA) =ˆ{ RA ∗ period } , ..$ So sp if x in$ RE ( cap A open ˆ{\ parenthesiscirc } ) RA plus ) . $ \quad Since $RA$ i s bounded $ , RA$ i s compactSubspaces . \quad andHence quotient $RA spaces of + locally RL$ convex spaces i s 7 RL closing parenthesis◦ ∗ ◦ then x minus Rl in RE line-intersection RA for some l in L period From RE cap RA = RA we get c lRE o s eσ cap d( sp . open (A\quad) parenthesis, spBy (A the)). RA bipolarSince plusRA RL closingi theorem s bounded parenthesis ,,RA we have =i sRA compact plus $ RE RL .\ Hencecap RARA+ RL =i s closed RA . . $ By the\quad So i f $ x \ in RE andbipolar\ socap this theorem set( is closed RA , we have in + RE $RE period∩RA ....= RA. Also RSo to if thex ∈ powerRE ∩( ofRA minus+ RL 1) thenopenx parenthesis−Rl ∈ REline 0 closing−intersectionRA parenthesis cap E subset epsilon A for any$RL epsilonfor some greater )$l ∈ thenL. 0 periodFrom $xRE .... Therefore∩ RA− = RARl we\ getin RE l i n e −intersection RA$ for some $ l \ in L . $ From $ RER to the\cap powerRA of minus = 1 open RA$ parenthesis weget 0 closing parenthesis cap open parenthesis E plus A closing parenthesis = A and A plus L subset E cap R to the power of minus 1 open parenthesisRE ∩ RA(RA plus+ RL) closing = RA + parenthesisRL subset A plus L period \ [ RE \cap ( RA + RL ) = RA + RL \ ] If Aand is almost so this bounded set is closed and in U inRE. U open parenthesisAlso E closingR−1(0) parenthesis∩ E ⊂ εA for comma any ε then > 0. by what we haveTherefore shown U to the power of circ subset rho A to the power of circ plus L for some delta greater 0 and L in F open parenthesis E to the power of prime closing parenthesis period .... This i s what motivates the next result \noindentwhich i s purelyand so algebraic this in set character is closed period ..in HereR $RE− and1(0) throughout∩ ( .E $+ A\)h =Capital fA i land l Also Gamma $ open R ˆ parenthesis{ − 1 } D closing( 0 parenthesis ) \cap .. denotesE the\subset \ varepsilon A $ f o r any $ \ varepsilon > 0 . $ \ h f i l l Therefore absolutely convex hull of the set D periodA + L ⊂ E ∩ R−1(RA + RL) ⊂ A + L. 1 period 2 period Lemma period Let A and B be two absolutely convex subsets of a lcs E which is \ begin { a l i g n ∗} not necessarily Hausdorff period ..If SupposeA is almost A is boundedclosed and and boundedU ∈ andU(E) B, then only boundedby what we period have shown U ◦ ⊂ R ˆ{ − 1 } ( 0 ) \cap ( E + A ) = A and \\ A + L \subset E \cap R ˆ{ − LetρA E◦ =+ spL openfor some parenthesisδ > 0 and A closingL ∈ F(E parenthesis0). oplus G where the sumThis is only i s algebraicwhat motivates period the .. If next there result are L in F open parenthesis E 1 } ( RA + RL ) \subset A + L . closingwhich parenthesis i s purely .. and algebraic in character . Here and throughout Γ(D) denotes the absolutely convex hull of \end{ a l i g n ∗} rhothe greater set D. 0 such that B subset1 . rho 2 . ALemma plus L . Let A and B be two absolutely convex subsets of a lcs E which is not necessarily \ hspace ∗{\ f i l l } If $A$ is almost bounded and $U \ in U ( E ) , $ thenbywhatwehaveshown thenHausdorff we can find . z Suppose sub 1 commaA is period closed and period bounded period and commaB only z sub bounded n in G . so Let thatE = sp (A) ⊕ G where the sum $ U ˆ{\ circ }\subset $ B subsetis only delta algebraic open .parenthesis If there A are plusL Capital∈ F(E) Gammaand openρ > 0 bracesuch z that sub 1 comma period period period comma z sub n closing brace closing parenthesis \noindent $ \rho A ˆ{\ circ } + L$ forsome $ \ delta > 0 $ and $ L \ in F ( E ˆ{\prime } for some delta greater 0 period B ⊂ ρA + L )P r. o $ o f\ periodh f i l l .. TheThis assumption i s what implies motivates the next result B subsetthen we rho can A plusfind spz1, open ..., zn parenthesis∈ G so that A plus B closing parenthesis line-intersection L period \noindentWe let L subwhich 1 = sp i open s purely parenthesis algebraic A closing in parenthesis character cap L . and\quad find LHere sub 2 and in F open throughout parenthesis $ E\ closingGamma parenthesis( D with ) L $ sub\ 1quad denotes the oplusabsolutely L sub 2 = convexsp open parenthesis hull of Athe plus set B closing $B D⊂ parenthesisδ(A .+ $ Γ{z1, cap ..., z Ln} period) .. We decomposefor some L subδ > 10. further as L sub 1 = M oplus F where M cap open brace 0 closing brace = open brace 0 closing brace and F subset open 1 . 2 . Lemma . Let $A$ and $B$ be two absolutely convex subsets of a lcs $ E $ which is brace 0 closing brace period .. Since P r o o f . The assumption implies notA i necessarilys absolutely convex Hausdorff and closed . comma\quad weSuppose have open $brace A $ 0 closing is closed brace subset and Abounded and therefore and for $ some B $ rho only 1 greater bounded 0 . Let $E =$ sp $( A ) \oplus G $ where the sum is only algebraic . \quad If there are $ L we obtain B ⊂ ρA + sp(A + B)line − intersectionL. \ inB subsetF rho ( 1 to E the ) power $ \ ofquad A plusand M plus L sub 2 period $If\ xWerho in M let commaL>1 = sp0 y in( $A L) ∩ suchsubL and 2 and that find x plusL2 ∈ y F in(E open) with braceL1 ⊕ 0 closingL2 = sp brace (A + commaB) ∩ L. thenWe x plus decompose y = a inL1 Afurther period asSo y = x minus a in sp open parenthesisL1 = AM closing⊕ F where parenthesisM ∩ {0} cap= { L0} suband 2F = ⊂ {0}. Since A i s absolutely convex and closed , we have {0} ⊂ A \ [Bopenand brace\subset therefore 0 closing for\ bracesomerho ρ period1 A> 0 ..we + Hence obtain L y\ =] 0 and therefore x = 0 period .. This gives open parenthesis M plus L sub 2 closing parenthesis cap open brace 0 closing brace = open brace 0 closing braceA and so the topology is Hausdorff on M plus L sub 2 periodB ⊂ ..ρ1 Now+ theM + aboveL2. inclusion implies that there \noindent then we can find $ z { 1 } , . . . , z { n }\ in G $ so that i s aIf boundedx ∈ M, y subset∈ L2 and D withx + y ∈ {0}, then x + y = a ∈ A. So y = x − a ∈ sp (A) ∩ L2 = { 0 } . Hence y = 0 B subsetand therefore rho 1 tox the= 0 power. This of A gives plus (M D cap+ L open) ∩ { parenthesis0} = {0} and M so plus the L topology sub 2 closing is Hausdorff parenthesis on M period+ L . Now \ [B \subset \ delta ( A + 2 \Gamma \{ z { 1 } , . . . ,2 z { n }\} ) \ ] Thisthe means above that inclusion thereare implies a finite that number there i of s a points bounded x sub subset i in ED sowith that B subset rho 1 to the power of A plus Capital Gamma open brace x sub 1 comma period period period comma x sub n closing brace period A We finish by writing each x sub i as y i plus zB sub⊂ ρ i1 where+ D y∩ i(M in sp+ L open2). parenthesis A closing parenthesis and z sub i in G period \noindentThe followingf o r is some an immediate $ \ delta consequence> of0 the .preceding $ result period This means that there are a finite number of points xi ∈ E so that \ centerline {P r o o f . \quad The assumption implies } A B ⊂ ρ1 + Γ{x1, ..., xn}. \ [B \subset \rho A + sp ( A + B ) line −intersection L . \ ] We finish by writing each xi as yi + zi where yi ∈ sp (A) and zi ∈ G. The following is an immediate consequence of the preceding result . \noindent We l e t $ L { 1 } = $ sp $ ( A ) \cap L$ and find $L { 2 }\ in F(E ) $ with $ L { 1 }\oplus L { 2 } = $ sp $ ( A + B ) \cap L . $ \quad We decompose $ L { 1 }$ further as $L { 1 } = M \oplus F $ where $ M \cap \{ 0 \} = \{ 0 \} $ and $ F \subset \{ 0 \} . $ \quad Since $ A $ i s absolutely convex and closed , we have $ \{ 0 \}\subset A $ and therefore for some $ \rho 1 > 0 $ we obtain
\ [B \subset \rho 1 ˆ{ A } + M + L { 2 } . \ ]
\noindent I f $ x \ in M , y \ in L { 2 }$ and $ x + y \ in \{ 0 \} , $ then $ x + y = a \ in A . $ So $ y = x − a \ in $ sp $ ( A ) \cap L { 2 } = $ \{ 0 \} . \quad Hence $y = 0$ andtherefore $x = 0 .$ \quad This gives $( M + L { 2 } ) \cap \{ 0 \} = \{ 0 \} $ and so the topology is Hausdorff on $M + L { 2 } . $ \quad Now the above inclusion implies that there i s a bounded subset $ D $ with
\ [B \subset \rho 1 ˆ{ A } + D \cap ( M + L { 2 } ). \ ]
\noindent This means that there are a finite number of points $ x { i }\ in E $ so that
\ [B \subset \rho 1 ˆ{ A } + \Gamma \{ x { 1 } , . . . , x { n }\} . \ ]
\noindent We finish by writing each $ x { i }$ as $ y i + z { i }$ where $ y i \ in $ sp $ ( A ) $ and $ z { i }\ in G . $
\ centerline {The following is an immediate consequence of the preceding result . } 8 .. S period O-dieresis nal and T period Terzio caron-g lu \noindent1 period 38 period\quad .. CorollaryS $ . period\ddot .. Let{O} A$ .. and nal B ..andT be .. absolutely . Terzio .. convex $ \check subsets{g} .. of$ a lu.. lcs E comma A c losed and B bounded period .. Suppose we have E = sp open parenthesis A closing parenthesis oplus G where the sum is purely 1 . 3 . \quad C o r o l l a r y . \quad Let $ A $ \quad and $ B $ \quad be \quad a b s o l u t e l y \quad convex subsets \quad o f a \quad l c s algebraic period¨ .. If there are L in F open parenthesis E closing parenthesis .. and rho greater 0 with $ EB subset8 , S rho A.O $ Anal plus and L T . Terzio gˇ lu cthen losed there1 . and 3 are . z $Corollary sub B 1 $ comma bounded . periodLet . period\Aquadand periodSupposewehaveB commabe z absolutely sub n in G$E and convex delta =$ subsets greater sp 0 $( of with a A lcs E,A ) \oplusc G $ where the sum is purely a lB g esubsetlosed b r a i andc delta . B\ openquadbounded parenthesisIf . there Suppose A plus are Capital we $ have L GammaE\ in= sp open (AF) ⊕ braceG (where z subE the 1 )comma sum $ is\ periodquad purely periodand algebraic $period\rho . comma If there> z sub0 $ n closing with brace closing parenthesisare L period∈ F(E) and ρ > 0 with \ [BP r o o\subset f period .. We\rho equip EA with + the topology L \ ] tau = sigma open parenthesis E comma sp open parenthesis A to the power of circ closing parenthesis closing parenthesis period .. Then A is tau hyphenB ⊂ ρA closed+ L
andthen tau therehyphen are boundedz1, ..., z periodn ∈ G and B is tauδ > hyphen0 with bounded period .. So the result follows from Lemma 1 period 2 period \noindentAn .. absolutelythen .. there convex are .. and $ bounded z { 1 ..} subset, ..B . .. will . be . .. called , .. z a ..{ discn }\ commain .. andG ..$ if and $ \ delta > 0 $ with the normed space E open square bracket B closingB ⊂ δ square(A + Γ{ bracketz , ..., z =}) ... sp open parenthesis B closing parenthesis .. i s complete comma B \ [B \subset \ delta ( A + \Gamma 1\{ nz { 1 } , . . . , z { n }\} ). \ ] will be called a Banach .. disc period ◦ From nowP r ono o if f we . write We equip E simeqE with F oplus the G topology with noτ additional= σ(E, sp remarks (A )). commaThen weA is meanτ− closed E i s and τ− bounded i somorphic.B is τ as− bounded a lcs to the . direct So the sum result of lcs follows F and from G period Lemma .. 1 By . 2 K . to the power of open parenthesis I closing parenthesis .. we denote the P rdirect o o sumAn f . of absolutely\ Iquad copiesWe of K equip convex period $ and E $ bounded with the subset topologyB will $ be\tau called= a \sigmadisc , and( E if the , $ sp $ ( A ˆ{\ circ } )1 period )normed . 4 $ period space\quadE Theorem[B]Then = period sp $A (B Let) $ E i i be s complete a $open\tau parenthesis,B−will$ ultrabe c lcalled o closings e d a Banach parenthesis disc .. bornological. From now lcs on period if we .. Suppose there is a c los andwrite $ \tauE ' F ⊕−G$with bounded no additional $. remarks B$ , we is mean $ \Etaui s − $ bounded . \quad So the result follows from Lemma 1 . 2 . ed comma (I) absolutelyi somorphic convex as subseta lcs to A the of directE with sum the of following lcs F and propertyG. By : ....K for everywe denote open parenthesis the direct sum Banach of I closingcopies of parenthesisK. An disc\quad B ina E b there s o l u1 tare e . l 4y L . in\Theoremquad F openconvex parenthesis . Let\quadE Ebe closing aand( ultra bounded parenthesis) bornological\ andquad rhosubset lcsgreater . 0\ Supposequad with B there subset$ B is $ rho a c\ Aquad los plus ed Lw, iperiod l l be ..\ Thenquad Ec simeq a l l e d F \quad a \quad d i s c , \quad and \quad i f thenormedspaceabsolutely convex subset $E A of [ E with B the ] following =$ property\quad sp: $ ( B ) $ \quadfor everyi s( completeBanach ) $ , B$ will be called a Banach \quad d i s c . oplus K to the power of open parenthesis I closing parenthesis (I) Fromfordisc some nowB index onin ifE setthere we I and write are A capL ∈ F $ F is E(E a) neighborhoodand\simeqρ > 0 with inF FB period\⊂oplusρA + L. G$Then withE ' noF ⊕ additionalK for some remarks index set , we mean $ E $ i s P rI oand o f periodA ∩ F ..is Let a neighborhood F = sp open in parenthesisF. A closing parenthesis and decompose E algebraically as E = F oplus G period .. Let P \noindentb e theP proj ri o ectiono somorphic f . with Let kernelF as= sp a F ( andA lcs) and range to decompose the G period directE Wealgebraically will sum show of that lcs as E P $F$= i sF continuous⊕ G. andLet period $GP b e the . $proj\ ectionquad By $ K ˆ{ (I ) }Let$ with B\quad be kernel a openweF parenthesis denoteand range theG. BanachWe will closing show parenthesis that P i s disccontinuous period ... Let ByB Corollarybe a ( Banach 1 period ) disc3 there . are By z Corollary sub i in G and delta greater 0 with directB subset1 . 3 sum there delta of are openz $i parenthesis∈ IG $and copiesδ > A0 plus with of Capital $K Gamma . $ open brace z sub 1 comma period period period comma z sub n closing brace closing parenthesis period \ hspaceHence∗{\ P openf i l l parenthesis}1 . 4 . B Theorem closing parenthesis . LetB ⊂ $E$ subsetδ(A + Capital Γ be{z1, a ..., Gamma ( zn} ultra). open ) \ bracequad zbornological sub 1 comma period lcs period . \quad periodSuppose comma z there sub n is a c los ed , closingHence braceP period(B) ..⊂ ThisΓ ..{z shows, ..., z that}. ..This every showsbounded that subset every .. of bounded G i s subset of G i s finite - dimensional \noindent absolutely convex1 n subset A of E with the following property : \ h f i l l for every ( Banach ) finiteand hyphenP : E → dimensionalE i s continuous and P : , E since rightE arrowis ( ultra E i ) s bornological continuous comma . since E is open parenthesis ultra closing parenthesis bornological period Since G is also ( ultra ) bornological , we have G ' K(I). \noindent disc $B$ in $E$ there are $L \ in F ( E ) $ and $ \rho > 0 $ with $ B Since GLet is alsoD ⊂ openF be parenthesis bounded . ultra Then closing there parenthesis are L ∈ bornological F(E) and ρ comma > 0 with we haveD ⊂ GρA simeq+ L. KThis to the gives power again of open parenthesis I closing \subset \rho A + L . $ \quad Then $ E \simeq F \oplus K ˆ{ (I) }$ parenthesisB ⊂ δA period+Γ{y1, ..., yn}, yi ∈ G. Applying I −P, we obtain B ⊂ δA. Since F i s also ( ultra ) bornological for some index set $ I $ and $A \cap F$ is a neighborhood in $F . $ Letas D a subset quotient F be space bounded of E,A period∩ F i .. sThen a neighborhood there are L b in ecause F open it absorbs parenthesis every E ( closing Banach parenthesis ) disc of F. and rho greater 0 with D subset rho A plus L period Before we proceed any further we would like t o examine the stability of the P r o o f . \quad Let $F =$ sp $( A )$ anddecompose $E$ algebraically as $E = F Thisclass gives of again almost B bounded subset delta set s A under plusCapital certain operations Gamma open . We brace will y first 1 comma prove that period the period period comma y n closing brace comma y i in \oplus G . $ \quad Let $ P $ G periodclosure .. Applying of an absolutely I minus P convex comma almost we obtain bounded subset i s also almost bounded . b e the proj ection with kernel $ F $ and range $G . $ We will show that $ P$ i s continuous . B subsetThe proof delta of A this period i s not .. Since trivial F and i s also we needopen a parenthesis modification ultra of Mazur closing ’ parenthesis s method for bornological extracting as basic a quotient sequences space of E comma A cap F i Let $B$ be a ( Banach ) disc . \quad By Corollary 1 . 3 there are $ z { i }\ in G $ and $ \ delta s a ( cf . [ 7 ] ; Chap . V ) . > 0 $ with neighborhood b ecause1 . 5it . absorbsLemma every . openLet A parenthesisand B Banachbe two closing absolutely parenthesis convex disc subsets of F period of a lcs E such Beforethat weB proceedis c losed any,A further⊂ sp we (B) wouldand likeA ∩ tL o⊥ examineis not absorbed the stability by B offor the each L ∈ F(E0). Suppose F ∈ F( sp \ [B \subset \ delta ( A + \Gamma \{ z { 1 } , . . . , z { n }\} ). \ ] class(B of)) almostis such bounded that qB, set s underthe gauge certain of operationsB, is a normperiod on WeF. will firstThen prove for everythat theε > 0 and ρ > 0 there closureis x of∈ anA \ absolutelyρB such that convexqB almostis still bounded a norm subset on sp i s{x, also F } almostand bounded for each periody ∈ F and each α ∈ K we have The proof of this i s not trivial and we need a modification of Mazur quoteright s method for \noindent Hence $ P ( B ) \subset \Gamma \{ z { 1 } , . . . , z { n }\} . $ extracting basic sequences open parenthesis cfqB period(y) ≤ open(1 + squareε)qB(y bracket+ αx). 7 closing square bracket semicolon Chap period V closing parenthesis period\quad This \quad shows that \quad every bounded subset \quad o f $ G $ i s f i1 n period i t e − 5dimensional period Lemma period and .. $P Let A ..: and E B ..\ berightarrow two .. absolutelyE convex $ i subsetss continuous of a lcs E such, since $ E $ is ( ultra ) bornological . that B is c losed comma A subset sp open parenthesis B closing parenthesis and A cap L to the power of bottom is not absorbed by B for each\noindent L in F openSince parenthesis $G$ E to is the also power ( of ultra prime closing ) bornological parenthesis period , we have $G \simeq K ˆ{ (I) } . $ Suppose F in F open parenthesis sp open parenthesis B closing parenthesis closing parenthesis .. is such that q B comma .. the gauge of B commaLet $ .. D is a norm\subset on F periodF $ .. Then be bounded . \quad Then there are $ L \ in F ( E ) $ and $ \rho > 0 $for everywith epsilon $ D greater\subset 0 .. and rho\rho greaterA 0 there + is L x in .A $backslash rho B such that q B .. is still a norm on Thissp open gives brace again x comma $ F B closing\subset brace .. and\ delta for each yA in F + and each\Gamma alpha in\{ K we havey 1 , . . . , y n \} ,q B y open i parenthesis\ in G y closing . $ parenthesis\quad Applying less or equal open$ I parenthesis− P 1 plus , $ epsilon weobtain closing parenthesis q B open parenthesis y plus alpha x closing$ B parenthesis\subset period\ delta A . $ \quad Since $ F $ i s also ( ultra ) bornological as a quotient space of $ E , A \cap F $ i s a neighborhood b ecause it absorbs every ( Banach ) disc of $ F . $
\ hspace ∗{\ f i l l } Before we proceed any further we would like t o examine the stability of the
\noindent class of almost bounded set s under certain operations . We will first prove that the
\noindent closure of an absolutely convex almost bounded subset i s also almost bounded .
\noindent The proof of this i s not trivial and we need a modification of Mazur ’ s method for extracting basic sequences ( cf . [ 7 ] ; Chap . V ) .
\ hspace ∗{\ f i l l }1 . 5 . Lemma . \quad Let $ A $ \quad and $ B $ \quad be two \quad absolutely convex subsets of a lcs $ E $ such
\noindent that $B$ isclosed $ , A \subset $ sp $( B )$ and $A \cap L ˆ{\bot }$ is not absorbed by $B$ for each $ L \ in F ( E ˆ{\prime } ) . $ Suppose $ F \ in F ($sp$( B ) )$ \quad issuchthat $q B ,$ \quad the gauge o f $ B , $ \quad is anormon $F .$ \quad Then f o r every $ \ varepsilon > 0 $ \quad and $ \rho > 0$ there is $x \ in A \setminus \rho B$ suchthat $q B$ \quad is still a norm on sp $ \{ x , F \} $ \quad and for each $ y \ in F $ and each $ \alpha \ in K $ we have
\ [ q B ( y ) \ leq ( 1 + \ varepsilon ) q B ( y + \alpha x ) . \ ] Subspaces and quotient spaces of locally convex spaces .. 9 \ hspaceP r o o∗{\ f periodf i l l } ..Subspaces Assume 0 less and epsilon quotient less 1 .. andspaces let z subof 1locally comma period convex period spaces period\ commaquad 9 z sub n in F cap S b e an epsilon slash 4 hyphen net with \ hspacerespect∗{\ t of q i lB l for}P the r o compact o f . set\quad F capAssume S comma where $ 0 S< = open\ varepsilon brace y : q B open< parenthesis1 $ \quad y closingand parenthesis l e t $ z = 1{ closing1 } brace, period...... We . choose , z { n }\ in F \cap SSubspaces $ b e and an quotient $ \ varepsilon spaces of locally convex/ 4 spaces− $ 9 net with v sub 1 comma period periodP r o o period f . Assume comma 0v< sub ε n
\ [ \ parallel y \ parallel \ leq ( 1 + \ varepsilon { n } ) \ parallel \alpha x { n + 1 } + y \ parallel \ ]
\noindent f o r every $ y \ in $ \quad sp $ \{ x { 1 } , . . . , x { n }\} , \alpha \ in K , x { n }\ in A \setminus 2 ˆ{ n + 1 } U $ and \quad $ \ parallel \cdot \ parallel $ \quad i s a norm on sp $ \{ x { i }\} . $ \quad The sequence $ ( x { n } ) $ i s a $ \ parallel \cdot \ parallel − $ basic sequence with basis constant 3 / 2 . \quad To see this let $ \alpha { 1 } ,..., \alpha { n }$ begiven $, m \ leq n . $ \quad Then
\ [ \ begin { a l i g n e d }\Arrowvert \sum ˆ{ i = 1 } { m }\alpha { i } x { i }\Arrowvert \ leq ( 1 + \ varepsilon { m } ) \Arrowvert \sum ˆ{ i = 1 } { m + 1 }\alpha { i } x { i } \Arrowvert \\ \ leq ( 1 + \ varepsilon { m } ) . . . ( 1 + \ varepsilon { n } ) \Arrowvert \sum ˆ{ i = 1 } { n }\alpha { i } x { i }\Arrowvert \ leq 3 { 2 }\Arrowvert \sum ˆ{ i = 1 } { n }\alpha { i } x { i }\Arrowvert . \end{ a l i g n e d }\ ]
\noindent This shows that $ ( x { n } )$ isabasisofsp $( x { i } ) $ and \ h f i l l $ \ parallel \cdot \ parallel $ i s a norm on this space . \ h f i l l We
\noindent will now perturb this sequence by choosing $ y n \ in A$ such that $ \ parallel x { n } − y n \ parallel \ leq 2 ˆ{ − n − 3 } . $ Let $ ( u { n } ) $ b e the associated sequence of coefficient functionals for $ ( x { n } ) . $ \quad We may
\ begin { a l i g n ∗} \ tag ∗{$ assume u { n }\ in 2 ( 3 / 2 ) U ˆ{\ circ } . $} Since \\\sum \ parallel u { n } \ parallel \ parallel x { n } − y n \ parallel < 1 \end{ a l i g n ∗} 1 0 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentthe image1 open 0 \ parenthesisquad S y $ n . open\ parenthesisddot{O} $ U closing nal andT parenthesis . Terzio closing parenthesis $ \check of{g open} $ parenthesis lu y n closing parenthesis in the associated Banach space E-tildewide sub open parenthesis U closing parenthesis is a basic sequence \noindent theimage $( y n ( U ) )$ of $( y n )$ intheassociatedBanachspace with basis constant¨ K and it i s equivalent t o open parenthesis x sub n open parenthesis U closing parenthesis closing parenthesis period .. From$ \ widetilde this1 0 it follows S .O{Enal} that and{ (U) T . Terzio gˇ lu}$ is a basic sequence barthe times image bar ( iyn s( alsoU)) a of norm (yn) inon the sp open associated brace Banach y n : n space in N closingEe(U) is brace a basic period sequence .. Note that y n in A but y n negationslash-element 2 to \noindentwith basiswith constant basisK constantand it i s equivalent $K$ t ando (xn( itU)) i. sFrom equivalent this it follows t o that $k ( · k xi s{ alson a} norm( on U ) ) . $ \quad From this it follows that the power of n U period .. Since A n 0 $i s\ almostspparallel{yn bounded: n ∈ N}\. wecdotNote can find that\ parallel Lyn in∈ FA openbut$ parenthesisyn i6∈ s2 U. also ESince to anormon theA poweri s almost of sp prime bounded $ closing\{ wey parenthesis can findnL : and∈ F(n rhoE ) greater and\ in 0N with \} . $ \quad Note that $ yCapitalρ n > 0 Gamma with\ in openA $ brace but y n $: n y in N n closing\not brace\ in cap L2 to ˆ{ then power} U of bottom . $ subset\quad ASince cap L to $ the A power $ of bottom subset line-rho i s almost bounded we can find $ L \ in F ( E ˆ{\prime } ) $ and $ \rho > 0 $ with U period ⊥ ⊥ Let x = sum sub n greater k alpha subΓ{yn n: yn n∈ commaN} ∩ L and⊂ alphaA ∩ L sub⊂ nline = 0− afterrhoU. some index comma bar x bar less or equal 1 period .... Then alpha\ [ \Gamma sub n y nP in\{ 2 KUy comma n : n \ in N \}\cap L ˆ{\bot }\subset A \cap L ˆ{\bot }\subset Let x = n>k αnyn, and αn = 0 after some index , k x k ≤ 1. Then αnyn ∈ 2KU, l i n e −rho U . \ ] n n wherewhere K iK s thei s the basis basis constant constant of open of (yn parenthesis). Since yn y n6∈ closing2 U, we parenthesis have | αn | period< 2K/ ..2 Since. This y n means negationslash-element 2 to the power of n U comma we have bar alpha sub n bar less 2 K slash 2 to the power of n period K This means sp{yn : n > k} ∩ U ⊂ 22 kA \noindentsp open braceLet y n $ : x n greater = k\ closingsum { bracen cap> U subsetk }\ 2alpha sub 2 sub{ kn to the} powery n of K , A $ and $ \alpha { n } = 0 $ afterandand therefore some therefore index $ , \ parallel x \ parallel \ leq 1 . $ \ h f i l l Then $ \alpha { n } y n \ insp open2 brace KU y n , : $n greater k closing brace cap U cap L to the power of bottom subset K 2 to the power of k minus sub 1 to the power of sp{yn : n > k} ∩ U ∩ L⊥ ⊂ Kρ U ∩ L⊥. rho U cap L to the power of bottom period 2k−1 \noindent where $K$ i sthe basis constant of $( y n ) .$ \quad Since $ y n \not\ in Since bar times bar .... is a norm also on sp open brace y n : n in N closing brace comma this implies sp open⊥ brace y n : n greater k closing 2 ˆ{ Sincen } kU · k , $ we have $ \ismid a norm\alpha also on sp{ {ynn }\: n ∈midN}, this< implies2 sp K{yn : /n > k 2} ∩ ˆ{L n=}{0}. $ brace cap L to the power of bottom = open brace0 0 closing brace Thisfor meansk sufficiently large , but since L ∈ F(E ) this cannot be true . for k sufficientlyR e m a r large k . comma The absolutely but since convex L in F openhull of parenthesis an almost E bounded to the power subset of need prime not closing b e almost parenthesis bounded this . cannot be true period R e m a r k period .. The absolutely convex hull of an almost bounded subset need not Here i s an extreme example . Let (en) b e the canonical basis of `1 and \ [b sp e almost\{ boundedy n period : Here n i s> an extremek \}\ examplecap periodU Let open\subset parenthesis2 e{ sub2 n} closingˆ{ K } parenthesis{ k } A b e\ the] canonical basis of l sub 1 and ∞ Line 1 infinity Line 2 A = union of sp open brace e sub n[ closing brace period Line 3 n = 1 A = sp{e }. \noindentFor e = openand parenthesis therefore 1 comma 1 comma period period periodn closing parenthesis in l sub infinity comma we have A cap e to the power of minus 1 open parenthesis 0 closing parenthesis = open brace 0n closing= 1 brace and so A i s almost bounded period \ [ sp \{ y n : n > k \}\cap U \cap L ˆ{\bot }\subset K { 2 ˆ{ k } − }ˆ{\rho } { 1 } However comma Capital Gamma open parenthesis−1 A closing parenthesis = phi i s not almost bounded comma b ecause it i s dense in l sub U1 period\Forcape =L (1, ˆ1{\, ...)bot∈ `∞}, we. have\ ] A ∩ e (0) = {0} and so A i s almost bounded . However , Γ(A) = φ i s not Wealmost now give bounded a dual , characterization b ecause it i s dense of absolutely in `1. convex almost .. bounded subsetsWe period now give a dual characterization of absolutely convex almost bounded subsets . \noindent1 . 7Since . Corollary $ \ parallel . An absolutely\cdot convex\ parallel subset A$of a\ h lcs f i lE l isis almost a norm bounded also if on and sp only $ if\{ for y n : n \ in 1 period 7 period Corollary period .. An absolutely0 convex subset A of a lcs E is almost bounded N if and\}every only,U if$∈ for U this(E every) impliesthere U in Uare openL sp∈ parenthesis F $(E\{) andy E closingρ n > 0 with parenthesis : n > .. therek are\}\ L in Fcap open parenthesisL ˆ{\bot E to} the= power\{ of prime0 closing\} $ parenthesis .. and rho greater 0 with ◦ ◦ \noindentU to the powerfor of $ circ k $subset sufficiently rho A to the power large ofU circ ,⊂ but plusρA + sinceLL. period $ L \ in F ( E ˆ{\prime } ) $ this cannot be true . P r o oP f r period o o f . .. If If AA isis almost almost bounded bounded comma , we may we assume may assume by the by theorem the theorem that it that i s itclosed i s . Hence from R eclosed m a period r k . ..\ Hencequad fromThe absolutely convex hull of an almost bounded subset need not b e almost bounded . Here i s an extreme example . Let $ ( e { n } ) $ b e the canonical basis A intersection-line L to the power of bottomAintersection subset rho U− lineL⊥ ⊂ ρU o fwe get$ \ bye l takingl { 1 polars}$ and and using Lemma 1 period 1 U towe the get power by taking of circ polars subset and open using parenthesis Lemma 1 . rho 1 A to the power of circ plus L closing parenthesis = rho A to the power of circ plus L \ [ \ begin { a l i g n e d }\ infty \\ period ◦ ◦ ◦ AIf on = the other\bigcup hand commasp we\{ start withe { theUn above}\}⊂ (ρA inclusion+ L.) =\\ commaρA + L. taking polars in E we n = 1 \end{ a l i g n e d }\ ] haveIf on the other hand , we start with the above inclusion , taking polars in E we have A cap L subset A to the power of circ circ cap L subset rho U to the power of circ circ = rho U period 1 period 8 period .. Proposition period .. TheA ∩ imageL ⊂ A under◦◦ ∩ L a⊂ continuousρU ◦◦ = ρU. operator of an absolutely \noindentconvex almostFor$e=( bounded subset is almost 1 bounded , 1 period , .. . The .sum of . two ) absolutely\ in convex\ e l l {\ infty } , $ we have $ A \capalmoste1 bounded . ˆ{ 8 − . subsetsProposition1 } is( also 0 almost . ) boundedThe = image\{ period under0 a\} continuous$ and operator so $A$ of an absolutely i s almost convex bounded almost . Howeverbounded $ subset , is\Gamma almost bounded( A . The ) sum = of\ twophi absolutely$ i s convex not almost almost bounded bounded subsets , b is ecause also almost it i s dense in $ \ e l l { 1 } . $ bounded .
We now give a dual characterization of absolutely convex almost \quad bounded s u b s e t s .
1 . 7 . Corollary . \quad An absolutely convex subset $ A $ of a lcs $ E $ is almost bounded if and only if for every $U \ in U ( E ) $ \quad the re are $ L \ in F ( E ˆ{\prime } ) $ \quad and $ \rho > 0 $ with
\ [ U ˆ{\ circ }\subset \rho A ˆ{\ circ } + L . \ ]
P r o o f . \quad If $ A $ is almost bounded , we may assume by the theorem that it i s c l o s e d . \quad Hence from
\ [ A intersection −l i n e L ˆ{\bot }\subset \rho U \ ]
\noindent we get by taking polars and using Lemma 1 . 1
\ [ U ˆ{\ circ }\subset ( \rho A ˆ{\ circ } + L ) = \rho A ˆ{\ circ } + L . \ ]
\noindent If on the other hand , we start with the above inclusion , taking polars in $ E $ we have
\ [A \cap L \subset A ˆ{\ circ \ circ }\cap L \subset \rho U ˆ{\ circ \ circ } = \rho U. \ ]
1 . 8 . \quad Proposition . \quad The image under a continuous operator of an absolutely convex almost bounded subset is almost bounded . \quad The sum of two absolutely convex almost bounded subsets is also almost bounded . Subspaces and quotient spaces of locally convex spaces .. 1 1 \ hspaceP r o o∗{\ f periodf i l l } ..Subspaces Let T : E right and arrow quotient F b e continuous spaces and of Alocally subset E convex absolutely spaces convex\ andquad 1 1 almost bounded period .. Given V in U open parenthesis F closing parenthesis we find U in U open parenthesis E closing parenthesis so that PT open r o parenthesis o f . \quad U closingLet parenthesis $ T : subset E V\ andrightarrow by F $ b e continuous and $A \subset E $ absolutely convex and almostCorollary bounded 1 period . 7 we\quad determineGiven L in $ F V open\ parenthesisin USubspaces E ( to the and F power quotient )$ of spacesprime wefind ofclosing locally $U parenthesis convex\ spacesin commaU 1 1rho ( greater E 0 )$ with sothat $ T (P r U o o f ) . Let\subsetT : E →VF $b ande continuous by and A ⊂ E absolutely convex and almost bounded . U to the power of circ subset rho A to the power of circ plus L period 0 CorollarySinceGiven T toV the 1∈ U. power( 7F ) we we of finddetermineprimeU open∈ U(E parenthesis) $ so L that \T Vin(U to) ⊂ theFV powerand ( by of CorollaryE circ ˆ{\ closingprime 1 .parenthesis 7 we} determine), subsetL U\∈rho to F( theE ) power,> ρ > 00 of $circ withcomma we have V towith the power of circ subset rho T to the power of prime minus 1 open parenthesis A to the power of circ closing parenthesis plus T to the \ [ U ˆ{\ circ }\subset \rho A ˆ{\ circ } + L . \ ] power of prime minus 1 open parenthesis L closing parenthesis◦ ◦ period Let L sub 1 = T to the power of prime minus 1 openU parenthesis⊂ ρA + L. 0 closing parenthesis cap T to the power of prime minus 1 open parenthesis L closingSince parenthesisT 0(V ◦) ⊂ andU ◦, Lwe sub have 2 be any algebraic complement of L sub 1 in T to the power of prime minus 1 open parenthesis L closing parenthesis\noindent periodSince $ T ˆ{\prime } ( V ˆ{\ circ } ) \subset U ˆ{\ circ } , $ we have Trivially we have L sub 1 subset T to the powerV ◦ ⊂ ρT of prime0−1(A◦ minus) + T 0− 11( openL). parenthesis A to the power of circ closing parenthesis = T open \ [ V ˆ{\ circ }\subset \rho T ˆ{\prime − 1 } ( A ˆ{\ circ } ) + T ˆ{\prime − 1 } parenthesis A closing0−1 parenthesis0−1 to the power of circ and L sub 2 in F open parenthesis0−1 F to the power of prime closing parenthesis period .. (L).Let L1 = T \ ](0) ∩ T (L) and L2 be any algebraic complement of L1 in T (L). Trivially we have L1 ⊂ Therefore0−1 ◦ ◦ 0 V toT the(A power) = T of(A circ) and subsetL2 delta∈ F(F T) open. Therefore parenthesis A closing parenthesis to the power of circ plus L sub 2
for some delta greater 0 period .. This proves the first◦ statement◦ period .. The second follows by applying \noindentthe first t oLet the map $ L open{ parenthesis1 } = x T comma ˆ{\prime yV closing⊂ δT ( parenthesis−A) +1L}2 mapsto-arrowright( 0 ) \cap x plus yT period ˆ{\prime − 1 } ( L ) $ and $ L { 2 }$ be any algebraic complement of $ L { 1 }$ in $ T ˆ{\prime − 1 } ( L ) . $ Ourfor next some resultδ > i0 s. commaThis in proves a certain the firstsense statement comma the . dual The of second Theorem follows 1 period by applying 4 period the .. It first gives t uso the map Trivially we have $ L { 1 }\subset T ˆ{\prime − 1 } ( A ˆ{\ circ } ) = T ( A ) ˆ{\ circ }$ means(x, y of) 7→ distinguishingx + y. b e-line tween boundedness and almost boundedness period .. In the and $ L { 2 }\ in F ( F ˆ{\prime } ) . $ \quad Therefore proof commaOur next Grothendieck result i s , in quoteright a certain s sense completeness , the dual theorem of Theorem is used 1 . period 4 . It gives us means of distinguishing b 1 periode − line 9 period tween boundednessTheorem period and Let almost E be boundedness a complete lcs . and In the A an proof absolutely , Grothendieck convex comma ’ s completeness .. almost theorem \ [ V ˆ{\ circ }\subset \ delta T ( A ) ˆ{\ circ } + L { 2 }\ ] boundedis used subset . period .. Then there is a subspace F of E an index set J with E simeq K to the power of J oplus F such that the image of A in1 . F 9 is . boundedTheorem period . Let E be a complete lcs and A an absolutely convex , almost P rbounded o o f period subset .. We. consider Then there a proj is a ection subspace Q : EF toof theE poweran index of prime set J rightwith arrowE ' K EJ to⊕ theF such power that of theprime image with kernel E to the power of \noindent f o r some $ \ delta > 0 . $ \quad This proves the first statement . \quad The second follows by applying prime openof A squarein F bracketis bounded A circ . closing square bracket .. and set thefirsttothemap $( x , y ) \mapsto x + y . $ G = Q open parenthesis E toP the r o power o f . of We prime consider closing a parenthesis proj ection Q period: E ....0 → OnE0 Ewith to the kernel powerE0[A of◦ prime] and we set put the topology gamma to the powerG = ofQ t(E comma0). where we follow the notationOn E0 ofwe open put squarethe topology bracketγt 1, where 0 closing we square follow the bracket notation period of [ 1 0 ] . Our next result i s , in a certain sense , the dual of Theorem 1 . 4 . \quad I t g i v e s us ForFor each each U inU U∈ openU(E), parenthesiswe find ρ > E0 closing and L ∈ parenthesis F(E0) so thatcomma we find rho greater 0 and L in F open parenthesis E to the power of prime closingmeans parenthesis of distinguishing so that b $ e−line $ tween boundedness and almost boundedness . \quad In the proof , Grothendieck ’ s completeness theorem is used . U to the power of circ subset rho A to the power ofU circ◦ ⊂ plusρA◦ L+ L open parenthesis Corollary 1 period 7 closing parenthesis period .. By Corollary 1 period 3 we obtain \ hspaceU to( Corollary the∗{\ powerf i l l 1} of .1 7circ . ) . 9 subset . By Theorem Corollary delta A to . 1 the . Let 3 powerwe obtain$ E of $circ be plus a Capital complete Gamma lcs open and brace $ wA sub $ 1 an comma absolutely period period convex period , comma\quad walmost sub n closing brace ◦ ◦ \noindentfor some deltabounded greater subset 0 and w . sub\quad i in GThen periodU ⊂ there ..δA Therefore+ is Γ{w a1, subspace ..., wn} $F$ of $E$ an index set $ J $ with $E \simeq K ˆ{ J }\oplus F $ Q openfor some parenthesisδ > 0 and U towi the∈ G. powerTherefore of circ closing parenthesis subset Capital Gamma open brace w sub 1 comma period period period comma wsuch sub n thatclosing the brace image period of $A$ in $ F $ is bounded . Suppose open parenthesis u sub alpha closing parenthesisQ(U ◦) ⊂ Γ{ iw s a, ..., net w in}. U to the power of circ which converges weakly t o u in U to the power \ hspace ∗{\ f i l l }P r o o f . \quad Weconsider a1 projn ection $Q : Eˆ{\prime }\rightarrow E ˆ{\prime }$ of circ period .. Let u = delta a plus◦ w ◦ ◦ withwhereSuppose kernel a in A (u to $α) Ethe i s ˆ a power{\ netprime in ofU circwhich} comma[A converges w in Capital\ circ weakly Gamma t o]u $ ∈ openU\quad. braceLetandu w= sub s eδa t 1+ commaw where perioda ∈ periodA , w period∈ comma w sub n closing Γ{w1, ..., wn}. Hence Qu = w. Now (Quα) i s a net in a finite - dimensional bounded subset . If it has brace period .. Hence Qu = w period ..0 Now open parenthesis Qu sub alpha closing parenthesis i s a net in a \noindenta subnet$ which G converges = Q t ( o w E∈ ˆ{\Γ{wprime1, ..., wn}}, then) (u .α $− Qu\ αh) f hasi l l aOn subnet $ E which ˆ{\ convergesprime } t$ o some we put the topology $ \gamma ˆ{ t } finite0 hyphen0 dimensional◦ bounded subset0 0 period .. If it has a subnet which converges0 t o w to the power of prime in , $Capitalδa where, a Gamma∈ weA . follow openSince braceu the w= subnotationδa 1 comma+ w of= period [δa 1 period+ 0w, ] . periodwe have commaQu w sub= nw closing= bracew. commaHence then (Qu openα) parenthesis u sub alpha converges weakly to w = Qu. This implies that Q i s γt− continuous and therefore γ− continuous ( [ 1 0 ] ; minus Qu sub alpha closing parenthesis has a subnet which converges t o some delta a to0 the0 power of prime comma a to the power of prime in \noindentA to9 .the 3 . power 4For ) . ofeach Since circ periodE $i U s complete .. Since\ in u , by =U delta the ( Grothendieck a to E the power) completion ,$wefind of prime theoremplus w to ( $E the\[γrho power]) = E of>and prime therefore0 $= delta and there a plus $ L w comma\ in ..F( we have .. E ˆ{\existsprime a σ }− σ )continuous $ so that proj ection P : Qu = w to the power of prime =0 w period .. Hence .. open parenthesis QuJ sub alpha closing parenthesis convergesE → E weaklywhose toadj w oint = QuP period= Q. ..We This next implies want that to show Q iP s gamma(E) ' K to. the power of t hyphen continuous and therefore \ [ U ˆ{\ circ }\subset \rho A ˆ{\ circ } + L \ ] gamma hyphen continuous open parenthesis open squareIfu ∈ bracketU ◦, then 1 0 closing square bracket semicolon 9 period 3 period 4 closing parenthesis period .. Since E i s complete comma by the Grothendieck completion theorem open parenthesis E to the power of prime open squaren bracket gamma closing square bracket closing parenthesis to the power of \noindent ( Corollary 1 . 7 ) . \quad By CorollaryX 1 . 3 we obtain prime = E and therefore there exists a sigma hyphen sigmaQu = continuousξjwj proj ection P : E right arrow E whose adj oint P to the power of prime = Q period .. We next want to show P open parenthesis E closing parenthesis simeq K\ [ to U the ˆ{\ powercirc of J}\ periodsubset \ delta A ˆ{\ circ j}= 1+ \Gamma \{ w { 1 } , . . . , w { n } \}\Line] 1 If u in U to the power of circ comma then Line 2 n Line 3 Qu = sum xi sub j w sub j Line 4 j = 1
\noindent f o r some $ \ delta > 0 $ and $ w { i }\ in G . $ \quad Therefore
\ [ Q ( U ˆ{\ circ } ) \subset \Gamma \{ w { 1 } , . . . , w { n }\} . \ ]
\noindent Suppose $ ( u {\alpha } )$ isanetin $Uˆ{\ circ }$ which converges weakly t o $ u \ in U ˆ{\ circ } . $ \quad Let $ u = \ delta a + w $ where $ a \ in A ˆ{\ circ } , w \ in \Gamma \{ w { 1 } , . . . , w { n }\} . $ \quad Hence $Qu = w .$ \quad Now $ ( Qu {\alpha } )$ isanetina f i n i t e − dimensional bounded subset . \quad If it has a subnet which converges t o $ w ˆ{\prime }\ in $ $ \Gamma \{ w { 1 } , . . . , w { n }\} , $ then $ ( u {\alpha } − Qu {\alpha } ) $ has a subnet which converges t o some $ \ delta a ˆ{\prime } , a ˆ{\prime }\ in $ $ A ˆ{\ circ } . $ \quad Since $ u = \ delta a ˆ{\prime } + w ˆ{\prime } = \ delta a + w , $ \quad we have \quad $ Qu = w ˆ{\prime } = w . $ \quad Hence \quad $ ( Qu {\alpha } ) $ converges weakly to $w = Qu . $ \quad This implies that $Q$ i s $ \gamma ˆ{ t } − $ continuous and therefore $ \gamma − $ continuous ( [ 10 ] ; 9 . 3 . 4) . \quad Since $ E $ i s complete , by the Grothendieck completion theorem $ ( Eˆ{\prime } [ \gamma ] ) ˆ{\prime } = E $ and therefore there exists a $ \sigma − \sigma $ continuous proj ection $ P : $
\noindent $ E \rightarrow E$ whose adj oint $Pˆ{\prime } = Q . $ \quad We next want to show $ P ( E ) \simeq K ˆ{ J } . $
\ [ \ begin { a l i g n e d } I f u \ in U ˆ{\ circ } , then \\ n \\ Qu = \sum \ xi { j } w { j }\\ j = 1 \end{ a l i g n e d }\ ] 1 2 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentwhere open1 brace 2 \quad w subS 1 comma $ . period\ddot period{O} $ period nal comma andT w sub. Terzio n closing $ brace\check i s as{g above} $ and lu sum bar xi sub j bar less or equal 1 period .. So we get \noindentq U openwhere parenthesis $ \{ Px closingw { parenthesis1 } , less . or equal . . 1 tothe , power w { ofn maximum}\} $ less i or s equal as above j less orand equal $ n\sum bar w sub\mid j open\ xi { j } \mid 1 2\ leq S .O¨ nal1 and . T $ . Terzio\quadgˇ luSo we get parenthesis Px closing parenthesis bar P wherewhere q U{w is1 the, ..., gauge wn} i ofs as U above in U open and parenthesis| ξj |≤ 1. E closingSo we get parenthesis period .. This means that the topology on the complete \ [ q U ( Px ) \ leq 1 ˆ{\max }\ leq j \ leq n \mid w { j } ( Px ) \mid \ ] lcs P open parenthesis E closing parenthesis is equivalentmax t o the weak t opology and so P open parenthesis E closing parenthesis simeq K to the power of J for some set J period qU(P x) ≤ 1 ≤ j ≤ n | wj(P x) | Wewhere put FqU = Pis to the the gauge power of U of∈ minus U(E) 1. openThis parenthesis means that 0 closingthe topology parenthesis on the period complete .. From lcs P the(E) inclusion is equivalent U to the power of circ subset \noindent where $q U$ is the gauge of $U \ in U ( E ) . $ \quad This means that the topology on the complete delta At circ o the plus weak Capital t opology Gamma and open so braceP (E) w' subKJ 1 commafor some period set J. periodWe put periodF = commaP −1(0) w. subFrom n closing the brace inclusion we obtain lcs $P◦ ( E )$ is equivalent totheweakt opologyandso $P ( E ) \simeq K ˆ{ J }$ openU parenthesis⊂ δA ◦ +Γ{ Iw minus1, ..., w Qn} closingwe obtain parenthesis U to the power of circ subset delta A to the power of circ forand some hence set open $J parenthesis . $ I minus P closing parenthesis open parenthesis A closing parenthesis subset delta U period .... This means that We put $ F = P ˆ{ − 1 } ( 0 ) . $ \quad From the inclusion $ U ˆ{\ circ }\subset \ delta open parenthesis I minus P closing parenthesis open parenthesis(I − Q)U ◦ ⊂ AδA closing◦ parenthesis i s a bounded subset of A F\ periodcirc .. Since+ P\Gamma open parenthesis\{ Ew closing{ 1 parenthesis} , . simeq . Kto . the , power w of{ Jn comma}\} it follows$ we that obtain P is continuous with respect to and hence (I − P )(A) ⊂ δU. This means that (I − P )(A) i s a bounded subset of the original J \ [(It opologyF. Since of− EP period(QE) ' )K , it U follows ˆ{\ circ that P}\is continuoussubset with\ delta respect toA the ˆ{\ originalcirc t opology}\ ] of E. R e mR a r e k m period a r k ... If If EE isis a complete a complete lcs lcs which which admits admits a continuous a continuous norm norm comma , it follows it follows from Theorem 1 . 9 fromthat Theorem any absolutely .. 1 period convex 9 that almost.. any absolutely bounded convex subset almost of E ..i boundeds either bounded subset .. or of the E i sum s of a bounded set \noindenteitherand bounded a finiteandhence - or dimensional the sum $ of ( subspace a bounded I − . setP)(A) and a finite hyphen dimensional\subset subspace\ perioddelta U . $ \ h f i l l This means that $ ( I − P )Let ( us A now see )$ how isaboundedsubsetof we can differentiate b etween bounded and almost boun d − line ed Let us now see how we can differentiate b etween bounded and almost boun0 d-line ed subsetssubsets period . We.. We keep keep the the notation notation of the of the preceding preceding proof proof . period Since Q ..= SinceP , it Q is = weakly P to the continuous power of and prime therefore comma it is weakly \noindentit s kernel$ sp F (A◦ .) i $ s σ(\Equad0,E)−Sinceclosed . If$A Pi s not ( bounded E ) , \simeq K ˆ{ J } , $ it follows that $ P $ is continuous with respect to the original continuous and therefore0 it s kernel sp open parenthesis A to the power of circ closing parenthesis i s sigma open parenthesis E to the power t opologywe find u of∈ E $Ewhich does . $ not b elong to sp (A◦). Now we determine x ∈ E such that u(x) 6= 0 but ea(x) = 0 of prime comma E closing◦ parenthesis hyphen◦◦ closed period If A i s not bounded comma wefor find every u in Eea ∈ toA the. powerThis of means primeλx which∈ A does= A notfor b every elongλ to∈ spK. openSo parenthesis we have proved A circ the closing following parenthesis result . period .. Now we determine x Rin E e such m a1 r . k 10 . . \Propositionquad If $ . E $An absolutely is a complete convex almost lcs which bounded admits subset A a of continuous a com - plete norm lcs is , either it follows fromthatbounded Theorem u open or parenthesis\Aquadcontains1 x . closinga non9 that - parenthesis trivial\quad subspace negationslash-equalany . absolutely 0 convex but tildewide-a almost open\quad parenthesisbounded x closing subset parenthesis\quad o f= 0 $ for E every $ i s a-tildewideeitherSuppose bounded in A toV theis or power an the almost of sum circ bounded periodof a boundedneighborhood .. This means set of lambda and a complete a x infinite A lcs toE the−and powerdimensional sup of- pose circU circ∈ subspace = U( AE) for is every such . that lambdaqU(·) in i Ks a period norm on.. SoE. weWe have choose provedW the∈U following(E) with resultW period⊂ V ∩ U. Then ,W i s an absolutely convex \ hspace1 periodclosed∗{\ 10 almostf periodi l l } Let bounded Proposition us neighnow period see - borhood how An absolutely we . Furthercan differentiateconvex,W contains almost bounded no bnon etween - subset trivial Abounded subspace of a com . and hyphen Therefore almost from boun $ d−l i n e $ ed pleteProposi lcs is either - bounded or A contains a non hyphen trivial subspace period \noindentSupposet ion 1 Vs . is u 10 b an s we e almost t sget . \ boundedquad We neighborhood keep the of notation a complete of lcs theE and preceding sup hyphen proof . \quad Since $ Q = P ˆ{\prime } , $pose it U1 in is . U 1 weakly open1 . parenthesisCorollary E closing . If parenthesis a complete is lcs suchE thathas q an U almost open parenthesis bounded neighborhood times closing then parenthesis either iE s a norm on E period .. We choosecontinuousadmits W in U no open and continuous parenthesis therefore norm E or closing itE sis parenthesiskernel a Banach sp space with $ . ( A ˆ{\ circ } ) $ i s $ \sigma ( E ˆ{\prime } , E)W subsetA− continuous V$ cap closed U period operator . .. If ThenT : $A$E comma→ F i i W s ssaid i s not an t o absolutely bounded be almost convex bounded , closedif there almost i s bounded a neighborhood neigh hyphenU in E such borhoodthat T period(U) i s .. an Further almost commabounded W subset contains of F. no nonUsing hyphen the properties trivial subspace of almost period bounded .. Therefore subsets we from have Proposi already hyphen \noindentt ionestablished 1 periodwe 10 f , i n we d we get can $ u easily\ in prove theE ˆ following{\prime result}$ . which does not b elong to sp $ ( A \ circ ) . $ \quad Now we determine $ x1 period\ in1 . 1 1 1 2 periodE . Proposition $ ..such Corollary . periodThe ..almost If a complete bounded operatorslcs E has anform almost an operator bounded ideal neighborhood on the category of locally thatthenconvex either $ u spaces E admits ( . x no continuous ) \not norm= 0or$ E is but a Banach $ \ spacewidetilde period{a} ( x ) = 0$ forevery $ \ widetilde {a} \ in A ˆ{\ circ } . $ \quadEveryThis bounded means operator $ \lambda is of coursex almost\ in boundedA ˆ .{\ Oncirc the other\ circ hand} , = A$ for every A continuous operatorJ T : E right arrow F i s said t o be almost bounded if there i s a $neighborhood\everylambdaT ∈ L U(\E,K inin E such)K i s almostthat . T $ boundedopen\quad parenthesis .So In fact we U , have if closingT : E proved→ parenthesisF is almostthe i sfollowing boundedan almost and bounded resultF i s complete subset . of , F period then .. Using by Theorem 1 . 9 there i s a continuous proj ection P : the properties of almost boundedJ subsets we have already established comma .. we can 1 .easily 10F → prove .F Propositionsuch the thatfollowingP (F ) . result' AnK periodabsolutelyand (I − P )T convexis a bounded almost operator bounded . subset A of a com − Therefore plete1 period(I − lcsP 1)T 2 is periodfactors either Proposition over boundeda Banach period space or .. The. $ A almost $ contains bounded Conversely operators a non ,− formtrivial an operator subspace any idealoperator on . which factors the category of locally convex spaces period SupposeEvery bounded $ V $ operator is an is almost of course bounded almost bounded neighborhood period .. On of the a other complete hand comma lcs $ E $ and sup − poseevery T $ in U L open\ in parenthesisU ( E comma E )$ K to the issuchthat power of J closing $q parenthesis U ( i s almost\cdot bounded)$ period isanormon In fact comma if T$E : E right .$ arrow\quad We choose F$ isW almost\ in boundedU ( E ) $ with $and W F i\ ssubset complete commaV \ ..cap then byU Theorem . $ ..\ 1quad periodThen 9 there $ i s , a continuous W$ i projs an ection absolutely P : convex closed almost bounded neigh − borhoodF right arrow . \quad F suchFurther that P open $ , parenthesis W$ contains F closing parenthesis no non − simeqtrivial K to thesubspace power of . J\ andquad openTherefore parenthesis from I minus Proposi P closing− parenthesis T is a bounded operator period .... Therefore \noindentopen parenthesist ion I 1 minus . 10 P closingwe get parenthesis T factors over a Banach space period .... Conversely comma .... any operator which factors 1 . 1 1 . \quad C o r o l l a r y . \quad If a complete lcs $ E $ has an almost bounded neighborhood then either $ E $ admits no continuous norm or $ E $ is a Banach space .
A continuous operator $T : E \rightarrow F $ i s said t o be almost bounded if there i s a neighborhood $U$ in $E$ such that $T ( U ) $ i s an almost bounded subset of $F . $ \quad Using the properties of almost bounded subsets we have already established , \quad we can easily prove the following result .
1 . 1 2 . Proposition . \quad The almost bounded operators form an operator ideal on the category of locally convex spaces .
\ hspace ∗{\ f i l l }Every bounded operator is of course almost bounded . \quad On the other hand ,
\noindent every $ T \ in L ( E , K ˆ{ J } ) $ i s almost bounded . In fact , if $T : E \rightarrow F $ is almost bounded and $F$ i s complete , \quad then by Theorem \quad 1 . 9 there i s a continuous proj ection $P : $
\noindent $ F \rightarrow F$ suchthat $P ( F ) \simeq K ˆ{ J }$ and $ ( I − P ) T$ is a bounded operator . \ h f i l l Therefore
\noindent $ ( I − P ) T$ factors over a Banach space . \ h f i l l Conversely , \ h f i l l any operator which factors Subspaces and quotient spaces of locally convex spaces .. 1 3 \ hspaceover K∗{\ to thef i l power l } Subspaces of J times and Banach quotient i s almost spaces bounded of by Propositionlocally convex 1 period spaces 1 2 period\quad .. We1 collect 3 these in the following \noindent1 period 13over period $ Proposition K ˆ{ J }\ periodtimes .. A ..$ continuous Banach operator i s almost from a boundedlcs into a complete by Proposition lcs is 1 . 1 2 . \quad We collect these in the following Subspaces and quotient spaces of locally convex spaces 1 3 almost boundedJ if and only if it factors over a space of the form K to the power of J times Banach period In openover K square× Banach bracketi 2 s closing almost square bounded bracket by Proposition .. Bessaga 1 and . 1 2 Pe . l-suppress We collect sub these czy in to the the following power of acute-n ski proved that a Fr acute-e \ hspace ∗{\ f i l l }1 . 131 . . Proposition13 . Proposition . \ .quadAA \ continuousquad continuous operator from operator a lcs into from a complete a lcs lcs intois a complete lcs is chet space which i s not i so hyphen J morphicalmost t obounded K to the if and power only of ifJ timesit factors Banach over has a space a subspace of the form whichK is i× somorphicBanach . t o a nuclear K o-dieresis the \noindent almost boundedIn if [ 2 and ] Bessaga only if and it Pe factorslczyn´ ski proved over that a space a Fre ´ chet of thespace form which i $ s not K ˆ i{ soJ - }\times $ Banach . space period .. InJ open square bracket 1 6 closing square bracket it was proved that if T : E right arrow F i s an unbounded operator and E commamorphic F t are o K Fr acute-e× Banach chethas spaces a subspace where Fwhich is assumed is i somorphic to b e i t somorphic o a nuclear t oK ao ¨ subspacethe space of . In [ 1 6 ] it was \ hspacea spaceproved∗{\ which thatf i l hasl if} InT a: basisE [→ 2 andF ] i\ asquad ancontinuous unboundedBessaga norm operator and comma Pe and then $E,F\ El hasare{ a Frczy subspacee ´ chet ˆ{\ spaces whichacute where{n}}}F is$ assumed ski proved to b e that a Fr $ \acute{e} $ cheti s isomorphici space somorphic which t t o o a a nucleari subspace s not K ofo-dieresis i a so space− the which space has .. a lambda basis and open a continuous parenthesis norm A closing , then parenthesisE has a subspace .. and the which restriction of T to .. lambda open parenthesisi s isomorphic A closing t o a nuclear parenthesis Ko ¨ the space λ(A) and the restriction of T to λ(A) i s an i somorphism ( \noindenti s ancf . i also somorphismmorphic [ 1 5 ] open ) t . o parenthesis Our $Kˆ final{ resultJ cf period}\ in thistimes also section .. openBanach combines square $ bracketand has generalizes 1 a 5 closingsubspace both square of these which bracket theorems is closing i somorphic . parenthesis t period o a ..nuclear Our final K result$ \ddot in this{o} section$ the combines1 and . 14 . Proposition . Let T : E → F be continuous and not almost bounded . spacegeneralizesSuppose . \quad bothE is ofIn a these Fr [e 16´ theoremschet ] space it period wasproved . Then that there if is $T a nuclear : K Eo¨ the\ spacerightarrowλ(A) F $which i s an unbounded operator and $E1 periodis , 14isomorphic period F$ Proposition areFr to a $ period subspace\acute .. Let{e} ofT :$ E right chetsuch arrow spaces that F the .. be where continuous restriction $ F andof $ T not isto almost assumedλ(A) boundedis to b periodan e is i somorphic t o a subspace of aSuppose spaceomorphism Ewhich is a .Fr has acute-e a basis chet space and period a continuous .... Then there norm is a nuclear , then K o-dieresis $ E $ the has space a subspace lambda open which parenthesis A closing parenthesis ....i which s isomorphicP r o o f . t We o a follow nuclear quite closely K $ the\ddot proof{o} of$ the themain space theorem\quad in [ 1 6$ ] .\lambda Using the continuity( A of ) $ \quad and the restriction of $ Tis $..T isomorphicand to \ thequad assumption .. to$ ..\ alambda subspace that T is ..( not of E almost A .. such ) bounded that $ the alt .. restriction ernately , of we T find to ..V lambdan ∈ U(F open),Un parenthesis∈ U(E) with A closing the parenthesis .. is .. an iis s omorphismfollowing an i somorphism properties period : ( cf . also \quad [ 1 5 ] ) . \quad Our final result in this section combines and generalizes both of these theorems . P r o o f period .. We follow quite closely(i) theU proof⊃ U of⊃ the..., V main⊃ V theorem⊃ V ⊃ in... open square bracket 1 6 closing square bracket period .. Using the continuity of T and the assumption that1 T is not2 almost1 bounded2 3 alt ernately comma \ hspacewe find∗{\ V subf i l ln} in1 U . open 14 .parenthesis Proposition F closing . \ parenthesisquad Let(ii) commaT $(U Tk U) ⊂ sub :Vk, n Ein U open\rightarrow parenthesis EF closing $ \ parenthesisquad be continuouswith the following and not almost bounded . properties : ⊥ 0 ( iii )T (Uk) ∩ L i s not absorbed by Vk+1 for any L ∈ F(F ), \noindentLine 1 openSuppose parenthesis $E$ i closing is parenthesis aFr $ U\ subacute 1 supset{e} $ U sub chet 2 supset space period . \ periodh f i l l periodThen comma there V is sub a 1 nuclear supset V sub K 2 $ supset\ddot V{o} $ ( iv )(Uk) i s a base of neighborhoods for E. subthe 3 space supset period $ \lambda period period( Line A 2 open ) $ parenthesis\ h f i l l iiwhich closing parenthesis T open parenthesis U sub k closing parenthesis subset V sub k Let | · | k and k · k k b e the gauges of Uk and Vk respectively . As in [ 1 6 ] and Theorem 1 . 6 comma we construct a sequence (xn) in E such that \noindentopen parenthesisi s \quad iii closingisomorphic parenthesis\quad T opento parenthesis\quad a U subspace sub k closing\quad parenthesiso f $ cap E L $ to the\quad powersuch of bottom that i the s not\ absorbedquad restriction by V of T to \quad $ \lambda ( A ) $ \quad i s \quad an sub k plus 1 for any L in F open parenthesisn F to the power of prime closing parenthesis comma 2 | xn | k
\noindent and $ ( Tx { n } ) $ i s a $ \ parallel \cdot \ parallel k − $ basic sequence for each $ k . $ \quad We s e t $ n { a }ˆ{ k } = \mid x { n }\mid k $ an $ l i n e −d $
\ [ A = \{ ( n { a }ˆ{ k } ):k=1,2,... \} . \ ]
\noindent $ \lambda ( A )$ isanuclearK $ \ddot{o} $ the space and the restriction of $ T $ t o sp $ ( x { n } ) \simeq \lambda ( A ) $ i s an i somorphism .
\ centerline {2 . \quad Eidelheit ’ s theorem }
In \quad [ 8 ] \quad Eidelheit proved that \quad any proper Fr $ \acute{e} $ chet \quad space $E$ has $ \omega \simeq K ˆ{ N }$ \quad as a quotient space . \quad I f $ Q : E \rightarrow \omega $ i s a quotient map , since $ \omega $ i s a nuclear space , for every bounded subset $ B $ \quad o f $ \omega $ there is even a precompact \quad subset \quad $ D $ \quad o f $ E $ with $ B \subset Q ( D ) . $ \quad This means that \quad $ \omega ˆ{\prime }{ b }\simeq \phi $ i s \quad isomorphic via $ Q ˆ{\prime }$ \quad to \quad a subspace of
\noindent $ E ˆ{\prime } [ \tau ] , $ where $ \tau $ is any topology b etween the t opology of uniform convergence on 14 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentprecompact14 subsets\quad ofS E and $ beta. open\ddot parenthesis{O} $ nal E to andT the power . Terzioof prime comma $ \check E closing{g} parenthesis$ lu period .. A lcs F is said t o be a faithful quotient \noindentof E if thereprecompact i s a continuous subsets operator of Q : $ E E right $ arrow and F $ with\beta the following( E properties ˆ{\prime : } , E ) . $ \quad A l c s $ F $ is said14 t S o.O¨ benal a and faithful T . Terzio quotientgˇ lu open parenthesis i closing parenthesis0 Q open parenthesis E closing parenthesis = F and Q is open comma ofopenprecompact $E$ parenthesis if subsets iithere closing of iE parenthesisand s aβ continuous(E ,E for). eachA lcsbounded operatorF is said B subset t o $Q beF a faithful there : is quotient Ea bounded\rightarrowof subsetE if there D subset i sF a continuous $E with with B subset the following Q open parenthesis properties : D closingoperator parenthesisQ : E period→ F with the following properties : \ centerlineIn this case{ we(i call Q $) a faithful Q surjection ( E and( i ) )Q of(E course=) = F F$ Qand to theQ andis power open $Q$ ,of prime isopen, i s an isomorphism} of F sub b to the power of prime ( ii ) for each bounded B ⊂ F there is a bounded subset D ⊂ E with B ⊂ Q(D). onto a subspace of E sub b to the power of period to the power0 of prime .... One could say0 that a surj ection which lift s bounded set s i s \ hspaceIn this∗{\ casef i l l we}( call iiQ )a forfaithful each surjection boundedand of $ course B \Qsubseti s an isomorphismF $ there of Fb is a bounded subset $ D \subset E $ a faithful surj ection period0 For example comma a surj ection of a Fr acute-e chet space onto a Fr e-acute chet endash onto a subspace of E . One could say that a surj ection which lift s bounded set s i s withMontel $ B .. space\subset .. is ..b faithfulQ ( period D .. Another ) . $result .. due t o .. Fakhoury .. open parenthesis open square bracket 5 closing square bracketa semicolon faithful surj .. Prop ection period . For .. example VI period , a surj3 period ection 5 closing of a Fr parenthesise ´ chet space onto a Fre ´ chet – Montel space is \noindentstatesfaithful thatIn .a surjection this Another case whose result we kernel call due i t s oa $ Fr Q Fakhoury acute-e $ a chetfaithful ( space [ 5 ] ; always surjection Prop lift . s compact VI and . 3 . setsof 5 ) statescourse period that $ a Q surjection ˆ{\prime }$ i s an isomorphism of $ FOn ˆwhose{\ theprime other kernel hand} i s{ acomma Frb e ´}$chet .. one space could always construct lift s compact various examplessets . On theof surj other ections hand which , one could construct various areexamples not faithful of surjperiod ections .. For which example are not comma faithful let F . be For a lcs example whose bounded, let F be subsets a lcs whose are not bounded all subsets are not \noindentfiniteall hyphen finiteonto - dimensional dimensional a subspace period . A of .. theorem A $theorem E of ˆ{\ Dierolf ofprime Dierolf [ 6 ] open} states{ squareb that ˆ{ bracketthere. }}$ i s 6 a\ closing continuoush f i l l squareOne open bracket could map states say of a lcs that thatE, there a surj i s a continuous ection which open lift s bounded set s i s mapwhose of a lcs bounded E comma subsets whose are bounded all finite subsets - dimensional are all finite , onto hyphenF. There dimensional are also comma examples onto ofF period Montel .. – There Ko ¨ the \noindentarespaces also examples whicha faithful fail .. of t o Montel b surj e Schwartz endash ection K ( o-dieresis cf . . For [ 1 example the 0 ] ;spaces 1 1 . 6 which, . a 4 ) surj fail. t Such o ection b e a Schwartz space of has a ..` Fr1 openas $ parenthesis a\ quotientacute{e spacecf} period$ chet .. open space square onto bracket a Fr 1$ 0\ closingacutebut of{ squaree course} $ bracket` chet1 here semicolon−− is not a faithful quotient . Montel1 1 periodOur\quad 6 purpose periodspace 4is closing t\ oquad determine parenthesisi s \ whichquad period locallyf a i .. t h Suchconvex f u l a . spacespaces\quad hasE canAnother l sub have 1 ..ω as resultas a aquotient faithful\quad space quotientdue but . of t course o If\ everyquad l subFakhoury 1 here is not\quad a ( [ 5 ] ; \quad Prop . \quad VI . 3 . 5 ) statesfaithfulcontinuous that quotient a operator period surjectionT : E → whoseω is bounded kernel , then i s of a course Fr $ω \cannotacute b{e } a quotient$ chet space space of E. alwaysIn case lift of s compact sets . OnOur thewebbed purpose other spaces is t hand o we determine shall , \quad see which thisone is locally the could only convex obstruction construct spaces E . can various have omega examples as a of surj ections which are notA sequencefaithful of absolutely. \quad For convex example subsets A ,1 let⊃ A2 $ F⊃ $A3 be⊃ a lcs... of whose a lcs E boundedi s called a subsetscompleting are s not all faithful quotient period .. If every continuous operatorP T : E right arrow omega is bounded comma then of f icourse n iequence t e omega− dimensionalif cannotxn ∈ A bn eimplies a. quotient\quad that space theA theorem series of E periodxn ofis .. convergent Dierolf In case of [webbed in 6E. ]states spacesLet (An we) that be shall a completing see there i sequence s a continuous in open mapthisE. of isIf the aT lcsonly: E → obstruction $F Ei s continuous , period $ whose , then bounded subsets are all finite − dimensional , onto $ F . $ \quad There are( alsoT (An)) examples is a completing\quad sequenceo f Montel in F. −−ForKU ∈ $ U(\Eddot), there{o} i s$ a k thesuchspaces that An which⊂ U failfor all tn o≥ b ek. Schwartz \quad ( c f . \quad [ 1 0 ] ; A sequence of absolutelyT convex subsets A sub 1 supset A sub 2 supset A sub 3 supset period period period of a lcs E i s 1called 1In . particular6 a completing . 4 ), . \ squad equenceAn =Such if x{0 sub a} space nsince in AE sub hasi s n Hausdorff implies $ \ e l thatl and{ the1 series}$ sum\quad x subas n a is convergentquotient space but of course $ \ e l l { 1 }$ E0 = S∞ E0[A◦ ]. For more information on completing sequences we refer to [ 5 ] herein E is period notn=1 .. a Let openn parenthesis A sub n closing parenthesis be a completing sequence in E period If T : E right arrow F i s continuous commafaithfuland then [ 1 quotient 7 ] . . Let W = {C } be an absolutely convex C− web in E([5], [11]). For any open parenthesis T open parenthesis A sub n closingn1...nk parenthesis closing parenthesis is a completing sequence in F period .. For U in U open N parenthesisOur purpose(nk) E∈ closingN is, let t parenthesis oρk determine > 0 comma b e the which there associated i locally s a k sequence such convex that , which spaces we may assume $ E $ t o can b e decreasing have $ without\omega loss$ as a faithfulA subof generality n subset quotient U . for If all we . n\ nowquad greater set If equal every k period continuous .. In particular operator comma intersection $ T : of E A sub\ nrightarrow = open brace 0\ closingomega brace$ ..is since bounded E i s , then of course $ \omega $ cannot b e a quotient space of $E . $ \quad In case of webbed spaces we shall see Hausdorff and C thisE to theis powerthe only of prime obstruction = union of sub . n = 1 to theAk power= ρk n of1...n infinityk E to the power of prime open square bracket A sub n to the power of circ closingwe get square bracket a completing period For moresequence information on ( completingA ), sequenceswhich we we refer will tocall open a completing square bracket sequence 5 closing square bracket A sequence of absolutely convex subsets $ A k { 1 }\supset A { 2 }\supset A { 3 }\supset . andderived open square from the bracket web 1W 7. closing square bracket period Our first lemma i s really a rewording of the classical . .$ ofalcs $E$ is Letopen W = mapping open brace theorem C sub of n Banachsub 1 period . period period n sub k closing brace be an absolutely convex C hyphen web in E open parenthesis called a completing s equence if $ x { n }\ in A { n }$ implies that the series $ \sum x { n }$ open square bracket2 . 15 .closingL e − squareline mma bracket . commaLet (A open) squarebe a completing bracket 1 sequence 1 closing of squareE and bracket let T closing: E → parenthesisF be period .. For any is convergent n opencontinuous parenthesis . n sub Then k closingthe fo llowing parenthesis are equivalent in N to the: power of N comma let .. rho k greater 0 b e the associated sequence comma which in $ E . $ \quad Let $ ( A { n } )$ beacompleting sequence in $E . $ If $T : E we may assume t o b e ( i )T (A parenright − line is a neighborhood for each n. \rightarrow F $ i s continuousn , then decreasing without loss of generality( period ii )T (A ..n) If weis now a neighborhood set for each n. A sub k = rho k to the power of( C iii n ) sub (T 1(A periodn)) periodis a base period of neighborhoods n sub k in F. \noindentwe get .... a$ completing ( T sequence ( A ....{ openn } parenthesis) ) $ A subis a k closing completing parenthesis sequence comma .... in which $F we will . $ call\ aquad completingFor sequence $ U \ in Uderived ( E from )the web ,$ W period thereisa .... Our first $k$ lemma i suchthat s really a rewording of the classical $open A mapping{ n }\ theoremsubset of BanachU $ period f o r a l l $ n \geq k . $ \quad In particular $ , \bigcap A { n } = 2 period\{ 0 1 period\} L$ e-line\quad mmasince period .. $ Let E $ open i parenthesis s Hausdorff A sub and n closing parenthesis .. be a completing sequence of E and let T : E right arrow F .. be \noindentcontinuous period$ E ˆ ..{\ Thenprime the fo} llowing= \ arebigcup equivalentˆ{\ : infty } { n = 1 } E ˆ{\prime } [A { n }ˆ{\ circ } ]open . $ parenthesis For more i closing information parenthesis on T open completing parenthesis sequences A sub n parenright-line we refer .. to is a [ neighborhood 5 ] for each n period open parenthesis ii closing parenthesis T open parenthesis A sub n closing parenthesis .. is a neighborhood for each n period \noindentopen parenthesisand [ iii 1 closing 7 ] . parenthesis open parenthesis T open parenthesis A sub n closing parenthesis closing parenthesis .. is a base of neighborhoods in F period \ hspace ∗{\ f i l l } Let $ W = \{ C { n { 1 } . . . n { k }}\} $ be an absolutely convex $ C − $webin$E ( [ 5 ] , [ 1 1 ] ) .$ \quad For any
\noindent $ ( n { k } ) \ in N ˆ{ N } , $ l e t \quad $ \rho k > 0 $ b e the associated sequence , which we may assume t o b e decreasing without loss of generality . \quad I f we now s e t
\ [A { k } = \rho k ˆ{ C } n { 1 } . . . n { k }\ ]
\noindent we get \ h f i l l a completing sequence \ h f i l l $ ( A { k } ) , $ \ h f i l l which we will call a completing sequence
\noindent derived from the web $W . $ \ h f i l l Our first lemma i s really a rewording of the classical
\noindent open mapping theorem of Banach .
\ hspace ∗{\ f i l l }2 . 1 . L $ e−l i n e $ mma . \quad Let $ ( A { n } ) $ \quad be a completing sequence of $E$ andlet $T : E \rightarrow F $ \quad be
\noindent continuous . \quad Then the fo llowing are equivalent :
\ centerline {( i $ ) T ( A { n } parenright −l i n e $ \quad is a neighborhood for each $ n . $ }
\ centerline {( i i $ ) T ( A { n } ) $ \quad is a neighborhood for each $ n . $ }
\ centerline {( i i i $ ) ( T ( A { n } ) ) $ \quad is a base of neighborhoods in $ F . $ } Subspaces and quotient spaces of locally convex s line-p aces .. 1 5 \ hspaceP r o o∗{\ f periodf i l l } ..Subspaces It .. i s .. necessary and quotient t o .. prove spaces .. open of parenthesis locally iconvex closing parenthesis s $ line double−p $ stroke aces right\quad arrow1 open 5 parenthesis ii closing parenthesis .. only period .. We .. set .. B sub 1 = A sub 1 comma B sub n = \ hspace2 to the∗{\ powerf i l lof}P minus r o n oplus f 1 . A\ subquad n commaI t \quad n greateri s equal\quad 2 periodnecessary .. Then open t o parenthesis\quad prove B sub\ nquad closing( parenthesis i $ ) is\ aRightarrow completing sequence( $ i i and ) \ Bquad sub nonly subset . 2 B\quad sub nWe minus\quad 1 periodSubspacess e t ..\ Itquad and i s quotient$ B spaces{ 1 of} locally= convex A { s 1line} −,Bp aces{ 1 5n } = $ P r o o f . It i s necessary t o prove ( i ) ⇒ ( ii ) only . We set B1 = A1,Bn = enough−n+1 to prove the assertion for T open parenthesis B sub n closing parenthesis period \noindentFix2 .. k ..A andn$, n 2 ..≥ ˆ let{2. −.. VThen ..n b ( eB +..n) a is .. 1 a neighborhood completing} A { n sequence ..} of zero, and .. n inB ..n\geq F ..⊂ with2Bn2− ..1. V . subsetIt $ i s\ Tquadenough openThen parenthesis to prove $ the ( B sub B k{ closingn } parenthesis) $ is a completing sequence and period$ B assertion{ .. Fixn }\ forsubsetT (Bn). 2 B { n − 1 } . $ \quad I t i s enoughy in VFix period to provek .. Chooseand the let xassertion subV k plusb e 1 for in a B neighborhood $T sub k plus ( 1 B .. withof{ zeron y} minus in) TxF . sub $with k plusV 1 in⊂ T openT (Bk parenthesis). Fix B sub k plus 1 closing parenthesisy ∈ periodV. ..Choose By inductionxk+1 find∈ Bk+1 with y − T xk+1 ∈ T (Bk+1). By induction find FixEquation:\quad x$ sub k i $ in B\quad sub i commaand \quad .. i greaterl e t equal\quad k plus$ V 1 comma $ \quad withb m ey minus\quad suma Tx\quad sub ineighborhood in T open parenthesis\quad B subo f mzero plus\ 1quad in \quad closing$ F $ parenthesis\quad with period\quad i = k plus$ V 1 \subset T(B { k } ) . $ \quad Fix i ≥ k + 1, with x ∈ B , $If ywe set\ in V . $ \quad Choose $ x { k + 1 }\ in B { k + 1 }$ i \quadi with $ y − Tx { k +Line 1 }\ 1 infinityin LineT(B 2 x = sum x{ subk i Line + 3 i1 =} k plus) 1 . $ \quadm By induction find X then x in B sub k period .. Also y minus Txy in− T open parenthesisT xi ∈ T (Bm B+1 sub). m closing parenthesis for each m and this implies y = Tx period \ beginExistence{ a l i g of n a∗} completing sequence consisting of unbounded sets i s essential \ tagin∗{ what$ x follows{ i period}\ in .. OurB next{ lemmai } i, s crucial $} i for\ thegeq inductivei =kk + construction+ 1 1 , used with in \\ m \\ y − \sum Tx { i } \ inthe proofT(B of the main{ resultm + of this 1 section} ). period\\ i = k + 1 If we set \end2 period{ a l i g n 2∗} period .. Lemma period .. Let open parenthesis A sub n closing parenthesis .. be .. a completing sequence in E such that each A sub n .. is \noindent I f we s e t ∞ unbounded period .. Then for every n and L in F open parenthesisX E to the power of prime closing parenthesis .. the s et A sub n cap L to the power of bottom .. is unbounded period x = xi \ [ \ begin { a l i g n e d }\ infty \\ P r o o f period .. First note that we may assume withouti = anyk + loss 1 of generality that xE i s = complete\sum periodx ....{ Supposei }\\ A sub k cap L to the power of bottom i s bounded for some L in F open parenthesis E to the power of i = k + 1 \end{ a l i g n e d }\ ] prime closingthen x ∈ parenthesisBk. Also periody − T .... x ∈ ThenT (Bm for) for any each m and this implies y = T x. U in U open parenthesis E closing parenthesisExistence weof a have completing U to the sequence power of consisting circ subset of rho unbounded A sub k sets to the i s power essential of circ plus L forin some what rho follows greater . 0 period Our next .. Therefore lemma i for s crucial some G for in the F open inductive parenthesis construction E to the used power in the of prime proof closingof the main parenthesis we have \noindentE toresult the power ofthen this of section prime $ x . =\ Ein to theB power{ k of} prime. $ open\quad squareAlso bracket $ A y sub k− to theTx power\ in of circT(B closing square{ bracketm } oplus) $ G f period o r each $m$ andthis implies $y = Tx .$ Let Q : E to the power2 . 2 . of primeLemma right . arrowLet E(A ton) thebe power a completingof prime be sequence the corresp in E ondingsuch thatproj each ectionA onton is G period .. There i s a proj 0 ⊥ ection unbounded . Then for every n and L ∈ F(E ) the s et An ∩ L is unbounded . \ hspaceP : E right∗{\ f arrow i l l } Existence E with P to the ofP power a r ocompleting o of f .prime First = Q note sequence open that parenthesis we consisting may assume see the without proof of unbounded of any Theorem loss of 1 sets generality period i 9 s closing that essential parenthesis period open ⊥ 0 parenthesisE i s I complete minus P . closing parenthesis Suppose openA parenthesisk ∩ L i s boundedA sub k closing for some parenthesisL ∈ F(E ) is. bounded Then for any \noindentand P openin parenthesis what follows E closing . \ parenthesisquad Our i next s finite lemma hyphen i dimensional s crucial period for ..the So inductive the t opology construction of P open parenthesis used in E closing parenthesisthe proof is generated of the bymain a norm result of this section . bar times bar period .. We find V in U open parenthesisU ∈ U( EE)wehave closing parenthesis such that bar Px bar less or equal rho q V open parenthesis x \ hspace ∗{\ f i l l }2 . 2 . \quad Lemma . \quad◦ Let◦ $ ( A { n } ) $ \quad be \quad a completing sequence in closing parenthesis and m greater equal k so that A subU m subset⊂ ρAk V+ L period $E$Then P such open that parenthesis each A sub$A m closing{ n }$ parenthesis\quad subseti s P open parenthesis V closing parenthesis which means that P open parenthesis 0 A sub mfor closing some ρ parenthesis > 0. Therefore is bounded for some periodG ∈ .. F Since(E ) A we sub have m subset A sub k comma we \noindentconclude thatunbounded A sub m i . s a\quad boundedThen subset for of everyE period $ n $ and $L \ in F ( E ˆ{\prime } ) $ \quad the s et $ A { n }\cap L ˆ{\bot }$ \quad is0 unbounded0 ◦ . We are now ready for the main result of this sectionE = periodE [Ak] ⊕ G. 2 periodLet Q 3: E period0 → E Theorem0 be the corresp period onding.. Let open proj ectionparenthesis onto G. A subThere n closing i s a parenthesis proj ection P ..: beE a→ completingE with P 0 s= equenceQ( in E .. and let T : E right\ hspace arrow∗{\ F f i l l }P r o o f . \quad First note that we may assume without any loss of generality that see the proof of Theorem 1.9). (I − P )(Ak) is bounded and P (E) i s finite - dimensional . So the t opology be suchof P ( thatE) is T generated open parenthesis by a norm A subk · n kclosing. We parenthesis find V ∈ U ..(E is) unbounded such that k forP x eachk≤ nρqV period(x) and .. Thenm ≥ therek so thatis a faithful surjection \noindentQ : F right arrow$E$ omega i s so complete that QT : E . right\ h f arrowi l l Suppose omega is also $ A a faithful{ k }\ surjectioncap periodL ˆ{\bot }$ i s bounded for some $ L \ in AFm ⊂ V. (Then EP ˆ{\(Amprime) ⊂ P (V}) which) means . $ that\ h fP i( lA l mThen) is bounded f o r any . Since Am ⊂ Ak, we conclude that Am P ri o s o a f bounded period .. subset Without of lossE. of generality we assume A sub n subset 2 to the power of 1 A sub n minus 1 period .. Since T open parenthesis A sub 1 closing parenthesis i s We are now ready for the main result of this section . \ beginunbounded{ a l i g n we∗} find v sub 1 in F to the power of prime with 2 . 3 . Theorem . Let (An) be a completing s equence in E and let T : E → F be such that U supremum\ in openU brace (E bar v ) sub 1 we open parenthesishave \\ U Tx ˆ{\ closingcirc parenthesis}\subset bar : x in\ Arho sub 1A closing ˆ{\ bracecirc =} infinity{ k } period+ L \end{Ta( lA i gn n) ∗}is unbounded for each n. Then there is a faithful surjection Q : F → ω so that QT : E → ω is also a faithful surjection . 1 P r o o f . Without loss of generality we assume An ⊂ 2 An−1. Since T (A1) i s \noindent f o r some $ \rho0 > 0 . $ \quad Therefore for some $ G \ in F ( E ˆ{\prime } ) $ we haveunbounded we find v1 ∈ F with
\ [ E ˆ{\prime } = E ˆ{\prime }sup{|[v1(T A x) ˆ{\|: x ∈circA1} =} ∞{ . k } ] \oplus G. \ ]
\noindent Let $ Q : E ˆ{\prime }\rightarrow E ˆ{\prime }$ be the corresp onding proj ection onto $ G . $ \quad There i s a proj ection $ P : E \rightarrow E $ with $ P ˆ{\prime } = Q ($ see the proof of Theorem $1 . 9 ) . ( I − P)(A { k } ) $ is bounded and $P ( E )$ isfinite − dimensional . \quad So the t opology of $P ( E ) $ is generated by anorm $ \ parallel \cdot \ parallel . $ \quad We f i n d $ V \ in U ( E )$ suchthat $ \ parallel Px \ parallel \ leq \rho q V ( x )$and$m \geq k $ so that $ A { m }\subset V . $ Then $ P ( A { m } ) \subset P ( V )$ whichmeansthat $P ( A { m } ) $ is bounded . \quad Since $ A { m }\subset A { k } , $ we conclude that $ A { m }$ i s abounded subset of $E . $
\ centerline {We are now ready for the main result of this section . }
2 . 3 . Theorem . \quad Let $ ( A { n } ) $ \quad be a completing s equence in $ E $ \quad and l e t $ T : E \rightarrow F $ besuchthat $T ( A { n } ) $ \quad is unbounded for each $ n . $ \quad Then there is a faithful surjection $ Q : F \rightarrow \omega $ sothat $QT : E \rightarrow \omega $ is also a faithful surjection .
\ hspace ∗{\ f i l l }P r o o f . \quad Without loss of generality we assume $ A { n }\subset 2 ˆ{ 1 } A { n − 1 } . $ \quad Since $ T ( A { 1 } ) $ i s
\noindent unbounded we find $ v { 1 }\ in F ˆ{\prime }$ with
\ [ \sup \{\mid v { 1 } ( Tx ) \mid : x \ in A { 1 }\} = \ infty . \ ] 1 6 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentWe determine1 6 x\ subquad 1 inS A sub $ . 1 with\ddot v sub{O 1} open$ parenthesis nal andT Tx . sub Terzio 1 closing $ parenthesis\check{g =} 1$ period lu T open parenthesis A sub 2 closing parenthesis is unbounded period .... We claim we can \noindent We determine $ x { 1 }\ in A { 1 }$ with $ v { 1 } ( Tx { 1 } ) = 1 . T find v sub 2 in¨ F to the power of prime with the following two properties : (Asupremum1 6{ 2 S open}.O nal) brace $ and baris T . Terziounbounded v sub 2gˇ openlu .parenthesis\ h f i l l We Tx closing claim parenthesis we can bar : x in A sub 2 comma v sub 1 open parenthesis Tx closing We determine x1 ∈ A1 with v1(T x1) = 1.T (A2) is unbounded . We claim we can parenthesis = 0 closing0 brace = infinity comma v sub 2 open parenthesis Tx sub 1 closing parenthesis = 0 period \noindentIf thisfind werev2 ∈f i impossible nF dwith $the v comma following{ 2 }\ then two thisin properties wouldF ˆ mean{\ : prime that whenever}$ with v open the parenthesis following Tx twosub 1 properties closing parenthesis : = 0 then v i s bounded on T open parenthesis A sub 2 closing parenthesis cap v sub 1 to the power of minus 1 open parenthesis 0 closing parenthesis comma\ [ \sup v in F\{\ to the powermid ofv primesup{ {|2 periodv2}(T x() ..|: Sox Tx∈ weA would2, ) v1( obtainT\ xmid) = 0} =:∞, x v2(T\ xin1) = 0A. { 2 } , v { 1 } ( Tx ) = 0 \} = \ infty , v { 2 } ( Tx { 1 } ) = 0 . \ ] openIf brace this were Tx sub impossible 1 closing ,brace then this to the would power mean of bottom that whenever subset Fv to(T the x1) power = 0 then of primev i s open bounded square on bracketT (A2) B∩ sub 2 to the power of circ −1 0 closingv square1 (0), bracket v ∈ F where. So B we sub would 2 = obtain T open parenthesis A sub 2 closing parenthesis cap v sub 1 to the power of minus 1 open parenthesis 0 closing parenthesis period Let F to the power of prime = open square bracket Tx sub 1 closing square bracket to the power of bottom oplus G \noindent If this were impossible , then this would mean that whenever $ v ( Tx { 1 } ) = 0 $ for some G in F open parenthesis F to the power of prime closing⊥ parenthesis0 ◦ period We find n greater 2 with thenG subset $ v F $ to the power of prime open square bracket open{T x1 parenthesis} ⊂ F [B2 T] open parenthesis A sub n closing parenthesis cap v sub 1 to the i sboundedon $T ( A { 2 } ) \cap v− ˆ1{ − 1 } { 1 } ( 0 ) , v \ in F ˆ{\prime } power of minus 1 open parenthesis 0 closing parenthesiswhereB2 = closingT (A2) parenthesis∩ v (0).Let to the power of circ closing square bracket period . $ \quad So we would obtain 1 This means comma however comma 0 ⊥ F = [T x1] ⊕ G F to the power of prime = F to the power of prime open square bracket open parenthesis T open parenthesis A sub n closing parenthesis cap\ begin v subfor{ a some1 l toi gn theG∗}∈ power F(F 0 of). We minus find 1n open > 2 parenthesis with 0 closing parenthesis closing parenthesis to the power of circ closing square bracket \{and henceTx { T1 open}\} parenthesisˆ{\bot A sub}\ n closingsubset parenthesisF ˆ{\ capprime v sub} 1 to[ the B power ˆ{\ ofcirc minus} 1{ open2 } parenthesis] \\ where 0 closing B parenthesis{ 2 } i= s T(A { 2 } ) \cap v ˆ{ − 1 0} { 1 } −(1 0◦ ) . Let \\ F ˆ{\prime } = [ Tx { 1 } bounded period .. However comma this contradictsG ⊂ LemmaF [(T (A 2n period) ∩ v 2(0)) period]. ] ˆ{\bot }\oplus G 1 ProceedingThis means in this , however fashion , comma we choose sequences v sub n in F to the power of prime comma x sub n in E with the \endfollowing{ a l i g n properties∗} : open parenthesis i closing parenthesis x sub n0 in A sub0 n comma−1 ◦ F = F [(T (An) ∩ v (0)) ] \noindentopen parenthesisf o r some ii closing $ G parenthesis\ in supF open ( brace F ˆ bar{\ vprime sub1 n open} ) parenthesis .$ Wefind Tx closing parenthesis $n > bar2 : x$ in with A sub n comma v sub −1 i openand parenthesis hence T Tx(An closing) ∩ v1 parenthesis(0) i s bounded = 0. for i However = 1 comma , this period contradicts period Lemma period comma 2 . 2 . n minus 1 closing brace = infinity comma 0 \ [Gopen parenthesis\Proceedingsubset iii inF closing this ˆ{\ fashion parenthesisprime , we} choose v[(T(A sub sequences n open parenthesisvn ∈ TxF{, subn x}n m closing)∈ E\cap parenthesiswith thev following ˆ ={ delta− properties sub1 } n{ comma1 : } m( period 0 ) ) ˆ{\ circ } ].We then\ ] define Q : F right arrow omega simply by Qy = open parenthesis v sub n open parenthesis y closing parenthesis closing parenthesis period .... So QTx = open parenthesis u sub n open parenthesis(i)xn ∈ A xn, closing parenthesis closing parenthesis where
u sub n = T to the power( iiof ) prime sup {| vv subn(T xn) period|: x ∈ An, vi(T x) = 0 for i = 1, ..., n − 1} = ∞, \noindentFor open parenthesisThis means xi sub , however n closing parenthesis , in omega comma .. we find z sub 1 in A sub 1 .. with v sub 1 open parenthesis Tz sub 1 closing parenthesis = xi 1 .. and for i greater equal(iii) 2 commavn(T xm z) sub = δ in,m in. A sub i comma \ [v F sub ˆ{\ j openprime parenthesis} = Tz F sub ˆ{\ i closingprime parenthesis} [(T(A = 0 comma j = 1 comma{ n period} ) period\cap periodv comma ˆ{ − i minus1 } 1{ comma1 } with( n 0 minus ) 1 We then define Q : F → ω simply by Qy = (v (y)). So QT x = (u (x)) where v) sub ˆ{\ n opencirc parenthesis} ] \ ] Tz sub n closing parenthesisn = xi sub n minus sum v sub n open parenthesis Tzn sub i closing parenthesis period i = 1 We set z = sum z sub n in E and get QTz = open parenthesis xi sub n closing parenthesis period .. Hence QT and Q are onto omega period 0 Recall that A sub n subset 2 to the power of 1 A subu n= minusT vn 1. period .. Given xi in omega with bar xi i bar less or equal 1 slash k for i = 1\noindent comma periodandhence period period $T comma ( k for A { n } ) \cap v ˆ{ − 1 } { 1 } ( 0 )$ isbounded. \quad However , this contradicts Lemma 2 . 2 . some k commaFor we (ξ writen) ∈ xi =ω, eta pluswe find muz where1 ∈ muA1 = openwith parenthesisv1(T z1) =0 commaξ1 and 0 comma for i ≥ period2, period zi ∈ periodAi, comma 0 comma xi k plus 1Proceeding comma xi k plus in 2 this comma fashion period period , we period choose closing sequences parenthesis $ period v { ..n Then}\ wein find F ˆ{\prime } , x { n }\ in E $ with the z sub i in A sub i comma i greater equal kv plus(T z 1) comma = 0, j = so 1, that ..., i − if 1, with followingLine 1 infinity properties Line 2 line-z : = sum z sub i Linej 3i i = k plus 1 then QTz = mu period .. We note z in A sub k period .. Since sumn − lambda1 sub i x sub i i s convergent in E for every open parenthesis \ [ ( i ) x { n }\ in A { n } , \ ] X lambda sub i closing parenthesis vn(T zn) = ξn − vn(T zi). which satisfies sum bar lambda sub i bar less or equal 1 we see that the set Line 1 infinity Line 2 L = open brace sum lambda sub i x sub i : sumi = bar 1 lambda sub i bar less or equal 1 closing brace Line 3 i = 1 \ centerline {(P i i ) sup $ \{\mid v { n } ( Tx ) \mid : x \ in A { n } , v { i } ( TxWe set )=0$for$iz = zn ∈ E and get QT z = ( =1ξn). Hence ,QT .and .Q are . onto ,ω. n − 1 \} = \ infty , $ } 1 Recall that An ⊂ 2 An−1. Given ξ ∈ ω with | ξi | ≤ 1/k for i = 1, ..., k for \ [ (some i ik, iwe )write vξ ={η +n µ}where(µ Tx= (0,{0, ...,m 0}, ξk)+ 1, ξk = + 2,\ ...delta). Then{ wen find ,zi ∈ mAi}, i ≥. k \+] 1, so that if ∞ X \noindent Wethen define $Q : F line\−rightarrowz = zi \omega $ simplyby $Qy = ( v { n } ( y ) ) . $ \ h f i l l So $ QTx = ( u { n } ( x ) ) $ where i = k + 1 P P \ beginthen{ a lQT i g n z∗}= µ. We note z ∈ Ak. Since λixi i s convergent in E for every (λi) which satisfies | λi |≤ 1 u {wen see} that= the T set ˆ{\prime } v { n } . \end{ a l i g n ∗} ∞ X X \ hspace ∗{\ f i l l }For $ ( \ xi { n L} = {) λ\ixini : \|omegaλi |≤ 1} , $ \quad we f i n d $ z { 1 }\ in A { 1 }$ \quad with $ v { 1 } ( Tz { 1 } ) = \ xi 1 $ \quad and f o r $ i \geq 2 , z { i }\ in A { i } , $ i = 1
\ begin { a l i g n ∗} v { j } ( Tz { i } )=0,j=1,...,i − 1 , with \\ n − 1 \\ v { n } ( Tz { n } ) = \ xi { n } − \sum v { n } ( Tz { i } ). \\ i = 1 \end{ a l i g n ∗}
\noindent We s e t $ z = \sum z { n }\ in E$ andget $QTz = ( \ xi { n } ) . $ \quad Hence $QT$ and $Q$ are onto $ \omega . $
\ hspace ∗{\ f i l l } Recall that $ A { n }\subset 2 ˆ{ 1 } A { n − 1 } . $ \quad Given $ \ xi \ in \omega $ with $ \mid \ xi i \mid \ leq 1 / k$for$i =1 , . . . , k$ f o r
\noindent some $k ,$ wewrite $ \ xi = \eta + \mu $ where $ \mu = ( 0 , 0 , . . . , 0 , \ xi k + 1 , \ xi k + 2 , . . . ) . $ \quad Then we f i n d $ z { i }\ in A { i } , i \geq k + 1 ,$ sothatif
\ [ \ begin { a l i g n e d }\ infty \\ l i n e −z = \sum z { i }\\ i = k + 1 \end{ a l i g n e d }\ ]
\noindent then $ QTz = \mu . $ \quad We note $ z \ in A { k } . $ \quad Since $ \sum \lambda { i } x { i }$ i s convergent in $E$ for every $ ( \lambda { i } ) $ which satisfies $ \sum \mid \lambda { i }\mid \ leq 1 $ we see that the set
\ [ \ begin { a l i g n e d }\ infty \\ L = \{\sum \lambda { i } x { i } : \sum \mid \lambda { i }\mid \ leq 1 \}\\ i = 1 \end{ a l i g n e d }\ ] Subspaces and quo to the power of line-t ient spaces of loca line-l ly convex spaces .. 1 7 \ hspacei s compact∗{\ f i in l l E} Subspaces open parenthesis and cf period $ quo .... ˆ{ alsol i n .... e − opent }$ square ient bracket spaces 1 7 closing of loca square $ bracket line −l semicolon $ ly ....convex Prop period spaces 5 closing\quad 1 7 parenthesis period .... So eta in QT open parenthesis Capital Gamma open brace x sub 1 comma period period period comma x sub k closing \noindent i scompactin $E ($ cf . \ h f i l l a l s o \ h f i l l [ 1 7 ] ; \ h f i l l Prop . 5 ) . \ h f i l l So $ \eta brace closing parenthesis subset QT open parenthesis L closingline parenthesis−t period \ inThisQT shows that( for\Gamma each k comma\{ QTx openSubspaces{ 1 parenthesis} and,quo . 2 to .theient power . spaces of , minus of xloca k{line Lk plus}\}− l Aly sub convex k) closing spaces\subset parenthesis 1 7 QT i s a neighborhood ( L ) in . $ i s compact in E( cf . also [ 1 7 ] ; Prop . 5 ) . So η ∈ QT (Γ{x1, ..., xk}) ⊂ QT (L). omega period .. We have −k parenleftbiggThis shows sub that 2 subfor each n tok, the QT power(2 L of+ 1A Lk) plus i s aA neighborhood sub n parenrightbigg in ω. We = 2 have to the power of 1 sub n L plus A sub n \noindent Thisshows that for each $k , QT ( 2ˆ{ − k } L + A { k } ) $ i s a neighborhood in and als o-line open parenthesis 2 to the power1 of minus n L plus1 A sub n closing parenthesis i s a completing sequence in E period .. Hence by$ \ Lemmaomega 2 period. $ 1\ commaquad We QT have (2nL + An) = 2nL + An andand therefore als o − Qline(2 are open−nL + mapsA ) period i s a completing sequence in E. Hence by Lemma 2.1, QT and therefore Q are \ [( { 2 }ˆ{ 1 } { n } nL + A { n } ) = 2 ˆ{ 1 } { n } L + A { n }\ ] Sinceopen omega maps is . a Fr acute-e chet endash Montel space comma .. a bounded subset S i s contained in the absolutelySince convexω is a closed Fre ´ chet hull – of Montel a sequence space open , a parenthesis bounded subset eta nS closingi s contained parenthesis in the which absolutely converges convex t o zero closed period .. Setting 2 tohull the of power a sequence of minus (ηn n) L which plus A converges sub n = tD o sub zero n .comma Setting .. we 2− choosenL + A i sub n= upwardsD , arrowwe choose infinityi .. such↑ ∞ that for i sub k less or equal \noindent and a l s $ o−l i n e ( 2 ˆ{ − n } L + An { n } n) $ i s a completingn sequence in $ E m lesssuch i sub that k plus for 1i ..k we≤ havem < ik+1 we have ηm ∈ QT (Dk). As in the proof of Lemma 2 . 1 we find . $ \mquad HencebyLemma $2 . 1 , QT$ etax ml in∈ QTDl such open that parenthesis D sub k closing parenthesis period .. As in the proof of Lemma 2 period 1 we find x to the power of m l in D suband l such therefore that $ Q $ are open maps . Line 1 infinity Line 2 eta m = QT open parenthesis sum x sub l to∞ the power of m closing parenthesis Line 3 l = 1 Since $ \omega $ i s a Fr $ \acute{e} $ chetX−− Montel space , \quad a bounded subset $ S $ i s contained in the for i sub 1 less or equal m less i sub 2 period .. Weηm have= QT ( xm) absolutelyLine 1 infinity convex Line 2 z closed sub 1 = sum hull x to of the a power sequence of m l in $ 2 ( Dl sub\eta 1 periodn Line ) 3 $ l = which 1 converges t o zero . \quad S e t t i n g $We 2 proceed ˆ{ − inn this} fashionL + and find A { x ton the} power= D of m{ ln in D} l sub=, 1 l $ for\ iquad sub k lesswe orchoose equal m $ less i i sub{ n k plus}\ 1uparrow so that \ infty $ \quadinfinitysuch z sub that m = for sum x $ to i the{ powerk }\ of mleq l in Dm sub k< minusi 1 l{ =k k Equation: + 1 and}$ QTz\quad subwe m = have eta m period .. Then F sub k = open for i1 ≤ m < i2. We have brace$ \ xeta to the powerm \ ofin m kQT : i sub (1 less D or equal{ k } m less) i sub . $k plus\quad 1 closingAs bracein the subset proof D sub of k Lemmaminus 1 2 . 1 we find $ x ˆ{ m }{ l } \ in D { l }$ such that and open brace x to the power of m l closing brace subset union of F∞ sub k period .. The sequence open parenthesis z sub m closing parenthesis certainly converges to zero period .. Since Capital GammaX openm parenthesis F sub k closing parenthesis \ [ \ibegin s .. a ..{ a compact l i g n e d ..}\ subsetinfty .. and\\ Capital Gammaz1 open= parenthesisx l ∈ 2D1. F sub k closing parenthesis subset D sub k minus 1 .. we .. can .. define \eta m = QT ( \sum x ˆ{ m } { l } ) \\ .. a continuous function l = 1 l = 1 \end{ a l i g n e d }\ ] Capital Theta : product sub k = 1 to them power of infinity Capital Gamma open parenthesis F sub k closing parenthesis right arrow E by CapitalWe Theta proceed open in parenthesis this fashion open and parenthesis find x l ∈ tD subl for ni closingk ≤ m < parenthesis ik+1 so that closing parenthesis = sum t sub n and so open parenthesis z sub m closing parenthesis is a sequence in the compact \noindentset Capitalf oTheta r $ open i { parenthesis1 }\ leq productm Capital< Gammai { 2 open} parenthesis. $ \quad F subWe k have closing parenthesis closing parenthesis period Hence ∞ for every alpha in l sub 1 comma the series sum alpha sub n z sub n is convergent and the \ [ \ begin { a l i g n e d }\ infty \\ X m absolutely convex closed hull of open parenthesis z subz nm closing= x parenthesisl ∈ Dk−1 i s equal to zinfinity{ 1 M} == open\ bracesum sumx m ˆ{ =m 1 lambda}{ l }\ subin m z sub2 m :D sum{ bar1 } lambda. \\ sub m bar less or equal 1 closing brace period We have S l = k subsetl QT = open 1 \ parenthesisend{ a l i gn M e dclosing}\ ] parenthesis period Although we shall deal with webbed spaces in the context of this resultThen in andQT zm = ηm. m detail subsequently comma let us first indicateFk = {x ank immediate: i1 ≤ m < generalization ik+1} ⊂ Dk−1 of Eidelheit quoteright s \noindenttheorem periodWe proceed .. If E i s in a Fr this acute-e fashion chet space and with find a base $ of x neighborhoods ˆ{ m }{ l open}\ parenthesisin D { U subl }$ n closing f o r parenthesis $ i { k then}\ settingleq m < i m{ kS + 1 }$ so that A suband k{x = 2l} to ⊂ theFk power. The of sequence minus k U(zm sub) certainly k we have converges a completing to zero sequence . Since open Γ(Fk) parenthesis i s a compact A sub k closing subset parenthesis in E period .. Q∞ P This simpleand Γ( observationFk) ⊂ Dk−1 we can define a continuous function Θ : k=1 Γ(Fk) → E by Θ((tn)) = tn \ begin { a l i g n ∗} Q P yieldsand the so ( followingzm) is a sequence result period in the compact set Θ( Γ(Fk)). Hence for every α ∈ `1, the series αnzn is convergent \ infty \\ z { m } = \sum x ˆ{ m }{ l }\ in D { k − 1 }\\ l = k \\\ tag ∗{$ and QTz { m } 2 periodand the 4 period absolutely .. Corollary convex closed period hull .. If of there (zn) .. i sis equal .. an to.. unbounded .. continuous .. linear map from .. a = Fr\ acute-eeta chetm space . $} intoThen a lcs\\ F commaF { thenk } there= is\{ a faithfulx ˆ surjection{ m }{ k of} F onto: omega i { period1 }\ leq m < i { k + 1 } \}\subset D { k − 1 } \end{ a l i g n ∗} ∞ X X M = { λ z : | λ |≤ 1}. \noindent and $ \{ x ˆ{ m }{ l }\}\m subsetm m\bigcup F { k } . $ \quad The sequence $ ( z { m } m=1 ) $ certainly converges to zero . \quad Since $ \Gamma (F { k } ) $ WehaveS ⊂ QT (M). i s \quad a \quad compact \quad subset \quad and $ \Gamma (F { k } ) \subset D { k − 1 }$ \quad weAlthough\quad wecan shall\quad deal withd e f i webbed n e \quad spacesa in continuous the context of function this result in detail subsequently , let us first $ \indicateTheta an: immediate\prod generalizationˆ{\ infty of} Eidelheit{ k = ’ s theorem 1 }\Gamma . If E i(F s a Fre ´ {chetk space} ) with\rightarrow a base of E $ by $ \Theta ( ( t { n } ) ) = \sum t { n }$ and so $ ( z { m } ) $ is a sequence in the compact neighborhoods (Un) then setting s e tA $ =\Theta 2−kU we have( a\prod completing\Gamma sequence (A(F) in E.{ kThis} simple) ) observation .$ Henceforyields the following every result $ \ .alpha \ in \ e l l { 1 } , $ thek seriesk $ \sum \alpha { n } k z { n }$ is convergent and the 2 . 4 . Corollary . If there is an unbounded continuous linear map from a Fr absolutelye´ chet space convex into a lcs closedF, then hull there of is a $ faithful ( surjectionz { n } of )F $onto iω. s equal to \ begin { a l i g n ∗} \ infty \\ M = \{\sum { m = 1 }\lambda { m } z { m } : \sum \mid \lambda { m } \mid \ leq 1 \} . \\ We have S \subset QT ( M ) . \end{ a l i g n ∗}
Although we shall deal with webbed spaces in the context of this result in detail subsequently , let us first indicate an immediate generalization of Eidelheit ’ s theorem . \quad If $E$ isaFr $ \acute{e} $ chet space with a base of neighborhoods $ ( U { n } ) $ then setting
\noindent $ A { k } = 2 ˆ{ − k } U { k }$ we have a completing sequence $ ( A { k } ) $ in $ E . $ \quad This simple observation yields the following result .
2 . 4 . \quad C o r o l l a r y . \quad I f there \quad i s \quad an \quad unbounded \quad continuous \quad linear map from \quad a Fr $ \acute{e} $ chet space into a lcs $ F , $ then there is a faithful surjection of $ F $ onto $ \omega . $ 1 8 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentBy a web we1 8 shall\quad alwaysS mean $ . an absolutely\ddot{O} convex$ nal C hyphen andT web . period Terzio .. We $ say\check a web{g} $ lu W = open brace C sub n sub 1 period period period n sub k closing brace i s .. eventually bounded if for any .. open parenthesis n sub k \ hspace ∗{\ f i l l }By a web we shall always mean an absolutely convex $ C − $ web . \quad We say a web closing parenthesis¨ in N to the power of N there i s some j .. so that1 C8 sub S n.O subnal 1 and period T . periodTerzio periodgˇ lu n sub j .. i s a bounded set period .. In this case if open parenthesis A sub k closing parenthesis i s any\noindent completing$ sequence W = \{ CBy a{ webn we{ shall1 } always.mean . . an absolutely n { k convex}}\}C−$web i s.\quad We sayeventually a web bounded if for any \quad $ (W n ={ k {}C ) } \iin s eventuallyN ˆ{ N bounded}$ thereif for i any s some (n ) $∈ j $ N there\quad i sso some j so that C derived from Wn comma1...nk then each A sub k i s bounded from somek k sub 0 onN period .... For example when n1...nj thatE ii s s a $a sequentially bounded C { n set complete{ .1 In} this DF. case hyphen . if (A . spacek) i nscomma any{ completingj the}}$ natural\quad sequence webi is s eventually a bounded bounded set period . \quad In this case if $( A { k } ) $Wederived iwould s any like from completingt oW apply, then Theorem each sequenceAk 2i s period bounded 3 to from the identity some k0 mapon . on a webbed space E period For example when If inE thisi s a case sequentially the condition complete of TheoremDF − space 2 period , the 3 does natural not web hold is comma eventually then bounded the web .W We i s would like t o apply \noindenteventuallyTheoremderived bounded 2 . 3 to period the from identity .. $WTherefore map , on let $ a us webbed then discuss each space thisE. case $A in some{ k detail}$ iperiod s bounded from some $ k { 0 }$ on . \ h f i l l For example when ForIf a inweb this W casewhich the i s condition eventually of bounded Theorem and 2 for . 3 a does strand not open hold parenthesis , then the n websub iW closingi s eventually parenthesis bounded in N to .the power of N we say \noindentk iTherefore s the terminal$ let E us $ index discuss i s if Ca this sub sequentially case n sub in some 1 period detail complete period . period $ n DF sub k− minus$ space 1 .. i s unbounded , the natural but C sub web n issub eventually 1 period period bounded period n . N subWe k would .. i sFor bounded likea web tperiodW owhich apply .. We i s eventually Theorem bounded 2 . 3 to and the for a identity strand (ni) map∈ N onwe a say webbedk i s the spaceterminal $ index Eif . $ C i s unbounded but C i s bounded . We collect all C over all strands (n ) which collectn1...n allk C−1 sub n sub 1 period periodn1 period...nk n sub k over all strands open parenthesisn1...nk n sub i closing parenthesisi which have k as t erminal index\noindent andhave denotek asIf t in erminal this index case and the denote condition the resulting of set Theorem by Wk. 2W .k 3i s does countable not . hold If we , take then the the union web of $W$ i s eventuallytheall resultingWk ’ sbounded setover by all W t erminalsub . \ kquad period indicesTherefore Wk, subwe k get i s a countable let countable us discuss period collection .. If thisW web takeof case bounded the unionin subsets some of all detailof WE. subWe k .order quoterightWb s over into a sequence (Bk) of bounded set s . For x ∈ E given we can find a strand (ni) such that x ∈ Cn ...n for all t erminal indices k comma we get a countableS collection W sub b of bounded subsets of E period 1 i ForWe aeach order webi ∈ W $W$N sub. So b intox which∈ aB sequencek for i some s openeventuallyk. parenthesisSo E = boundedB Bk suband k thus andclosing the for tparenthesis opology a strandτm ofof bounded uniform $ ( set convergence n s period{ i } For on) ( xB ink)\ Ein givenN we ˆ can{ N find}$ we say $a strandki s $ a locally openi s theparenthesis convex terminal metrizable n sub index i t closing opology if parenthesis satisfying $C { suchn that{ 1 x} in C. sub . n sub . 1 period n { periodk − period1 n}} sub$ i\ ..quad for eachi s i in unbounded N period .. but $ C { n { 1 } . . . n { k }}$ \quad i s bounded . \quad We So x in B sub k for some k period .. So 0 0 collectE = union all of B $Csub k and{ n thus{ the1 } t opology. .σ tau(E sub,E .) m≤ n ofτm uniform{≤ βk(E}},E convergence$). over all on open strands parenthesis $ ( B sub n k{ closingi } parenthesis) $ which i s a locally have $ k $ as t erminal index and denote convexSuppose metrizable further t thatopologyW is satisfying a strict web . This i s automatically satisfied if each C i s also closed . the resulting set by $W { k } .W { k }$ i s countable . \quadn1...nk If we take the union of all $W { k }$ sigmaLet openT b e parenthesis a sequentially E to closed the power linear of map prime of acomma Fre ´ chet E closing parenthesis less or equal tau sub m less or equal beta open parenthesis E ’ s over to thespace powerF ofinto primeE. comma E closing parenthesisThe localization period theorem of De Wilde ( [ 5 ] , [ 1 1 ] ) states that there all t erminal indices $ k , $ we(k) get a countable collection $W { b }$ of bounded subsets of $ E Supposei s a strand further (n thati) W∈ N isN aand strict a sequence web periodU .. Thisof neighborhoods i s automatically of satisfiedzero in F ifsuch each that for every k . $C sub n sub 1 period period period n sub k .. i s also closed period .. Let T b e a sequentially closed linear map of a Fr acute-e chet Wespace order F into $ E W period{ b ....}$ The into localization a sequence theorem of $(k De) ( Wilde B open{ k parenthesis} ) $ openof bounded square bracket set s5 closing . For square $ x bracket\ in commaE $ open given we can find T (U ) ⊂ Cn1...nk . squarea strand bracket $1 1 ( closing n square{ i } bracket) $ closing such parenthesis that $x states\ thatin thereC { n { 1 } . . . n { i }}$ \quad f o r each $ ii s aThis strand\ in implies openN that parenthesis .T $ maps\quad n a subneighborhoodSo i closing $ x parenthesis into\ in someBB inj N.{ tok the}$ power forsome of N and a $k sequenceHence .$ U , if to\E thequadis apower strictlySo of open parenthesis k closing parenthesis$ Ewebbed = .. of space\ neighborhoodsbigcup whose webB of i s{ zero eventuallyk in}$ F such and bounded thus , we the have tL opology(F,E) = LB $ (F,E\tau) for{ everym }$ Fre ´ ofchet uniform space F. convergence on $ ( B that{ k for}A every corollary) $ k i of s the a localization locally theorem states that every Banach disc D of E i s contained in some αjBj [11]. convexT openThis metrizableparenthesis means that U to t the opology power of satisfying open parenthesis k closing parenthesis closing parenthesis subset C sub n sub 1 period period period n sub k period 0 0 \ [ This\sigma implies that( T E maps ˆ{\ aprime neighborhood} ,E) intob( someE ,E) B≤ sub\τleqm j≤ periodβ(E\tau,E ....) Hence{ m comma}\ leq if E is\ abeta strictly ( E ˆ{\prime } ,E ). \ ] webbedif W spaceis a whose strict web web is which eventually i s eventually bounded comma bounded we . have Here L openb(E parenthesis0,E) denotes F comma the t opology E closing of parenthesis uniform = LB open parenthesis 0 F commaconvergence E closing on parenthesis Banach discs for of E. In particular , if E i s also locally complete , then τm = β(E ,E) and everytherefore Fr acute-e the strongchet space dual FE period0 i s a metrizable locally convex space . \noindent Suppose furtherb that $W$ is a strict web . \quad This i s automatically satisfied if each A corollaryWe recall of the that localization a lcs theoremE has the statescountable that every Banach boundedness disc D property of E if for every sequence (B ) of $ C { n { 1 } . . . n { k }}$ \quad i s also closed . \quad Let $ Tn $ b e a sequentially closed linear map of a Fr i s containedbounded subsets in some there alpha are subµn j B> sub0 soj open that squareS∞ µn bracketBn i s 1 also 1 closing square bracket period .. This means that $ \acute{e} $ chet n=1 b openbounded parenthesis . Every E to Fr thee ´ chet power space of prime enj oys comma this property E closing . parenthesis less or equal tau sub m less or equal beta open parenthesis E to the power2 of . prime 5 . Corollary comma E closing . If parenthesisE is a webbed c − barrelled lcs then one and only one of the fo l lowing is \noindent space $F$ into $E .$ \0 h f i l l The localization theorem of De Wilde ( [ 5 ] , [ 1 1 ] ) states that there if Wtrue is ..: a strict .. web which i s .. eventually bounded period .. Here .. b open parenthesis E to the power of prime comma E closing parenthesis .. denotes the \noindentt opology ofi uniform sastrand convergence $( on Banach n { i discs} of) E period\ in ..N In particularˆ{ N }$ comma and a .. ifsequence E i s $Uˆ{ ( k ) }$ \quad of neighborhoods of zero in $ Falso $ locally such complete comma then tau sub m = beta open parenthesis E to the power of prime comma E closing parenthesis and therefore thethat strong for dual every E sub b $ to k the $ power of prime i s a metrizable locally convex space period \ [We T recall ( that U ˆ ..{ a( lcs .. k E has ) the} ..) countable\subset .. boundednessC { n property{ 1 } if for. every . . n { k }} . \ ] sequence open parenthesis B sub n closing parenthesis .. of bounded subsets there are mu n greater 0 so that union of sub n = 1 to the power of infinity mu n to the power of B n i s also \noindentbounded periodThis .. implies Every Fr acute-ethat chet $ T space $ maps enj oys a this neighborhood property period into some $ B { j } . $ \ h f i l l Hence , if $E$ is2 a period strictly 5 period Corollary period .. If E is a webbed c sub 0 hyphen barrelled lcs then one and only one of the fo l lowing is true : \noindent webbed space whoseweb i s eventually bounded , wehave $L ( F , E ) = LB ( F , E ) $ f o r every Fr $ \acute{e} $ chet space $F . $
A corollary of the localization theorem states that every Banach disc $ D $ of $ E $ i s contained in some $ \alpha { j } B { j } [ 1 1 ] . $ \quad This means that
\ [ b ( E ˆ{\prime } ,E) \ leq \tau { m }\ leq \beta ( E ˆ{\prime } ,E) \ ]
\noindent i f $ W $ i s \quad a s t r i c t \quad web which i s \quad eventually bounded . \quad Here \quad $ b ( E ˆ{\prime } , E ) $ \quad denotes the t opology of uniform convergence on Banach discs of $ E . $ \quad In particular , \quad i f $ E $ i s also locally complete , then $ \tau { m } = \beta ( E ˆ{\prime } , E ) $ and therefore the strong dual $ E ˆ{\prime } { b }$ i s a metrizable locally convex space .
We recall that \quad a l c s \quad $ E $ has the \quad countable \quad boundedness property if for every sequence $ ( B { n } ) $ \quad of bounded subsets there are $ \mu n > 0 $ so that $ \bigcup ˆ{\ infty } { n = 1 }\mu n ˆ{ B } n $ i s a l s o
\noindent bounded . \quad Every Fr $ \acute{e} $ chet space enj oys this property .
2 . 5 . Corollary . \quad If $E$ is awebbed $c { 0 } − $ barrelled lcs then one and only one of the fo l lowing is true : Subspaces and quotient spaces of locally convex spaces .. 1 9 \ hspaceopen parenthesis∗{\ f i l l } Subspaces a closing parenthesis and quotient .. There is spaces a faithful of surjection locally Q : convex E right arrow spaces omega\quad period1 9 open parenthesis b closing parenthesis L open parenthesis E comma omega closing parenthesis = LB open parenthesis E comma omega closing\ centerline parenthesis{( aperiod ) \quad There is a faithful surjection $Q : E \rightarrow \omega . $ } If E is a webbed comma barrelled space then open parenthesisSubspaces b closingand quotient parenthesis spaces of ..locally can be convex rep laced spaces by 1 9 \ [open ( parenthesis b ) L b closing ( parenthesis E( , a ) \ toomegaThere the power is a) faithful of prime = surjectionL LB open ( parenthesisQ : EE → ω. , E comma\omega F closing). parenthesis\ ] = LB open parenthesis E comma F closing parenthesis for every lcs F which has the countable boundedness property period (b)L(E, ω) = LB(E, ω). \noindent If $ E $ is a webbed , barrelled space then ( b ) \quad can be rep laced by P rIf o oE f periodis a webbed .. Either, barrelled there is space a completing then ( b sequence ) can beopen rep parenthesis laced by A sub n closing parenthesis in E such that each A sub n i s unbounded or the web is eventually( b )0L( boundedE,F ) = periodLB(E,F ..) Infor the every former lcs caseF which we have has open the countable parenthesis boundedness a closing parenthesis \ hspace ∗{\ f i l l }( b $ ) ˆ{\prime } L ( E , F ) = LB ( E , F )$ foreverylcs fromproperty Theorem . 2 period 3 period .. So it remains t o prove open parenthesis b closing parenthesis or open parenthesis b closing parenthesis $ F $ which has the countable boundedness to the powerP r of o primeo f . in Either case the there web is ais completing eventually sequence (An) in E such that each An i s unbounded or the web boundedis eventually period bounded . In the former case we have ( a ) \noindent property . Letfrom T : E Theorem right arrow 2 . 3 omega-line . So it remains b e continuous t o prove and ( b open) or ( parenthesis b )0 in case B the sub web k isclosing eventually parenthesis bounded the . seq to the power of u-line ence of bounded subsets of E Let T : E → omega − line b e continuous and (B ) the sequ−line ence of bounded subsets of E P r o o f . \quad Either there is a completing sequencek $ ( A { n } ) $ in $E$ such that each wewe have have constructed constructed comma, (Vk) open a base parenthesis of neighborhoods V sub k closingof ω. We parenthesis deter m − aline base ine ofρk neighborhoods > 0 so that of omega period We deter m-line ine rho k$ greater A { n 0 so}$ i s unbounded or the web is eventually bounded . \quadV In the former case we have ( a ) that T (B1) ∪ ... ∪ T (Bk) ⊂ ρk k. T open parenthesis B− sub1 1V closing parenthesis cup period period−1 V period cup T open parenthesis B sub k closing0 parenthesis subset rho k to the\noindent powerHence of VBfromm k period⊂ TheoremT (ρn 2n) .for 3n .≥ \m.quadEachSoline it− remainsT (ρn tn) o i s prove an absolutely ( b ) convex or ( , b closed $ )σ( ˆE,E{\prime)− }$ in case the web is eventually neighborhood and each x ∈ E is in T∞ T −1(ρnV n) for some boundedHence B sub . m subset T to the power of minusn=m 1 open parenthesis rho n to the power of V n closing parenthesis for n greater equal m period .. Eachm. line-TTherefore to the power , if E ofi s minusc0− barrelled 1 open parenthesis then rho n to the power of V n closing parenthesis i s an absolutely convex comma \ hspace ∗{\ f i l l } Let $ T : E \rightarrow omega−line $ b e continuous and $ ( B { k } ) $ the closed sigma open parenthesis E comma E to the power of prime closing∞ parenthesis hyphen neighborhood and each x in E is in intersection of$ sub seq n ˆ ={ mu− tol i the n e power}$ ofence infinity of T bounded to the power subsets of minus of 1 open $ E parenthesis $ rho n to the power of V n closing parenthesis for some \ m period .. Therefore comma if E i s c sub 0 hyphenU = barrelledT −1(ρn thenV n) \noindent we have constructed $ , ( V { k } ) $ a base of neighborhoods of $ \omega . $ We det er Line 1 infinity Line 2 U = intersection of T to the power of minusn = 1 1 open parenthesis rho n to the power of V n closing parenthesis Line 3 n =$ 1m−l i n e $ i n e $ \rho k > 0 $ so thati s ai s neighborhood a neighborhood open ( [ parenthesis 1 0 ] ; 1 2 . open 1 . 6 square ) and so bracketT (U) 1 is 0 bounded closing square . bracket semicolon 1 2 period 1 period 6 closing parenthesis and so T openSuppose parenthesis now UE closingi s barrelled parenthesis and T is bounded: E → periodF i s continuous . Using the as - sumption that F has \ [T(BSupposethe countable now E{ i sboundedness1 barrelled} ) and\ propertycup T : E right (... cf . arrow [ 3 ] ) F , i we s continuous\ determinecup T(B period .. Using{ thek } as hyphen) \subset \rho k ˆ{ V } k . \ ] S∞ T sumptionµn > 0 that so that F hasn=1 theµn countable(Bn) i s boundedness a bounded subset property of F. open parenthesis cf period openIf square we denote bracket by A 3 closingthe square bracket closing parenthesisabsolutely comma convex we determine , closed hull of this set , we conclude that T −1(A) i s a barrel . Hence T i s bounded . \noindentmu n greaterHence 0 so thatR $ e Bm union a{ r km of s sub}\ . nAsubset = lcs 1 toE thehas powerT ˆ the{ of − infinitycountable1 } mu( n to neighborhood\ therho powern of property ˆ T{ openV } parenthesisn( cnp ) ) $ B if sub forf o r n closing $ n parenthesis\geq m i s a . $ \quad Each $ l i n e −T ˆ{ − 1 } ( \rho n ˆ{ V } n ) $ i sT an∞ absolutelyU convex , boundedany subset sequence of F periodUn ....∈ If U we(E) denote there by A is the a sequence ρn > 0 such that n=1 ρn n i s c labsolutely o s ea d neighborhood $ convex\sigma comma [ 5( ] ... closed E It i s hull ,known of E this that ˆ{\ setprime commaE has the} we cnp conclude) if− and$ that only neighborhood T if toL( theE,F power) = of andLB minus(E,F each 1) open for $ every parenthesis x Fr\ in A closingE $ iparenthesis s in $ \bigcap ˆ{\ infty } { n = m } T ˆ{ − 1 } ( \rho n ˆ{ V } n ) $ f o r some 0 i s a barrele´ chet period space F [4]. It can b e shown that L(E, ω) = LB(E, ω) if and only if for every sequence Un of σ(E,E )− Henceneighborhoods T i s bounded there period i s a sequence ρn > 0 so that T ρnU n i s a neighborhood . This property i s equivalent \noindentR eto m what a r k Bonet$ s periodm calls . .. $ theA lcs\countablequad .. E hasTherefore linear .. the form .. countable ,property if ..$E$ neighborhood( [ 3 ] ; Propi s . property$c 14 ) .{ From..0 open} Theorem − parenthesis$ barrelled 2 . 3 cnp we closing obtain then parenthesis .. if for anythat sequence if E has .... thisU sub property n in U openand (A parenthesisk) i s a completing E closing sequence parenthesis , then .... thereAk i s is bounded .... a sequence after some .... rhok0. n greater 0 such that .... intersection \ [ \ begin { a l i g n e d }\ infty \\ of sub n =Let 1 toE theb e poweran infinite of infinity - dimensional rho n to Banach the power space of and U n (u ....n) i a s linearly independent sequence of bounded linear U = \bigcap T ˆ{ − 10 } ( \rho n ˆ{ V } n ) \\ a neighborhoodforms on E. ..Then open squareS : E[σ bracket(E,E )] 5 closing→ ω defined square by bracketSx = period (un(x)) .. i It s i continuous s known that . .. Moreover E has the , since cnp ifS and(E) only if L open parenthesis E commanis = infinite F closing 1 \ -end dimensional parenthesis{ a l i g n e,S d =}\ ]i s unbounded , although E, and therefore E[σ(E,E0)], is a webbed space whose LBweb open i sparenthesis eventually E bounded comma . F From closing this parenthesis simple example for every it follows Fr acute-e that thechet barrelledness space F open square bracket 4 closing square bracket period It can bcondition e shown on thatE Lin open Corollary parenthesis 2 . 5 cannot E comma b e droppedomega closing altogether parenthesis . = LB open parenthesis E comma omega closing parenthesis \noindentif and onlyi if for saneighborhood every sequence U sub ( n of[ sigma 10 ] open ; 12 parenthesis . 1 . E 6)andso comma E to the $T power of ( prime U closing )$ parenthesis is bounded hyphen . neighborhoods there i s a sequence Supposenowrho n greater 0 $E$ so that intersection i s barrelled of rho n and to the $T power of : U n E i s a\ neighborhoodrightarrow periodF $.. This i s property continuous i s equivalent . \quad to whatUsing the as − sumptionBonet calls that the countable $ F $ linear has form the propertycountable open boundedness parenthesis open property square bracket ( cf 3 . closing [ 3 ] square ) , we bracket determine semicolon Prop period 14 closing parenthesis period .. From Theorem 2 period 3 \noindentwe obtain that$ \ ifmu E hasn this> property0 $ and so open that parenthesis $ \bigcup A sub kˆ{\ closinginfty parenthesis} { n i s a = completing 1 }\ sequencemu n comma ˆ{ T } then(B A sub k { n } ) $i s bounded i s abounded after some subset k sub 0 period of $F . $ \ h f i l l If we denote by $A$ the Let E b e an infinite hyphen dimensional Banach space and open parenthesis u sub n closing parenthesis a linearly independent \noindentsequence ofabsolutely bounded linear convex forms on , E\quad periodclosed .. Then S hull : E open of square this bracketset , sigmawe conclude open parenthesis that E $ comma T ˆ{ −E to the1 } power( of A prime ) $ closingi s a parenthesis b a r r e l . closing square bracket right arrow omega defined by HenceSx = open $T$ parenthesis i s bounded u sub n open . parenthesis x closing parenthesis closing parenthesis i s continuous period .. Moreover comma since S open parenthesis E closing parenthesis .. is infinite hyphen dimensional comma S i s \ hspaceunbounded∗{\ f comma i l l }R although e m a r E kcomma s . and\quad thereforeA l c Es open\quad square$ bracket E $ has sigma\quad open parenthesisthe \quad E commacountable E to the\quad power ofneighborhood prime closing property \quad ( cnp ) \quad i f f o r parenthesis closing square bracket comma is a webbed space whose web \noindenti s eventuallyany bounded sequence period\ h From f i l l this$ simple U { examplen }\ itin followsU that ( the E barrelledness ) $ \ h f i l l the re i s \ h f i l l a sequence \ h f i l l $ \conditionrho n on E> in Corollary0 $ such 2 period that 5 cannot\ h f i l b l e dropped$ \bigcap altogetherˆ{\ periodinfty } { n = 1 }\rho n ˆ{ U } n $ \ h f i l l i s \noindent a neighborhood \quad [ 5 ] . \quad It i s known that \quad $ E $ has the cnp if and only if $ L ( E , F ) = $ $LB ( E , F )$ foreveryFr $ \acute{e} $ chetspace $F [ 4 ] .$ Itcanbeshownthat $ L ( E , \omega ) = LB ( E , \omega ) $ if and only if for every sequence $ U { n }$ o f $ \sigma ( E , E ˆ{\prime } ) − $ neighborhoods there i s a sequence $ \rho n > 0 $ so that $ \bigcap \rho n ˆ{ U } n $ i s a neighborhood . \quad This property i s equivalent to what Bonet calls the countable linear form property ( [ 3 ] ; Prop . 14 ) . \quad From Theorem 2 . 3 we obtain that if $E$ has this property and $ ( A { k } ) $ i s a completing sequence , then $ A { k }$ i s bounded after some $ k { 0 } . $
Let $E$ b e an infinite − dimensional Banach space and $ ( u { n } ) $ a linearly independent sequence of bounded linear forms on $ E . $ \quad Then $ S : E [ \sigma ( E , E ˆ{\prime } )] \rightarrow \omega $ d e f i n e d by $ Sx = ( u { n } ( x ) )$ iscontinuous. \quad Moreover , since $S ( E )$ \quad i s i n f i n i t e − dimensional $ , S $ i s unbounded , although $ E , $ and therefore $ E [ \sigma ( E , E ˆ{\prime } ) ] , $ is a webbed space whose web i s eventually bounded . From this simple example it follows that the barrelledness
\noindent condition on $ E $ in Corollary 2 . 5 cannot b e dropped altogether . 20 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentAssume E20 open\quad squareS bracket $ . tau\ closingddot{O square} $ bracket nal andT is only .webbed Terzio and there $ \check is a continuous{g} $ lulinear map T from E onto a proper Fr acute-e chet space F period .. Let tau sub b be the associated barrelled topology period .. If Assume $ E [ \tau ] $ is only webbed and there is a continuous linear map $ T $ from $ E $ I : E open square¨ bracket tau sub b bracketright-line right arrow E open square bracket tau closing square bracket .. i s the identity then TI : Eonto open20 a square proper S .O bracketnal Fr and tau T$ . sub\ Terzioacute b closinggˇ{elu} square$ chet bracket space right arrow$F F .is line-c$ \quad ontinuousLet and $ open\tau period{ b }$ be the associated barrelled topology . \quad I f $Hence IAssume by : Corollary EE[τ [] 2 is period only\tau webbed 5 comma{ b and} E there openbracketright issquare a continuous bracket−l i tau linearn e sub map\ brightarrow closingT from squareE onto bracketE[ a proper has Fromega\taue ´ chet as space a] quotient $ \quad spacei period s the .. More identity then particularly$ TIF. :Let comma Eτb be there [ the associated\tau { barrelledb } ] topology\rightarrow . If I : E[τFbbracketright $ i s $− l iline n e −c→ $E[τ ontinuous] i s the identity and open . i sthen a continuousTI : E[τb linear] → mapF is Qline from− c Eontinuous open square and bracketopen . tau sub b closing square bracket onto omega and a completing sequence open parenthesis\noindentHence A byHencebyCorollary sub Corollary n closing 2 parenthesis.5,E[τb] has $2ω as a quotient . 5 space , . E More [ particularly\tau { ,b there} i] s $a continuous has $ \ linearomega $ as a quotient space . \quad More particularly , there iin s Emap a open continuousQ squarefrom E bracket[τb] linear onto tauω and sub map ba closingcompleting $Q$ square from sequence bracket $E (A ..n so) that [ open\tau parenthesis{ b } Q] open $ ontoparenthesis $ \ toomega the power$ and of A-tildewide a completing sub n sequence $ ( A { n } ) $ Ae closingin parenthesisE[τb] so closing that ( parenthesisQ(n )) i s a i base s a base of neighborhoods of neighborhoods in ω. in omegaHere periodAe denotes .. Here the tildewide-A closure with denotes resp ect the to closureτb. with(An) resp i s of ect course to tau a subcompleting b period sequence open parenthesis in E[τ] A and sub n closing parenthesis i s of course a completing sequence in E open square \noindent in $ E [ \tau { b } ] $ \quad sothat $( Q (ˆ{\ widetilde {A}} { n } ) ) $ i s a base of neighborhoods in bracketAe taun ⊂ closingAn, the square closure bracket with resp .. and ect t o τ. So each Q(An) i s a neighborhood in ω and therefore each An is $ \A-tildewideomegaunbounded. sub $ . n subset\ Soquad we AhaveHere sub proved n comma $ \ thewidetilde thefollowing closure{ resultA with} $ . resp denotes ect t o tau the period .. So each Q open parenthesis A sub n closing parenthesis i sclosure a neighborhood2 . with 6 . Corollary inresp omega ect . toIf there $ \tau is a continuous{ b } linear.(A map from{ an webbed} ) space $ iE sonto of a course proper Fr ae´ completingchet sequence in $ Eandspace therefore [ ,\thentau eachω Ais] sub a $ faithful n\ isquad unbounded quotientand of periodE. .. So we have proved the following result period 2 period 6 period Corollary period3 If . there isNuclear a continuous K linearo¨ the map quotients from a webbed space E \noindentonto aIn proper$ [ 1 Fr\ 3widetilde acute-e ] it chet was{A space proved} { comman that}\ then ifsubsetE omegaand isFA a faithful{aren Fr} quotiente ´ chet, $ of spaces the E period closure,T : withE resp→ F ecti s t o $ \tau . $ \quad3 periodunboundedSo ..each Nuclear and $F QKsatisfies o-dieresis ( aA thecertain{ quotientsn normability} ) $ i condition s a neighborhood ( y ) which will in b e given $ \omega $ andIn ..b therefore elowopen , square then eachthere bracket i s $ 1 a A 3 nuclear closing{ n K} square$o ¨ the is bracketspace unboundedλ( ..A) it and .. was . a\ quotientquad provedSo thatmap we ..Q haveif: F E→ and provedλ( ..A) F such .. arethe that Fr following acute-eQT chet: result .. spaces . comma T : E right arrowE → F i s λ(A) i s also a quotient map . A simple consequence of this theorem i s that either 2 .unbounded 6every . Corollary continuous and F satisfies linear . If a map there certain from normability is a given continuous conditionFre ´ chet linearopen space parenthesisE mapinto from y closing any a webbed nuclear parenthesis K spaceo ¨ the which $ space will E $ b e is given ontob elowboun a comma properline − thend Fred there $ or \ iacute sE ahas nuclear{e} a Knuclear$ dieresis-o chet Ko ¨ space the spacequotient , thenlambda [ 1 3 open$] . \omega parenthesis$ is A closing a faithful parenthesis quotient and a quotient of $E map Q .: F $ right arrow lambda open parenthesisWe recallA closing that parenthesis a lcs F satisfies ( y ) if there is V1 ∈ U(F ) such that \ centerlinesuch that ..{3 QT . :\ Equad rightNuclear arrow lambda K $ open\ddot parenthesis{o} $ A the closing quotients parenthesis} .. i s .. also .. a quotient .. map period .. A simple 0 [ 0 ◦ consequence of F = F [V1 ] ∩ V ◦ In \quad [ 1 3 ] \quad i t \quad was proved that \quad i f $ E $ and \quad $ F $ \quad are Fr $ \acute{e} $ this theorem i s that .. either every continuous linear map ..V from∈ U a(F given) Fr acute-e chet chetspace\quad .. E ..spaces into .. any $, nuclear T K o-dieresis : E the\ ..rightarrow space .. is .. bounF $ line-d i ed s .. or .. E has .. a nuclear K o-dieresis the unboundedquotientwhere open closure and square is $ taken F bracket $ in any satisfies 1 3 admissible closing square a locally certain bracket convex normability period t opology of the condition dual pairing (hF,F y0 )i which[13]. A will lcs which b e given Wesatisfies recall that ( a y lcs ) F satisfiesadmits a open continuous parenthesis norm y .closing Conversely parenthesis , any if there lcs which is V sub has 1 the in Ubounded open parenthesis approximation F closing parenthesis such that \noindentLineproperty 1 F tob the and elow power a continuous , ofthen prime there norm = union , isatisfies ofs Fa to nuclear ( they ) power [ 13 K of] . $ prime\ Weddot open also{o refer} square$ the the bracket reader space V t subo [ 14 $1 to ]\ andlambda the [ power 2 1 ] forof( circ A closing ) square $ and bracket a quotient map cap$ Q V circfurther : Line F facts 2 V\ about inrightarrow U open this parenthesis condition\lambda F and closing it s( parenthesisrelation A t o ) $ other normability conditions . We also note suchwherethat that closure by\quad a is nuclear taken$ in K QT anyo ¨ the admissible : space E we locally\rightarrow mean convex a nuclear t opology Fr\lambdae ´ chet of the space dual( pairing which A has ) $ a \quad i s \quad a l s o \quad a quotient \quad map . \quad A simple consequence of thisangbracketleftbasis theorem and admits F i comma s a that continuous F to\ thequad power normeither of . prime Our every right first and angbracket continuous main result open linear is square a Ko ¨ bracketthe map space\quad 13 closingversionfrom square of Theorem a given bracket 2 Fr. period 3 $ ..\acute A lcs which{e} satisfies$ chet ..space openand parenthesis\quad it generalizes$ y closingE $the theorem\ parenthesisquad i in n t [ o.. 13 admits\ ]quad which aany wecontinuous have nuclear just norm discussed K period $ \ . ddot .. Conversely{o} $ comma the \quad space \quad i s \quad boun $ l i n e −d $ ed any\quad lcs3 which .or 1 .\ hasquadTheorem the bounded$ E . $Let approximation hasT :\Equad→ Fa propertybe nuclear continuous and K a and continuous $ assume\ddot{ norm thato} $ commaF satisfies the ( y ) . If there is a quotient [ 1 3 ] . satisfiescompleting open parenthesis sequence (A yn closing) in parenthesisE such that ..each openT ( squareAn) is bracket unbounded 13 closing then square there is bracket a nuclear period K o¨ ..the We space also refer the reader t o open squareλ bracket(B) and 14 closing a faithful square surjection bracketQ and: F open→ λ( squareB) such bracket that QT 2 1 closing: E → λ square(B) is bracket als o afor faithful further surjection facts about . \ centerline {We recall that a lcs $F$ satisfies ( y ) if there is $V { 1 }\ in U ( F ) $ this ..P condition r o o f . .. andLet V it1 sb relation e the neighborhood t o .. other normabilityin condition .. ( y conditions ) . Using period alternately .. We ..the also continuity note of T and the such that } thatassumption by .. a nuclear on T ( KAn o-dieresis), we determine the .. space 1 = n we1 < mean n2 < a ... nuclearand Vk Fr acute-e∈ U(F ) chet so .. that spaceVk+1 which⊂ hasV ..k,T a (Ank ) ⊂ 0 0 ◦ ◦ basisVk andand admitsT (F a[V continuous1 ] ∩ Vk+1) norm i s not period .. Our first and main result is a K dieresis-o the space \ [ \versionbegin { ofa Theorem l i g n e d } 2F period ˆ{\ 3prime and it generalizes} = \bigcup the theoremF in ˆ open{\prime square bracket} [ 13 V closing ˆ{\ circ square} bracket{ 1 } which] we\ havecap justV \ circ \\ Vdiscussed\ in periodU(F) \end{ a l i g n e d }\ ] 3 period 1 period Theorem period Let T : E right arrow F be continuous and assume that F satisfies open parenthesis y closing parenthesis period \noindentIf there iswhere a completing closure sequence is taken open parenthesis in any admissibleA sub n closing locally parenthesis convex .. in E t such opology that each of T the open dual parenthesis pairing A sub n closing parenthesis$ \ langle .. is unboundedF , F ˆ{\prime }\rangle [ 13 ] . $ \quad A lcs which satisfies \quad ( y ) \quad admits a continuous norm . \quad Conversely , anythen lcs there which is a nuclear has the K dieresis-o bounded the approximation space lambda open property parenthesis and B closing a continuous parenthesis norm .. and ,a faithful surjection Q : F right arrow lambdasatisfies open parenthesis ( y ) \quad B closing[ 13 parenthesis ] . \quad We also refer the reader t o [ 14 ] and [ 2 1 ] for further facts about t hsuch i s \ thatquad QTc : o E n d right i t i o arrow n \quad lambdaand open it parenthesis s relation B closing t o \ parenthesisquad other .. is normabilityals o a faithful surjection\quad conditions period . \quad We \quad a l s o note thatP r o by o f\ periodquad ..a Let nuclear V sub 1 K b e $the\ neighborhoodddot{o} $ in the condition\quad openspace parenthesis we mean y closing a nuclear parenthesis Fr period $ \acute .. Using{e} alternately$ chet the\quad space which has \quad a continuity of T and the assumption on T open parenthesis A sub n closing parenthesis comma we determine 1 = n sub 1 less n sub 2 less period\noindent periodbasis period and admits a continuous norm . \quad Our first and main result is a K $ \ddot{o} $ the space versionand V sub of k inTheorem U open parenthesis 2 . 3 and F closing it generalizes parenthesis .. the so that theorem V sub k in plus [ 1 13 subset ] which V sub k we comma have T just open parenthesis A sub n sub k closingd i s c u parenthesis s s e d . subset V sub k .. and T to the power of prime open parenthesis F to the power of prime open square bracket V sub 1 to the power of circ closing square bracket cap V sub k plus 1 to the power of circ closing parenthesis i s not 3.1.Theorem. Let $T : E \rightarrow F $ be continuous and assume that $ F $ satisfies ( y ) . If there is a completing sequence $ ( A { n } ) $ \quad in $E$ suchthateach $T ( A { n } ) $ \quad is unbounded then there is a nuclear K $ \ddot{o} $ the space $ \lambda ( B ) $ \quad and a faithful surjection $ Q : F \rightarrow \lambda ( B ) $ such that $QT : E \rightarrow \lambda ( B ) $ \quad is als o a faithful surjection .
P r o o f . \quad Let $ V { 1 }$ b e the neighborhood in condition ( y ) . \quad Using alternately the continuity of $T$ and the assumption on $T ( A { n } ) ,$ wedetermine $1 = n { 1 } < n { 2 } < . . . $ and $ V { k }\ in U ( F ) $ \quad so that $ V { k + 1 }\subset V { k } ,T( A { n { k }} ) \subset V { k }$ \quad and $ T ˆ{\prime } ( F ˆ{\prime } [ V ˆ{\ circ } { 1 } ] \cap V ˆ{\ circ } { k + 1 } ) $ i s not Subspaces and quotient spaces of locally convex spaces .. 2 1 \ hspaceabsorbed∗{\ byf i A l l sub} Subspaces n sub k to the and power quotient of circ period spaces .... Noteof locally that if there convex i s some spaces m such\ thatquad T to2 the1 power of prime open parenthesis F to the power of prime open square bracket V sub 1 to the power of circ closing square bracket cap V to the power of circ closing parenthesis ....\noindent i s absorbed by $ A ˆ{\ circ } { n { k }} . $ \ h f i l l Note that if there i s some $m$ such that $ Tabsorbed ˆ{\prime by .. A} sub( m to F the ˆ{\ powerprime of circ} ..[ for V every ˆSubspaces{\ ..circ V in Uand} open{ quotient1 parenthesis} spaces] of\ Fcap locally closing convexV parenthesis ˆ{\ spacescirc comma} 2 1 ) .. $ then\ h by f i l.. l conditioni s .. absorbed by A◦ . Note that if there i s some m such that T 0(F 0[V ◦] ∩ V ◦) i s nk 1 open parenthesis y closing◦ parenthesis comma .. we would have 0 0 \noindentabsorbedabsorbed by Am byfor\quad every $V A ∈{ m U(F})ˆ,{\thencirc by}$ condition\quad f o( r y ) every , we would\quad have$T V(F )\⊂insp U ( F ) , $ T to the◦ power of prime open parenthesis F to the power of prime closing parenthesis subset sp open parenthesis A sub m to the power \quad(Athenm), which by \ wouldquad implyc o n d that i t i o nT (A\quadm) is bounded( y ) ., \quad Denotewe (A wouldnk ) by (A haven) t o simplify the notation . Let of circ closing parenthesis comma which would imply that T open◦ parenthesis◦ A sub m closing parenthesis is bounded period .. Denote open $ Tk ˆ{\ · kprimek and |} · | k(denote F ˆ resp{\prime ectively} the gauges) \subset of Vk and$Ak spdefined $ ( on the A ˆ spans{\ circ of these} { setsm .} We) , $ which would imply that parenthesis A sub n sub k closing parenthesis0 ◦ by $ Topenchoose ( parenthesis Axn ∈{ E,m A zn} sub∈ A) nn closing, $ un is∈ F parenthesis bounded[V1 ] and L t .n o∈\ simplifyquad F(E)Denote such the notationthat $ ( period A Let{ barn times{ k }} bar k) and $ bar by times bar k denote resp ectively $ ( A { n } ) $ t o simplify the notation . Let $ \ parallel \cdot \ parallel k $ and $ \mid the gauges (i)T 0u (x ) = δ , \cdotof V sub\mid k to thek power $ denote of circ and resp A sub ectively k to the power the of gauges circ definedn m on then,m spans of these sets period .. We choose x sub n in E comma o f $ V ˆ{\ circ } { k }$ and $ A ˆ{\ circ } { k }0$ defined⊥ on the spans of these sets . \quad We choose z sub n in A sub n comma (ii)T un+1 ∈ Ln , $ x { n }\ in E , z { n }\0 in A n+1{ n } , $ u sub n in F to the power of prime open(iii) square| T u bracketn | k > 2 V subk 1un tok thek + power 1forn of> circk, closing square bracket and L sub n in F open parenthesis E closing$ u { parenthesisn }\ in such thatF ˆ{\prime } [ V ˆ{\ circ } { 1 } ] $ and $ L { n }\ in F ( E )$ suchthat 0 1 0 Line 1 open parenthesis( iv i closing ) there parenthesis is y ∈ Ln ∩ TAk tosuch the that powerT ofun prime(y) > 2 u sub| T u nn open| k for parenthesisn > k, x sub m closing parenthesis = delta sub n comma\ [ \ begin m comma{ a l i g n Linee d } 2( open i parenthesis ) T iiˆ{\ closingprime parenthesis} u { T ton the} power( x of{ primem } u sub) n = plus 1\ delta in L sub{ n ton the , power m } of bottom, \\ comma( Line i i 3 open ) parenthesisT ˆ{\prime iii closing} u parenthesis{ n(v) +x barn ∈ T 1sp to{}\z the1, ...,in power zn}. L of prime ˆ{\bot u sub} n{ barn k} greater, \\ 2 to the power of n plus 1 bar u sub n bar( k plus i i 1 i for n ) greater\mid k commaST∞ ˆ{\prime } u { n }\mid k > 2 ˆ{ n + 1 }\ parallel u { n }\ parallel We set M = sp ({xn} ∪ n=1 Ln) and consider the quotient space M/L where kopen + parenthesis 1 f o r iv closing n > parenthesisk , there\end{ isa y l i ing n L e d sub}\ ] n cap A sub k such that T to the power of prime u sub n open parenthesis y closing parenthesis greater 2 to the power of 1 bar T to0 the power of⊥ prime u sub n bar k for n greater k comma L = {T un : n ∈ N} ∩ M. open parenthesis v closing parenthesis x sub n in sp open brace z sub 1 comma period period period comma z sub n closing brace period \ centerlineWeThe set M image ={ sp( of open iv (xn) ) parenthesis spans thereM/L. is openLet $yB bracek be the x\ in sub image n closingL of Γ({ {xn brace1,}\ ..., xcupncap−1 union}∪AkA) of∩M sub{ ink nM/L. =}$ 1 to such theEach powerB thatk is of absorbent infinity $Tˆ L{\ subprime n closing} parenthesisu { n } and( consider yand )it the s gauge> quotient defines2 ˆ{ space1 a}\ seminorm M slashmid L on whereTM/L. ˆ{\primeOn M/L} weu consider{ n }\ the metrizablemid k $t opology f o r determined $ n > byk , $ } L =the open sequence brace T (B tok). theThen power the of canonical prime u maps sub n : n in N closing brace to the power of bottom cap M period \ [ ( v ) x { n }\ in sp \{ z { 1 } , . . . , z { n }\} . \ ] The image of open parenthesis x sub n closing parenthesis0 spans⊥ M slash⊥ L period Let B sub k be the image of Capital Gamma open parenthesis open brace x sub 1 comma periodG = periodM/L period→ E/(T commaun) → x subF/( nun minus) 1 closing brace cup A sub k closing parenthesis cap M in Mare slash all continuous L period .. . Each We B have sub k is absorbent and it s gauge defines a seminorm on M slash L period .. On \noindentM slash LWeset we consider $M the metrizable =$ sp t opology $( determined\{ x by{ then }\}\ sequence opencup parenthesis\bigcup B subˆ{\ k closinginfty parenthesis} { n period = 1 Then} L { n } ) $the and canonical consider maps the quotient space0 ∼ $Mn / L $ where | T un |k > 2 k un k k + 1 forn > k G = M slash L right arrow E slash open parenthesis T to the power of prime u sub n closing parenthesis to the power of bottom right arrow \ [ L = \{ ∼ T ˆ{\prime } ◦u { n } :0 n \ in N \} ˆ{\bot }\cap M. \ ] F slashwhere open parenthesis| · |k is u the sub gauge n closing of B parenthesisk. If vn to= theT u powern, then of bottomvn(xm) = δn,m and G = sp {xn(L): aren all= continuous 1, 2, ...}. periodBy a .. result We have of Bellenot and Dubinsky [ 1 ] there i s a subsequence n ↑ ∞ such that the image of (x ) i s a basis of G/(u )⊥. By passing to a subsequence we bar T to the powerk of prime u sub n bar sub k to thenk power of thicksim greaternk 2 to the power of n bar u sub n bar k plus 1 for n greater k 0 \noindentwherecan .. assume barTheimage times (xn( barL)) sub is of a k basis to $ the ( in powerG xand{ of ifn thicksimu ∈} G )$ .. is spansthe gauge $M of B sub / k to L the power .$ of Let period $B to the{ powerk }$ of circbe the.. If v image sub n = of T to$ the\Gammathen power there of( prime are\{k u∈ subNxand n comma{ρ >1 0} so.. then that, v . sub n . open . parenthesis , x x{ subn m closing− 1 parenthesis}\}\ =cup delta subA n{ commak } m) .. and\cap G = M $ insp open $ M brace / xsub L n . open $ parenthesis\quad Each L closing $ B parenthesis{ k }$ : is n = absorbent 1 comma 2 commaand it period s gauge period defines period closing a seminorm brace period on .. $M By a / L . $ \quad On ∼ result of Bellenot and Dubinsky .. open square bracketu(xn 1(L closing)) | vn square|k < ρ bracket .. there i s a $M / L $ we consider the metrizable t opology determined by the sequence $ ( B { k } ) . $ subsequencefor all n. Let n subbk k= upwardsk u k arrow −k1 and infinityB = such{(bk ): thatk = the 1, 2 image, ...}. of λ open(B) i parenthesis s a nuclear Kx subo ¨ the n subspace k .closing We parenthesis define i s a basis of G slash Then n n n open parenthesisQ : F → λ( uB sub) by nQy sub= k (u closingn(y)). parenthesisThe maps to the power of bottom period .. By thepassing canonical to a subsequence maps we can assume open parenthesis x sub n open parenthesis L closing parenthesis closing parenthesis is a basis in G 0 ⊥ ⊥ and if u in G to the power of prime G → E/(T un) → F/(un) → λ(B) \ [then G there = are M k in / N and L rho\rightarrow greater 0 so thatE / ( T ˆ{\prime } u { n } ) ˆ{\bot }\rightarrow F 0 0 0 0 /u open (are continuous parenthesisu { n } and x sub)T ˆQ n{\ open(λbot(B) parenthesis)}\ = G] . L closing parenthesis closingThe parenthesis image of barG in v subλ(B n) bar i s dense sub k and to the so power of thicksim less rho 0 0 forT allQ ni period s one - Let to - b one sub . n By to the the closedpower range of k = theorem bar u sub the n completion bar minusGb kof toG theand powerλ(B) ofare 1 isomorphic and B = open and also brace open parenthesis b sub 0 0 n to theT powerQ i s an of isomorphism k closing parenthesis . This : gives k = that 1 commaQT : G 2 comma→ λ(B) period i s a dense period imbedding period closing . So the brace closure period of the lambda image open parenthesis B closing \noindent are all continuous . \quad We have parenthesiso f − iline s aB nucleark in λ(B K) i dieresis-o s a neighborhood the . Hence QT (Ck +Ak) is also a neighborhood where Ck = Γ{z1, ..., zk−1}. spaceSince period (Ck ..+ WeAk) define is also Q a : F completing right arrow sequence lambda, open(QT ( parenthesisCk + Ak)) Bis closing a base ofparenthesis neighborhoods by Qy by = open Lemma parenthesis 2 . 1 . u sub n open parenthesis 0 0 0 y\ [ closing\midNow parenthesis asT in ˆ part{\ closingprime ( ii ) of parenthesis} Theoremu { 2periodn .}\ 3 it .. canmid The be mapsˆ shown{\sim that} Q{ kand} T >Q are2 imbeddings ˆ{ n }\ withparallel respect to theu { n }\ parallel kG + rightstrong 1 arrow topologies f Eo r slash . n open> parenthesisk \ ] T to the power of prime u sub n closing parenthesis to the power of bottom right arrow F slash open parenthesis u sub n closing parenthesis to the power of bottom right arrow lambda open parenthesis B closing parenthesis are continuous and T to the power of prime Q to the power of prime open parenthesis lambda open parenthesis B closing parenthesis to the power\noindent of primewhere closing\quad parenthesis$ \ =mid G to the\cdot power of\mid primeˆ{\ periodsim ....} The{ k image}$ of\quad G in lambdais the open gauge parenthesis of $Bˆ B closing{\ parenthesiscirc } { ik s ˆ{ . }}$ dense\quad andI f so $ v { n } = T ˆ{\prime } u { n } , $ \quad then $ v { n } ( x { m } ) = \ delta { n ,T m to} the$ power\quad ofand prime $ Q G to the = power $ of prime i s one hyphen to hyphen one period .. By the closed range theorem the completion G-hatwidesp $ \{ of G andx { lambdan } open(L):n=1,2,... parenthesis B closing parenthesis \} . $ \quad By a result of Bellenot and Dubinsky \quad [ 1 ] \quad ther e i s a are isomorphic and also T to the power of prime Q to the power of prime i s an isomorphism period .. This gives that QT : G right arrow lambda\noindent open parenthesissubsequence B closing $ n parenthesis{ k }\uparrow \ infty $ such that the image of $ ( x { n { k }} ) $ isabasisofi s a dense imbedding $G period / So the ( closure u { ofn the{ imagek }} o f-line) ˆ B{\ subbot k in} lambda. $ open\quad parenthesisBy B closing parenthesis i s a neighborhood periodpassing to a subsequence we can assume $ ( x { n } ( L ) )$ isabasisin $G$ andif $u \ inHenceG QT ˆ{\ openprime parenthesis}$ C sub k plus A sub k closing parenthesis is also a neighborhood where C sub k = Capital Gamma open brace z sub 1 comma period period period comma z sub k minus 1 closing brace period .. Since \noindentopen parenthesisthen thereC sub k are plus A $ sub k k closing\ in parenthesisN $ and is also$ \rho a completing> 0 sequence $ so comma that open parenthesis QT open parenthesis C sub k plus A sub k closing parenthesis closing parenthesis is a base of neighborhoods \ [by u Lemma (2 x period{ n 1} period(L)) Now as in part open\mid parenthesisv { iin closing}\mid parenthesisˆ{\sim of Theorem} { k } 2 period< \ 3rho it can\ ] be shown that Q to the power of prime and T to the power of prime Q to the power of prime are imbeddings with respect to the strong topologies period \noindent forall $n .$ Let $bˆ{ k } { n } = \ parallel u { n }\ parallel −{ k }ˆ{ 1 }$ and $ B = \{ ( b ˆ{ k } { n } ):k=1,2,... \} . \lambda ( B )$ isanuclearK $ \ddot{o} $ the space . \quad Wedefine $Q : F \rightarrow \lambda ( B )$by$Qy = ( u { n } ( y ) ) . $ \quad The maps
\ [G \rightarrow E / ( T ˆ{\prime } u { n } ) ˆ{\bot }\rightarrow F / ( u { n } ) ˆ{\bot }\rightarrow \lambda (B) \ ]
\noindent are continuous and $ T ˆ{\prime } Q ˆ{\prime } ( \lambda ( B ) ˆ{\prime } ) = G ˆ{\prime } . $ \ h f i l l The image of $G$ in $ \lambda ( B )$ isdenseandso
\noindent $ T ˆ{\prime } Q ˆ{\prime }$ i s one − to − one . \quad By the closed range theorem the completion $ \widehat{G} $ o f $ G $ and $ \lambda ( B ) $ are isomorphic and also $ T ˆ{\prime } Q ˆ{\prime }$ i s an isomorphism . \quad This gives that $ QT :G \rightarrow \lambda ( B ) $ i s a dense imbedding . So the closure of the image o $ f−l i n e B { k }$ in $ \lambda ( B ) $ i s a neighborhood . Hence $ QT ( C { k } + A { k } ) $ is also a neighborhood where $ C { k } = \Gamma \{ z { 1 } , . . . , z { k − 1 }\} . $ \quad Since $ ( C { k } + A { k } ) $ is also a completing sequence $ , ( QT ( C { k } + A { k } ) ) $ is a base of neighborhoods by Lemma 2 . 1 . Now as in part ( ii ) of Theorem 2 . 3 it can be shown that $Q ˆ{\prime }$ and $ T ˆ{\prime } Q ˆ{\prime }$ are imbeddings with respect to the strong topologies . 22 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentThere are several22 \quad immediateS $ corollaries . \ddot of{ thisO} theorem$ nal comma andT similar . Terzio to the corol $ \check hyphen{g} $ lu laries of Theorem 2 period 3 period .. We shall only give one period .. We may also get some more by There are several immediate corollaries of this theorem , similar to the corol − elaborating on¨ the condition open parenthesis y closing parenthesis period .. Here let E be a webbed space period .. We have seen lariesthat22 if its of S web Theorem.O inal s eventually andT 2 . . Terzio 3 bounded . gˇ\quadlu commaWe then shall E i only s equal give t o the one union . \ ofquad a sequenceWe may also get some more by elaboratingof boundedThere subsets are on several the period condition immediate corollaries ( y ) . of\quad this theoremHere , let similar $ toE the$ corolbe a - webbed laries of Theorem space . 2\ .quad 3 . We have seen that3 periodWe if shall 2 its period only web give Corollary i one s eventually . period We may Let also E bounded be get a webbed some , more then space by which elaborating $ E satisfies $ i on s open the equal condition parenthesis t o ( the y y ) closing . union Here parenthesis oflet E abe sequence period .. Then either E ofhas boundeda a webbed faithful subsetsspace nuclear . K We .dieresis-o have seen the thatquotient if its lambda web i s open eventually parenthesis bounded A closing , then parenthesisE i s equal or t o E the is equal union to of the a union of a sequence of boundedsequence subsets of bounded period subsets . \ hspace4 period∗{\ ..f Nuclear i l l }3 K . dieresis-o 23 . . 2 .CorollaryCorollary the subspaces . . LetLet andE completing $be E a $ webbed be sequences a space webbed which space satisfies which( y ) . satisfies Then either ( yE ) . \quad Then e i t h e r $ EIn $ Shas 1 .. a we faithful have nuclear shown that K o¨ ifthe T : quotient E right arrowλ(A) For i sE notis equal .. almost to the bounded union of and a sequence E i s of bounded subsets . Fr acute-e chet comma4 . .. Nuclear then E has K .. a nuclearo¨ the subspaces K dieresis-o the and subspace completing so that the sequences restriction of T to \noindentthis subspaceIn §has1 i we s a an have faithful i somorphism shown that nuclear open if T parenthesis: KE $ \→ddot PropositionF{oi} s not$ 1 the period almost quotient 14 bounded closing and parenthesis $ \Elambdai s Fr periode ´ chet( .., We A then willE ) now $ generalize or $E$ this is equal to the union of a sequence offurther boundedhas by a using nuclear subsets an K ideao ¨ the . ofsubspace Valdivia open so that square the restriction bracket 1 of7 closingT to this square subspace bracket i s an period i somorphism ( Proposition 4 period1 . 14 1 ) period . We Proposition will now generalize period Let this T : E right arrow F be continuous period .. Suppose open parenthesis A sub n closing parenthesis is\ centerline a completfurther hyphen by{4 using . \quad an ideaNuclear of Valdivia K [ 1 $ 7\ ]ddot . {o} $ the subspaces and completing sequences } ing s equence in E such4 that . 1. TProposition open parenthesis . Let sp openT : E parenthesis→ F be continuous B closing . parenthesis Suppose cap(A An) subis a n complet closing parenthesis - .... is not almost boundedIn \Sing1 for s\ equencequad all n andwehave in E such shown that T that( sp (B if) ∩ A $Tn) : E \rightarrowis not almostF $ bounded i s not for all\quadn andalmost bounded and $ E $ i sforfor some some Banach Banach disc disc B periodB. .... Then thereThen is there a nuclear is a nuclear K dieresis-o K o¨ thethe subspace subspace lambdaλ(A) open parenthesisof E Aso closing parenthesis .... of E Frthat $ \acute the restriction{e} $ of chet T to ,λ(\Aquad) is anthen isomorphism $ E $ . has \quad a nuclear K $ \ddot{o} $ the subspace so that the restriction of so −n $ Tthat $ the to restrictionP r of o oT f to . lambda Let Bn open= 2 parenthesisB + An. A(B closingn) is a parenthesis completing .. sequence is an isomorphism in E. Define period this subspace i s an i somorphism ( Proposition 1 . 14 ) . \quad We will now generalize this P r o o f period .. Let B sub n = 2 to the power of minus n B plus A sub n period∞ open parenthesis B sub n closing parenthesis is a completing sequence in E period .. Define \noindent further by using an ideaX of Valdivia [ 1 7X ] . Line 1 infinity Line 2 C sub k = openCk brace= { sumξjx xij : subxj ∈ j xB subk+j− j1 :, x sub j| inξj B|≤ sub1} k plus j minus 1 comma sum bar xi sub j bar less or equal 1 closing brace Line 3 i = 1 i = 1 \ hspaceand E∗{\ sub 0f i = l l intersection}4 . 1 . of Proposition sub k = 1 to the . power Let of $T infinity : sp open E parenthesis\rightarrow C sub kF closing $ be parenthesis continuous period . On\quad E sub 0Suppose the set s $ ( A { n T}∞ ) $ is a complet − −1 r to theand powerE0 = of minusk=1 sp 1 ( openCk). parenthesisOn E0 the set E sub s r 0( capE0 ∩ CC subr) define r closing a complete parenthesis locally define con a - complete locally con hyphen vexvex metrizable metrizable t opology t opology which which i s finer i s finer than than the one the inducedone induced by E by openE([17]; parenthesis Prop . open6 ) . square bracket 1 7 closing square bracket semicolon \noindent ingsequencein $E$ suchthat $T ($ sp−1 $( B ) \cap A { n } ) $ \ h f i l l is not almost bounded for all Prop periodTherefore 6 closing the canonical parenthesis imbedding period j : E0 → E is continuous . If T j(r (E0 ∩ Cr)) i s almost bounded for $ n $ and Thereforesome r, thesince canonicalB ⊂ Bn imbedding⊂ Cn for all j :n, E T sub(E[B 0] right∩ Ar) arrow would E b is e continuous almost bounded period . If From Tj open this parenthesis contradiction r to by the power of minus 1 open parenthesisProposition E sub 0 cap 1 . C 14 sub we rconclude closing parenthesis that T j : λ( closingAe) → F parenthesisi s an isomorphism for some nuclear Ko ¨ the space λ(Ae) \noindent for some Banach disc $B . $ \ h f i l l Then there is a nuclear K $ \ddot{o} $ the subspace i s almostwhich bounded for some r comma since B subset B sub n subset C sub n for all n comma T open parenthesis E open square bracket B $ \lambda ( A ) $ \ h f i l l o f $ E $ so Ae closingi square s isomorphic bracket t cap o a subspaceA sub r closing of E0. parenthesisWe let λ( wouldA) = j(λ( )). b e almost bounded period .. From thisLet contradictionA ⊂ E b e such by Propositionthat for some ..U 1 period∈ U(E) 14 and we for conclude each L ∈ F(E0) the set \noindent⊥ that the restriction of T to $ \lambda ( A ) $ \quad is an isomorphism−n . thatA Tj∩ L : lambdai s not absorbed open parenthesis by U. By to Mazur the power ’ s method of A-tildewide , as in the closing proof parenthesis of Theorem right 1 . 6, arrow we choose F i sx ann ∈ isomorphism4 A for some nuclear K dieresis-osuch the that spacex lambda6∈ U and open (x ) parenthesis i s a qu−basic to thesequence powerof with tildewide-A basis constant closingδ parenthesis > 0. Let whichD = {x : n ∈ } and \ centerline {Pn r o o f . n\quad Let $ B { n } = 2 ˆ{ − n } B + An { n N} .(B { n } ) $ i s isomorphicsuppose for t some o a subspaceL ∈ F(E of0), Eρ > sub0 we 0 period have .. We let lambda open parenthesis A closing parenthesis = j open parenthesis lambda open parenthesisis a completing to the power sequence of A-tildewide in $ closing E parenthesis . $ \quad closingDefine parenthesis} period Let A subset E b e such that for some U in U opensp(D parenthesis) ∩ A ∩ L⊥ E⊂ closingρU. parenthesis and for each L in F open parenthesis E to the power of prime\ [ \ begin closing{ a parenthesisl i g n e d }\ theinfty set \\ CA capThen{ k L to} the= power\{\ of bottomsum i s\ notxi absorbed{ j } byx U period{ j } .. By: Mazur x { quoterightj }\ in s methodB comma{ k as + in the j proof− of1 Theorem} , 1 period\sum 6\mid comma \ xi { j }\mid \ leq 1 \}\\ ⊥ 1k ⊥ iwe choose = 1x sub\end n in{ a 4 l ito g n the e d power}\ ]U ∩ ofL minus∩ sp{ nx An such: n ≥ thatk} ⊂ x2 subδA n∩ negationslash-elementL ⊂ ρδ2kU. U and open parenthesis x sub n closing parenthesis i s a q u to the power of hyphen basic sequence with basis constant delta greater 0 period .. Let D = open brace x sub n : n in N closing brace and suppose for some L in F open parenthesis E to the power\noindent of primeand closing $ Eparenthesis{ 0 } comma= \ rhobigcap greaterˆ 0{\ infty } { k = 1 }$ sp $ ( C { k } ) . $ On $ E { 0 }$ thewe s have e t s $ r ˆ{ − 1 } (E { 0 }\cap C { r } ) $ define a complete locally con − sp open parenthesis D closing parenthesis cap A cap L to the power of bottom subset rho U period \noindentThen vex metrizable t opology which i s finer than the one induced by $E ( [ 1 7 ] ; $ PropU cap . 6 L to) the. power of bottom cap sp open brace x sub n : n greater equal k closing brace subset 2 to the power of 1 k delta A cap L to the power of bottom subset rho delta 2 sub k U period \noindent Therefore the canonical imbedding $ j : E { 0 }\rightarrow E$ is continuous . If $ Tj ( r ˆ{ − 1 } (E { 0 }\cap C { r } ) ) $ i s almost bounded for some $ r , $ since $B \subset B { n }\subset C { n }$ f o r a l l $ n ,T(E[B] \cap A { r } ) $ would b e almost bounded . \quad From this contradiction by Proposition \quad 1 . 14 we conclude that $ Tj : \lambda ( ˆ{\ widetilde {A}} ) \rightarrow F $ i s an isomorphism for some nuclear K $ \ddot{o} $ the space $ \lambda ( ˆ{\ widetilde {A}} ) $ which
\noindent i s isomorphic t o a subspace of $ E { 0 } . $ \quad We l e t $ \lambda ( A ) = j ( \lambda ( ˆ{\ widetilde {A}} ) ) . $
\ hspace ∗{\ f i l l } Let $ A \subset E$ b e such that for some $U \ in U ( E )$ andforeach $ L \ in F ( E ˆ{\prime } ) $ the s e t
\noindent $ A \cap L ˆ{\bot }$ i s not absorbedby $U . $ \quad By Mazur ’ s method , as in the proof of Theorem 1 . 6 , we choose $ x { n }\ in 4 ˆ{ − n } A$ such that $x { n }\not\ in U $ and $ ( x { n } ) $ i s a $ q u ˆ{ − b a s i c }$ sequence with basis constant $ \ delta > 0 . $ \quad Let $ D = \{ x { n } : n \ in N \} $ and suppose for some $ L \ in F ( E ˆ{\prime } ), \rho > 0 $ we have
\ [ sp ( D ) \cap A \cap L ˆ{\bot }\subset \rho U. \ ]
\noindent Then
\ [U \cap L ˆ{\bot }\cap sp \{ x { n } : n \geq k \}\subset 2 ˆ{ 1 { k }}\ delta A \cap L ˆ{\bot }\subset \rho \ delta { 2 } { k } U. \ ] Subspaces and quotient spaces of locally convex spaces .. 23 \ hspaceThis would∗{\ f then i l l } implySubspaces that sp and open quotientparenthesis D spaces closing ofparenthesis locally cap convex L to the spaces power of\ bottomquad has23 a precompact neighborhood period .. With \noindentthis contradictionThis would we have then proved imply the following that sp t echnical $ ( result D period ) \cap L ˆ{\bot }$ has a precompact neighborhood . \quad With this contradiction we have proved the followingSubspaces t and echnical quotient spaces result of locally . convex spaces 23 4 period 2 period Lemma period .. If A subset⊥ E is not almost bounded comma .. there is a countable subset D of AThis such would that thensp open imply parenthesis that sp (D ) closing∩ L has parenthesis a precompact cap A neighborhood is still not almost . With bounded this periodcontradiction we have 4 .Our 2proved immediate . Lemma the following goal . \quad is to t examineechnicalI f $resultthe A relation\ . subset b etweenE the $ existence is not of almost an bounded , \quad there is a countable subset $ Dunbounded $ 4 . 2 continuous . Lemma . operatorIf A from⊂ E ais Fr not acute-e almost chet bounded space into, athere given is lcs a countable E and the subset D of A such that ofexistencesp $A$ (D) of∩ aA suchthatsp completingis still not sequencealmost $( bounded open D parenthesis . ) \ Acap sub nA closing $ is parenthesis still not in E suchalmost that bounded each A sub . n i s unbounded or not almostOur immediatebounded period goal is .. to Before examine this the we relation need another b etween concept the existence period .. of Following an unbounded Bonet continuous open square operator bracket 3 closing square bracket commaOur immediatefrom a Fre ´ chet goal space is into to aexamine given lcs E theand relation the existence b etweenof a completing the existence sequence (An of) in anE such that each unboundedweA sayn i that s unbounded continuous a lcs E has or the not operator individual almost bounded countable from . a Frboundedness Before $ this\acute we property need{e} another$ open chet parenthesis concept space . icbp into Following closing a given Bonet parenthesis [lcs 3 ] , if $ for E $ and the existencewe say that of a a lcs completingE has the individual sequence countable $ ( boundedness A { n property} ) $( in icbp $E$) if for each such sequence that (x eachn) in E $A { n }$ i s unbounded or each sequence open parenthesis x subx n closing parenthesis in E we can find rho n greater 0 so that open brace rho n to the power of x n : n innot N closingwe almost can brace find bounded iρn s a> bounded0 so . that\quad{ρn Beforen : n ∈ N} thisi s a webounded need subset another of E. concept . \quad Following Bonet [ 3 ] , wesubset say4 of that . E 3 period . aTheorem lcs $ E . $ hasSuppose theE individualis a countable locally complete boundedness space property which ( has icbp the ) if for each4 periodic sequence bp . 3 period Suppose .. $ TheoremT (: E x→ periodF{ isn such} .. Suppose that)$ for in someE .. $E$is completing .. a .. locally wecanfind sequence .. complete(An) $ ..\ spaceinrhoE, .. whicheachn > ..T ( hasAn0) .. $ theis not so.. ic that bp period $ \{\rho n ˆSuppose{ almostx } Tn bounded : E right : . arrow n Then\ Fin is there suchN is that a nuclear\} for$ some K i completing so¨ the abounded space sequenceλ(B) which open parenthesis is is omorphic A sub n to closing a subspace parenthesis .. in E comma .. each subsetT openof E of parenthesissuch $E A that . sub $ the n closing restriction parenthesis of T to .. isλ( notB) almostis boundedan is omorphism period .. . Then there is a nuclear K dieresis-o the space lambda open parenthesisP r o o B f . closing By Lemma parenthesis 4 . 2 .. , which for each n there is a countable set Dn such that T ( sp (Dn) ∩ An) 4 .is 3isi omorphics . not\quad almost ..Theorem to bounded .. a subspace .\ Byquad .. the ofSuppose icbpE .. such and the ..$ that assumption E the $ ..\quad restriction that iE s i\ s ofquad locally T to lambdaa complete\quad open ,l we o parenthesis c cana l l y find\quad a B Banach closingcomplete parenthesis\quad .. isspace .. an \quad which \quad has \quad the \quad i c bp . Supposeis omorphismdisc D $Twith periodDn :⊂ sp E (D) for\rightarrow each n. So theF result $ is follows such from that Proposition for some 4 . 1 completing . sequence $ ( A { n } ) $ \quadP r oin o f period $ E .. By , $ Lemma4\quad . 4 .4Theorem periodeach 2 comma . Let ..E forbe each a locally n there complete is .. a countablelcs . Consider set .. D the sub following n .. such state that - $T T open (parenthesis A { spn open} ) parenthesis $ \quad Dis sub not n closing almost parenthesis bounded cap A . sub\quad n closingThen parenthesis there is i s a not nuclear almost bounded K $ \ periodddot{ ..o} By$ the the space icbp$ \lambda and the assumption( B that ) $E \quad which ments : isi s is locally omorphic complete\quad commato we\ canquad finda a Banachsubspace disc\ Dquad with Do f sub $ n E subset $ \ spquad opensuch parenthesis\quad Dthat closing the parenthesis\quad forrestriction each n period of .. T to So$ \lambda ( B ) $ \quad i s \quad an is omorphism( a ) There . is continuous operator from a Fr e´ chet space into E which is not almost thebounded result follows . from Proposition 4 period 1 period 4 period 4 period Theorem( b )E periodhas a subspace Let E be which a locally is is complete omorphic lcs to period a nuclear .. Consider K o¨ the the space following . state hyphen P rments o o : f . \quad By Lemma 4 . 2 , \quad for each $ n $ there is \quad a countable set \quad $ D { n }$ \quad such( c ) thatThere is a completing s equence (An) in E such that each An is not almost bounded . openWe parenthesis have ( a ) a⇔ closing( b ) ⇒ parenthesis( c ) . .. If ThereE has .. is the .. ic continuous bp then all .. three operator statements from a Frare acute-e equivalent chet . space into E .. which is not $almost T bounded ( $ sp period $ ( D { n } ) \cap A { n } ) $ i s not almost bounded . \quad By the icbp and the assumption that $ E $ P r o o f . The equivalence of ( a ) and ( b ) follows from Proposition 1 . 14 . ( b ) ⇒ ( c ) i s trivial . openThat parenthesis ( c ) ⇒ ( b b closing ) if E has parenthesis the icbp E follows has a subspacefrom Theorem which 4 is . is 3 .omorphic to a nuclear K dieresis-o the space period iopen s locally parenthesis complete c closing parenthesis , we can .. find There a is Banach a completing disc s equence $ D $ open with parenthesis $ D { An sub}\ n closingsubset parenthesis$ sp in $ E ( such D that )each $ R e m a r k . Here is a simple application of this result : Suppose each En i s an infinite - dimensional Aforeach sub n is not $n almost .$ \quad So theFr resulte ´ chet space follows which from admits Proposition a continuous norm 4 . and 1 . let boundedE = Q periodE . Let U (n) ⊃ U (n) ⊃ ... b e a base of neighborhoods of E where the We have openn parenthesis a closing parenthesis Leftrightarrowk k+1 open parenthesis b closing parenthesisn double stroke right arrow open paren- \ hspace ∗{\ f i l l }4(n .) 4 . Theorem . Let $ E $ be a locally complete lcs . \quad Consider the following state − thesis cgauge closing of parenthesiseach Uk i speriod a norm .. Ifline E− hason theEn. ic bpWe then set all three statements are equivalent period
P r o o f period .. The equivalence of open parenthesis(1) a closing( parenthesisk) and open parenthesis b closing parenthesis follows from Proposition \ begin { a l i g n ∗} 1k 1 period 14 period .. open parenthesis b closingAk = 2 parenthesis(Uk × ... double× Uk stroke× Ek+1 right× ... arrow). open parenthesis c closing parenthesis ments : i s trivialIt is period simple .. That t o verify open parenthesis that each cA closingi sparenthesis not almost double bounded stroke and right that arrow ( openA ) parenthesis i s a completing b closing parenthesis if E has the \end{ a l i g n ∗} k k icbp followssequence from in TheoremE. As 4a periodFre ´ chet 3 period space E enj oys the icbp and therefore it has a nuclear Ko ¨ the subspace . R e m a r k period .. Here is a simple application of this result : .. Suppose each E sub n i s ( aan ) infinite\quad hyphenThere dimensional\quad i s Fr\ acute-equad continuous chet space which\quad admitsoperator a continuous from norm a and Fr let $ \acute{e} $ chet space into $E$ \quadE =which product iE s sub not n period .... Let U sub k to the power of open parenthesis n closing parenthesis supset U sub k plus 1 to the power of openalmost parenthesis bounded n closing . parenthesis supset period period period b e a base of neighborhoods of E sub n where the gauge of each U sub k to the power of open parenthesis n closing parenthesis i s a norm line-o sub n E sub n period .. We set \ centerlineA sub k = 2{ to( theb power $ ) of E 1k $ open has parenthesis a subspace U sub which k to the is power isof omorphic open parenthesis to a nuclear 1 closing parenthesis K $ \ddot times{o} period$ the period space period . } times U sub k to the power of open parenthesis k closing parenthesis times E sub k plus 1 times period period period closing parenthesis period ( cIt) .. is\quad .. simpleThere t o verify is a that completing .. each A sub s k equence .. i s not .. $ almost ( Abounded{ n and} that) $ .. inopen $E$ parenthesis such A sub that k closing each parenthesis $A { ..n i} s$ a iscompleting not almost sequence in E period .. As a Fr acute-e chet space E enj oys the icbp and therefore it boundedhas a nuclear . K o-dieresis the subspace period \noindent We have ( a $ ) \Leftrightarrow ( $ b $ ) \Rightarrow ( $ c ) . \quad If $ E $ has the ic bp then all three statements are equivalent .
P r o o f . \quad The equivalence of ( a ) and ( b ) follows from Proposition 1 . 14 . \quad ( b $ ) \Rightarrow ( $ c ) i s t r i v i a l . \quad That ( c $ ) \Rightarrow ( $ b ) if $E$ has the icbp follows from Theorem 4 . 3 .
R e m a r k . \quad Here is a simple application of this result : \quad Suppose each $ E { n }$ i s an i n f i n i t e − dimensional Fr $ \acute{e} $ chet space which admits a continuous norm and let
\noindent $ E = \prod E { n } . $ \ h f i l l Let $ U ˆ{ ( n ) } { k }\supset U ˆ{ ( n ) } { k + 1 }\supset . . . $ be a base of neighborhoods of $E { n }$ where the
\noindent gauge of each $ U ˆ{ ( n ) } { k }$ i s anorm $ line −o { n } E { n } . $ \quad We s e t
\ [A { k } = 2 ˆ{ 1 { k }} ( U ˆ{ ( 1 ) } { k }\times ... \times U ˆ{ ( k ) } { k } \times E { k + 1 }\times ...). \ ]
\noindent I t \quad i s \quad simple t o verify that \quad each $ A { k }$ \quad i s not \quad almost bounded and that \quad $ ( A { k } ) $ \quad i s a completing sequence in $ E . $ \quad As a Fr $ \acute{e} $ chet space $ E $ enj oys the icbp and therefore it has a nuclear K $ \ddot{o} $ the subspace . 24 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentLet us now24 relax\quad the conditionS $ . open\ddot parenthesis{O} $ b closing nal andT parenthesis . Terzio of Theorem $ \ 4check period{g 4} so$ that lu we search for proper Fr acute-e chet subspaces only period \ hspace ∗{\ f i l l } Let us now relax the condition ( b ) of Theorem 4 . 4 so that we search for proper 4 period 5 period¨ Theorem period Let E be a complete lcs period .. Consider the following statements : open24 parenthesis S .O nal a and closing T . Terzio parenthesisgˇ lu .. There is an unbounded continuous operator from a Fr acute-e chet space into E period \noindentopen parenthesisFr $ b\ closingacute{ parenthesise}Let$ us chet now E has relax subspaces a proper the condition Fr only acute-e ( b. chet ) of subspaceTheorem period4 . 4 so that we search for proper openFr parenthesise ´ chet subspaces c closing only parenthesis . .. There is a completing s equence open parenthesis A sub n closing parenthesis in E such that each A\ centerline sub n is unbounded{44 . 5 5 period. Theorem . Theorem . Let . LetE be $E$ a complete be lcs a . complete Consider lcs the following . \quad statementsConsider: the following statements : } We have( a open ) There parenthesis is an aunbounded closing parenthesis continuous Leftrightarrow operator from open a Fr parenthesise´ chet space b closinginto E. parenthesis( b ) E has double a proper stroke right arrow open paren- thesis( a ) cFr closing\quade´ chet parenthesisThere subspace is . period an unbounded .. If E has the continuous ic bp then all operator three statements from aare Fr equivalent $ \acute period{e} $ chet space into $E . $ (bP r o $) o f period E$ .. open hasaproperFr parenthesis( c ) There b closing is a $ parenthesiscompleting\acute{ sdoublee} equence$ stroke chet(An right) subspacein arrowE such open. that parenthesis each An is a closing unbounded parenthesis . and open parenthesis b closingWe parenthesis have ( a ) double⇔ ( b )stroke⇒ ( c right ) . arrow If E openhas parenthesis the ic bp then c closing all three parenthesis statements are are trivial equivalent period . .. Suppose now there i s an unbounded \ hspacecontinuous∗{\ f operator i l l }(P cr T o )from o\ fquad . a Fr ( b acute-eThere ) ⇒ ( a chet is ) and aspace (completing b F ) ⇒ into( c E ) period are s trivial equence If T . i s not Suppose $ almost ( now A bounded there{ n i} comma s an) unbounded $ in $E$ such that each $ Athencontinuous{ En has}$ a nuclear operatoris unbounded KT o-dieresisfrom a . Fr thee ´ subspacechet space byF Theoreminto E. If 4 periodT i s not 4 period almost .. bounded If T i s almost , then E boundedhas a nuclear comma K theno¨ the T open subspace parenthesis by Theorem U closing 4 . 4 parenthesis . If T i s is almost an almost bounded bounded , subset for some U in U open parenthesis F closing parenthesis period \noindentthen TWe(U) ishave an almost ( a bounded $ ) \ subsetLeftrightarrow for some U ∈ U(F(). $ b $ ) \RightarrowBy Theorem( $ c 1 .) 9 . , \quad If $ E $ has the ic bp then all three statements are equivalent . .... By Theorem 1J period 9 comma E simeqE ' K toK the⊕ powerG and of J the oplus image G .... and of T the(U) image in .... ofG Tis open bounded parenthesis . U This closing means parenthesis that .... in ....T G is .... bounded period ....\ hspace Thiswould ....∗{\ means bef i bounded l l } ....P that r if o ....J owere T f . a\ finitequad set( . b $ ) \Rightarrow Since($ this a)and(b i s not the case E $)contains\Rightarrow a ( $ c ) are trivial . \quad Suppose now there i s an unbounded wouldcomplemented be bounded copy if J were of ω. a finiteSo ( seta ) ⇒ period( b ) .... . Since this i s not the case E contains a \noindentcomplementedcontinuous copy of omega operator periodSuppose .. $ each TSo $ openAn fromis parenthesis unbounded a Fr a .$ closing\ Thenacute parenthesis by{e a} simpler$ chetspace double version stroke of Lemma right $F$ 4 arrow . 2into we open parenthesis$E .$ b closing If $T$ parenthesisi s notcan find almost period a countable bounded subset , Dn such that each sp (Dn) ∩ An i s unbounded ( cf . also thenSupposeLemma $E$ each 2 . A 2 has sub ) . By n a is the nuclearK unbounded icbp and periodthe $ assumption\ddot .. Then{o} by that$ a simplerE theis complete subspace version , of we Lemma by find Theorem a Banach4 period 4 2 . we 4 . \quad If $T$ i s almost bounded , candisc findD asuch countable that D subsetn ⊂ sp D (D sub) for n all suchn. thatThen each sp sp(D) open∩ An parenthesisi s unbounded D sub . We n setclosing parenthesis cap A sub n i s unbounded open \noindent then $T ( U ) $ is an almost bounded subset for some $U \ in U ( F ) . $ parenthesis cf period also 1 \ hLemma f i l l By 2 period Theorem 2 closing 1 . parenthesis 9 , periodC Byn = the (2n icbpD + andAn) the∩ sp( assumptionD). that E is complete comma we find a Banach discCertainly D such that each DC subi n ssubset unbounded sp open and parenthesis the sequence D closing (C ) defines parenthesis a metrizable for all n locally period convex .. Then t sp opology open parenthesis on sp D closing parenthesis \noindent $ E \nsimeq K ˆ{ J }\oplus Gn $ \ h f i l l and the image \ h f i l l o f $ T ( U ) $ \ h f i l l in \ h f i l l cap A sub(D). n iIt s unbounded s completion periodF i s We a Fr sete ´ chet space and the exten - sion of the imbedding of sp (D) into E i s an $ G $ i s \ h f i l l bounded . \ h f i l l This \ h f i l l means \ h f i l l that \ h f i l l $ T $ C subunbounded n = parenleftbigg continuous sub operator 2 sub n from to the power of 1 D plus A sub n parenrightbigg cap sp open parenthesis D closing parenthesis period Certainly each C sub n i s unbounded and the sequence open parenthesis C sub n closing parenthesis defines a metrizable locally \noindentconvex t opologywould on be sp bounded open parenthesis if $ D J closing $ were parenthesis a finite period set .. It . s\ completionh f i l l Since F i s a this Fr acute-e i s chetnot space the andcase the $exten E $ hyphen contains a sion of the imbedding of sp open parenthesis D closingF parenthesisintoE. into E i s an unbounded continuous operator from \noindentF into E periodcomplemented copy of $ \omega . $ \quad So ( a $ ) \Rightarrow ( $ b ) . Let .. ELet be ..E abe lcs .. a which lcs has which .. the has .. bounded the bounded .. approximation approximation .. property property period .. . That That .. is commais , there is an \ hspace ∗{\ f i l l }Suppose each $ A { n }$ is unbounded . \quad Then by a simpler version of Lemma 4 . 2 we thereequicontinuous is an equicontinuous net {Tα} netα ∈ openI of brace finite T rank sub operators alpha closing on E bracesuch sub that alphax = inlim IT ofα finitex for everyrank operatorsx ∈ E. onFor E a such that x =subset lim TA sub⊂ E alphawe define x for every x in E period .. For a subset A subset E we define \noindentA-hatwidecan = Capital find Gamma a countable open brace subset T sub alpha $ D open{ n parenthesis}$ such A closing that each parenthesis sp $: alpha ( Din I{ closingn } brace) period\cap A { n }$ i s unbounded ( cf . also We start with a t line-e chnical result period Ab = Γ{Tα(A): α ∈ I}. 4 periodWe start 6 period with aLemma t line − periode chnical .. open result parenthesis . a closing parenthesis A subset A-hatwide sub period \noindentopen parenthesisLemma b 2 closing . 2 ) parenthesis . By the If Eicbp = sp and open the parenthesis assumption B closing that parenthesis $ E $ for is some complete absolutely , convex we find subset a B Banach of E comma 4 . 6 . Lemma . ( a )A ⊂ Ab. then ( b ) If E = sp (B) for some absolutely convex subset B of E, then \noindentE-hatwidedisc = sp open $D$ brace T such sub alpha that open $D parenthesis{ n }\ E closingsubset parenthesis$ sp : $( alpha in D I closing )$ brace forall = sp open $n parenthesis .$ to\quad the powerThen sp $ ( D ) \cap A { n }$ i s unbounded . We set of hatwide-B closing parenthesis period Bb Eb = sp{Tα(E): α ∈ I} = sp( ). open parenthesis c closing parenthesis A is bounded Leftrightarrow A-hatwide is bounded period \ [Copen{ parenthesisn } = d closing ( { 2 parenthesis}ˆ{ 1 } If({ c A-hatwide )nA is} boundedD is almost +⇔ AAb boundedis{ boundedn } then .) A is\ almostcap boundedsp ( period D ) . \ ] open parenthesis e closing parenthesis( d ) If Ab Ifis E almost is complete bounded and then openA parenthesisis almost boundedA sub n .closing parenthesis is a completing s equence with 2 A sub n plus( 1 e subset ) If E Ais sub complete n comma and then(A ) is a completing s equence with 2A ⊂ A , then (Ab) is also a completing \noindent Certainly each $ C n { n }$ i s unbounded andn+1 the sequencen n $ ( C { n } ) $ defines a metrizable locally opens equence parenthesis . to the power of A-hatwide sub n closing parenthesis is also a completing s equence period convex t opologyonsp $( D ) . $ \quad It s completion $F$ i s aFr $ \acute{e} $ chet space and the exten − sion of the imbedding of sp $ ( D ) $ into $ E $ i s an unbounded continuous operator from
\ begin { a l i g n ∗} F i n t o E . \end{ a l i g n ∗}
Let \quad $ E $ be \quad a l c s \quad which has \quad the \quad bounded \quad approximation \quad property . \quad That \quad i s , there is an equicontinuous net $ \{ T {\alpha }\} {\alpha }\ in I $ of finite rank operators on $ E $ such that $ x = $ lim $ T {\alpha } x$ for every $x \ in E . $ \quad For a subset $ A \subset E $ we d e f i n e
\ [ \widehat{A} = \Gamma \{ T {\alpha } (A): \alpha \ in I \} . \ ]
\noindent We start with a t $ line −e $ chnical result .
\ centerline {4 . 6 . Lemma . \quad ( a $ ) A \subset \widehat{A} { . }$ }
\ centerline {(b) If $E =$ sp $( B )$ forsomeabsolutely convex subset $B$ of $E ,$ then }
\ [ \widehat{E} = sp \{ T {\alpha } (E): \alpha \ in I \} = sp ( ˆ{\widehat{B}} ). \ ]
\ centerline {(c $) A$ isbounded $ \Leftrightarrow \widehat{A} $ is bounded . }
\ centerline {( d ) I f $ \widehat{A} $ is almost bounded then $ A $ is almost bounded . }
(e) If $E$ is completeand $( A { n } ) $ is a completing s equence with $ 2 A { n + 1 }\subset A { n } , $ then $ ( ˆ{\widehat{A}} { n } ) $ is also a completing s equence . Subspaces and quotient spaces of locally convex spaces .. 25 \ hspaceP r o o∗{\ f periodf i l l } ..Subspaces open parenthesis and a quotient closing parenthesis spaces follows of locally from x = convex lim T sub spaces alpha x\ periodquad Now25 Line 1 infinity Line 2 sp open parenthesis B closing parenthesis = union of nB = M Line 3 n = 1 \ centerlineand so by continuity{P r o o T fsub . alpha\quad open(a) parenthesis follows M closing from parenthesis $x =$ is a limdense subspace $T {\ ofalpha T sub alpha} x open . parenthesis $ Now } E closing parenthesis period Hence T sub alpha open parenthesis E closingSubspaces parenthesis and quotient = T spaces sub alpha of locally open convex parenthesis spaces M closing 25 parenthesis period \ [ \Sobegin { a l i g n e d }\ infty \\P r o o f . ( a ) follows from x = lim Tαx. Now sp ( B ) = \bigcup nB = M \\ Line 1 infinity Line 2 T sub alpha open parenthesis E closing parenthesis∞ = union of nT sub alpha open parenthesis B closing parenthesis periodn Line = 3 1 n =\end 1 { a l i g n e d }\ ] [ This means that union of sub alpha in I T subsp( alphaB) =open parenthesisnB = M B closing parenthesis spans E-hatwide sub comma but hatwide-B i s the absolutely convex hull of this n = 1 \noindentset periodand .... This so byproves continuity open parenthesis $ T b closing{\alpha parenthesis} ( period M .... ) $ To prove is a open dense parenthesis subspace c closing of $Tparenthesis{\alpha comma} let( A E ) .$ Hence $T {\alpha } ( E ) = T {\alpha } ( M ) . $ subsetand E bso e bounded by continuity periodTα ....(M For) is U a indense U open subspace parenthesis of Tα(E E). closingHence parenthesisTα(E) = Tα we(M find). So SoV in U open parenthesis E closing parenthesis such that T sub alpha open parenthesis V closing parenthesis subset U for every alpha in I period .. If A subset rho V then A-hatwide subset rho U period .. This∞ \ [ \ begin { a l i g n e d }\ infty \\ [ and open parenthesis a closing parenthesis proveTα open(E) = parenthesisnTα(B c) closing. parenthesis period TIf A-hatwide{\alpha i s} almost( bounded E ) comma = by\bigcup Theorem 1nT period{\ 6 commaalpha hatwide-A} (B). is also almost\\ bounded period .. Now the nresult = follows 1 \ againend{ froma l i g openn e d }\ parenthesis] a closing parenthesis periodn = 1 .. If 2 A sub n plus 1 subset A sub n comma then of course 2 A-hatwide sub n plus 1 subset hatwide-AS sub n period .. Again This means that α∈I Tα(B) spans Eb, but Bb i s the absolutely convex hull of this forset U in . U open This parenthesis proves ( E b closing ) . parenthesis To prove we ( findc ) , V let inA U⊂ openE b eparenthesis bounded . E closing For parenthesisU ∈ U(E) with we find T sub alpha open parenthesis V\noindent closing parenthesisThis means subset Uthat for every $ \ alphabigcup in I period{\alpha .. Since open\ in parenthesisI } T A{\ subalpha n closing} parenthesis( B i s ) $ spans $ \widehat{E} { , }$ V ∈ U(E) such that T (V ) ⊂ U for every α ∈ I. If A ⊂ ρV then A ⊂ ρU. This and ( a ) prove ( c ) . buta completing $ \widehat sequence{B} $ commaα i s there the i absolutelys n sub 0 in N convexwith A sub hull n subset ofb this V for n greater equal n sub 0 period .. Hence If A i s almost bounded , by Theorem 1.6, A is also almost bounded . Now the result follows again from ( A-hatwideb sub n subset U b \noindentfora n ) greater .s If e 2t equalAn . +1\ n⊂h sub fA in l, l0then periodThis of ..course proves Completeness 2Abn (+1 b⊂ )A ofbn. . E\ implieshAgain f i l l now forToproveU open∈ U( parenthesisE) ( we c find ), toV let∈ the U( powerE $A) with ofT A-hatwide\α(subsetV ) ⊂ U subE n closing $ be parenthesis bounded i . \ h f i l l For s$ a U completingfor\ in every sequenceαU∈ I. ( periodSince E (A )$n) i s wefinda completing sequence , there i s n0 ∈ N with An ⊂ V for n ≥ n0. Hence As the final result .. of this .. section we .. examine the consequences .. of the .. ex hyphen \noindenti stence of an$ unboundedV \ in completingU ( sequence E )$ in a lcssuchthatA whichbn ⊂ U has the $T bounded{\alpha } (V) \subset U $ f o r every $ \alpha \ in I . $ \quad I f $ A \subset \rho V $ then $ \widehat{A}\subset \rho U approximationfor n ≥ n . propertyCompleteness period of.. WeE implies keep the now notation (Ab) i swe a completing have introduced sequence above . period . $ \quad This0 n 4 periodAs 7 the period final Theorem result of period this Let section E be we a complete examine lcs the which consequences has the bounded of the approximation ex - i stence of an unbounded and ( a ) prove ( c ) . propertycompleting period sequence .. Assume in a E lcs sub which b to hasthe thepower bounded of prime approximation admits a continuous property norm . We period keep .. the If notationthere is a we completing have sequence openintroduced parenthesis above A sub . n closing parenthesis in E such that each A sub n is unbounded open parenthesis respectively not almost bounded I f $ \widehat{A} $ i s almost bounded , byTheorem $1 . 6 , \widehat{A} $ is also almost bounded . \quad Now the closing parenthesis4 . 7 . comma thenLet E be a complete lcs which has the bounded approximation property . Assume result followsTheorem again . from ( a ) . \quad I f $ 2 A { n + 1 }\subset A { n } , $ then of course E hasE0 aadmits proper a Fr continuous acute-e chet norm subspace . If open there parenthesis is a completing respectively sequence a nuclear K o-dieresis the subspace closing parenthesis period $ 2 \b widehat{A} { n + 1 }\subset \widehat{A} { n . }$ \quad Again P r( oA o) fin periodE such .. Let that B eachbe anA absolutelyis unbounded convex( closedrespectively bounded not subset almost such bounded that the), then f o r $n U \ in U ( En )$ wefind $V \ in U ( E )$ with $T {\alpha } (V) gaugeE has of B a to proper the power Fr e´ ofchet circ subspace i s a continuous( respectively norm on a nuclearE sub b Kto theo¨ the power subspace of period) . to the power of prime .. Therefore sp open parenthesis \subset U$ for every $ \alpha \ in I . $ \quad Since $ ( A { n } ) $ i s B closing parenthesisP r o o f . i Let s denseB be in an E absolutely and so convex closed bounded subset such that the gauge of B◦ i s a continuous a completing0 sequence , there i s $ n { 0 }\ in N $ with $ A { n }\subset V $ f o r $ n \geq by ournorm lemma on E we. haveTherefore sp (B) i s dense in E and so n { 0 } . $b \quad Hence E-hatwideby our lemma = sp open we have parenthesis to the power of hatwide-B closing parenthesis period Also B-hatwide i s bounded period .. Now open parenthesis to the power of hatwide-A sub n closing parenthesis i s a completing sequence \ [ \widehat{A} { n }\subset U \ ] by Lemma 4 period 6 and each Eb = sp(Bb). A-hatwide sub n is unbounded or not almost bounded period .. Also hatwide-A sub n subset sp open parenthesis to the power of B-hatwide Ab closingAlso parenthesisBb i s bounded by definition . Now period (n ) .. i s Now a completing sequence by Lemma 4 . 6 and each Abn is unbounded or not \noindent f o r $ n \geq n B{b 0 } . $ \quad Completeness of $ E $ implies now $ ( ˆ{\widehat{A}} { n } thealmost assertion bounded follows . from Also PropositionAbn ⊂ sp ( 4) period by definition 1 in case . each Now A sub n is not almost bounded period ) $ i s a completing sequence . If eachthe assertion A sub n i follows s unbounded from Proposition but some A 4 sub . 1 in n sub case 0 each .. isA almostn is not bounded almost it bounded follows . now If each fromAn i s unbounded
Theorembut some 1 periodAn0 9is that almost omega bounded is isomorphic it follows to now a subspace from Theorem of E period 1 . 9 that ω is isomorphic to a subspace of E. As5 the period final .. Applications result \quad o f t h i s \5quad .s eApplications c t i o n we \quad examine the consequences \quad o f the \quad ex − iOur stence first t of opic an will unbounded b e LF hyphenOur completing first spaces t opic period will sequence .. b Let e LF E− = inspaces limint a lcs . sub Let whichrightE arrow=lim has open( theE , parenthesis i bounded) be E an sub induc n comma - i sub n plus 1 comma approximation property . \quad We keep the notation we have introduced→ n n+1,n above . n closingt ive parenthesis limit of a .. sequence be an induc of Fr hyphene ´ chet spaces (En) which we always assume t o b e t ive limit of a sequence of Fr acute-e chet spaces .... open parenthesis E sub n closing parenthesis .... which we always assume t o b e 4 . 7 . Theorem . Let $ E $ be a complete lcs which has the bounded approximation property . \quad Assume $ E ˆ{\prime } { b }$ admits a continuous norm . \quad If there is a completing sequence
\noindent $ ( A { n } ) $ in $E$ such that each $A { n }$ is unbounded ( respectively not almost bounded ) , then
\noindent $E$ has a proper Fr $ \acute{e} $ chet subspace ( respectively a nuclear K $ \ddot{o} $ the subspace ) .
P r o o f . \quad Let $ B $ be an absolutely convex closed bounded subset such that the gauge o f $ B ˆ{\ circ }$ i s a continuous norm on $ E ˆ{\prime } { b ˆ{ . }}$ \quad Therefore sp $ ( B )$ i sdensein $E$ andso
\noindent by our lemma we have
\ [ \widehat{E} = sp ( ˆ{\widehat{B}} ). \ ]
\noindent Also $ \widehat{B} $ i s bounded . \quad Now $ ( ˆ{\widehat{A}} { n } ) $ i s a completing sequence by Lemma 4 . 6 and each $ \widehat{A} { n }$ is unbounded or not almost bounded . \quad Also $ \widehat{A} { n }\subset $ sp $ ( ˆ{\widehat{B}} ) $ by definition . \quad Now
\noindent the assertion follows from Proposition 4 . 1 in case each $ A { n }$ is not almost bounded . I f each $ A { n }$ i s unbounded but some $ A { n { 0 }}$ \quad is almost bounded it follows now from Theorem 1 . 9 that $ \omega $ is isomorphic to a subspace of $E . $
\ centerline {5 . \quad Applications }
\ hspace ∗{\ f i l l }Our first t opic will b e $ LF − $ spaces . \quad Let $ E = \lim {\rightarrow } (E { n } , i { n + 1 , n } ) $ \quad be an induc −
\noindent t ive limit of a sequence of Fr $ \acute{e} $ chet spaces \ h f i l l $ ( E { n } ) $ \ h f i l l which we always assume t o b e 26 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentreduced period26 \ ..quad We assumeS $ E. i s\ Hausdorffddot{O} and$ denote nal andT by i sub . n Terzio: E sub n right $ \check arrow{ Eg the} $ canonical lu inj ection period .. Let T : E right arrow F be continuous period .. Then either Ti sub n : E sub n right arrow F i s bounded \noindent reduced . \quad We assume $ E $ i s Hausdorff and denote by $ i { n } :E { n }\rightarrow for each n or¨ it i s unbounded for some n period .. If the latter holds we have again two E $possibilities26 the Scanonical. :O ..nal either and TTi . sub Terzio n ..gˇ ilu s not .. almost .. bounded or Ti sub n .. is .. almost bounded but i nunbounded jreduced e c t i o n period. . \quad We assumeLetE $i T s Hausdorff : E and\ denoterightarrow by in : EFn $→ beE continuousthe canonical inj . \ ectionquad .Then Let either $ Ti { n } :EAfterT : this{E n→ summary}\F berightarrow continuous of the logical . Then frameworkF $ either i s weT bounded i proceedn : En → withF i the s bounded results period for each n or it i s unbounded for some for5 periodn. eachIf 1 the period $ lattern $ Theorem holds or it we period havei s ..again unbounded Let two .. T possibilities be afor continuous some : eitheroperator $ nT in . from $i s the not\quad LF almost hyphenIf the space bounded latter E into or holdsT .. in a is we have again two possibilitiescompletealmost lcs bounded F period : but\quad .. unbounded Thene i one t h e of r . the $ following Ti { n holds}$ : \quad i s not \quad almost \quad bounded or $ Ti { n }$ \quad i s \quad almost bounded but unboundedopen parenthesis . a closingAfter thisparenthesis summary T isof factorable the logical over framework an LB hyphenwe proceed space with period the results . open parenthesis5 . 1 . Theorem b closing . parenthesisLet T E be has a continuous a subspace operator E sub 0 ..from which the is LF isomorphic - space E to into a nuclear a complete K dieresis-o lcs F the space and \ centerlinethe. restriction Then{ After one of T of to the this E following sub summary 0 is an holds is omorphism of: the logical period framework we proceed with the results . } open parenthesis c closing parenthesis( a .. )T Thereis factorable is a continuous over an LB projection− space P . : F right arrow F such that PT open parenthesis E closing parenthesis5 . 1 . Theorem = P open parenthesis. \quad (Let b F ) closing\Equadhas parenthesis aT subspace be a simeq continuousE0 omegawhich period is operator isomorphic from to a nuclear the LF K −o¨ thespace space E and i n t o \quad a completeP rthe o o restriction f period lcs F .. of . Suppose\Tquadto EThen each0 is an Ti one is sub omorphism of n : the E sub .following n right arrow holds F is bounded: period .. We will show that open parenthesis a closing parenthesis holds ( c ) There is a continuous projection P : F → F such that PT (E) = P (F ) ' ω. \ centerlinein this case{ period( a .. $ Let ) UP subT$ r o n o in f is . U open factorable Suppose parenthesis each T over in E: subE ann → n closingF $LBis bounded parenthesis− $ . space We be such will . show that} Ti that sub ( a n ) open holds parenthesis U sub n closing parenthesisin this i s case a bounded . Let subsetUn ∈ U of(E Fn) and be such that T in(Un) i s a bounded subset of F and set \ hspace ∗{\ f i l l }(b $) E$ hasasubspace $E { 0 }$ \quad which is isomorphic to a nuclear K $ \ddot{o} $ set n theLine space 1 n Line and 2 B sub n = Capital Gamma open parenthesis union of Ti sub j open parenthesis U sub j closing parenthesis closing parenthesis [ period Line 3 j = 1 Bn = Γ( T ij(Uj)). \noindent the restriction of $T$ to $E { 0 }$ is an is omorphism . Each B sub n .. i s .. a Banach .. disc period .. Let j sub n plusj = 1 1 comma n : F open square bracket B sub n closing square bracket right arrow F open square bracket B sub n plus 1 closing square bracket .. be the .. canonical \ hspace ∗{\ f i l l }( c ) \quad There is a continuous projection $ P : F \rightarrow F $ such that inclusionEach B periodn i s .. We a Banach have disc . Let jn+1, n : F [Bn] → F [Bn+1] be the canonical inclusion . $PT(E)=P(j subWe n have plus 1 comma n circ Ti sub n = F Ti sub ) n plus\simeq 1 circ i sub\omega n plus 1. comma $ n period So T : E right arrow limint sub right arrow F open square bracket B sub n closing square bracket is continuous period .. So T can b e \ hspace ∗{\ f i l l }P r o o f . \quad Suppose each $ Ti { n } :E { n }\rightarrow F $ is bounded . \quad We will show that ( a ) holds factored over the LB hyphen space jn+1, n ◦ T in = T in+1 ◦ in+1,n. limint sub right arrow F open square bracket B sub n closing square bracket since the inclusion of this space into F i s also continuous period \noindentSo T : inE → thislim→ caseF [Bn] . is\ continuousquad Let . $ So U T {cann b}\ e factoredin U(E over the LB− {spacen } lim→)F $[Bn be] since such the that $ Ti { n } ( Nowinclusion we assume of this Ti sub space m into i s unboundedF i s also continuous for some m . period .. We will prove that if Ti sub m i s U not{ n almost} ) bounded $ i s then a bounded open parenthesis subset b of closing $F$ parenthesis and .. holds but if it i s almost bounded then open parenthesis c closing s e t Now we assume T im i s unbounded for some m. We will prove that if T im i s not almost bounded then ( parenthesisb ) .. holds holds but period if it i s almost bounded then ( c ) holds . If T i i s not almost bounded , there i s a nuclear If Ti sub m i s not almost bounded comma .. there i s a nuclear Km o-dieresis the subspace lambda open parenthesis A closing parenthesis .. \ [ \ beginKo ¨ {thea l isubspace g n e d } nλ(A\\) of Em such that T im i s an i somorphism when restricted t o λ(A)( Proposition 1 . of E sub14 m ) . Let E = i (λ(A)). Then the restriction of T to E is an i somorphism and E i s i somorphic t Bsuch{ thatn } Ti= sub0 m\Gamma i sm an i somorphism( \bigcup when restrictedTi { tj o lambda} 0(U open{ parenthesisj } )). A closing0 parenthesis\\ open parenthesis Proposition 1 jo =λ(A) 1 via\endim. { a l i g n e d }\ ] period 14Suppose closing parenthesisT i : E period→ F is Let almost bounded . Then there is an index set J so that F ' G ⊕ KJ and E sub 0 = i sub mm openm parenthesis lambda open parenthesis A closing parenthesis closing parenthesis period .. Then the restriction of T to (I − Q)T im : Em → F is bounded where Q : F → F i s E sub 0the .. projis an ection i somorphism with range andK EJ suband 0 kernel i s G( Theorem 1 . 9 ) . This means QT i i s unbounded , since T i \noindenti somorphicEach t o lambda $ B open{ n parenthesis}$ \quad A closingi s \quad parenthesisa Banach via i sub\quad m periodd i s cm . \quad Let $ j {mn + } 1 , n :F[Bi s unbounded{ .n We} apply] Theorem\rightarrow 2 . 3 t o TF[B im and infer { n + 1 } ] $ \quad be the \quad c a n o n i c a l Suppose Ti sub m : E sub m right arrow FJ is almost bounded period .. Then there is an index set J so i n c lthat u s i o there n . i\ squad a faithfulWe havesurj ection q : K → ω so that qQT im i s also a faithful that F simeq G oplus K to the power of J and open parenthesis I minus Q closing parenthesis Ti sub m : EJ sub m right arrow F is bounded surj ection of Em onto ω. The kernel of q i s a complemented subspace of K and where Qso :q Fhas right a left arrow inverse F i sS : ω → KJ . We set P = SqQ : F → F. \ [the j proj{ n ection + with} 1 range , K to n the power\ circ of JTi and kernel{ n } G open= parenthesis Ti { n Theorem + 1 1}\ periodcirc 9 closingi parenthesis{ n + period 1 .. This , meansn } . \ ] We shall now apply this result t o the identity on E and therefore we will have QTi subt o m assume i s the completeness of E. unbounded comma since Ti sub m i s unbounded period .. We apply Theorem 2 period 3 t o Ti sub m and infer 5 . 2 . Corollary . Let E be a complete LF - space . The following are equivalent : that there i s a faithful surj ection q : K to the( i ) powerE is an of JLB right− space arrow . omega so that qQTi sub m i s also a faithful \noindentsurj ectionSo of E $ sub T m onto : omega E \ periodrightarrow .... The kernel\lim of q i{\ s arightarrow complemented} subspaceF[B of K to the{ powern } of] J $ and is continuous . \quad So $ Tso $ q has can a left b einverse factored S : omega over right the arrow $ K LB to the− power$ space of J period .. We set P = SqQ : F right arrow F period $We\lim shall now{\ applyrightarrow this result} t oF[B the identity on{ En and} therefore] $ since we will the have inclusion of this space into $ F $ i s also continuous . t o assume the completeness of E period Now5 periodwe assume 2 period $ Corollary Ti { m period}$ Let i sE be unbounded a complete for LF hyphen some space $m period . $ .. The\quad followingWe will are equivalent prove that : if $ Ti { m }$ i sopen parenthesis i closing parenthesis E is an LB hyphen space period not almost bounded then ( b ) \quad holds but if it i s almost bounded then ( c ) \quad holds . I f $ Ti { m }$ i s not almost bounded , \quad there i s a nuclear K $ \ddot{o} $ the subspace $ \lambda ( A ) $ \quad o f $ E { m }$ such that $ Ti { m }$ i s an i somorphism when restricted t o $ \lambda ( A ) ($ Proposition1. 14) . Let $ E { 0 } = i { m } ( \lambda ( A ) ) . $ \quad Then the restriction of $ T $ to $ E { 0 }$ \quad is an i somorphism and $ E { 0 }$ i s i somorphic t o $ \lambda ( A ) $ via $ i { m } . $
Suppose $ Ti { m } :E { m }\rightarrow F $ is almost bounded . \quad Then there is an index set $ J $ so that $ F \simeq G \oplus K ˆ{ J }$ and $ ( I − Q ) Ti { m } :E { m }\rightarrow F$ is boundedwhere $Q : F \rightarrow F $ i s
\noindent the proj ection with range $ K ˆ{ J }$ and kernel $G ( $ Theorem1 . 9 ) . \quad This means $ QTi { m }$ i s unbounded , since $ Ti { m }$ i s unbounded . \quad We apply Theorem 2 . 3 t o $ Ti { m }$ and i n f e r
\noindent that there i s a faithful surj ection $ q : Kˆ{ J }\rightarrow \omega $ so that $ qQTi { m }$ i s also a faithful
\noindent surj ection of $ E { m }$ onto $ \omega . $ \ h f i l l The kernel of $ q $ i s a complemented subspace of $ K ˆ{ J }$ and
\noindent so $q$ hasa left inverse $S : \omega \rightarrow K ˆ{ J } . $ \quad We s e t $ P = SqQ : F \rightarrow F . $
\ hspace ∗{\ f i l l }We shall now apply this result t o the identity on $ E $ and therefore we will have
\noindent t o assume the completeness of $ E . $
\ hspace ∗{\ f i l l }5 . 2 . Corollary . Let E be a complete LF − space . \quad The following are equivalent :
\ centerline {(i $) E$ isan $LB − $ space . } Subspaces and quotient spaces of locally convex spaces .. 27 \ hspaceopen parenthesis∗{\ f i l l } Subspaces ii closing parenthesis and quotient .. There is spaces .. a sequence of locally n sub k upwards convex arrow spaces infinity\quad .. such27 that each i sub n sub k plus 1 comma n sub k : E sub n sub k right arrow E sub n sub k plus 1 .. is \ hspacebounded∗{\ periodf i l l }( i i ) \quad There i s \quad a sequence $ n { k }\uparrow \ infty $ \quad such that each $ iopen{ parenthesisn { k iii + closing 1 } parenthesis, n E{ hask no}} proper:ESubspaces Fr acute-e{ n and chet{ quotientk subspace}}\ spacesrightarrow period of locally convexE spaces{ n { 27k + 1 }}$ \quad i s ( ii ) There is a sequence n ↑ ∞ such that each i : E → E is P r o o f period .. Suppose open parenthesis iii closingk parenthesis i s true periodnk+1,n ..k By Theoremnk 4 periodnk+1 5 each i sub n : E sub n right arrow\noindent Ebounded mustbounded b . e . bounded period .. Let U sub n in U( iii open )E has parenthesis no proper E Fr sube´ nchet closing subspace parenthesis . be such that i sub n open parenthesis U sub n closing parenthesis\ centerlineP i r s o bounded{ o( f . iii Suppose in $) E period ( iii E$ ).. i Now s true hasnoproperFr i sub . 1By open Theorem parenthesis 4 . $ 5 each\ Uacute subin 1{ closing:eE} n$ parenthesis→ chetE must subspace b i s e bounded . } . Let Un ∈ U(En) be such that in(Un) i s bounded in E. Now i1(U1) i s contained in some i sub n sub 2 open− parenthesis1 E sub n sub 2 closing parenthesis and i sub n sub 2 to the power of minus 1 open parenthesis P r ocontained o f . \ inquad someSupposein (En ) and ( iiii (i1 )(U i1)) s i trues bounded . \quad in En Bysince Theorem completeness 4 . 5 each $ i { n } :E { n }\rightarrow i sub 1 open parenthesis U2 sub 12 closingn parenthesis2 closing parenthesis2 i s bounded in E sub n sub 2 since completeness E $implies must b that e the LF − space E i s regular . So we see that i : E → E i s bounded . implies that .. the LF hyphen space E i s regular period .. So we see thatn2, ..1 i sub n sub1 2 comman2 1 : E sub 1 right arrow E sub n sub 2 .. i s boundedWe let . n\quad=Let 1. We $ U continue{ n }\ this wayin andU(E find n >{ nn } such) $ that bei such(U ) that is contained $ i { inn } (U { n } ) $ bounded period1 .. We let n sub 1 = 1 period .. We continue3 this way2 and find n sub 3n2 greatern2 n sub 2 .. such that i s boundedi (E ) in and $Ei .: $ E \quad→ NowE $i s i bounded{ 1 } .(U This gives{ (1 ii} ) . ) $ i s i subn3 n subn3 2 open parenthesisn3,n2 Un2 sub n sub 2n3 closing parenthesis .. is .. contained in i sub n sub 3 open parenthesis E sub n sub 3 closing Now assume ( ii ) and let U ∈ U(E ) be such that i (U ) i s bounded in E . Let parenthesis .. and i sub n sub 3 commak n subn 2k : E sub n sub 2n rightk+1,nk arrowk E sub n sub 3 ..n ik+1 s bounded period .. This \noindent contained in some $ i { n { 2 }} (E { n { 2 }} ) $ and $ i ˆ{ − 1 } { n { 2 }} gives open parenthesis ii closing parenthesis period k (Now i assume{ 1 } open(U parenthesis{ 1 ii} closing) parenthesis )$ i sboundedin and let U sub k in $E U open{ n parenthesis{ 2 }}$ E sub since n sub completeness k closing parenthesis be such that i [ sub n sub k plus 1 comma n sub k open parenthesisBk U= sub Γ ki closingn ,n ( parenthesisUj). i s bounded in \noindent implies that \quad the $ LF − $k space+1 j $E$ i s regular . \quad So we see that \quad $ i { n { 2 } E sub n sub k plus 1 period .. Let j = 1 ,Line 1 } 1 k Line:E 2 B sub{ 1 k}\ = Capitalrightarrow Gamma unionE of{ i subn n{ sub2 }} k plus$ \ 1quad commai n s sub j open parenthesis U sub j closing parenthesis period bounded . \quad We l e t $ n { 1 } = 1 . $ \quad We continue this way and find $ n { 3 } > n { 2 }$ Line 3B jk =i 1 s a Banach disc in Enk+1 and we have the commutative diagram \quadB subsuch k i s a that Banach disc in E sub n sub k plus 1 and we have the commutative diagram $ i { n { 2 }} (U... {−→n {E 2 }}−→) $E \quad−→i s E\quad −→contained... in $ i { n { 3 }} (E { n { 3 }} Line 1 period period period minus-arrowrightnk E sub n subnk+1 k arrowright-minusnk+2 E sub n sub k plus 1 minus-arrowright E sub n sub k plus 2) arrowright-minus $ \quad and period $ i period{ n period{ 3 Line} , 2 arrowvertex-arrowbt n { 2 }} :E nearrow{ ↓ arrowbt-arrowvertexn %{ ↓2 }}\ % rightarrow nearrow LineE 3 period{ n period{ 3 }} period$ \ minus-quad i s bounded . \quad This arrowrightg i v e s ( E i sub i ) n .sub k plus 1 open square bracket B sub k closing square bracket arrowright-minus E sub n sub k plus 2 open square bracket ... −→ Enk+1 [Bk] −→ Enk+2 [Bk+1] −→ ... B sub k plus 1 closing square bracket minus-arrowright period period period Now assume ( ii ) and letFrom $ this U we{ havek }\E 'inlim EU(E[B ]. So{ ( in ) i{ s truek }} . ) $ be such that $ i { n { k + From this we have E simeq limint sub right arrow E sub→ n subnk+1 k plusk 1 open square bracket B sub k closing square bracket period .. So open parenthesis1 } , i n closing{ k parenthesis}} (U iFinally s true{ k period ,} we assume) $ iE sis bounded an LB− space in and F a Fre ´ chet subspace of E. Hence $ E { n { k + 1 }} . $ \quad Let S∞ FinallyE = comma lim→ E[ weBn assume] E is an where LB hyphenBn ⊂ spaceE andis a F Banach a Fr acute-e disc and chetE subspace= n=1 ofnB En period. .. Hence Hence E =for limint some subm, rightthe set arrowF ∩ EmB openm is square a bounded bracket neighborhood B sub n closing and square so F must bracket be a .... Banach where space B sub . n subset E is a Banach disc and E = union of\ [ sub\ begin n ={ 1a to l i gthe n5 e power d .} 3 .kCorollary of\\ infinity nB . subFor n a period complete .... HenceLF − space E the following are equivalent : Bfor{ somek } m comma(= i ) \GammaE the is set F is cap omorphic\bigcup mB sub m to isi a{ bounded ann inductive{ k neighborhood + limit 1 }lim and→, soGn F n mustwhere{ bej a}} each(UGn '{ jω×} ). \\ jBanach = space 1 \ periodend{ a l i g n e d }\ ] 5 period 3 period Corollary period For a complete LF hyphen space E the following are equivalent : Banach. open parenthesis i closing parenthesis .. E is .. is omorphic .. to .. an inductive .. limit limint sub right arrow G sub n .. where .. each G sub\noindent n simeq omega$ B times{ k }$ i s a Banach disc in $E { n { k + 1 }}$ and we have the commutative diagram ( ii ) There is a s equence nk ↑ ∞ such that each in ,n is almost bounded but Banach period k+1 k \ [ \ beginunbounded{ a l i g n . e d } ... \longrightarrow E { n { k }}\longrightarrow E { n { k + 1 }} open parenthesis ii closing( iii ) E parenthesis has no subspace .. There which is a is s isomorphicequence n sub to a k nuclear upwards K arrowo¨ the infinity space . such that each i sub n sub k plus 1 comma n sub\longrightarrow k .. is almost boundedE but{ n { k + 2 }}\longrightarrow ... \\ \unboundeddownarrow period\nearrow \downarrow \nearrow \\ ... \longrightarrow E { n { k + 1 }} [B { k i}n ] \longrightarrow E { n { k open parenthesisPJ r iii closing parenthesis E has no subspace which is isomorphic to a nucleark+ K1,n dieresis-ok the space periodBanach ×Kk oof. SupposewhereKJ kis(aii)holds.Wecancomplementedsubspacefactorizeof eachE +1 (seeovertheGk = proofof Propo− +times 2 }} K to[B the power{ ofk P J + r sub 1 k} o o f] period\longrightarrow Suppose where sub K... to the power of\end J kn{k ia s l iopen g n ed parenthesis}\ ] a ii closing parenthesis holds periodsition We can 1 .complemented 13 or Theorem sub 1 .subspace 9 ) . Hence factorizeJk submust of be each countable E sub n . sub So k subGk plus' Banach 1 to the× powerω and of ( i sub) is ntrue sub . k plus sub open parenthesis see to theNow power we of assume 1 comma ( i ) n and sub prove k over ( iiithe ) G. sub Since k =E proof= S i of( toG the), if power a nuclear of Banach Ko ¨ the Propo space λ hyphen(A) i s i somorphic \ centerline {From this we have $ E \simeq \nlimn {\rightarrow } E { n { k + 1 }} [B { k } sitiont o 1 a period subspace 13 orof E Theoremvia an operator1 period 9T closingthen by parenthesis the lo cal period- ization .. theorem Hence J (sub cf . k must[10]; 5 be.6) countableT can b e period factored .. So G sub k simeq Banach ] . $ \quad So( i ) i s true . } times omegaover some in(Gn). Comparison of the t opologies of E,Gn and λ(A) gives that λ(A) i s isomorphic t o a andsubspace open parenthesis of Gn, but i closingGn ' ω parenthesis× Banach isand true this period is impossible . \ hspaceNow we∗{\ assumef i l l } openFinally parenthesis , we i assume closing parenthesis $E $ and is prove an $LB open parenthesis− $ spaceand iii closing parenthesis $F$ period aFr .. $ Since\acute E ={ unione} $ of i chet sub subspace of n$ open E parenthesis . $ \quad G subHence n closing parenthesis comma if a nuclear K o-dieresis the space lambda open parenthesis A closing parenthesis i s i somorphic t o a subspace of E via an operator T then by the lo cal hyphen \noindentization theorem$ E open = parenthesis\lim {\ cf periodrightarrow open square} bracketE[B 10 closing{ squaren } bracket] $ \ semicolonh f i l l where 5 period $ 6 B closing{ n parenthesis}\subset T can bE e $ factoredis a Banach over some disc i sub and n open $E parenthesis = \ Gbigcup sub n closingˆ{\ infty parenthesis} { periodn = .. Comparison 1 } nB { n } . $ \ h f i l l Hence of the t opologies of E comma G sub n and lambda open parenthesis A closing parenthesis gives that lambda open parenthesis A closing parenthesis\noindent i sfor isomorphic some t $m o a subspace ,$ the set $F \cap mB { m }$ is a bounded neighborhood and so $ F $ mustof G be sub a n comma but G sub n simeq omega times Banach and this is impossible period Banach space .
\ centerline {5 . 3 . Corollary . For a complete $ LF − $ space $ E $ the following are equivalent : }
\ hspace ∗{\ f i l l }( i ) \quad E i s \quad i s omorphic \quad to \quad an inductive \quad l i m i t $ \lim {\rightarrow } G { n }$ \quad where \quad each $ G { n }\simeq \omega \times $
\ begin { a l i g n ∗} Banach . \end{ a l i g n ∗}
\ hspace ∗{\ f i l l }( i i ) \quad There is a s equence $ n { k }\uparrow \ infty $ such that each $ i { n { k + 1 } , n { k }}$ \quad is almost bounded but
\noindent unbounded .
\ centerline {( iii ) E has no subspace which is isomorphic to a nuclear K $ \ddot{o} $ the space . }
\ [ \times K ˆ{ P { J } r } { k } o o f . Suppose { where } { K ˆ{ J } k i s } ( { a } ii ) holds . We can { complemented } { subspace } f a c t o r i z e { o f } each { E { n { k }}}ˆ{ i { n { k + }}} { + 1 }ˆ{ 1 , n { k }} { ( see } over { the } G { k } = { proof } o f ˆ{ Banach } Propo − \ ]
\noindent sition 1 . 13 or Theorem 1 . 9 ) . \quad Hence $ J { k }$ must be countable . \quad So $ G { k } \simeq $ Banach $ \times \omega $ and ( i ) is true .
Now we assume ( i ) and prove ( iii ) . \quad Since $ E = \bigcup i { n } (G { n } ) , $ if a nuclear K $ \ddot{o} $ the space $ \lambda ( A ) $ i s i somorphic t o a subspace of $E$ via an operator $T$ then by the lo cal − izationtheorem(cf $. [ 10 ] ; 5 . 6 ) T$ canbefactoredoversome $i { n } (G { n } ) . $ \quad Comparison of the t opologies of $E , G { n }$ and $ \lambda ( A )$ givesthat $ \lambda (A ) $ i s isomorphic t o a subspace o f $ G { n } , $ but $ G { n }\simeq \omega \times $ Banach and this is impossible . 28 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentFinally we28 will\ provequad ..S open $ parenthesis. \ddot iii{O closing} $ nal parenthesis andT .. . implies Terzio .. open $ \ parenthesischeck{g} ii$ closing lu parenthesis period .. By Theorem 4 period 4 each i sub n : E sub n right arrow E Finally we will prove \quad ( i i i ) \quad i m p l i e s \quad ( i i ) . \quad By Theorem 4 . 4 each $ i { n } i s almost bounded¨ period .. Since E i s complete comma .. we know by Theorem .. 1 period 9 that .. E simeq :EK to28 the{ n S power}\.O nal ofrightarrow andJ n Toplus . Terzio G subgˇ Elu n where$ P sub n i sub n : E sub n right arrow G sub n i s bounded period .... Here P sub n is the proj ectioni s of almost EFinally bounded we will prove . \quad ( iiiSince ) implies $E$ ( ii ) i . s By complete Theorem 4, .\quad 4 each wein know: En by→ TheoremE i s almost\quad 1 . 9 that \quad $ E bounded\simeq .$ Since E i s complete , we know by Theorem 1 . 9 that E ' onto GJ sub n with kernel K to the power of J n period K to the power of J n i s a Baire space period .... By a corollary of the localization K n ⊕ Gn where Pnin : En → Gn i s bounded . Here Pn is the proj ection of E theorem .. open parenthesisJ open squareJ bracket 10 closing square bracket semicolon .. 5 period 6 period 4 closing parenthesis .. there i s an imbedding\noindentonto ofGn K$with to K the ˆkernel{ powerJ }K ofn.n J n K ..\oplus inton i s some a Baire EG sub space{ kn sub .}$ n period where E i $ s also P { n By} a corollaryi { n of} the:E localization{ n }\rightarrow G { n }$ i s boundedtheorem . (\ [h 10 f i ] l ; l Here 5 . 6 . $4 ) P there{ n i} s$ an is imbedding the proj of K ectionJ n into of some $E$E .E i s also regular and regular and therefore there is an l sub n with P sub n i sub n : E sub n right arrow Ek subn l sub n .. continuous period .. We have therefore there is an l with P i : E → E continuous . We have thus shown that for each n there i s thus shown that for eachn n there in sn m subn n greaterln n so that i sub m sub n comma n : E sub n right arrow E sub m sub n is factorable \noindentmn > nontoso that $im Gn,n {: En →}$Emn withis factorable kernel $Kˆ{ J } n . K ˆ{ J } n $ i s a Baire space . \ h f i l l By a corollary of the localization through Banach times KJ to the power of J comma which means that it i s almost bounded period Finallythrough commaBanach we consider×K , which the inductive means that limit it of i as almost sequence bounded of Fr acute-e . chet spaces each \noindentof which admitstheorem a continuous\quad ( norm [Finally 10 period ] ; , .. we\ Wequad consider have5 a . dichotomythe 6 inductive . 4 ) in\ thislimitquad case ofthere a period sequence i s of an Fre ´imbeddingchet spaces each of $ K ˆ{ J } n $ \quad i n t o some $ E5 periodof{ whichk 4{ period admitsn }} Corollary a. continuous E period $ norm i Let s.E a l= s oWe limint have sub a dichotomy right arrow in E this sub case n .. . be complete and suppose each E sub n .. admits regulara continuous and norm therefore period5 . .... 4 there . ThenCorollary either is an E .is $ anLet l LBE{ hyphen=n lim}$→ spaceE withn orbe E complete $ has P a{ subspace andn } suppose whichi { each isn } En:Eadmits{ n }\rightarrow E a{ nuclearal continuous{ Kn dieresis-o}}$ norm\quad the . spacecontinuous period . \quad Then eitherWe haveE is an LB− space or E has a subspace which is thusP ra o shown nuclear o f period Kthat ..o¨ Supposethe for space each E . i s not $ n .. $an LB there hyphen i spaces $m period{ ..n Then} > by Corollaryn $ so 5 period that 2 $ there i i{ sm a { n } , n } : E proper{ n }\ Fr acute-erightarrow chet spaceP r FoE oand f{ . anm Suppose imbedding{ n }}E$i T s : notis F right factorable an arrowLB− Espace period . .. Then Now by T factors Corollary over 5 .some 2 there i s a E subproper n and Fr Te ´ chet : F right space arrowF and E an sub imbedding n cannotT b: eF almost→ E. boundedNow T commafactors because over some E subEn and n admitsT : F a→ continuousEn cannot \noindentnormb e open almostthrough parenthesis bounded Banach Theorem , because $ 1E\ periodntimesadmits 9 closing aK continuous ˆ{ parenthesisJ } norm, $ period ( Theorem which .. By means 1 Theorem . 9 ) . that 4 By period it Theorem i 3 s comma almost 4.3,F Fhas has bounded a a nuclear . K dieresis-o the nuclear Ko ¨ the subspace . subspace period n \ hspaceNext we∗{\Next willf wei deall l will} Finally with deal the with projective , the we projective consider limit oflimit a the sequence of a sequence inductive of certain of certain limit locally locally of aconvex sequence spaces . of Let Fr ( $En\,acute jn+1) {e} $ chet spaces each b e a proj ective sp ectrum where each En has an convex spaces period .. Let .. open parenthesis E sub n comma j sub n plus 1 to the power of n closing parenthesisn b e a proj ective sp ectrum\noindentincreasing whereof each fundamental which E sub n admits has sequence an a continuous of bounded set norm s Bn .consisting\quad We of Banach have discsa dichotomy . We let E in= lim this→(En case, jn+1) . increasingbe the proj fundamental ective limit sequence . The ofsituation bounded resembles set s B sub what n consisting of Banach discs period \ hspaceWeVogt let∗{\ E considers =f ilimint l l }5 in sub . [ 204 right ]. . Corollary arrow open parenthesis . Let $E E sub =n comma\lim j sub{\ n plusrightarrow 1 to the power} E of n{ closingn }$ parenthesis\quad be be complete the proj ective and suppose each $ E { n }$ \quad5 .admits 5 . Theorem . Let E be as above . The following are equivalent : limit period The situation resembles what 0 0 Vogt considers in open square bracket(20 a ) closingE [β( squareE ,E)] bracketis not period metrizable . \noindent5 period( b 5 )a period continuousThere Theorem is a completing period norm Let . s equence\ Eh bef i l as l ( aboveThenAn) in period eitherE such .. The that $E$following each A isn areis an unbounded equivalent $LB . : (− c$ ) ω spaceis a faithful or $ E $ has a subspace which is openquotient parenthesis of E. a closing parenthesis E to the power of prime open square bracket beta open parenthesis E to the power of prime comma \noindentP ra o o nuclear f . ( b K ) ⇒ $( c\ddot ) i s true{o} in$ general the space ( Theorem . 2 . 3 ) . ( c ) ⇒ ( a ) i s trivial because ( c ) E closing parenthesis closing square bracket .. is not metrizable0 0 period openimplies parenthesis that φ bis closing i somorphic parenthesis to a subspace .. There of isE a completing[β(E ,E)]. s equenceSo it remains open parenthesis to prove A( a sub ) ⇒ n( closing b ) . parenthesis Let in E such that each A\ hspace sub njn is∗{\ unbounded: fE i l→ l }PEn period rbe o the o canonical f . \quad projSuppose ection . There $E$ are two i s possibilities not \quad .an Either $ (LB i ) for− every$B spacen ∈ Bn . \quad Then by Corollary 5 . 2 there i s a openthere parenthesis i s a k such c closing that parenthesis omega is a faithful quotient of E period \noindent proper Fr $ \acute{e} $ chet space $F$ andan imbedding $T : F \rightarrow E P r o o f period .. open parenthesis b closing parenthesis doublek stroke right arrow open parenthesis c closing parenthesis i s true in general. $ \ openquad parenthesisNow $ T Theorem $ factors 2 period over 3 closing some parenthesis period .. open parenthesis c closing parenthesis double stroke right arrow open $ E { n }$ and $ T : F \rightarrow\ −1 E { n }$ cannot b e almost bounded , because $ E { n }$ parenthesis a closing parenthesis i s trivial because ji (Bi) admits a continuous open parenthesis c closing parenthesis implies that phi is i somorphic= 1 to a subspace of E to the power of prime open square bracket beta opennorm parenthesis ( Theorem E to 1 the . power 9 ) of. prime\quad commaByTheorem E closing parenthesis $4 . closing 3 square , F$ bracket hasanuclearK period .. So it remains $ to\ddot{o} $ the subspace . provei s ..bounded open parenthesis in E or ( ii a ) closing there i parenthesiss Bn ∈ Bn such double that stroke for every rightk, arrowthe set open defined parenthesis above b closing parenthesis period .. Let j sub n : E rightNext arrowi we s not will E bounded sub deal n be . the with In canonical the the firstprojective proj case ection we will period show limit .. that There ofE0[ aβ are(E sequence0 two,E)] i s metrizable of certain . locally convex spaces . \quad Let \quad $ ( E { n } , j ˆ{ n } { n + 1 } ) $ b e a proj ective sp ectrum where each possibilitiesSuppose periodD i.. s a Either bounded open subset parenthesis of E. iThen closing for parenthesis each n we findfor everyBn ∈ B Bn subwith n injn B(D sub) ⊂ nB theren. For i this s a kchoice such that $ ELinewe{ 1n now k Line}$ find 2 hask intersectionsuch an that of j sub i to the power of minus 1 open parenthesis B sub i closing parenthesis Line 3 i = 1 i s bounded in E or open parenthesis ii closing parenthesis there i s B sub n in B sub n such that for every k comma the set defined above \noindenti s not boundedincreasing period .. fundamental In the first case sequence we will show of that bounded E tok the set power s of $ prime B { openn }$ square consisting bracket beta of open Banach parenthesis discs E to . the powerWe l of e t prime $ E comma = E closing\lim parenthesis{\rightarrow closing square} (E\ bracket−1 { i sn metrizable} , period j ˆ{ n } { n + 1 } ) $ be the proj ective limit . The situation resembles what B = jn (Bn) Suppose D i s a bounded subset of E period .. Then for each n we find B sub n in B sub n with \noindentj sub n openVogt parenthesis considers D closing in parenthesis[ 20 ] . subset B sub nn period= 1 For this choice we now find k such that Line 1 k Line 2 B = intersection of j sub n to the power of minus 1 open parenthesis B sub n closing parenthesis Line 3 n = 1 \ centerline {5 . 5 . Theorem . Let $E$ be as above . \quad The following are equivalent : }
\ centerline {( a $ ) E ˆ{\prime } [ \beta ( E ˆ{\prime } , E ) ] $ \quad is not metrizable . }
( b ) \quad There is a completing s equence $ ( A { n } ) $ in $E$ such that each $A { n }$ is unbounded . ( c $ ) \omega $ is a faithful quotient of $E . $
P r o o f . \quad ( b $ ) \Rightarrow ( $ c ) i s true in general ( Theorem 2 . 3 ) . \quad ( c $ ) \Rightarrow ( $ a ) i s trivial because ( c ) implies that $ \phi $ is i somorphic to a subspace of $ E ˆ{\prime } [ \beta ( E ˆ{\prime } , E ) ] . $ \quad So it remains to prove \quad ( a $ ) \Rightarrow ( $ b ) . \quad Let $ j { n } :E \rightarrow E { n }$ be the canonical proj ection . \quad There are two possibilities . \quad Either ( i ) for every $B { n }\ in B { n }$ there i s a $k$ such that
\ [ \ begin { a l i g n e d } k \\ \bigcap j ˆ{ − 1 } { i } (B { i } ) \\ i = 1 \end{ a l i g n e d }\ ]
\noindent i s bounded in $E$ or ( ii ) there i s $B { n }\ in B { n }$ such that for every $ k , $ the set defined above
\noindent i s not bounded . \quad In the first case we will show that $ E ˆ{\prime } [ \beta ( E ˆ{\prime } , E ) ]$ ismetrizable.
Suppose $D$ i s a bounded subset of $E . $ \quad Then for each $ n $ we find $B { n }\ in B { n }$ with $ j { n } (D) \subset B { n } . $ For this choice we now find $ k $ such that
\ [ \ begin { a l i g n e d } k \\ B = \bigcap j ˆ{ − 1 } { n } (B { n } ) \\ n = 1 \end{ a l i g n e d }\ ] Subspaces and quotient spaces of locally convex spaces .. 29 \ hspacei s bounded∗{\ f i period l l } Subspaces .. Hence Dand subset quotient B period .. spaces So if we deof n-line locally ote by convex B the collection spaces of\quad all bounded29 subsets .. of E we .. obtain by the process .. open parenthesis i closing parenthesis comma .. we .. see that .. E has .. a fundamental \noindentsequence ..i of s bounded bounded sets . period\quad .. ThereforeHence $ .. D open\ parenthesissubset iiB closing . parenthesis $ \quad ..So must i f .. we be detrue period $ n−line .. Then $ for ote each by n we $B$ thechoose collection B sub n in of B sub all n bounded so that for every k the set Subspaces and quotient spaces of locally convex spaces 29 s uA bsub si e s t s boundedk =\quad 2 to the .o f power Hence $ E ofD $ 1⊂ k weintersectionB. \quadSo if we ofobtain fromde n − n by =line 1 theto ote k by j process subB the n to collection the\quad power of( of all i minus ) bounded , 1\quad open subsets parenthesiswe \quad of E Bsee sub n that closing\quad parenthesis$ E $ hasi s\ unboundedwequad obtaina fundamental period by the .. process Since j k open ( i ), parenthesis we see A that sub kE closinghas parenthesis a fundamental subset sequence 2 to the power of bounded of minus sets k B. sub k where B sub k is a BanachsequenceTherefore disc in\quad E sub ( ii kof ) comma bounded must open be parenthesissets true . . \ Thenquad A sub forTherefore k each closingn we parenthesis choose\quadBn( is∈ a Bi in so ) that\quad for everymust k\thequad set be true . \quad Then for each $ ncompleting $ we sequence in E period choose $ B { n }\ in B { n }$ so thatn=1 for every $ k $ the set We would like to point out the similarity b etween this1k proof\ and−1 the argument Ak = 2 jn (Bn) given in S 2 when we discussed webs which are eventuallyk bounded period \ [AWe ..{ alsok } .. note= .. 2 that ˆ{ ..1 this{ ..k result}}\ .. maybigcap b e ..ˆ applied{ n ..= to .. 1 the} { .. projk } ectivej ˆ ..{ limit− ..1 of} a{ n } (B { n } ) \ ] i s unbounded . Since jk(A ) ⊂ 2−kB where B is a Banach disc in E , (A ) is a completing sequence in E. DFS hyphen spectrum of Vogt openk squarek bracket 20k closing square bracketk periodk .. Let We would like to point out the similarity b etween this proof and the argument given in §2 when we discussed Line 1 E sub k comma m = open brace xi = open parenthesis xi sub j closing parenthesis : bar xi bar k comma m = sum bar xi sub j bar k webs which are eventually bounded . j\ anoindent to the poweri sof unboundedcomma m less . infinity\quad closingSince brace $ Line j 2 k j ( A { k } ) \subset 2 ˆ{ − k } B { k }$ where We also note that this result may b e applied to the proj ective limit of a DFS $ Bwhere{ k open}$ parenthesis is a Banach k j a to disc the power in of $E comma{ k m} closing,(A parenthesis{ i sk an} infinite) $ matrix i s a with - spectrum of Vogt [ 20 ] . Let completingk j a to the power sequence of comma in m $ greater E . equal $ k j a to the power of comma m plus 1 greater 0 and k j a to the power of plus 1 comma m greater equal k j a to the power of comma m X E = {ξ = (ξ ): k ξ k k, m = | ξ | kj ,m < ∞} Wefor would all j comma liketo k comma point m out periodk,m the ....similarity Set E subj k = b union etween of sub this mj E sub proofa k comma and them and argument equip it with the natural inductive t opology periodgiven in \S 2 when we discussed webs which are eventually boundedj . Let E = intersection of E sub k be the projective limit period .. We also assume that for every k comma m there where (kj ,m) i s an infinite matrix with We i\ squad l with a l s oa \quad note \quad that \quad t h i s \quad r e s u l t \quad may b e \quad appl ied \quad to \quad the \quad pr oj e c t i v e \quad l i m i t \quad o f a DFSlimint− spectrum j k j a to the of power Vogt of comma [ 20 ] m . slash\quad k j aLet to the power of comma l = 0 period kj ,m ≥ kj ,m+1 > 0 and kj +1,m ≥ kj ,m Thus .. the .. set hyphen up .. i s ..a what .. Vogta .. open parenthesisa open squarea bracket 20 closing square bracket closing parenthesis .. \ [ \ begin { a l i g n e d } E { k , m } = \{\S xi = ( \ xi { j } ): \ parallel \ xi \ parallel calls ..for the all ..j, sequence k, m. .. space .. case .. of a ..Set DFSEk hyphen= Ek,m and equip it with the natural inductive t opology . k , mT = \sum \mid \ xi { j }\mmid k{ j { a }}ˆ{ , m } < \ infty \}\\ spectrumLet E = periodEk ..be Although the projective we have limit .. only . We introduced also assume the case that p for =every 1 .. fork, the m there .. sake i s ofl with jsimplicity\end{ a comma l i g n e d what}\ ] follows i s valid for any 1 less or equal p less or equal infinity period ,m ,l 5 period 6 period Proposition period Let E be thelim projectivekja /kja limit= 0 of. a DFS hyphen spectrum of sequence spaces period .. Then either the strong dual E subj b to the power of prime .. is metrizable or E has a nuclear K dieresis-o the \noindentsubspaceThus andwhere the a faithful set $ - up( nuclear k{ i sj K what dieresis-o{ a }} Vogtˆ the{ , quotient ( [ m20} ] )period) calls $ i the s an sequence infinite space matrix case with of a DFS P r- o spectrum o f period . .. Since Although E has we .. have a continuous only introduced norm and the E i scase .. completep = comma 1 for .. the every sake almost of simplicity , what \ [ kbounded{ followsj { ..a i absolutely s}} validˆ{ for, .. any convex m 1}\≤ p ..≤geq subset ∞. k ..{ ofj ..{ E ..a i}} s ..ˆ{ the, .. sum m .. + of a .. 1 bounded} > ..0 set .. and and .. k a{ j { a }}ˆ{ + 1 , m } \geqfinitek hyphen{5 .j 6 .{ dimensionalPropositiona }}ˆ{ , subspace . mLet}\ ..]E openbe the parenthesis projective Theorem limit of a .. DFS 1 period - spectrum 9 closing of sequence parenthesis spaces period . .. We Then .. apply Theorem 5 period 5 0 t o .. deduceeither the strong dual Eb is metrizable or E has a nuclear K o¨ the subspace and a faithful nuclear K o¨ thatthe either quotient E sub . b to the power of prime i s metrizable or E has a completing sequence consisting of set s \noindentwhich areP r notoforall o f almost . Since bounded $jE hasperiod , a continuous k .. Since , E m norm subb .$and toE the\ih s power f i complete l l ofSet prime , $ admits every E { almost ak continuous} bounded= \ normbigcup absolutely comma{ wem can} apE hyphen{ k , m }$ andply equipconvex Theorem it subset 4 with period the of7 in naturalthisE casei s t the inductiveo conclude sum that of t a Eopology has bounded a nuclear . set K dieresis-o and a the finite subspace - dimensional comma subspace 0 provided( Theorem E sub b 1 to . the 9 ) power . We of prime apply i s Theorem not metrizable 5 . 5 t period o deduce that either Eb i s metrizable or E has a \noindentR ecompleting m a r kLet period sequence $ .. E In consisting investigating = \bigcap of set the s projE ective{ k limit}$ .. be of a the .. DFS projective hyphen sp ectrum limit comma . \quad .. VogtWe also assume that for every $ k , m $ t here 0 provedwhich a fundamentalare not almost theorem bounded about . Since the completenessEb admits a of continuous E sub b to norm the , power we can of ap prime - ply open Theorem parenthesis 4 . 7 in open this square bracket 20 closing i s $ l $ with 0 squarecase bracket t o concludesemicolon that TheoremE has 4 a period nuclear 2 K closingo ¨ the parenthesis subspace , provided period Eb i s not metrizable . If E subR b e to m the a r power k . of In prime investigating is complete the and proj metrizable ective limit then E of sub a bDFS to the power- sp ectrum of prime , = Vogt E sub proved k to the a power of prime .. and by \ [ \lim { j } k{ j { a }}ˆ{ , m } /0 k{ j { a }}ˆ{ , l } = 0 . \ ] reflexivityfundamental E = E sub theorem k for about the completeness of Eb([20]; Theorem 4 . 2 ) . 0 0 0 someIf E kb periodis complete and metrizable then Eb = Ek and by reflexivity E = Ek for some k. \noindent Thus \quad the \quad s e t − up \quad i s \quad what \quad Vogt \quad ( [ 20 ] ) \quad c a l l s \quad the \quad sequence \quad space \quad case \quad o f a \quad DFS − spectrum . \quad Although we have \quad only introduced the case $ p = 1 $ \quad f o r the \quad sake o f simplicity , what follows i s valid for any $ 1 \ leq p \ leq \ infty . $
5 . 6 . Proposition . Let $ E $ be the projective limit of a DFS − spectrum of sequence spaces . \quad Then either the strong dual $ E ˆ{\prime } { b }$ \quad is metrizable or $ E $ has a nuclear K $ \ddot{o} $ the subspace and a faithful nuclear K $ \ddot{o} $ the quotient .
P r o o f . \quad Since $E$ has \quad a continuous norm and $ E $ i s \quad complete , \quad every almost bounded \quad a b s o l u t e l y \quad convex \quad subset \quad o f \quad $ E $ \quad i s \quad the \quad sum \quad o f a \quad bounded \quad s e t \quad and \quad a f i n i t e − dimensional subspace \quad ( Theorem \quad 1 . 9 ) . \quad We \quad apply Theorem 5 . 5 t o \quad deduce that either $ E ˆ{\prime } { b }$ i s metrizable or $ E $ has a completing sequence consisting of set s
\noindent which are not almost bounded . \quad Since $ E ˆ{\prime } { b }$ admits a continuous norm , we can ap − ply Theorem 4 . 7 in this case t o conclude that $ E $ has a nuclear K $ \ddot{o} $ the subspace , provided $ E ˆ{\prime } { b }$ i s not metrizable .
R e m a r k . \quad In investigating the proj ective limit \quad o f a \quad DFS − sp ectrum , \quad Vogt proved a fundamental theorem about the completeness of $ E ˆ{\prime } { b } ( [ 20 ] ;$ Theorem4.2).
\noindent I f $ E ˆ{\prime } { b }$ is complete and metrizable then $ E ˆ{\prime } { b } = E ˆ{\prime } { k }$ \quad and by reflexivity $E = E { k }$ f o r some $ k . $ 30 .. S period O-dieresis nal and T period Terzio caron-g lu \noindent6 period ..30 Spaces\quad of continuousS $ . functions\ddot{O} $ nal andT . Terzio $ \check{g} $ lu In this final section we shall apply our results t o spaces of continuous functions period \ centerline {6 . \quad Spaces of continuous functions } Throughout X¨ will denote a completely regular topological space and C open parenthesis X closing parenthesis the space30 .. of S real.O nal hyphen and T valued . Terzio .. continuousgˇ lu .. functions .. on X .. with the t opology .. of uniform Inconvergence this final on compact section subsets we shall6 of . X period apply Spaces .. Weour of shall results continuous identify t Y o subset spaces functions X with of the continuous set of functions . Throughoutpoint evaluationsIn this final $ X open section $ brace will wedelta shall denote apply sub x a our: x completelyin results Y closing t o spaces brace regular of subset continuous topological C open functions parenthesis . space Throughout X closing and parenthesisX $will C denote ( to Xthe power ) $ of the prime where ofspace coursea completely delta\quad subo xfregular open r e a l parenthesistopological− valued space f closing\quadand parenthesisCcontinuous(X) the space= f open\quad of parenthesis realf u - n valued c t i o x n closing s continuous\quad parenthesison functions $ periodX $ ..\quad on HenceX with we the t opology \quad o f uniform convergencewith the t opologyon compact of uniform subsets convergence of $ X on compact . $ \ subsetsquad We of X. shallWe identify shall identify $Y Y⊂ X\withsubset the X$ with the set of take the lib erty of setting 0 pointY toset the evaluations of power point evaluations of circ = $ open\{\{δx brace: x ∈delta fY in} ⊂CC open{(Xx) parenthesis}where: of x course X closing\ inδx(f parenthesis)Y = f(x\}\). :Hence barsubset f open we take parenthesisC the lib ( erty y closing X of ) parenthesis ˆ{\prime bar} less$ orwhere of course equal$ \ delta 1setting for all{ yx in Y} closing( fbrace ) period = f ( x ) . $ \quad Hence we take the lib erty of setting We refer the reader to open square bracket◦ 9 closing square bracket for t opological concepts not defined here period Let phi n : R right arrow R b e definedY by= {f ∈ C(X):| f(y) |≤ 1forally ∈ Y }. \ [ Y ˆ{\ circ } = \{ f \ in C(X): \mid f ( y ) \mid \ leq 1 f o r a l l phiWe n open refer parenthesis the reader tox closing [ 9 ] for parenthesis t opological = concepts open brace not x defined comma here from . nLet commaφn : R to→ minusR b e n defined to the bypower of comma minus n from n less toy x less\ in less orY equal\} x from. \ x] comma to minus n comma less or equal n comma n, , n< x, and for A subset C open parenthesis X closingφn(x) = parenthesis{x,−n − definenx< ≤ thex− setn, ≤ ofn, truncations as phi open parenthesis A closing parenthesis = open brace phi n circ f : f in A comma n in N closing brace period \noindentWeand first for haveWeA ⊂ a refer line-tC(X) echnical define the reader the result set of period to truncations [ .... 9 Our] for as approach t opological resembles the concepts treatment not of spaces defined here . Letwith $ the\phi boundedn approximation : R \ prrightarrow to the power ofR$ line-o p b erty e definedopen parenthesis by cf period Lemma 4 period 6 closing parenthesis period 6 period 1 period Proposition period Letφ A(A subset) = {φn C◦ openf : f parenthesis∈ A, n ∈ XN} closing. parenthesis .. and Y subset X period .. The following are \ [ \phi n ( x ) = \{ x , ˆ{ n , } { − n }ˆ{ , } − n ˆ{ n < } { x < }\ leq x ˆ{ x true : We first have a line − t echnical result . Our approach resembles the treatment of spaces , } { − n } ,{\ leq } n , \ ] openwith parenthesis the bounded a closing approximation parenthesis pr Aline subset−o p erty phi open( cf . parenthesis Lemma 4 . A6 ) closing . parenthesis period open parenthesis6 b . closing1 . Proposition parenthesis . ALet subsetA ⊂ rhoC Y(X to) theand powerY ⊂ ofX. circ ..The if and following only if are phi true open: parenthesis A closing parenthesis subset rho Case 1 circ Case 2 period \noindent and f o r $ A \subset C ( X ) $ define the set of truncations as open parenthesis c closing parenthesis phi open parenthesis(a)A ⊂ φ A(A closing). parenthesis subset sp open parenthesis X to the power of circ closing parenthesis period \ [ \phi ( A ) = \{\phi n \ circ f : f ◦ \ in A , n \ in N \} . \ ] open parenthesis d closing parenthesis( b ) ForA ⊂ KρY subset◦ if X and compact only if commaφ(A) K⊂ toρY the power of circ subset X to the power of circ plus intersection of sub epsilon greater 0 epsilon K to the power of circ period . P r o o f period .. Let K subset X b e compact period(c)φ(A ..) ⊂ Tosp( proveX◦). open parenthesis a closing parenthesis we observe that open parenthesis phi\noindent m circ f closingWe first parenthesis have open a $parenthesis line −t x $ closing echnical parenthesis result = . \ h f i l l Our approach resembles the treatment of spaces ◦ ◦ T ◦ f open parenthesis x closing parenthesis( d ) For ifK x in⊂ XX andcompact m greater,K equal⊂ X n+ whereε>0 εK f open. parenthesis K closing parenthesis subset open square bracket\noindent minusP rwith o n o comma f . the Let n bounded closingK ⊂ X squareb approximation e compact bracket . period To prove open $ pr parenthesis( a ˆ ){ wel i observen e − bo closing} that$ parenthesisp (φm erty◦ f)(x () follows =cff( .x) Lemma4from if x ∈ theX observationand . 6 ) . thatm ..≥ barn where open parenthesisf(K) ⊂ [−n, phi n]. m circ( b ) f followsclosing fromparenthesis the observation open parenthesis that | x(φm closing◦ f)( parenthesisx) | ≤ | barf(x less) | orand equal bar f open parenthesis ◦ x\ centerline closingfrom parenthesis ( a{ )6 . . bar( 1 c ). .. is Propositionand evident from . open Now parenthesis . let Letf ∈ $AK a closing. By\subset theparenthesis TietzeC extension period ( .. theorem open X parenthesis ) we$ find\quadg c closing∈andC parenthesis(X $) Y \subset is evident periodX . .. $ ◦ ◦ Now\quad letwithThe f in K following| tog(x the) | power ≤ are1 of for circ trueall periodx and : ..}g( Byy) = f(y) for all y ∈ K. Thus g ∈ X and f − g ∈ εK for any ε > 0. the TietzeWe note extension that bytheorem ( b ),A weis find bounded g in C in openC(X parenthesis) if and only X closing if φ(A) parenthesis is bounded .. . with Also ..{φ bar(A): g openA ⊂ parenthesisC(X) x closing parenthesis bar\ [ less (bounded or a equal )} 1i for s A a all fundamental x\subset and system\phi of bounded(A). subsets \ ] g open parenthesis y closing parenthesis = f open parenthesis y closing parenthesis for all y in K period .. Thus g in X to the power of circ and f minus g in epsilon K to the power of circ for any epsilon greater 0 period ofC(X). \ centerlineWe note that{( by b open $ parenthesis) A \ bsubset closing parenthesis\rho commaY ˆ{\ Acirc is bounded}$ \ inquad C openif parenthesis and only X if closing $\ l parenthesis e f t . \phi if and(A) only if phi \subset \rho Y\ begin { a l i g n e d } & \ circ \\ open parenthesis A closing parenthesis is bounded period 0 &. \end{ a l i g n e d }\ right6 . 2. $ . Corollary} .C(X)b satisfies condition ( y ) . Also open braceP rphi o o open f . Let parenthesisB ⊂ C(X A) closingbe absolu parenthesisline − t ely : A convex subset an C opend − line parenthesis bounded X . By closing Proposition parenthesis 6 . 1 bounded closing brace i s a fundamentalwe have system of bounded subsets \ [of ( C open c parenthesis ) \phi X closing(A) parenthesis\ periodsubset sp ( X ˆ{\ circ } ). \ ] 6 period 2 period Corollary period C open parenthesisB ⊂ φ(B) X= closingsp(X◦) parenthesis∩ φ(B). sub b to the power of prime satisfies condition open parenthesis y closing parenthesis period \ centerline {( d ) For $ K 00\subset X$ compact $ , Kˆ{\ circ }\subset0 X00 ˆ{\ circ } + \bigcap {\ varepsilon P rWe o o carry f period this Let t o the B subset bidual CC( openX) by parenthesis ta line−k Xing closingpolars parenthesisand closure withbe absolu respect line-t to the ely duality convexhC an(X d-line) ,C( boundedX) i period By Proposition >6 period0and}\ 1 obtainvarepsilon K ˆ{\ circ } . $ } we have P r o o f . \quad Let $ K \subset X$ b e compact◦ ◦ . \quad To prove ( a ) we observe that $ ( \phi B subset phi open parenthesis B closing parenthesisB◦◦ ⊂ sp(X =◦ sp ◦) open∩ φ(B parenthesis) X circ closing parenthesis cap phi open parenthesis B closing mparenthesis\ circ periodf ) ( x ) = $ . $We f carry ( this x t o the ) $ bidual i f C $ open x parenthesis\ in X X $ closing and parenthesis $ m \geq to the powern$ of where prime prime $f by ( ta line-k K sub ) ing\subset polars and closure[ − withn respect , n to ] . ($ b) follows fromthe observation thatthe duality\quad angbracketleft$ \mid C( open\phi parenthesism X\ closingcirc parenthesisf ) to ( the x power ) of prime\mid comma\ leq C open\mid parenthesisf X ( closing x parenthesis ) \mid $ to\quad the powerand of from prime ( prime a ) right. \quad angbracket( c ) and is obtain evident . \quad Now l e t $ f \ in K ˆ{\ circ } . $ \quad By theB to Tietze the power extension of circ circ subset theorem sp open we parenthesis find $ g X circ\ in circ closingC parenthesis ( X )cap $ phi\quad open parenthesiswith \quad B Case$ 1\ circmid circ Caseg 2 ( period x ) \mid \ leq 1$ for all $x$ and $g ( y ) = f ( y )$forall$y \ in K . $ \quad Thus $ g \ in X ˆ{\ circ }$ and $ f − g \ in \ varepsilon K ˆ{\ circ }$ f o r any $ \ varepsilon > 0 . $
Wenotethatby(b $) , A$ isboundedin $C ( X )$ if andonly if $ \phi ( A ) $ i s bounded . Also $ \{\phi (A):A \subset C ( X )$ bounded \} i s a fundamental system of bounded subsets
\ begin { a l i g n ∗} o f C ( X ) . \end{ a l i g n ∗}
\ centerline {6.2.Corollary $. C ( X )ˆ{\prime } { b }$ satisfies condition ( y ) . }
\ hspace ∗{\ f i l l }Proof .Let $B \subset C ( X )$ beabsolu $line −t $ ely convex an $ d−l i n e $ bounded . By Proposition 6 . 1
\noindent we have
\ [B \subset \phi ( B ) = sp ( X \ circ ) \cap \phi (B). \ ]
\noindent Wecarry this t othe bidual $C ( X ) ˆ{\prime \prime }$ by ta $ l i n e −k { ing }$ polars and closure with respect to the duality $ \ langle C ( X ) ˆ{\prime } , C ( X ) ˆ{\prime \prime }\rangle $ and obtain
\ [ \ l e f t . B ˆ{\ circ \ circ }\subset sp ( X \ circ \ circ ) \cap \phi (B)\ begin { a l i g n e d } & \ circ \ circ \\ &. \end{ a l i g n e d }\ right . \ ] Subspaces and quotient spaces of locally convex spaces .. 31 \ hspaceSince open∗{\ f brace i l l } Subspaces phi open parenthesis and quotient B closing parenthesisspaces of : B locally subset C convex open parenthesis spaces X\quad closing31 parenthesis bounded closing brace is a fundamental sequence of bounded sets comma \noindentthis showsSince that C open $ \{\ parenthesisphi X closing(B):B parenthesis sub b to the\subset power of primeC ( satisfies X open )$ parenthesis bounded y closing\} is parenthesis a fundamental period sequence of bounded sets , thisshowsthatThe space C open parenthesis $C ( X closing X parenthesis )ˆ{\prime it selfSubspaces} satisfies{ b } the and$ openness quotient satisfies spaces condition ( of y locally open ) . convexsquare spaces bracket 3 31 closing square bracket period Since {φ(B): B ⊂ C(X) bounded } is a fundamental sequence of bounded sets , this shows that C(X)0 satisfies .. This follows from b TheProposition space( y ) . $C 6 period ( 1 open X parenthesis ) $ it d self closing satisfies parenthesis the period openness .. In particular condition comma [ .. 3 a quotient] . \quad spaceThis of C followsopen parenthesis from X closingProposition parenthesisThe space 6 i s.C either( 1X) ( it da self normed ) satisfies . \quad theIn openness particular condition , [ 3\quad ] . Thisa quotient follows from space Proposition of $C 6 . 1 ( d( ) . X ) $ i s either anormed spacespaceIn orparticular or it it admits admits , no a continuous quotient no continuous space norm of periodC(X norm) i s either . a normed space or it admits no continuous norm . We recallWe that recall .. that a t opological a t opological space X space is pseudocompactX is pseudocompact .. if everyif continuous every continuous real - valued function on Wereal recallX hypheni s bounded that valued\quad [ function 9 ] .a At on completely opological X i s bounded regular space open topological square $ X bracket $ space is 9X closing pseudocompacti s pseudocompact square bracket\quad if period andif only .. everyA if completely for every continuous regular topological space r e a ldescending− valued sequence function (Vn) ofon nonempty $X$ i s bounded [ 9 ] . \quad A completely regular topological space X i s pseudocompact if andT only if for every descending sequence open parenthesis V sub n closing parenthesis◦ of nonempty $open Xopen $ subsets subsets i s we pseudocompact wehave have intersectionVn 6= ∅ ifof[9] V. and subIf nonlyX negationslash-equalis a ifcompletely for every regular varnothing descending pseudocompact open square sequence space bracket , then $ 9X closing(i s V a square barrel{ n bracket} ) $ period of .. nonempty If X is a completelyin C( regularX). pseudocompact \noindentspace commaIn generalopen thenC subsets( XX to) the is wenot power have complete of circ $ i\ s ,bigcap a barrel but if inX CisV open a completely{ n parenthesis}\not regular= X closingk−\ varnothingspace parenthesis then C period(X [) can 9 ]b .$ \quad I f $ X $ isIn a generale completely expressed C open parenthesis regular as the pseudocompact X proj closing ective parenthesis limitof .. is notthe .. familycomplete{ commaC(K): ..K but⊂ X ifcompact X is a completely} and therefore regular k hyphen space spacethenin .. this , C then open case parenthesisC( $XˆX) i s{\ complete Xcirc closing (} [$ parenthesis 1 0 isabarrelin ] ; 3 . 6 .. . can 4 ) ... b Wee .. now expressed$C state ( ..one as X of .. our the ) main .. proj .$ results ective about .. limit spaces .. of .. of the .. family .. open brace C opencontinuous parenthesis functions K closing . parenthesis : IngeneralK subset6 . X 3 compact. $CTheorem (closing X . brace )$Let andX\ thereforequadbei sa in not this completely\ casequad C opencomplete regular parenthesisk ,− \spacequad X closing .but The parenthesis if following $ X i $ s complete areis a completely open parenthesis regular open square$ k bracket− $ 1space 0 closing square bracket semicolon 3 period 6 period 4 closing parenthesis period .. We thennow state\quad one$ of Cour main ( results X ) about $ \ spacesquad ofcan continuous\quad functionsb e \quad periodexpressed \quad as \quad the \quad pr oj e c t i v e \quad l i m i t \quad o f \quad the \quad family \quad equivalent : $ \{6 periodC 3 period ( K .. Theorem ) : period $ .. Let X .. be .. a .. completely .. regular k hyphen space period .. The following .. are $equivalent K \subset : X $ compact \} and therefore in this case $C ( X )$ i s complete ( [ 10 ] ; 3 . 6 . 4) . \quad We now state one of our main results( i about ) X is not spaces pseudocompact of continuous . functions . open parenthesis i closing parenthesis( ii X )ω isis not isomorphic pseudocompact to a subspace period of C(X). open parenthesis ii closing parenthesis( iii omega)C(X) ishas isomorphic a proper Fr to ae´ subspacechet subspace of C open. parenthesis X closing parenthesis period \ hspaceopen parenthesis∗{\ f i l l }6 iii closing. 3 . parenthesis\quad Theorem C open parenthesis . \quad Let X closing $ X parenthesis $ \quad hasbe a\ properquad Fra acute-e\quad chetcompletely subspace period\quad r e g u l a r $ k − ($ iv ) spaceThere is . a\ completingquad The s followingequence (An)\quadin Care(X) such that each An is un - bounded . open parenthesis iv closing parenthesis( There v )ω is a a completing faithful quotient s equence of C open(X). parenthesis A sub n closing parenthesis .. in C open parenthesis X closing parenthesisP r o o f . .. The such that implications each A sub ( n ii .. ) is⇒ un( iii hyphen ) ⇒ ( iv ) ⇒ ( v ) are true in general by our \ beginbounded{ a l i period g n ∗} equivalentprevious results : and therefore it remains to prove ( v ) ⇒ ( i ) ⇒ ( ii ) . open parenthesis v closing parenthesis omega is a faithful quotient◦ of C open parenthesis X closing parenthesis period \end{ a l iIf g nX∗}i s pseudocompact , the closure of the image of X in a barrelled quotient space of C(X) would b e a P rneighborhood o o f period . .. Hence The ..every implications barrelled quotient .. open ofparenthesisC(X) i s aii normed closing spaceparenthesis . double stroke right arrow open parenthesis iii closing parenthesis double strokeTo prove right arrow( i ) ⇒ open( ii ) parenthesis we will construct iv closing directly parenthesis an imbedding double stroke of ω into rightC( arrowX). open parenthesis v closing parenthesis .. are\ centerline .. true .. in{ ..( general i ) X .. is by .. not our pseudocompact . } If X is not pseudocompact , we find a sequence of closed proper subsets Xn of X such that each Xn i s a previous results and therefore it remains to prove open parenthesis v closing parenthesis double stroke right arrow open parenthesis i closing proper subset of Xn+1 and parenthesis\ centerline double{( stroke i i $ right ) arrow\omega open$ parenthesis is isomorphic ii closing parenthesis toasubspace period of $C ( X ) . $ } If X i s pseudocompact comma the closure of the image of X to the∞ power of circ in a barrelled quotient \ centerline {(iii $) C ( X )$ hasaproperFr $ \acute{e} $ chet subspace . } space of C open parenthesis X closing parenthesis would b[ e a neighborhood period Hence every barrelled quotient of C open parenthesis X X = intX . closing parenthesis n ( iv ) There is a completing s equence $ ( A { n } ) $ \quad in $ C ( X ) $ \quad such that each i s a normed space period n = 1 $ ATo prove{ n } open$ \ parenthesisquad i s iun closing− parenthesis double stroke right arrow open parenthesis ii closing parenthesis we will construct directly an We find n ↑ ∞ such that imbeddingbounded of . omegak into C open parenthesis X closing parenthesis period If X is not pseudocompact comma we find a sequence of closed proper subsets X sub n of Y = intX ∩ (X \ X ) 6= \ centerlineX such that{ each( v X $sub ) n i s\omega a proper$ subset isafaithfulk of X subnk+1 n plus quotient 1 andnk ∅ of $C ( X ) .$ } Lineand 1 infinity choose Linexk from 2 X this = union set . Weof int define X sub continuous n period Linefunctions 3 n =fk 1 : X → [0, 1] such P rWe othat find o ffk n sub.(x\)quad k = upwards 1 andThefk arrow\vanishesquad infinityimplications on X such\ Y that. \quad ( i i $ ) \Rightarrow ( $ i i i $ ) \Rightarrow ( $ i v $ ) \Rightarrowk ( $ v ) \quadk are \quad true \quad in \quad g e n e r a l \quad by \quad our Y sub kGiven = int a X compact sub n sub subset k plusK ⊂ 1 capX, we open find parenthesism with K X⊂ backslashint Xm. For X subnk > n msubwe k haveclosing parenthesis negationslash-equal varnothing previousand choose results x sub k from and this therefore set period itWe remainsdefine continuous to prove functions ( v f k $ : ) X right\Rightarrow arrow open square( $ bracket i $ 0comma ) \Rightarrow 1 closing square ( $ i i ) . bracket such fk | Xm = 0. that f k open parenthesis x sub k closing parenthesis = 1 and f k vanishes on X backslash Y sub k period IfGiven $ X a $ compact i s pseudocompact subset K subset X comma, the we closure find m with of the K subset image int X of sub $ m X period ˆ{\ Forcirc n sub}$ k greater in a barrelled m we quotient spacehave of $C ( X ) $ would b e a neighborhood . Hence every barrelled quotient of $C ( X ) $f k bar X sub m = 0 period i s a normed space .
\ centerline {Toprove ( i $ ) \Rightarrow ( $ ii ) we will construct directly an imbedding of $ \omega $ into $C ( X ) .$ }
If $ X $ is not pseudocompact , we find a sequence of closed proper subsets $ X { n }$ o f $X$ such that each $X { n }$ i s a proper subset of $X { n + 1 }$ and
\ [ \ begin { a l i g n e d }\ infty \\ X = \bigcup i n t X { n } . \\ n = 1 \end{ a l i g n e d }\ ]
\noindent We f i n d $ n { k }\uparrow \ infty $ such that
\ [Y { k } = i n t X { n { k + 1 }}\cap (X \setminus X { n { k }} ) \not= \ varnothing \ ]
\noindent and choose $ x { k }$ from this set . We define continuous functions $ f k : X \rightarrow [ 0 , 1 ] $ such
\noindent that $ f k ( x { k } ) = 1$ and $f k$ vanisheson $X \setminus Y { k } . $
Given a compact subset $ K \subset X ,$ wefind $m$ with $K \subset $ i n t $ X { m } . $ For $ n { k } > m $ we have
\ [ f k \mid X { m } = 0 . \ ] 32 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentEquation:32 Let\ openquad parenthesisS $ . xi\ iddot closing{O} parenthesis$ nal andT in omega . period Terzio .. Then $ \check m minus{g} 1 open$ lu parenthesis sum xi i f i closing paren- thesis vextendsingle-vextendsingle-vextendsingle sub K = open parenthesis sum i = 1 xi i f i closing parenthesis vextendsingle-vextendsingle- \ begin { a l i g n ∗} vextendsingle sub¨ K \ tagand∗{32 so$ the Let S series.O nal ( sum and\ xi Txi i. Terziof i isi Cauchygˇ )lu in\ in the complete\omega lcs C. open $} parenthesisThen \\ m X closing− parenthesis1 \\ ( period\sum .. Therefore\ xi i the map f i ) \arrowvertT : omega right{ K arrow} = C open ( parenthesis\sum { Xi closing = parenthesis 1 }\ xi obtainedi by f setting i ) \arrowvert { K } \endLine{ a 1l i infinityg n ∗} Line 2 T open parenthesis open parenthesis xi i closing parenthesis closing parenthesis = sum xi i f i Line 3 i = 1 Then Let(ξi) ∈ ω. i s well defined period .. Also if bar xi i bar less or equal 1 slash m for i = 1 comma period period period comma m comma for x in K we have\noindent and so the series $ \sum \ xi i fm i$− 1 isCauchyinthecomplete lcs $C ( X ) . $ \quad Therefore the map X X Line 1 infinity m Line 2 vextendsingle-vextendsingle-vextendsingle( ξifi)| = ( ξifi)| sum xi i f i open parenthesis x closing parenthesis vextendsingle- $ T : \omega \rightarrow C (K X ) $ obtainedK by setting vextendsingle-vextendsingle = vextendsingle-vextendsingle-vextendsinglei=1 sum xi i f i open parenthesis x closing parenthesis vextendsingle- vextendsingle-vextendsingle less or equal 1 Line 3 i = 1 i = 1 P \ [ \andbeginand so T{ soa is thel icontinuous g n series e d }\ infty periodξifi is .. Cauchy\\ We also in note the complete that f i open lcs C parenthesis(X). Therefore x sub j the closing map parenthesisT : ω → C( =X 0)if obtained i negationslash-equal by j comma which impliesT((setting in \ xi i ) ) = \sum \ xi i f i \\ iparticular = 1that\end T i{ sa one l i g nhyphen e d }\ ] t o hyphen one period .. Let U subset omega b e the neighborhood ∞ U = open brace open parenthesis xi i closing parenthesis : bar xi i bar less or equal 1 for i = 1 comma period period period comma l closing X brace T ((ξi)) = ξifi \noindentfor some li in s N well period defined .. Let K = . open\quad braceAlso x sub i f 1 comma $ \mid period\ xi periodi period\mid comma\ xleq sub l closing1 / brace m and $ suppose f o r $ f ini T= open i = 1 parenthesis1 , . omega . closing . parenthesis ,m cap ,$for$x K to the power of\ circin periodK $ .. So we f =have sum xi i f i andi for s well 1 less defined or equal .j Also less ifor| equalξi |≤ l1/m we havefor i bar= 1, f ..., open m, parenthesisfor x ∈ K we x sub have j closing parenthesis bar = bar xi sub j bar less or equal 1 period ..\ [ Therefore\ begin { a l i g n e d }\ infty m \\ \Tarrowvert open parenthesis\sum omega\ closingxi parenthesisi f icap K ( to the x power ) of\∞ circarrowvertm subset T open= parenthesis\arrowvert U closing\ parenthesissum \ xi periodi f i ( x ) \arrowvert \ leq 1 \\X X Next we will deal with certain almost bounded| ξifi subsets(x)| of= | C openξifi parenthesis(x)| ≤ 1 X closing parenthesis period .. As we shall iuse the = proj 1 ection i considered = 1 \ inend the{ a proof l i g n e of d Theorem}\ ] 1 period 9 comma we need completeness period 6 period 4 period Lemma period Let X be a completely regulari = k hyphen1 i = 1 space comma .. and let A subset X be c losed period If A to the power of circ .. is .. an almost bounded subset of C open parenthesis X closing parenthesis .. then A .. is .. also .. open and X \noindentand so andT is continuous so $ T .$ We is also continuous note that fi . (x\jquad) = 0 ifWealsonotethati 6= j, which implies in particular $f i that (T i s x one{ - tj } ) = 0 $ i f backslasho - oneA .. . is Let U ⊂ ω b e the neighborhood $ idiscrete\not period= j , $ which implies in particular that $ T $ i s one − t o − one . \quad Let $ U \subset \omega $ b e the neighborhood P r o o f period .. Set .. E = C open parenthesisU = {(ξi X): closing| ξi |≤ parenthesis1fori = 1, .. ..., and l} .. L = .. sp open parenthesis X backslash A closing parenthesis .. which .. is .. a .. subspace .. of E to the power of prime period \ [ U = \{ ( \ xi i ) : \mid \ xi i ◦ \mid P\ leq 1for i=1 , . . . Supposefor some u inl sp∈ N open. Let parenthesisK = {x1, A ..., to xl the} and power suppose of circf circ∈ T ( closingω) ∩ K parenthesis. So f = capξifi L periodand for .... 1 Then≤ j ≤ ul =we sum have sub i = 1 to the power of n , l \}\ ] alpha sub| f(x ij x) sub|=| ξ ij with|≤ 1. x subTherefore i in X backslash A period Note that we are identifying as usual x in X with the corresp onding point evaluation functional period We find f i in C open parenthesis X closing parenthesisT (ω) ∩ withK◦ ⊂ fT i open(U). parenthesis x sub i closing parenthesis = 1 comma \noindentf i open parenthesisf o r some x closing $ l parenthesis\ in N = 0 for . $ x in\ Aquad cup openLet brace $ K x sub = j : j\{ equal-negationslashx { 1 } i, comma . 1 less . or . equal , j less x or{ equall } \} $ andNext suppose we will deal $ with f certain\ in almostT( bounded\omega subsets of)C(X\).capAs weK shall ˆ{\ usecirc the proj} ection. $ considered\quad So $ f = \sum n closingin thebrace proof of Theorem 1 . 9 , we need completeness . \ xiand seti f = f sum i sub $ i = 1 to the power of n alpha sub i f i period .. Then 6 . 4 . Lemma . Let X be a completely regular k− space , and let A ⊂ X be c losed . If A◦ is andLine f o 1 r n Line $ 1 2 u open\ leq parenthesisj \ leq f closingl parenthesis $ we have = sum $ bar\mid alpha subf i bar ( to x the{ powerj } of) 2 Line\mid 3 i = 1 = \mid \ xi { j } \mid an\ almostleq bounded1 . subset $ \ ofquadC(XTherefore) then A is also open and X \ A is discrete . but u in spP open r o o parenthesis f . Set AE circ= circC closing(X) parenthesis and L = implies sp ( uX open\ A) parenthesis which is f closing a subspace parenthesis of =E 00. period .. Hence each alpha sub i = 0 and so ◦◦ Pn \ [T(Suppose \uomega∈ sp (A ))∩ L.\cap K ˆ{\ circ }\Thensubsetu = i=1T(U).αixi with xi ∈ X \ A. Note\ ] that we are sp openidentifying parenthesis as usual A tox the∈ X powerwith of the circ corresp circ closing onding parenthesis point evaluation cap L = functional open brace . 0 We closing find fi brace∈ C period(X) with Nowfi we(x ) find = 1 an, algebraic complement G of sp open parenthesis A to the power of circ circ closing parenthesis in E to the power of prime which containsi L period .... The Next we will deal with certain almost bounded subsets of $C ( X ) . $ \quad As we s h a l l proj ection P : E to the power offi prime(x) = right 0 for arrowx ∈ A E∪ to {x the: j power6= i, of1 ≤ primej ≤ n with} P to the power of minus 1 open parenthesis 0 closing parenthesisuse the = proj .. sp ectionopen parenthesis considered A circ circ in closingthe proof parenthesis ofj Theorem .. and range 1 . G 9 is gamma, we need hyphen completeness continuous . Pn openand parenthesis set f = Theoremi=1 αifi. 1Then period 9 closing parenthesis period .. Since X is a k hyphen space comma C open parenthesis X closing parenthesis6 . 4 . Lemma is complete . Let and so $X$ by the Grothen be a completely hyphen regular $ k − $ space , \quad and l e t $ A \subset X $ bedieck c l o completion s e d . theorem P i s sigma open parenthesis E to the powern of prime comma E closing parenthesis hyphen continuous comma .. I f $ A ˆ{\ circ }$ \quad i s \quad an almostX bounded subset of $C ( X ) $ \quad then $ A $ \quad i s \quad a l s o \quad open and which implies that .. G i s a u(f) = | α |2 $ X \setminus A $ \quad i s i d i s c r e t e . i = 1 but u ∈ sp (A ◦ ◦) implies u(f) = 0. Hence each α = 0 and so \ hspace ∗{\ f i l l }P r o o f . \quad Set \quad i $ E = C ( X ) $ \quad and \quad $ L = $ \quad sp $ ( X \setminus A ) $ \quad which \quad i s \quad a \quad subspace \quad o f $ E ˆ{\prime } . $ sp(A◦◦) ∩ L = {0}. \noindentNow weSuppose find an algebraic $ u complement\ in $ spG of $ sp ( (A◦◦ A) in ˆ{\E0 whichcirc contains\ circL. } ) \cap L . $ The\ h f i l l Then $ u = \sum projˆ{ n ection} { Pi: =E0 1 →}\Ealpha0 with P −{1(0)i } =x sp{ (Ai ◦} ◦$) with and range $ xG is{γ−i continuous}\ in (X Theorem\setminus 1 A . $ Note that we are . 9 ) . Since X is a k− space ,C(X) is complete and so by the Grothen - dieck completion theorem P i s \noindentσ(E0,Eidentifying)− continuous , as which usual implies $ x that \ inG i s aX $ with the corresp onding point evaluation functional . We f i n d $ f i \ in C ( X )$with$f i ( x { i } ) = 1 , $
\ [f i ( x )=0 for x \ in A \cup \{ x { j } : j \ne i , 1 \ leq j \ leq n \}\ ]
\noindent and s e t $ f = \sum ˆ{ n } { i = 1 }\alpha { i } f i . $ \quad Then
\ [ \ begin { a l i g n e d } n \\ u ( f ) = \sum \mid \alpha { i }\mid ˆ{ 2 }\\ i = 1 \end{ a l i g n e d }\ ]
\noindent but $ u \ in $ sp $ ( A \ circ \ circ )$ implies $u ( f ) = 0 .$ \quad Hence each $ \alpha { i } = 0 $ and so
\ [ sp ( A ˆ{\ circ \ circ } ) \cap L = \{ 0 \} . \ ]
\noindent Nowwe find an algebraic complement $G$ of sp $ ( A ˆ{\ circ \ circ } ) $ in $ E ˆ{\prime }$ which contains $ L . $ \ h f i l l The
\noindent proj ection $P : Eˆ{\prime }\rightarrow E ˆ{\prime }$ with $ P ˆ{ − 1 } ( 0 ) = $ \quad sp $ ( A \ circ \ circ ) $ \quad and range $G$ is $ \gamma − $ continuous ( Theorem 1 . 9 ) . \quad Since $X$ isa $k − $ space $ , C ( X ) $ is complete and so by the Grothen − dieck completion theorem $ P $ i s $ \sigma ( E ˆ{\prime } ,E) − $ continuous , \quad which implies that \quad $ G $ i s a Subspaces and quotient spaces of locally convex spaces .. 33 \ hspacesigma∗{\ hyphenf i l l closed} Subspaces subset of and E to quotient the power of spaces prime period of locally .. It is easy convex to see spaces that X\ backslashquad 33 A = X cap G comma and hence X backslash A \noindenti s closed period$ \sigma X backslash− A$ i s closed ther e-line subset fore a k of hyphen $E space ˆ{\ periodprime ..} If we. show $ \ thatquad it sIt compact is easy subsets to are see that $X \setminus A = X \cap G ,$ andhence $X Subspaces\setminus and quotientA $ spaces of locally convex spaces 33 always finite comma then0 it will follow that X backslash A i s disc r-line ete period If K subset X backslash A is compact comma isclosedthenσ− Pclosed open parenthesis subset $. of XE K. closing\setminusIt is easyparenthesis to seeA that = $ KX isubset\ sA ther= CapitalX ∩ $G, Gamma eand−line hence open $X brace fore\ A i u s a sub closed $k 1 comma.X−\ period$A i s space ther period . period\quad commaIf we u sub show n that it s compact subsets are closingalwayse brace− line finite .. fore as in a , thek− then proofspace it of . Theorem will If we follow show .. 1 period that that it 9 s period compact $ X .. Therefore subsets\setminus are alwaysA finite $ i ,s then d i sit c will $ follow r−line that $ ete . If $K \subset X sp\ openXsetminus\ A parenthesisi s disc rA$− Kline to the is ete power compact . If K of⊂ circX , \ circA is closing compact parenthesis , then P in(K E) to the = powerK ⊂ of primeΓ{u1 has, ..., finite un} dimensionas in the comma which implies that then$Pproof of Theorem ( K 1 ). 9 . = Therefore K \subset sp (K◦◦) in \EGamma0 has finite\{ dimensionu { , which1 } implies, . that .K is . finite , . u { n }\} $ \quad as in the proof of Theorem \quad 1 . 9 . \quad Therefore K is finite period ◦ spSuppose $ (Suppose again K ˆ A again{\ subsetcircA ⊂ XX i si\ such scirc such that that} A)A to $ thei s almost in power $ ofbounded E circ ˆ{\ i sprime . almost If X bounded}i$ s k− hasspace period finite we .. have If dimension X shown i s k hyphen that X , space\ whichA we implies that $ K $ i shave f iis n ishown open t e . , that closed X and backslash discrete A .is Hence open comma if either closedX is connected and discrete orperiod if X has Hence no i solatedif either points X is connected , we have X = A. or ifThis X has proves no i the solated following points result comma . we have X = A period .. This proves the following result period Suppose6 period again 5 period $ Corollary A6 . 5\subset . periodCorollary LetX$ X be. Let a i completely sX suchbe a completely regularthat k $Aˆ hyphen regular{\ spacekcirc− space and}$ A and subset iA s⊂ almost XX. periodIf bounded ..X Ifhas X has . \quad I f $ X $ i sno $ k is olated− $ points space or if weX is connected , the following are equivalent : no is olated points or if X is connected comma the following◦ are equivalent : haveopen shown parenthesis that i closing $ X parenthesis\setminus A to theA( power i $ )A is ofis circ open bounded .. is , bounded closed . period and discrete . Hence if either $ X $ is connected oropen if parenthesis $X$ hasnoii closing parenthesisi solated A points circ( ii .. )A is◦ almost ,wehaveis almost bounded bounded $X period . = A . $ \quad This proves the following result . ( iii ) A is dense in X . open parenthesis iii closing parenthesis AJ is dense in X period \ hspaceIf X i s∗{\If discreteXfi i s l l discrete} comma6 . 5 , then then . Corollary CC( openX) ' parenthesisR for . Let some XJ. $X$closingIf X parenthesis bei s compact a completely simeq then R of to course the regular powerC(X) of is $ J a for k Banach some− $ space J period space . .. If and X i s $A compact\subset then Xof course .Now $ if\quadX = Y I∪ fZ,Y $ X∩ $Z = has∅ where Y i s compact and Z discrete and closed then
C open parenthesis X closing parenthesis is a Banach spaceJ period .. Now if X = Y cup Z comma Y cap Z = varnothing where Y i s compact and\noindent no is olated points or ifC $(X X) $' R is× connectedBanach. , the following are equivalent : Z discreteThe converse and closed of this then observation i s also true . \ centerline {( i $ ) A ˆ{\ circ }$ \quad i s bounded . } C open parenthesis X closing6 . 6parenthesis . Theorem simeq . For R to a thecompletely power of regular J timesk− Banachspaceperiod the fo llowing are equivalent : The converse of this observation i s also true period \ centerline {( i i $ ) A \ circ $ \quad is almost boundedJ . } 6 period 6 period Theorem period For a completely(a)C(X regular) ' Banach k hyphen× R space. the fo llowing are equivalent : open parenthesis a closing parenthesis C open parenthesis X closing parenthesis simeq Banach times R to the power of J period \ centerlineopen parenthesis{( iii b( bclosing )A )X = is parenthesisY ∪ denseZ where inXXY =is Y . compact cup} Z where,Z discrete Y is compact, c losedcomma and Z discreteY ∩ Z = comma∅. .. c losed and Y cap Z = varnothing period P r o o f . We need t o prove ( a ) ⇒ ( b ) . If B i s the closed unit ball of the Banach If $X$space , then isdiscreteB × RJ i s an ,then almost bounded $C neighborhood ( X ) of C\(simeqX). R ˆ{ J }$ forsomeThis $J means .$ \quad I f $ X $ P r o o f period .. We need t o prove open parenthesis◦ a closing parenthesisJ ◦ double stroke right arrow open parenthesis b closing parenthesis periodi s compactthat .. If thereB i s then the i s a closed compact of courseunit subset ball ofK theof BanachX with K ⊂ B × R . So K is almost bounded and by Lemma $Cspace6.4 comma,Z (= X X then\ K )$i B s times closed isaBanachspace R and to the discrete power . of J i s . an\quad almostNow bounded i f neighborhood $ X = Yof C open\cup parenthesisZ,Y X closing\ parenthesiscap Z period = ....\ varnothing $ Thiswhere means $Let YX $be i a completelys compact regular and k− space which i s locally connected . We consider the decomposition of X $that Zinto $there maximal discrete i s a compact connected and subset closed subsets K of{ then Xα with}; i . K e to . the power of circ subset B times R to the power of J period .. So K to the power of circ .. is almost [ \ [C(X)bounded and by Lemma\simeq 6 period 4R comma ˆX{ αJ∩ ZX}\ =β = Xtimes∅ backslashforα 6=Banach Kβ, i s closedXα = and. X\ ] discrete period Let X be a completely regular k hyphen space which i s locally connected period We consider and each Xα i s both open and closed [ 9 ] . We certainly have the decomposition of X into maximal connected subsets open brace X sub alpha closing brace semicolon i period e period \noindentX sub alphaThe cap converse X sub beta of = varnothing this observation for alpha equal-negationslash iY s also true beta . comma union of X sub alpha = X C(X) = C(Xα). and each X sub alpha i s both open and closed open square bracket 9 closing square bracket period .. We certainly have \ hspace ∗{\ f i l l }6 . 6 . Theorem . For a completely regular $ k − $ space the fo llowing are equivalent : C openIf each parenthesisXα i s pseudocompact X closing parenthesis and C(X =) product is barrelled C open , then parenthesis it i s i somorphic X sub alpha t o closing the product parenthesis of the period Banach ◦ If eachspaces X subC(X alphaα) since i s pseudocompacteach Xα is a bounded and C barrel open parenthesis in C(Xα). However X closing , parenthesis much more is is barrelled true . We comma have then again it a i s i somorphic t o the \ [product (dichotomy a of the ) . Banach C ( spaces X C open ) parenthesis\simeq XBanach sub alpha closing\times parenthesisR ˆ{ sinceJ } each. \ X] sub alpha to the power of circ is a bounded barrel in C6 open . 7 . Theorem parenthesis . XLet sub alphaX be closing a completely parenthesis regular period, locally connected k− space and assume C(X) Howeveris barrelled comma . much Then more either is trueC( periodX) has .. We a nuclear have again K o¨ athe dichotomy subspace period or \ centerline6 period 7 period{( b Theorem $ ) period X = Let XY .. be\cup a completelyZ$ regular where comma $Y$ .. locally is compact connected k $ hyphen , Z$ space anddiscrete , \quad c l o s e d and $ Yassume\cap C openZ parenthesis = \ varnothing X closing parenthesis . $ ..} is barrelled period .. Then either C open parenthesis X closing parenthesis .. has a J nuclear K dieresis-o the subspace .. or C(X) ' Banach × R . \ hspaceC open∗{\ parenthesisf i l l }P X r closing o o f parenthesis . \quad Weneed simeq Banach t o times prove R to ( the a power $ ) of J\ periodRightarrow ( $ b ) . \quad If $ B $ i s the closed unit ball of the Banach \noindent space , then $ B \times R ˆ{ J }$ i s an almost bounded neighborhood of $C ( X ) . $ \ h f i l l This means
\noindent that there i s a compact subset $K$ of $X$ with $Kˆ{\ circ }\subset B \times R ˆ{ J } . $ \quad So $ K ˆ{\ circ }$ \quad i s almost boundedandbyLemma $6 . 4 , Z = X \setminus K$ i s closed and discrete .
Let $ X $ be a completely regular $ k − $ space which i s locally connected . We consider the decomposition of $ X $ into maximal connected subsets $ \{ X {\alpha }\} ; $ i . e .
\ [X {\alpha }\cap X {\beta } = \ varnothing for \alpha \ne \beta , \bigcup X {\alpha } = X \ ]
\noindent and each $ X {\alpha }$ i s both open and closed [ 9 ] . \quad We certainly have
\ [ C ( X ) = \prod C(X {\alpha } ). \ ]
\noindent I f each $ X {\alpha }$ i s pseudocompact and $C ( X ) $ is barrelled , then it i s i somorphic t o the product of the Banach spaces $ C ( X {\alpha } ) $ since each $Xˆ{\ circ } {\alpha }$ is a bounded barrel in $ C ( X {\alpha } ) . $ However , much more is true . \quad We have again a dichotomy .
6 . 7 . Theorem . Let $X$ \quad be a completely regular , \quad locally connected $ k − $ space and assume $C ( X )$ \quad is barrelled . \quad Theneither $C ( X )$ \quad has a nuclear K $ \ddot{o} $ the subspace \quad or
\ begin { a l i g n ∗} C(X) \simeq Banach \times R ˆ{ J } . \end{ a l i g n ∗} 34 .. S period O-dieresis nal and T period Terzio caron-g lu \noindentP r o o f period34 \quad .. FirstS assume $ . each\ddot X sub{O alpha} $ i s nal pseudocompact andT . Terzio period .. Then $ \check C open{g parenthesis} $ lu X closing parenthesis is i somorphic t o the product of the Banach spaces C open parenthesis X sub alpha closing parenthesis period .. Suppose infinitely many of these P r o o f . \quad First assume each $ X {\alpha }$ i s pseudocompact . \quad Then $ C ( X ) $ are infinite hyphen¨ dimensional comma say C open parenthesis X sub alpha sub i closing parenthesis comma i = 1 comma 2 comma period periodis i34 somorphicperiod S By.O thenal remarkand T . Terziofollowinggˇ lu Theo hyphen trem o the 4 periodP rproduct o o 4 f the . Fr First of acute-e the assume Banach chet each spaceX spacesα producti s pseudocompact sub $C i = 1 ( to . the X Then power{\Calpha of(X infinity) is i} somorphic C) open . parenthesis t $ o the\quad product XSuppose sub of alpha the sub infinitely i closing parenthesis many of these areBanach infinite spaces− Cdimensional(X ). Suppose , say infinitely $C many ( of these X {\ arealpha infinite -{ dimensionali }} ) , say ,C(X i), i == 1, 12, ... , 2 , . . has a nuclear K o-dieresis theα subspace period αi . $If there By the the is remark.. remark a finite following .. following set .. Theo F such - Theo that− .. for .. alpha negationslash-element F the .. space .. C open parenthesis X sub alpha closing rem 4 . 4 the Fre ´ chet space Q∞ C(X ) has a nuclear Ko ¨ the subspace . parenthesis .. i s .. finite hyphen i=1 αi \noindentdimensionalrem4 thenIf product there . 4 theis sub a Fr alpha finite $ in\ F setacute C openF{suche parenthesis} $ that chet Xfor sub spaceα alpha6∈ $closing F\prodthe parenthesisˆ space{\ infty isC a(X Banachα}){ ii sspace finite = period - 1 ..} SoC(X C open parenthesis{\alpha { i }} dimensional then Q C(X ) is a Banach space . So C(X) ' J × Banach . X) closing $ has parenthesis a nuclearK simeqα∈F R $ to\ theddotα power{o} of$ J times the Banach subspace period . R AssumeAssume now some now X some subX alphaα i s i not s not pseudocompact pseudocompact . period We use .. theWe covering use the covering in Theo in - rem Theo 6 . hyphen 3 . That i s , we \ hspaceremfind 6∗{\ period a sequencef i l 3 l period} I f of there proper .. That closed i i s s comma\quad subsets weaW findfn i nof i a tX esequenceα \suchquad that ofs proper e t \quad closed subsets$ F $ W such sub n that of X sub\quad alphaf sucho r \ thatquad $ \alpha \not\ in F $W subW then nis\ is aquad propera properspace subset subset\ ofquad ofW Wn+1 suband$ Cn plus ( 1 and X {\alpha } ) $ \quad i s \quad f i n i t e − Line 1 infinity Line 2 X sub alpha = union of int open parenthesis∞ W sub n closing parenthesis period Line 3 n = 1 \noindentLet C subdimensional n = 2 to the power then of minus $ \prod n W sub{\ n toalpha the power\ in of circF period} C(X Evidently for{\ eachalpha K subset} X) sub $ alpha is a compact Banach we space have C . \quad So $ C ( X ) \simeq R ˆ{ J }\times $[ Banach . sub n subset K to the power of circ for all n Xα = int(Wn). after some n sub 0 period .. By Proposition 6 period 1 comman A= sub 1 n = phi open parenthesis C sub n closing parenthesis subset K to the powerAssume of circ now for some n greater $ X equal{\ nalpha sub 0 period}$ Iti follows s not that pseudocompact . \quad We use the covering in Theo − rem 6 . 3 . \−quadn ◦ That i s , we find a sequence of proper closed◦ subsets $W { n }$ o f $ X {\alpha }$ openLet parenthesisCn = 2 AW subn . Evidently n closing parenthesis for each K i s⊂ aX completingα compact sequence we have inC then ⊂ completeK for lcs all Cn openafter parenthesis some n0. XBy sub alpha closing parenthesis such that ◦ periodProposition Now by Proposition 6.1,An = 6φ period(Cn) ⊂ 1 K for n ≥ n0. It follows that (An) i s a completing sequence in the complete ◦ ◦ welcs haveC( AXα sub). Now n = by phi Proposition open parenthesis 6 . 1 we C have subA nn closing= φ(Cn parenthesis) ⊂ sp (Xα) subset and therefore sp open trivially parenthesis sp (X Xα) sub∩An alpha= An to. the power of circ closing \noindent $ W { n }$ is a proper subset of $W { n + 1 −}n$◦ and parenthesisSince andWn i therefore s not dense trivially in the sp connected open parenthesis space Xα, Xby sub Corollary alpha to 6 . the 5 , each powerCn of= circ 2 closingWn i s not parenthesis almost bounded cap A sub n = A sub n period .. Since . Hence by Proposition 4 . 1 this C(Xα) has a nuclear Ko ¨ the subspace . \ [ \Wbegin subWe n{ ia srecalll i not g n e dense thatd }\ ininftyC the(X) connected\\ i s barrelled space if X and sub only alpha if comma every closed by Corollary and σ(C 6(X period)0,C( 5X comma))− bounded each C subset sub n of= 2 to the power of minus n WX subX{\ n toi salpha the compact power} ( of= [ 1circ 0\ ]bigcup ; 1 1 . 7 .i n5 t ) . ( This W will{ ben used} in). the proof\\ of the following result which deals ni swith not = almost the 1 strong\end bounded{ duala l i g ofperiod n eC d(X}\ Hence)]. by Proposition 4 period 1 this C open parenthesis X sub alpha closing parenthesis has a nuclear K o-dieresis the6 . 8 . Theorem . Let X be a first countable completely regular space . Suppose C(X) is barrelled 0 subspace. If periodX is not locally compact , there is a completing sequence in C(X)b consisting of unbounded sets \noindentWeand recallC thatLet(X)0 ..has $ C C a open faithful{ parenthesisn } quotient= X is 2 closing omorphic ˆ{ − parenthesisn to} a nuclearW .. ˆ{\ i s K barrelledcirco¨ the} space{ if andn . } only. if $ every Evidently closed and for sigma each open $K parenthesis\subset C open X {\alpha }b$ compact we have $ C { n }\subset K ˆ{\ circ }$ f o r a l l $ n $ parenthesisP X r o closing o f . parenthesis Suppose tox the∈ powerX ofhas prime no comma compact neighborhood . Let (Wn) be a closed after some $ n { 0 } . $ \quad ByProposition−n ◦◦ $60 . 1 , A { n } = \phi (C { n } C openneighborhood parenthesis base X closing at this pointparenthesis and A closingn = 2 parenthesisW ⊂ hyphenC(X) bounded. If some subsetAn ofis Xbounded i s compact then sinceopenC parenthesis(X) open square bracket ) \subset K ˆ{\ circ }$ f o r $ n \geqn n { 0 } . $ It follows that 1 0 closingis barrelled square bracket , by the semicolon theorem quoted .. 1 1 period before 7 the period statement 5 closing of parenthesis this theorem period we deduce .. This that willW ben usedi s a in compact the $ ( A { n } ) $ i s a completing sequence in the complete lcs0 $C ( X {\alpha } ) . $ Nowby Proposition 6 . 1 proofneighborhood of the following of x. resultFrom which this contradiction deals with the we strong see that dual each ofA Cn openis unbounded parenthesis in XC( closingX)b. parenthesis period we have $ A { n } = \phi (C { n } ) \subset $ sp $ (◦ X ˆ{\ circ } {\alpha } ) $ and therefore trivially sp 6 periodLet 8B period⊂ C( TheoremX) b e a bounded period Let subset X .. and be a suppose first countable for each n, completely An is not regular con - tained space in periodB . ..This Suppose means there $ ( X ˆ{\ circ } {\alpha } ) \cap A { nn } = A { n } . $ \quad Since C openare f parenthesisn ∈ B Xand closingxn parenthesis∈ Wn with .. isfn barrelled(xn) > period2 . But .. If limX isx notn locally= x compactand so K comma= ..{x theren, x is a: completing sequence in $ W { n }$ i s not dense in the connected space◦ $ X {\alpha } , $ by Corollary 6 . 5 , each $C { n } C openn ∈ parenthesisN} i s X a compactclosing parenthesis subset of subX. b toSo theB ⊂ powerρK for ofprime some ρ consisting > 0. Therefore of unbounded| fn(xn sets) |≤ andρ for C all open parenthesis X closing = 2 ˆ{ − n } W ˆ{\ circ } { n }$ ◦ parenthesisn. From sub b this to the contradic power of - t prime ion we has find a faithful that for quotient each B ⊂ isC omorphic(X) bounded , there i s an n0 such that An ⊂ B i s not almost bounded . Hence by Proposition 4 . 1 this $ C ( X {\alphaP} ) $ has a nuclear K to afor nuclear all n ≥ Kn dieresis-o0. Now theC(X space) is barrelled period . So if un ∈ An, from the above we see that the series un is $ \ddot{o} $ the 0 P rCauchy o o f period and therefore .. Suppose convergent .. x in X ..in hasC(X ..)b no. ..Thus compact (An) .. neighborhood period .. Let .. open parenthesis W sub n closing parenthesis .. subspace . 0 be .. ai s a completing sequence with the required properties . We know C(X)b satisfies 0 closed( y neighborhood ) base( Corollary at this 6 point . 2 ) and . A sub n = Therefore2 to the power by Theorem of minus 3 n.1,C W sub(X n)b tohas the a power faithful ofquotient circ circ subset C open parenthesis WeX closing recalli somorphic parenthesis that t o\ toquad some the nuclear power$ C of K prime (o ¨ the Xperiod space )λ ..( $A If). some\quad i s barrelled if and only if every closed and $ \sigma (A C sub n ( is bounded X ) then ˆ{\ sinceprime C open} parenthesis, $ X closing parenthesis is barrelled comma by the theorem quoted before the $statement C ( of X this theorem ) ) we− deduce$ bounded that W sub subset n i s a compact of $X$ neighborhood i s compact of x period ( [ 1 0 ] ; \quad 1 1 . 7 . 5 ) . \quad This will be used in the proofFrom this of contradictionthe following we see result that each which A sub deals n is unbounded with the in C strong open parenthesis dual of X closing $C parenthesis ( X sub ) b to. $the power of period to the power of prime 6 .Let 8 B . subset Theorem C open . parenthesisLet $X$ X closing\quad parenthesisbe a first b e a bounded countable subset completely and suppose regularfor each n commaspace A . sub\quad n is notSuppose con hyphen $tained C in ( B to X the )power $ \ ofquad circ periodis barrelled .. This means . \ therequad areIf f sub $X$ n in B and is x not sub locallyn in W sub compact n with f sub , \ nquad openthere parenthesis is xa sub completing n sequence in closing$ C parenthesis ( X greater ) ˆ{\ 2 toprime the power} { ofb n} period$ consisting of unbounded sets and $C ( X ) ˆ{\prime } { b }$ hasBut a lim faithful x sub n = quotient x and so K is = open omorphic brace x sub n comma x : n in N closing brace .. i s a compact subset .. of X period .. So toB a subset nuclear rho K K to the $ \ powerddot of{o circ} $ for the some space rho greater . 0 period .. Therefore bar f sub n open parenthesis x sub n closing parenthesis bar less or equal rho for all n period .. From this contradic hyphen P rt ion o o we f find . \ thatquad forSuppose each B subset\quad C open$ parenthesis x \ in XX closing $ \ parenthesisquad has bounded\quad commano \quad therecompact i s an n sub\quad 0 suchneighborhood that A sub n subset . \quad Let \quad B$ to ( the W power{ n of} circ) $ \quad be \quad a closedfor all n neighborhood greater equal n sub base 0 period at this .. Now point C open and parenthesis $ A X{ closingn } = parenthesis 2 ˆ{ is− barrelledn } periodW ˆ{\ .. Socirc if u sub\ circ n in A} sub{ nn comma}\subset Cfrom the ( above X we ) seeˆ{\prime } . $ \quad I f some $that A the{ seriesn }$ sum is u bounded sub n is Cauchy then and since therefore $ C convergent ( X in C ) open $ parenthesis is barrelled X closing , by parenthesis the theorem sub b to quoted the power before of period the to thestatement power of prime of this.. Thus theorem open parenthesis we deduce A sub that n closing $ W parenthesis{ n }$ i s a compact neighborhood of $ x . $ Fromi s a completingthis contradiction sequence with we the see required that properties each period $ A { ....n We}$ know isunboundedin C open parenthesis X $C closing ( parenthesis X )ˆ sub{\ bprime to the power} { b of ˆ{ . }}$ prime satisfies Letopen $ parenthesis B \subset y closingC parenthesis ( X .... ) open $ beabounded parenthesis Corollary subset 6 period and 2 suppose closing parenthesis for each period $n .... Therefore , A by{ n Theorem}$ i s 3 not con − periodtained 1 comma in $B C open ˆ{\ parenthesiscirc } X. closing $ \quad parenthesisThis sub means b to the there power are of prime $ f has{ an faithful}\ quotientin B $ and $ x { n }\ in W i{ somorphicn }$ with t o some $ nuclear f { n K dieresis-o} ( xthe{ spacen } lambda) open> parenthesis2 ˆ{ n } A closing. $ parenthesis period But lim $ x { n } = x$ andso $K = \{ x { n } , x : n \ in N \} $ \quad i s a compact subset \quad o f $ X . $ \quad So $ B \subset \rho K ˆ{\ circ }$ f o r some $ \rho > 0 . $ \quad Therefore $ \mid f { n } ( x { n } ) \mid \ leq \rho $ forall $n .$ \quad From this contradic − t ion we find that for each $ B \subset C ( X )$ bounded ,there isan $n { 0 }$ such that $ A { n }\subset B ˆ{\ circ }$ f o r a l l $ n \geq n { 0 } . $ \quad Now $C ( X )$ is barrelled . \quad So i f $ u { n } \ in A { n } , $ from the above we see that the series $ \sum u { n }$ is Cauchy and therefore convergent in $C ( X ) ˆ{\prime } { b ˆ{ . }}$ \quad Thus $ ( A { n } ) $
\noindent i s a completing sequence with the required properties . \ h f i l l Weknow $C ( X )ˆ{\prime } { b }$ s a t i s f i e s
\noindent ( y ) \ h f i l l ( Corollary 6 . 2 ) . \ h f i l l ThereforebyTheorem $3 . 1 , C ( X )ˆ{\prime } { b }$ has a faithful quotient
\noindent i somorphic t o some nuclear K $ \ddot{o} $ the space $ \lambda ( A ) . $ Subspaces and quotient spaces of locally convex spaces .. 35 \ hspaceFinally∗{\ commaf i l l we} Subspaces will deal with and the quotient problem of spacesfinding Fr of acute-e locally chet subspaces convex of spaces C open\ parenthesisquad 35 X closing parenthesis sub b to the power of period to the power of prime FinallyFirst we , give we a will simple deal technical with result the which problem may be of of independentfinding Fr interest $ \acute period{e} $ chet subspaces of $C ( X )ˆ{\prime } { b ˆ{ . }}$ First we give a simple technical result whichSubspaces may and be quotient of independent spaces of locally interest convex spaces . 35 6 period 9 period .. Lemma period .. If L open parenthesis lambda open parenthesis A0 closing parenthesis comma F closing parenthesis = Finally , we will deal with the problem of finding Fre ´ chet subspaces of C(X) . First we give a simple LB open parenthesis lambda open parenthesis A closing parenthesis comma F closing parenthesisb .. for every .. nuclear K dieresis-o the .. space 6 .lambda 9technical . \ openquad result parenthesisLemma which . may A\quad closing be ofI parenthesis independent f $ L comma ( interest\lambda then . L open( parenthesis A ) E comma , F F closing ) = parenthesis LB ( = LB\lambda open parenthesis(A E comma) , F6 closing F . 9 . ) parenthesis $Lemma\quad . forf oeveryIf rL every( Frλ(A acute-e),F\quad) chet = nuclear spaceLB(λ E(A period) K,F ) $ for\ddot every{o} $ nuclear the K\quado¨ thespace space λ(A), $P\ rthenlambda o o f periodL(E,F() .. = SupposeLB A(E,F ) T) :for E ,$then$L right every arrow Fr e´ Fchet is unbounded space ( E . E where , E i F s some ) Fr acute-e= LB chet space( E period , F )$ foreveryFr $ \Weacute .. choose{e} $ .. a chet base .. space of neighborhoodsP E r o. o f . .. Suppose U sub 1T supset: E → UF subis unbounded2 supset period where periodE i s period some Fr ..e of´ chet E .. space and neighborhoods . V subWe 1 supset choose V suba base 2 supset of neighborhoods period period periodU1 ⊃ in FU such2 ⊃ that... T openof E parenthesisand neighborhoods U sub n closingV1 ⊃ parenthesisV2 ⊃ ... subset V sub n but T open\ hspace parenthesisin F∗{\suchf i l that U l } subPT r( nUn closingo) o⊂ fV parenthesis .n but\quadT (UnSuppose is) notis not absorbed absorbed $T by by V :V subn+1 E n. Let plus\krightarrow 1 period · k n and | · |n Fb e $ the is gauges unbounded of Un where $ E $ i s some Fr $ \acuteand {Vne}respectively$ chet . space Choose . now xn ∈ E Let bar times bar n and−k bar times bar sub n b e the gaugesk of U sub n and V sub n respectively period .. Choose now x sub n in E withwith bark xx subn k k n < bar2 k less| T x 2n to| k the+ 1 power for n >of k minusand set k barna Tx=k subxn k nk. barDefine k plusS 1 for: λ( nA) greater→ E by k and set n a to the power of k = bar x sub n \noindent We \quad choose \quad a base \quad of neighborhoods \quad $ U { 1 }\supset U { 2 }\supset bar k period .. Define S : lambda open parenthesis A closing parenthesis∞ right arrow E .by . . $ \quad o f $ E $ \quad and neighborhoods $ V { 1 }\supset V { 2 }\supset .X . .$ in $F$ suchthat $T ( U { n } ) \subset Line 1 infinity Line 2 S open parenthesis open parenthesisS((ξn)) = xi subξnxn n. closing parenthesis closing parenthesis = sum xi sub n x sub n period V { n }$ but $ T ( U { n } ) $ is not absorbed by $V { n + 1 } . $ Line 3 n = 1 n = 1 LetS is continuous $ \ parallel and T circ\cdot S : lambda\ parallel open parenthesisn $ A andclosing $ parenthesis\mid \ rightcdot arrow\mid F is unbounded{ n }$ period b e the gauges of $U { n }$ and6 periodS $is V continuous 10{ periodn }$ and .. respectively TheoremT ◦ S : λ period(A) → F ...is\ Ifquad unbounded C openChoose parenthesis . now X $ closing x { parenthesisn }\ in .. isE .. $ barrelled comma .. then L open parenthesis 0 0 E comma6 C . open 10 . parenthesisTheorem X . closingIf C parenthesis(X) is sub barrelled b to the, powerthen ofL prime(E,C( closingX)b) parenthesis = LB(E,C( =X LB)b) openfor every parenthesis E comma C open parenthesis\noindentFr e´ Xchetwith closing space parenthesis$ E\ .parallel sub b tox the{ powern }\ of primeparallel closing parenthesisk < 2 ˆ{ − k }\mid Tx { n }\mid k + 1 $for every f o rP r Fr $ o acute-eo n f .> By chet thek space $ lemma and E period it s is e t sufficient $ n t{ o provea }ˆ{ thek result} = for E\ parallel= λ(A), ax nuclear{ n K}\o ¨ theparallel space . k . $ \quad Define $ S : \lambda (A)0 \rightarrow E $ P rLet o oT f period: λ(A ..) By→ theC lemma(X)b be it continuous is sufficient . t o If proveB the∈ B result(λ(A)) for, since E =T lambda(B) is openbounded parenthesis and C(X A) closing barrelled parenthesis comma .. a bynuclearwe have K o-dieresisT (B) ⊂ ρK the◦◦ spacefor some periodK ⊂ ..X Letcompact T : lambda . So open the restriction parenthesis of AT closing0 to C(X parenthesis) i s a continuous right arrow operator C open parenthesis X closing 0 0 parenthesisinto λ sub(A) bb. toλ( theA)b powersatisfies of prime ( y ) beand continuous thus it has period a continuous .. If B in norm B open.C parenthesis(X) satisfies lambda the openopenness parenthesis condition A closing parenthesis closing \ [ \ begin { a l i g n e d }\0 infty \\ 0 ◦ 0 parenthesisand therefore comma T i s bounded . So for some K ⊂ X compact T (K ) i s bounded in λ(A)b, which means S((sinceT 0 T(K open◦) parenthesis⊂ \ xiρV ◦ { Bforn closing} some)V parenthesis )∈ U =(λ(A is)) bounded\.sumThus and\Txi0( C( openX{ ))n⊂ parenthesis}λ(Ax)0[V{◦] Xn and closing} so.T \\ parenthesisis bounded barrelled . we have T open parenthesis n = 1 \end{ a l i g n e d }\ ] 0 B closing parenthesisR e m a r k subset . This rho means K to thatthe powerC(X)b ofhas circ no circ proper for some Fre ´ Kchet subset subspace X if C(X) i s barrelled . However 0 compact, from period Theorem .. So 6the . 8 restriction we know thereof T to are the spaces power ofC( primeX) for to C which openC parenthesis(X)b has X closing a completing parenthesis sequence i s a continuous operator into lambdaconsisting open parenthesis of unbounded A closing sets parenthesis . From this sub bit to follows the power that the of period individual to the countable power of boundedness prime property i s \noindentlambdaessential open$ in parenthesis S $ is Acontinuous closing parenthesis and sub$T b to\ thecirc powerS: of prime satisfies\lambda open parenthesis(A) y closing\rightarrow parenthesis andF thus $ it is has unbounded a . continuousTheorem norm 4 period . 5 . C open parenthesis X closing parenthesis satisfies the openness 6 .condition 10 . \ andquad thereforeTheorem T to . the\quad powerI f of prime $ CReferences i s ( bounded X period ) $ \ ..quad So fori s some\quad K subsetb a r r X e l lcompact e d , \ Tquad to thethen power $ of L prime ( open E parenthesis, C[1] ( K to X the S.BellenotandE.Dubinsky, power ) ˆ{\ ofprime circ closing} { parenthesisb } )=LB(E i s Fr e´ chet spaces with ,C(X)ˆ nuclear K o¨ the quotients{\,prime Trans . } { b } ) $ forboundedAmer every . in Math Fr lambda . $ Soc\ openacute . 273 parenthesis ({ 1e 9} 82$ ) , 579chet A closing – 594 space . parenthesis E . sub b to the power of comma to the power of prime .. which means T to the power of prime[ 2 ] open C . B parenthesis e s s ag a and K to A .the P e powerlczyn´ s ofk circ i , On closing bases andparenthesis unconditional subset convergence rho V to the of series power in of circBanach .. for spaces some V in U open parenthesis Plambda r o, Studia openo f parenthesis . Math\quad . 1 7By ( A 1 closing 9 the 58 ) lemma , parenthesis 1 5 1 – it 1 64 is closing . sufficient parenthesis t period o prove .. Thus the result for $ E = \lambda (A) , $T to\quad the[3] powera J.Bonet, of prime openOn the parenthesis identity CL( openE,F ) parenthesis = LB(E,F ) Xfor closing pairs ofparenthesis locally convex closing spaces parenthesisE and F, subsetProc lambda . open parenthesis A closingnuclearAmer parenthesis K . Math $ \ to .ddot Soc the . power{ 99o} ( 1$ of987 prime the ) , 249 space open – 255 square . . \quad bracketLet V to $ the T power : of\ circlambda closing square(A) bracket and\rightarrow so T is bounded periodC(X ) ˆR{\ e mprime a r k period} { b ..} This$ be means continuous that C open . parenthesis\quad I f X closing $ B parenthesis\ in B( sub b to\ thelambda power of prime( A has no ) proper ) Fr , acute-e $ chet subspacesince if $T C open (parenthesis B )$ X closing isboundedand parenthesis $C ( X )$ barrelledwehave $T ( B ) \subset \rhoi s barrelledK ˆ{\ periodcirc .. However\ circ comma}$ f o .. r from some Theorem $ K 6 period\subset 8 we knowX $ there are spaces .. C open parenthesis X closing parenthesis .. forcompact . \quad So the restriction of $ T ˆ{\prime }$ to $C ( X ) $ i s a continuous operator into $ \whichlambda C open( parenthesis A ) X ˆ{\ closingprime parenthesis} { b subˆ{ b. to}} the$ power of prime .. has .. a completing sequence consisting of unbounded set s period$ \lambda .. From ( A ) ˆ{\prime } { b }$ satisfies ( y ) and thus it has a continuous norm $ . C (this X it follows ) $ that satisfies the individual the opennesscountable boundedness property i s essential in conditionTheorem 4 periodand therefore 5 period $ T ˆ{\prime }$ i s bounded . \quad So for some $K \subset X $ compact $ TReferences ˆ{\prime } ( K ˆ{\ circ } ) $ i s boundedopen square in bracket $ \lambda 1 closing square( A bracket ) ˆ{\ .. Sprime period B} e{ lb l e ˆ n{ o, t and}}$ E\ periodquad Dwhich u b i n means s k y comma $ T ˆ Fr{\ acute-eprime chet} spaces( K with ˆ{\ circ } nuclear) \subset K dieresis-o\ therho quotientsV ˆ{\ commacirc Trans}$ period\quad f o r some $ V \ in U( \lambda ( A ) ) . $ \quad Thus $Amer T ˆ{\ periodprime Math} period(C(X)) Soc period 273 open parenthesis\subset 1 9 82 closing\lambda parenthesis( comma A 579 ) ˆ endash{\prime 594 period} [ V ˆ{\ circ } ] $ andopen so square $T$ bracket is 2 bounded closing square . bracket .. C period B e s s ag a and A period P e suppress-l sub c z y to the power of acute-n s k i comma On bases and unconditional convergence of series in R eBanach m a r spaces k . comma\quad StudiaThismeansthat Math period 1 7 open$C parenthesis ( X 1 9)ˆ 58{\ closingprime parenthesis} { b comma}$ has 1 5 no 1 endash proper 1 64 Fr period $ \acute{e} $ chetopen subspace square bracket if 3 $C closing (square X bracket )$ .. J period B o n e t comma On the identity L open parenthesis E comma F closing parenthesis =i LB s open barrelled parenthesis . \ Equad commaHowever F closing , parenthesis\quad from for pairs Theorem of locally 6 .convex 8 we spaces know E there and F comma are spaces \quad $ C ( X ) $ \quadProcf period o r Amer period Math period Soc period 99 open parenthesis 1 987 closing parenthesis comma 249 endash 255 period which $C ( X )ˆ{\prime } { b }$ \quad has \quad a completing sequence consisting of unbounded set s . \quad From this it follows that the individual countable boundedness property i s essential in
\noindent Theorem 4 . 5 .
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Math 1 8 .closing 345 square ( 1 983 bracket ) , .. 1 D 82period−− Vo2 g 0 t .comma Fr acute-e chetr dieresis-a ume comma zwischen denen jede stetige lineare$ [ Abbildung 1 9 beschr ] dieresis-a−−−{ , nkt}$ ist Kernels comma J of period Eidelheit matrices and related topics , Do $ \breve{g} $ aMat.10(1986) ,232 −− 256 . $Reine [ Angew 20 period ] −−−{ Math, period}$ \ 345quad openTopics parenthesis on projective1 983 closing parenthesis spectra comma of $ 1 ( 82 endash LB 2 ) 0 period− $ spaces , in : Advances in the Theory of Fr $ \openacute square{e} bracket$ chet 1 9 closing square bracket emdash comma Kernels of Eidelheit matrices and related topics comma Do g-breve a Mat periodSpaces 1 0 open , Kluwer parenthesis , 1 1 989 9 86 , closing 1 1 −− parenthesis1 8 . comma 2 32 endash 256 period open square bracket 20 closing square bracket emdash comma .. Topics on projective spectra of open parenthesis LB closing parenthesis hyphen\noindent spaces comma$ [ in 2 : Advances 1 ] in−−−{ the Theory, }$ of Remarks Fr acute-e on chet a paper of S $ . \ddot{O} $ nal andT . Terzio $ \breve{g} $ luSpaces , Do comma $ \breve Kluwer{g} comma$ aMat. 1 989 comma 15(1991) 1 1 endash 1 8 period ,202 −− 206 . open square bracket 2 1 closing square bracket emdash comma Remarks on a paper of S period O-dieresis nal and T period Terzio g-breve lu comma Do g-breve a Mat period 1 5 open parenthesis 1 99 1 closing parenthesis comma 202 endash 206 period