Uniquenessof Representationby Trigonometric Series
J.MARSHALL ASH', DePaul University,Chicago, IL 60614
J. MARSHALL ASH receivedhis S.B., S.M., and Ph.D. degreesat theUniver- I"k sityof Chicago where he was a studentof Antoni Zygmund. He wasa Joseph Fels Rittinstructor at Columbia University from 1966 to 1969.Since then he has beenon theDePaul Universityfaculty, where he becamea fullprofessor in 1974and chairedfrom 1985 to 1987.In 1977he wasvisiting professor at StanfordUniversity. His researchinterests have centeredon multiple trigonometricseries, singular integrals, and real analysis,with forays into functionaland numericalanalysis.
Abstract.In 1870Georg Cantor proved that a 2srperiodic complex valued function of a realvariable coincideswith the values of at mostone trigonometric series. We presenthis proof and thensurvey some of themany one dimensionalgeneralizations and extensionsof Cantor'stheorem. We also surveythe situationin higherdimensions, where a greatdeal lessis known.
1. Cantor'suniqueness theorem. In 1870Cantor proved THEOREMC (Cantor [5]). If, forevery real numberx N lim E ce inx = 0, N-?oo n=-N
thenall thecomplex numbers cn, n = 0, 1, - 1, 2, - 2,... are zero. Thisis calleda uniquenesstheorem because it has as an immediatecorollary the factthat a 2iTperiodic complex valued function of a realvariable coincides with the values of at most one trigonometricseries. (Proof: Suppose Yaneinx= Ybeeinxfor all x. Form the differenceseries E(an - bn)einxand apply Cantor's theorem.) This theoremis remarkableon twocounts. Cantor's formulation of theproblem in sucha clear,decisive manner was a majormathematical event, given the point of viewprevailing among his contemporaries.2 Equally enjoyable to beholdis therapid resolutionthat we willnow sketch. Cantor'stheorem is relativelyeasy to prove,if, as Cantordid, you have studied Riemann'sbrilliant idea of associating to a generaltrigonometric series T := ECneinx, the formalsecond integral, namely, F(x) = En,O(cn/(in)2)einX+ cO(x2/2). For some interestingremarks on the importanceof thisidea, see thevery enjoyable surveyarticle of Zygmund[27]. Definethe secondSchwarz derivative D of a
'The researchpresented here was supported in partby a grantfrom the University Research Council of DePaul University. The authoris gratefulto one refereefor proposing an expandedtreatment of multipletrigonometric seriesand to theother referee for making careful corrections and adding some historical remarks. 2In theeighteenth century, physicists just "did" Fourierseries (often quite successfully) without worryingabout convergence very much at all. Whendoubts about convergence began to arisein the nineteenthcentury, the first attempts at rigorwere rather heavy handed. See Dauben[9, pp. 6-31] foran interestingaiscussion of thehistorical context.
873 874 J. MARSHALL ASH [December functionG(x) by
- DG(x) = rn G(x + h) 2G(x) + G(x - h) h-0 h The stepsof theproof are: 1. Since T convergeseverywhere, it is immediatethat for every value of x, cneifX ? c_nei-nx 0 as n -* oo. By theCantor-Lebesgue theorem, ICnl + Ic_nl O as n -x oo. Appendix1 givesCantor's weak but easy versionof this.(For a statementof themore powerful Cantor-Lebesgue theorem see thesurvey article by RogerCooke [8]. The proof given there is muchshorter than the one in Appendix1, but requiressome of themachinery of modem analysis.) 2. By theWeierstrass M-test from
|c__inx icnx) SUp(|C | + I_e 2 + )2 nI (in in)2 n2 it followsthat F(x) - co(x2/2)is a continuousfunction and thatF is theuniform limitof itspartial sums. (See Theorems25.7 and 24.3in [14]for the M-test.) 3. An importantresult of Schwarz's states that if G is continuousand DG(x) = 0 forall x, thenG is a linearfunction. (See [5,pp. 82-83].)Before presenting a proof of this,Cantor remarks that Schwarz mailed this result to himfrom Zurich.) We givea proofin Appendix2. 4. Since
- ei(x+h) - 2eix + ei(x-h) sin h2 =-eix hI 2 (Checkthis.), we have nh 2 F(x + h)-2F(x) + F(x - h) sinm
n h2 n*0c0 nh 2 From a0 ? S2?=1an= 0 it followsthat limk OaO + n=1an(sinnk/nk)2 = 0. See Appendix3 forRiemann's summation by parts proof of this. Hence DF(x) = 0 so F(x) = ax + /3for some a and /3. 5. The rightside of theequation
co-2 + ax + n3 )2
is bounded.(From 2. aboveit is continuous,hence bounded on [0,2 ], and hence bounded everywhereby periodicity.)Letting x -x oo twice firstshows co = 0 and thena = 0. 6. Fromthe observation made in step2 above,we see thatthe sequence
SN(X) :-A + E ei O< InI AN (in) 1989] UNIQUENESS OF REPRESENTATION BY TRIGONOMETRIC SERIES 875
convergesuniformly to 0. Butfor each n # 0,
(in)2 2Nx)d = -| SN X) e ixdx Cn 2sr ( forall N > n bythe orthonormality of { einx/ xv }, so lettingN - oo givescn = 0 forall n * 0. (The uniformityof convergence allows the interchange of limitand integral.) Q.E.D. Cantor'sbeautiful theorem suggests a varietyof extensionsand generalizations. The remainingfour sections of thispaper will consider some of these. 2. Summabilityand uniqueness. Can we improveTheorem C by weakeningthe assumptionof convergenceto zeroto an assumptionof beingmerely summable to zero?As is oftenthe case in mathematics,the starting point is a counterexample whichdestroys the "obvious extension." The trigonometricseries Ec einx is said to be Abel summableto s if foreach r, O< r < 1, f(x, r) 2=Ec einxrInI convergesand iflimr f-f(x,r)= s. Let z reix = r(cos x + i sin x). Differentiatethe identity X0 00X A 1 1-z 1-rcosx (cos 9te / , z ) S E=O nx)r\= 1 -z 1 - z ) 1 -2rcos x + r withrespect to x to obtain
? (1-r2)r sin x E (n sin nx)r n 2_2 n=1 (1-2rcosx+r 2)2
If x * 0, as r -> 1-, theright side tends to 0; whileif x = 0, everyterm of f (O, r) is 0, so that f(O, r) = 0, whencelimr - f(O, r) = 0. Thus E?2=Insin nx is everywhere Abel summableto 0. Althoughthis example is unpleasant,it turnsout to be just aboutthe worst thing that can happen. THEOREM V (VERBLUNSKY [25,VOL. I, PP. 352, 383],[22], [23]). If cn/InI-> 0 as InI - oo and Ec einx is Abel summableto 0 at everyx, thenall c, are 0. 3. Higherdimensions. When we move fromone to severaldimensions, the picturebecomes much more cloudy. Here is the land of opportunity.Almost nothingis known;almost every question that the novice might ask turnsout to be an open question.The firstgoal is to mimicCantor's Theorem C. Even this apparentlymodest goal remainsto a largeextent unachieved. The hypothesisthat EC einx convergesto 0 has severaldifferent interpretations in higher dimensions. Mostof thesecan be illustratedin twodimensions, so to ease notationI willrestrict myselfto that case. The basic object will be the double trigonometricseries T be T(x, y) = E(m,n)EZxZcmnei(mx?nY) We definea rectangularpartial sum of to
Tmn(XI y) = E c,,ei(Pix+vY)I JL,tt Tn (XI y) := E cE,,^ei(x+y) I,ul+ Ivl and a circularpartial sum to be T(X, Y) = E cUe(F+) L2 + v2 Then formally F(X) = O I{2ei = S and formallyF( X) is a secondintegral of T( X). FromCooke's two-dimensionalCantor-Lebesgue theorem it followsthat equa- tion(1) holds.This guarantees the existence of F(X, t). That F(X) existsfor all X followsfrom the convergence and consequentcircular Abel summabilityof T [18, 1989] UNIQUENESS OF REPRESENTATION BY TRIGONOMETRIC SERIES 877 pp. 67-68]. We have ICM12 0ICI? 0 k CI M#O MI IMI=1 k=2 (k - 1k-< MI Fh(X) = ghh2/IF(X+H)dH. IH 6h This agreeswith the usual Laplacian (= d 2F/ax2 + d 2F/dy2) forC2 functionsas can be seen by expandingF into its Taylor series.We also have the representation F(X, t) = +-f2[3? FU) dU [l8, p.56]. 27rI2 [t2+ (X- U)2]/ Changingto polar coordinatesand integratingby partsgives 3 = FF(X, t ) 1otf r3F(X)+ dr [18, pp. 66-67]. (2) it (t2 +r212d Now - [ F(x, t)] - Le>2cMe MIt is the circularAbel meansof the originalseries, so thatfrom the hypothesis (1) it is immediate that limt O(d2/dt2)[F(x, t)] = 0. Shapiro then adapts a clever one- dimensionallemma of Rajchman[25, vol 1, pp. 353-354] to concludefrom this that AF(X) = 0 at each X. This implicationdepends heavilyon equation (2) [18, pp. 66-67]. By an argumentlike the one in appendix 2, zero generalizedLaplacian forces harmonicity.(The analog of the subtractedlinear functionof appendix 2 is the Poisson integralof F.) See Rado fordetails [12, p. 14]. It followsthat if F(X) can be shown continuouseverywhere, it will be harmoniceverywhere. The continuous functionF will be necessaiilybounded on the compact set [0,2s] x [0,g2] and 878 J.MARSHALL ASH [December consequentlyby periodicitybounded on theentire plane. But boundedharmonic functionsare constant. (Apply Liouville's Theorem to thebounded analytic function exp(F + iF), whereF is a harmonicconjugate of F.) So therewill be a constantd so that d+ M eiM-X = o forall X. By theuniqueness theorem for square integrable functions (an immediate consequenceof Parseval'sformula [25, vol. II, p. 301])it willfollow that all CM= 0. It remainsonly to establishthe continuity of F. This is doneby generalizinga Bairecategory argument employed by Verblunsky. Since T(x, t) the circularAbel mean of the originalseries is continuouson {(X, t): t > 0} and has finitelimit (namely 0) at eachpoint, by a basictheorem of Bairefor each discin T2 thereis a subdiscand a constantK so thatI T(X, t) I < K forevery X of thesubdisc and everypositive t [25,Vol. I, p. 29 (12.3i)]and [18,pp. 69-70]. Integrationin the t variablethen shows that in thatsubdisc F(X) is continuous,being the uniformlimit of F(X, t). Let Z be theset where F is notcontinuous. Let Z be theclosure of Z. Assume Z is non-empty.An inspiredidea thatappears already in theproof of TheoremV now producesthe desiredcontradiction, as follows.Applying the same Baire categoryargument that showed F(X, t) well-behavedon an opendense set to the set Z producesa pointX0 E Z withF(X, t) converginguniformly to F(X) with respectto Z throughouta neighborhood of X0.If X is anypoint very close to X0, let X1 be a pointof Z closestto X, say IX1- Xj = s. AssumeX ? X1,X1 = X0.(If eitherequality holds, the argumentis even easier.)Then (i) F(X) = Fs(X), (ii) Fs(X) is closeto Fs(X1),(iii) Fs(Xl) is closeto F(X1,s), (iv) F(X1,s) is closeto F(X1), and (v) F(X1) is closeto F(Xo); whenceF is continuousat X0,contrary to thedefinition of Z. (Reasons:(i) Continuityforces harmonicity as mentioned above, and harmonicityis equivalent to themean value property ([12], p. 7). (ii),(iii) These need technicallemmas whose proofs are straightforwardprovided one is awareof thedelicate estimate fl eiXe dX i3/2 forlarge t. This is equivalentto thefact that the Bessel functionJ1(s) = 0(s-1/2) as s -* + x0 [18,pp. 68-69]and [20,p. 199].(iv) Uniformityofconvergence gives this. (v) Uniformlimits are continuous.) Thiscompletes our discussion of theproof of Theorem SC. The othermajor result in twodimensions is this. THEOREMAW (J. M. Ash and G. Welland) [1], [2]. If T(x, y) is unrestrictedly rectangularlyconvergent to 0 everywhere,then all c ..n are 0. Fromthe hypothesis itis immediatethat at each X thepartial sums tend to 0 "in thenortheast," i.e., that = 0 limrln m n} 00 Tmn( X) One-dimensionalconvergent sequences of numbersare necessarilybounded, but two dimensionalones need not be. Neverthelessit can be provedthat at each X, 1989] UNIQUENESS OF REPRESENTATION BY TRIGONOMETRIC SERIES 879 sUpm, I Tmn( X) I < 00. In factthis is thehardest part of theproof of Theorem AW. It requiresa techniquethat first appeared in theunpublished thesis of P. J.Cohen [2, pp. 402,404-407]. Write Cmn(X) = Cm,nei(mx+nY) + C ,nei(-mx+ny) + Cm,-nei(mx ny) + C ei(-mx-ny). The "Mondrian"identity Cmn(X) = Tmn(X) - Tm-l,n(X) - Tm,n-l(X) + Tm-1,n-1(X) impliesthat at each X theCmn(X) also tendto 0 in thenortheast and arebound- ed [2, pp. 410-411].A Cantor-Lebesguetheorem finally gets back to the coef- ficientthemselves. The result is that { cmn} is a bounded sequence and limmin(I m,I~ n c- = 0 [2, p. 408 and p. 411]. These two facts allow one to readilydeduce that hypothesis (1) holds[2, p. 423].It is also truethat a unrestrictly rectangularlyconvergent double sequence of numbersis circularlyAbel summable to the same value, providedthe partial sums are bounded [2, pp. 413-416]. In particular,since T(X) convergesunrestrictly rectangularly to 0 everywhere,it is circularlyAbel summableto 0 everywhere.A careful examination of theproof of TheoremSC showsthat condition (1) togetherwith everywhere circular Abel summabilityto 0 are sufficienthypotheses for the proof to work. Remarks.When Grant Welland and I discoveredthis proof, I feltthat there was an elementof goodluck involved here. First, the rectangular convergence gave just enoughcontrol of the coefficient size to forcecondition (1) andcondition (1) is,in a sense,sharp. (The seriesTo(x, y) := E2nsin nx "almost"satisfies (1) and is circu- larlyAbel summableto 0 everywhere.The computationsare in section2 above.) More impertantly,there are usuallyno nontrivialconnections between various modesof convergenceand in particularunrestricted rectangular convergence of a trigonometricseries on a setdoes not force circular convergence of the series, even if one is willingto discarda subsetof measure0 [2, pp. 417-420].Thus theabove mentionedresult connecting unrestricted rectangular convergence and circularAbel summabilitycame as a happysurprise. The otherside of thecoin is thatthis proof of TheoremAW was "too lucky." Oftenrectangular results for two-dimensional series can be extendedto higher dimensionswithout much additional effort. The unexpecteddependence of Theo- remAW on TheoremSC meansthat a goodthree-dimensional uniqueness theorem for unrestrictedrectangular convergence awaits either a good three-dimensional sphericaluniqueness theorem or a completelynew method of proof.Shapiro has proveda three-dimensionalspherical uniqueness theorem, but it needsthe hypothe- sis S1S~~~~~ ra Icimni 0 as r -X 00 (Na) (r-1)2 < 12 + m2 + n2< r2 witha = 1. Buta three-dimensionalCantor-Lebesgue theorem can onlybe expected to produce condition(Na) witha = 2. (See [2, p. 425] fora relevantcounterexam- ple.) Thuseven if one assumes that T(x, y, z) is everywherecircularly convergent to 0, it is notknown whether all Clmn mustthen be 0. Thereis evengreater ignorance 880 J.MARSHALL ASH [December concerninguniqueness questions about the other 3 modesof convergence mentioned above.For example,here are 3 openquestions in twodimensions. Question1. If T(x, y) is everywhererestrictedly rectangularly convergent to 0, does thisforce all Cm,, to be 0? Question2. If T(x, y) is everywheresquare convergent to 0, does thisforce all Cmn to be 0? Question3. If T(x, y) is everywheretriangularly convergent to 0, doesthis force all Cmn to be 0? Again,it mustbe emphasizedthat easy counterexamplesshow that not much help willbe availablefrom Cantor-Lebesgue type theorems for square, restricted rectangular,or triangularconvergence [2, pp. 416-418]. 4. Fourierseries. Return to onedimension. A thirdtype of extension of Cantor's TheoremC occurswhen the limitof the trigonometricseries S c= cneinxis a Lebesgueintegrable function f. The questionnow is whetherS is necessarilythe Fourierseries of f, i.e.,whether for each integer n theremust hold the relation = cn 2n - j2TT f()-inx W dx. A typicalresult in thisdirection is thefollowing. THEOREM (de la Vallee-Poussin[25, Vol. I, pp. 326, 382],[21]). If S convergesto f at each x, and iff is finiteat each x and if fo` jf(x) I dx < oo, thenS is theFourier seriesof f. 5. Sets of uniqueness.If we replace the hypothesisEccneinx = 0 everywhereby Ecneinx = 0 forx not in a "thin"subset E of [0,2XT], we may stillbe able to concludethat all c,n= 0 if E is thinenough. Sufficiently thin sets E arecalled sets of uniqueness.A restatementof TheoremC is thatthe emptyset is a set of uniqueness.It was alreadyproved by W. H. Young[24] that every countable subset of [0,2 T] is a set of uniqueness.(Cantor's earlier work showing that closed countablesets weresets of uniquenessled Cantorto thecreation of set theory!) However,a set E ofpositive measure is too thickto be a setof uniqueness.(Proof from[25, Vol. I, p. 344]:Let E1 be a subsetof E whichis perfectand of positive measure,and let f(x) be thecharacteristic function of E1. The Fourierseries of f convergesto 0 outsideE1, and so also outsideE, but does not vanishidentically sinceits constant term is I ElI/2q 7> 0.) One ofthe most interesting theorems in all analysisproduces two classes of uncountable measure zero sets which are very much likeeach othermetrically, although the first are setsof uniquenessand thesecond are not. Let 0 < t < 1/2. Dissect[0, 2T] into2 closed"white" intervals, each of length 27T~,[0, 27rf] and [27T(1- t), 27T],and one open"black" interval (27Tt, 27r(1 - ())- Removethe black interval and repeatthe process by dissectingeach white interval into2 closedwhite intervals of length2 T42,and 1 centeredopen black interval of length2 iTr - 2 - 27T 2. Iteratingthis process k timesproduces 2k closed white intervals,each of length 27Tk. (See FIGURE 1.) Now let k -- oo rememberingto remove the black intervals at each stage.The resultingset E(() is said to be of Cantortype with constant ratio of dissection. 1989] UNIQUENESS OF REPRESENTATION BY TRIGONOMETRIC SERIES 881 stage 1 ------t e stage 2 stage 3 O "white" intervalsfor E(1/3) * "white" intervalsfor E(1/3) FIG. 1. Since 2k 2Xgk 0 as k oo, E(,) alwayshas measurezero. In particularE(1/3) is theclassical Cantor set. THEOREM SZ (R. Salemand A. Zygmund[16], [17], [25, Vol. II, p. 152],[15], [11]). If 1/t is an algebraicinteger all of whoseconjugates have modulusless than1 (i.e., (1/)n + a1(1/)n-1 + - - - ?an = 0 forsome integersa,, a2,..., a", n is minimal, and the othern - 1 rootsof X + alXn-1 + **. +an = 0 have absolutevalue less than 1), thenE(t) is a set of uniqueness.Otherwise E(t) is not a set of uniqueness. Now 1/3 = .33 ... > .3 = 3/10 so the process that formsE(1/3) leaves more materialat each stagethan does theprocess that forms E(3/10). HenceE(1/3) is intuitivelythicker than E(3/10). Nevertheless1/(1/3) satisfiesx - 3 = 0 and hence is an algebraicinteger without conjugates so that E(1/3) is a set of uniqueness;while 1/(3/10) is notan algebraicinteger at all3so thatE(3/10) is not a setof uniqueness. Thus the "wrong" one ofthe pair E(1/3), E(3/10) is theset of uniqueness. Even moredramatically counterintuitive is the factthat the "verythick" set UE(t), wheret variesover all reciprocalsof algebraicintegers with small conju- gates(t = 1/2 mustbe excluded),is also a setof uniqueness.This is truebecause N. Baryhas provedthat a countableunion of closedsets of uniqueness is stilla set of uniqueness.See ([25],Vol. I, p. 349) forthis fact as wellas furtherremarks on TheoremSZ. The otherside of thecoin from a setof uniqueness is a setof multiplicity.A set of multiplicityby definitionis a subsetof [0,2Xw] which is nota setof uniqueness. More directly,E is a set of multiplicityif there is a trigonometricseries which convergesto 0 outsideE but whichdoes not vanishidentically. Just as sets of 3Assume 10/3 satisfiesx" + *-. +an = 0 with all a, integer.Clearly n > 2, so (10`"1 + *-- +an_1311-1)10 + a,,3" = 0. Hence 3 divides lO"n- + 3(al + *-- +an13n-2), so 3 divides 10"-1, a contradiction. 882 J.MARSHALL ASH [December uniqueness become more interestingas theyget "thicker",so sets of multiplicity become more interestingas theyget "thinner."As was shown above, everyset of positive measureis a set of multiplicity.The firstmeasure zero set of multiplicity was produced by Men'shov[10] and TheoremSZ above giveslots of examples. In a later reviewI will surveysome of the extensiverecent work thathas been done on sets of uniqueness.At presentthis is the most activeand excitingarea of the four extensionsI have discussed.The interestedreader is encouragedto turn now to the books of Zygmund[25] or Bary [3] for much more comprehensive overviewsof all the one-dimensionaltopics I have highlightedhere. Note added in proof.A remarkablefacet of Cantor's proof of TheoremC has been its uniqueness.The manydifficulties encountered in attemptsto generalizeit suggestthat a differentproof of TheoremC could proveuseful. Because of a recent developmentin real analysisI can now givesuch a proof.Suppose the seriesEcnelnx convergesto zero everywhere.Form thefirst integral c0x + 2(cn/in)einx.Although this L2 functionis not easilyseen to be continuous(The examplesXsin n0/ln n = 0 but - Ecos nO/nln n divergentgive the flavorof the difficultyin workingwith a firstintegral.), it does followdirectly from a theoremof Rajchman and Zygmund [25, Vol. 1, p. 324] thatit has symmetricapproximate derivative 0 at everypoint. By the aforementionedrecent results {"A symmetricdensity property; monotonicity and the approximatesymmetric derivative," Proc. Amer. Math. Soc., 104 (1988) 1078-1102, and "A symmetricdensity property for measurable sets," Real Analysis Exchange, 14 (1988-89) 203-209, both by C. Freilingand D. Rinne}, thereis a constant c so that + e inx -C = 0 Cox in almost everywhere.The conclusionof TheoremC now followsfrom periodicity and Plancherel's Theorem.I will publish the details of this proof in the Proc. Amer. Math. Soc. Appendix1. THEOREM Al (Cantor [4]). If cneifx ? c__ne-i 0 as n -4 oo for everyx, then IcCn+ Ic n- 0. Proof. From the convergenceto zero of cneinx + C-ne-inX followsthe conver- gence to zero of its real and imaginaryparts. One may easily find real numbers a, bn, a, b"t,so that cnei ? c__eine = (a cos nx + bnsin nx) + (a' cos nx + b,nsin nx)i and directcalculation shows a2 + bn2+ a72 + bn,2= 2(I cn l12+Ic-n 12). Hence it sufficesto prove that an cos nx + bnsin nx -O 0 for every x implies an2+ b2 _0. Define Pn a= a2 + bn2and find Onso pncos On = an pnsin69bn = Then an cos nx + bnsin nx = pncos(nx - Osn),pn cos(nx - On) 0 for every x, and we need only show that pn 0. If pndoes not tendto 0, thereis a subsequence{ nk } and a number8 so thatfor everypositive integer k, pnk > 8 > 0. 1989] UNIQUENESS OF REPRESENTATION BY TRIGONOMETRIC SERIES 883 By discarding as many terms as necessaryfrom {nk} we may also assume (nk+l/nk) > 3 forall k. Then cos(nlx - 0n) > - forx E1= - +, 2 3j/f 3n Since 2,r n2 I111= and -> 3, 3n1 n, as x ranges over I1, (n2x - 0n2) rangesover n2Il - 01n2and 2 7T - = > 2,g. 1n2I1 1n21 = n2111 n2" 3n, Thus there is a closed intervalI2 c I1 with cos(n 2x - 02) > 1/2 for all x E I2 and II21 = 27T/3n2. Proceedinginductively produces a point t E ln 1Ik with cos(nk -n k) > 1/2 forevery k. Thus pnkcos(nkt -Onk) > 3/2 forevery k. This contradictsthe convergenceto zero of pncos(nx - On)at x = {. Appendix2. THEOREM A2 (Schwarz,Cantor [5], [9, p. 33]. A continuousfunction G witheverywhere 0 Schwarzderivative is necessarilya linearfunction. Proof. First suppose thatthere is a continuousnonconvex function H(x) satisfy- ing DH(x) > 0. Then thereare points a < b < d with H(b) > L(b), whereL is the linear functionwhose graph passes through(a, H(a)) and (d, H(d)). Let H1(x):= H(x) - L(x). Then H1(a) = H1(d) = 0, H1(b) > 0, H1 is continuous and DH1(x) = DH(x) > 0 for all x. Let c be a point of [a, d] where H1 is maximum. Note c E (a, d). (See Figure 2.) Then for all h < min{c - a, d - c},(1/2)[H1(c + h) + H1(c - h)] - H1(c) < 0, contraryto DH1(c) > 0. a b c d FIG. 2. Now fromD(x2) = 2 (an easy calculation)and DG = 0, foreach ? > 0 we have D(G + ex2)(X) > 0 for all x, so G + EX2 is convex. Let ? -> 0 to see that G is convex. Symmetrically,G - Ex2 is concave and hence G is also. Since the graphof 4Warning:Here we mustthink of [0,2 7] as R/27rZ,i.e. as the circumferenceof the unit circle.All thisreally means is thatthe set 11 shouldbe thoughtof as an intervalof length27r/3n,, even if tnllnl is less than 7r/3n,from 0 or 27r. 884 J. MARSHALL ASH [December G lies both above and below the chordpassing through any two of its points,it is a line. Appendix3. THEOREMA3 (Riemann[13, ?8, Theorem1]. Let Isin nk\ 2 Snk :=I k n >O, SOk := 1 and supposeEY Oan= 0. Let Gk = Y2flOafsfk. Thenlimk Gk = 0 Proof For each fixedk # 0, the seriesdefining Gk convergessince an 0 and ISnkl < c/n 2. Summationby partsyields N N-1 ansnkf = Sn(Snk - Sn+1k) + SNSNk' n=O n=O wheresn= a0+ an.Let N -oo toget 00 - Gk = Snf(Snk Sn +lk) n=O Now foreach integerN > 1 and each k + 0, N-1 00 IGkI < SUp ISnI NE (ISnk - 1I + 11 - Sn+lkl) + (sup snl) E ISnk Sn+lkl 1=O n>N n=N A + B. Since ISnk - Sn+lkI = fJn(k+l)kf(x)dxl wheref(x) {= [sinx/x]2}', the sum in B is boundedby J'Olf(x)I dx.This is a finiteconstant since f(x) = 2 ix x x~~~~~x xcosx - sinx x(1 - ) - (x - 5) _ near x = 0 so that f is bounded on (0,1], and lf(x)I < 2 1 (x . 1 + 1)/x3 < 4/x2 forx > 1. 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