James' space: a (counter-)example in Banach spaces Robert Hines December 3, 2014 1 Preliminaries In this section, we introduce some of the ideas needed to prove desired results about James' space. All of our vector spaces are real. We being with some definitions. A (Schauder) basis in a separable Banach space (X; k · k) is a sequence of vectors ek 2 X P1 such that every x 2 X can be uniquely written as a convergent series k=1 αkek with αk 2 R. P1 Pn We have canonical projections Pn : X ! X, Pn( k=1 αkek) = k=1 αkek and coordinate ∗ P1 functionals, en( k=1 αkek) = αn. [A few remarks. First, bases are ordered and the order matters; there is a notion of an unconditional basis, but James' Space does not have an unconditional basis, so we will not discuss it at any length. Second, not every separable Banach space has a basis (Enflo, 1973), although most familiar examples do. Every Banach space contains a basic sequence (a basis for its closed linear span), but not every Banach space contains an unconditional basic sequence (Gowers and Maurey, 1993).] We now show that the canonical projections and coordinate functionals are continuous, in fact uniformly bounded. We call K = supn kPnk the basis constant for the basis ek and say the basis is monotone if K = 1. Proposition 1. If (X; k · k) is a Banach space with basis ek, the the canonical projections are uniformly bounded P The proposition follows from the following lemma (noting that kPn( k αkek)k ≤ kfαkgkkY using the notation that follows). N P1 Lemma 1. If Y = fαkgk 2 R : k=1 αkek converges in (X; k · k) and Pn kfαkgkkY = supnfk k=1 αkekkg, then (Y; k · kY ) is a Banach space isomorphic to X. n (i)o Proof. It is clear that k · kY is a norm. To see that Y is complete, let αk be a Cauchy k (i) sequence in Y . Then for a fixed l, αl is Cauchy because ( ) l l−1 n (i) (j) X (i) (i) X (i) (i) X αl − αl kelk = αk − αk ek − αk − αk ek ≤ 2 sup αkek ! 0 n k=1 k=1 k=1 (n) (n) as i; j ! 1. Hence αk has a limit, limn αk =: αk. Let > 0, j such that for i; j ≥ j and for every n > 0 we have n X (i) (j) αk − αk ek < , k=1 1 and n such that for n; m ≥ n we have m X (j) αk ek < . k=n Then for j = j and i ! 1 we have (for all n) n X (j) αk − αk ek ≤ , k=1 and for n; m ≥ n m m m m X X X (j) X (j) αkek = αkek − αk ek + αk ek k=n k=n k=n k=n m m n−1 n−1 m X X (j) X X (j) X (j) ≤ αkek − αk ek + αkek − αk ek + αk ek k=1 k=1 k=1 k=1 k=n ≤ 3. P Hence k αkek converges and Y is complete. P1 The map S : Y ! X given by S(fαkgk) = k=1 αkek is clearly a linear bijection, and is also continuous since ( n ) X kS(fαkgk)k ≤ sup αkek = kfαkgkkY n k=1 Pn −1 (note that limn!1 k k=1 αkekk = kS(fαkgk)k). By the open mapping theorem, S is con- P1 tinuous and there is a K such that kfαkgkkY ≤ Kk k=1 αkekk. More generally we have the following (proof omitted). Proposition 2. A sequence ek of a Banach space (X; k · k) is a basis if and only if 1. ek 6= 0 for each k, 2. X = hekiR, and 3. there is an M such that for all n ≤ m and αk 2 R we have n m X X αkek ≤ M αkek : k=1 k=1 Next we have a characterization of reflexivity for a Banach space with basis. Theorem 1. A Banach space (X; k · k) with basis en is reflexive if and only if the basis is shrinking: for every f 2 X∗ we have ( 1 ) X lim sup jf(x)j : x = αkek; kxk = 1 = 0 n!1 k=n 2 and boundedly complete: if αk 2 R is such that ( n ) X sup αkek < 1; n k=1 P1 then there is an x 2 X such that x = k=1 αkek. Proof. Recall that X is reflexive if and only if the unit ball is weakly compact (a consequence of Alaoglu's theorem). ∗ If X is reflexive and has a non-shrinking basis ek, then there is an f 2 X , > 0, and sequence x = P1 α(n)e with kx k = 1 and p ! 1 such that jf(x )j > for every n k=pn k k n n n wk n. Because X is reflexive, fxng has a weakly convergent subsequence, xni ! x. Along this ∗ ∗ ∗ subsequence, we have ek(xni ) ! 0 and ek(xni ) ! ek(x) so that x ≡ 0 (all of its coordinates are zero), but < jf(xni )j ! jf(0)j = 0, a contradiction. Hence a basis for a reflexive space is shrinking. Pn Pn If X is reflexive and αn are such that supnfk k=1 αkekkg < 1, then xn = k=1 αkek has wk P1 a weakly convergent subsequence (using reflexivity), xni ! x. If x = k=1 βkek, then αk = βk ∗ ∗ for all k because αk = ek(xni ) ! ek(x) = βk. Hence a basis in a reflexive space is boundedly complete. We now assume that the basis is shrinking and boundedly complete and show that X is reflexive by proving that the unit ball is weakly compact. To this end, let 1 X (n) yn = βk ek 2 fkxk ≤ 1g: k=1 (n) (n) P1 For a fixed k the sequence βk is bounded, jβk j ≤ 2K=kekk (note the for any x = k=1 αkek 2 Pn+p X and n > 0; p ≥ 0 we have k k=n αkekk ≤ 2Kkxk where K is the basis constant). Hence (qn) there are qn such that for a fixed k, βk converges, and, using a diagonal argument, pn such (pn) that βk ! βk for some βk 2 R. For every n; N we have kPN (ypn )k ≤ Kkypn k ≤ K so that PN k k=1 βkekk ≤ K for all N. Because ek is boundedly complete, there is a y 2 X such that P1 y = k=1 βkek. ∗ We claim that ypn converges weakly to y. Let f 2 X and > 0. Since ek is shrinking, there P PN−1 is an N such that supfjf(x)j : x = k≥N αkek; kxk = 1g < . Choose M such that k k=1 (βk− (pn) (pn) PN+p (pn) β )ekk < for n ≥ M (remember β ! βk for all k). Recall that k β ekk ≤ 2K k k P k=N k for all p ≥ 0 and let p; pn tend to infinity to obtain k k≥N βkekk ≤ 2K. Hence for pn ≥ M, we have ! 0 1 0 1 N−1 X (pn) X X (pn) jf(y) − f(ypn )j ≤ f βk − β ek + f @ βkekA + f @ β ekA k k k=1 k≥N k≥N ≤ kfk + 2K + 2K, wk and ypn ! y. 1 For example, in our favorite non-reflexive spaces c0 and ` , the \standard basis" ek(l) = δkl Pn Pn fails to be boundedly complete in c0 (supnfk k=1 ekk1g < 1 but k=1 ek does not converge 3 1 P P in c0), and fails to be shrinking in ` (for instance, if f( k αkek) = k αk, then f(ek) = 1 for all k). We need a few more facts about shrinking bases. ∗ Proposition 3. If ek is a shrinking basis for a Banach space (X; k · k), then ek is a basis for ∗ ∗∗ ∗ X with coordinate functionals ek 2 X and the basis constant for ek is no more than the basis ∗ ∗ constant for ek. (In fact, ek is a basis for X if and only if ek is shrinking.) Proof. First note that for all n we have 1 n 1 1 X X X X αkek ≤ αkek + αkek ≤ (1 + K) αkek : k=n+1 k=1 k=1 k=1 If f 2 X∗, then ! ! ! n 1 X ∗ X X f − f(ek)ek αkek = f αkek k=1 k k=n+1 X ≤ kfnk(1 + K) αkek ! 0 k as n ! 1 because the basis is shrinking, where fn is the restriction of f to hek : k > niR. Hence ∗ ∗ X is the closed linear span of the coordinate functionals ek. P For n ≤ m, > 0, αk 2 R, and k k βkekk = 1 such that ! ! n m X ∗ X X ∗ αkek βkek ≥ αkek − , k=1 k k=1 we have ! ! n n X ∗ X ∗ X αkek ≤ αkek βkek + k=1 k=1 k ! ! m n X ∗ X = αkek βkek + k=1 k=1 m n X ∗ X ≤ αkek βkek + k=1 k=1 m X ∗ X ≤ K αkek βkek + k=1 k m X ∗ = K αkek + , k=1 ∗ ∗ where K is the basis constant for ek. Hence ek is a basis of X , with basis constant smaller ∗ than K. Finally, it is clear that the ek act as the coordinate functionals for the ek. Finally, we characterize X∗∗ for Banach spaces with a shrinking basis. 4 Proposition 4. Let (X; k · k) be a Banach space with shrinking basis ek, ( ( ) ) ( ) n n N X X Z = fαkgk 2 R : sup αkek < 1 ; and kfαkgkkZ = sup αkek : n n k=1 k=1 ∗∗ ∗ Then (Z; k · kZ ) is a Banach space and the map T : X ! Z given by T (φ) = fφ(ek)gk is an isomorphism, isometric if ek is monotone.