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 (X, k · k) is a sequence of vectors ek ∈ X P∞ such that every x ∈ X can be uniquely written as a convergent series k=1 αkek with αk ∈ R. P∞ Pn We have canonical projections Pn : X → X, Pn( k=1 αkek) = k=1 αkek and coordinate ∗ P∞ 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 ), 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 ≤ k{αk}kkY using the notation that follows).

 N P∞ Lemma 1. If Y = {αk}k ∈ R : k=1 αkek converges in (X, k · k) and Pn k{αk}kkY = supn{k k=1 αkekk}, then (Y, k · kY ) is a Banach space isomorphic to X.

n (i)o Proof. It is clear that k · kY is a . 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 → ∞. 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 → ∞ 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. P∞ The map S : Y → X given by S({αk}k) = k=1 αkek is clearly a linear bijection, and is also continuous since

( n ) X kS({αk}k)k ≤ sup αkek = k{αk}kkY n k=1

Pn −1 (note that limn→∞ k k=1 αkekk = kS({αk}k)k). By the open mapping theorem, S is con- P∞ tinuous and there is a K such that k{αk}kkY ≤ 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 ∈ 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 ∈ X∗ we have

( ∞ ) X lim sup |f(x)| : x = αkek, kxk = 1 = 0 n→∞ k=n

2 and boundedly complete: if αk ∈ R is such that

( n ) X sup αkek < ∞, n k=1 P∞ then there is an x ∈ 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 ∈ X ,  > 0, and sequence x = P∞ α(n)e with kx k = 1 and p → ∞ such that |f(x )| >  for every n k=pn k k n n n wk n. Because X is reflexive, {xn} 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  < |f(xni )| → |f(0)| = 0, a contradiction. Hence a basis for a reflexive space is shrinking. Pn Pn If X is reflexive and αn are such that supn{k k=1 αkekk} < ∞, then xn = k=1 αkek has wk P∞ 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

∞ X (n) yn = βk ek ∈ {kxk ≤ 1}. k=1

(n) (n) P∞ For a fixed k the sequence βk is bounded, |βk | ≤ 2K/kekk (note the for any x = k=1 αkek ∈ 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 ∈ 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 ∈ X such that P∞ y = k=1 βkek. ∗ We claim that ypn converges weakly to y. Let f ∈ X and  > 0. Since ek is shrinking, there P PN−1 is an N such that sup{|f(x)| : x = k≥N αkek, kxk = 1} < . 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 !     N−1   X (pn) X X (pn) |f(y) − f(ypn )| ≤ f βk − β ek + f  βkek + f  β ek 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 (supn{k k=1 ekk∞} < ∞ 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 ∈ 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

∞ n ∞ ∞ X X X X αkek ≤ αkek + αkek ≤ (1 + K) αkek . k=n+1 k=1 k=1 k=1

If f ∈ X∗, then ! ! ! n ∞ 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 → ∞ 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 ∈ 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 = {αk}k ∈ R : sup αkek < ∞ , and k{αk}kkZ = 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) = {φ(ek)}k is an , isometric if ek is monotone. Proof. The proof that Z is a Banach space is similar to that of Lemma 1, and T is clearly ∗ ∗ linear. If T (φ) = {φ(ek)}k = 0, then for every f ∈ X we have ! X ∗ X ∗ φ(f) = φ f(ek)ek = f(ek)φ(ek) = 0, k k ∗ so that T is injective. To see that T is surjective, let {αk}k ∈ Z, f ∈ X , and n ≤ m. We have

m m X X αkf(ek) ≤ kfnk αkek ≤ (1 + K)kfnkk{αk}kkZ → 0 k=n k=n

as n → ∞ because ek is shrinking (here fn is the restriction of f to hek : k > niR). Hence P ∗ P k αkf(ek) converges and we can define φ0 : X → R by φ0(f) = k αkf(ek), which is clearly linear with T (φ0) = {αk}k, and bounded because for any n

n n X X αkf(ek) ≤ kfk αkek ≤ Kkfkk{αk}kkZ . k=1 k=1 Finally, we have the bounded

( n ) ( n ) X ∗ X ∗ kT (φ)kZ = sup φ(ek)ek = sup φ(ek)f(ek) n k=1 n,kfk=1 k=1 ( n ! ) X ∗ = sup φ f(ek)ek ≥ sup{φ(f)} n n,kfk=1 k=1 = kφk,

Pn ∗ since k=1 f(ek)ek → f. So if K = 1, T is an . The last two propositions show that if X has a shrinking basis, we can identify X∗ and X∗∗ with sequence spaces.

2 James’ space

We now introduce the subject of this paper, the space

J = {x ∈ c0(R): kxkJ < ∞}

where k · kJ is defined by

( l !1/2 ) X 2 kxkJ = sup |x(pk+1) − x(pk)| : 2 ≤ l, 1 ≤ p1 < ··· < pl . k=1

5 We will also make use of the equivalent norm (for which the isometric isomorphism J =∼ J ∗∗ holds) given by

( l !1/2 ) 2 X 2 kxkcyc = sup |x(pl) − x(p1)| + |x(pk+1) − x(pk)| : 2 ≤ l, 1 ≤ p1 < ··· < pl . k=1 The space J has some interesting properties, such as

1 1. J is not reflexive and does not contain any subspaces isomorphic to c0 or ` , 2. J has one in J ∗∗ under the canonical injection ι : J → J ∗∗,

3. J (with the “cyclic quadratic variation” norm above) is isometrically isomorphic to J ∗∗ (but not via the canonical injection!),

4. J is not the underlying real space of a complex Banach space (we do not show this, but it ∗∗ follows from the fact that dimR(J /J) = 1 is odd since if X is the underlying real space ∗∗ ∼ ∗∗ of the complex space X0, then X /X = X0 /X0, real duals on the left, complex duals on the right),

5. J does not have an unconditional basis (we do not show this, but a Banach space with 1 an unconditional basis is either reflexive, contains c0, or contains ` ).

Properties 1,2, and 3 follow from the next series of propositions.

Proposition 5. (J, k · kJ ) is a Banach space and ek defined by ek(l) = δkl, is a monotone basis for J

Proof. That k · kJ is a norm on J is obvious and we now show completeness. If xj ∈ J is such P P∞ that j kxjkJ < ∞, then |xj(k)| = limn→∞ |xj(k) − xj(n)| ≤ kxjkJ , j=1 xj(k) =: αk exists P∞ and αk → 0 (since c0 is complete). For x = k=1 αkek we have

( l !1/2 ) X 2 kxkJ = sup |x(pk+1) − x(pk)| : 2 ≤ l, 1 ≤ p1 < ··· < pl k=1 ( ∞ l !1/2 ) X X 2 ≤ sup |xj(pk+1) − xj(pk))| : 2 ≤ l, 1 ≤ p1 < ··· < pl j=1 k=1 X ≤ kxjkJ < ∞. j

To see that ek is monotone, let p ≤ q and note that

p ( l !1/2 ) X X α e = sup |α − α |2 : 2 ≤ l, 1 ≤ p < ··· < p ≤ p k k pk+1 pk 1 l k=1 J k=1 ( l !1/2 ) X 2 ≤ sup |x(pk+1) − x(pk)| : 2 ≤ l, 1 ≤ p1 < ··· < pl ≤ q k=1 q

X = αkek k=1 J

6 (with equality when extending by zero for instance).

Finally, to see that J = hekiR, let x ∈ X,  > 0, 2 ≤ l, 1 ≤ p1 < ··· < pl such that

l !1/2 X 2 |x(pk+1) − x(pk)| ≥ kxkJ − . k=1

Pl Then kx − k=1 x(k)ekkJ < , and ek is a basis for J.

Proposition 6. (J, k · kJ ) is not reflexive (more specifically, ek is not boundedly complete). Pn Proof. Let sn =√ k=1 ek. Then ksnkJ = 1 for every n ≥ 1, but sn does not converge in J, e.g. ksm − snkJ = 2 for n 6= m.

Proposition 7. The basis ek is shrinking.

∗ Pni+1 (i) Proof. If not, there is an f ∈ X ,  > 0, and yi = β ek with ni increasing to infinity, P k=ni k P kyikJ = 1, and f(yi) >  for all i. If y = i yi ∈ J, then f(y) >  i 1/i = ∞, a contradiction. However y ∈ J. Let  > 0 and 2 ≤ l, 1 ≤ p1 < ··· < pl such that

l 2 X 2 kykJ −  ≤ |y(pk+1) − y(pk)| . k=1 Each term in the sum is either of the form

1 y (p ) y (p )2 y (p )2 y (p )2 (y (p ) − y (p ))2 or i+j k+1 − i k ≤ 2 i+j k+1 + 2 i k i2 i k+1 i k i + j i i + j i depending on whether or not we are in the same “block”. We get

∞ 2 X kyik kyk2 ≤ 5 J < ∞ J i2 i=1 and y ∈ J.

∗∗ Pn Proposition 8. J = ι(J) ⊕ Rs∞, where s∞ is the weak-∗ limit of sn = k=1 ek. ∗∗ Proof. From Proposition 4, J is isomorphic the the space Z of seqences {αk}k such that Pn k{αk}kkZ = supn{k k=1 αkekkJ } < ∞. If {αk}k is such a sequence, then λ = limk αk clearly P ∗∗ exists, and k αkek ∈ J if and only if λ = 0. Hence any φ = {αk}k ∈ J can be decomposed as φ = {αk − λ}k + λs∞.

∗∗ Proposition 9. J is isometrically isomorphic to J under k · kcyc.

∗∗ Proof. Define U : Z → (J, k · kcyc) (with Z isometrically isomorphic to J as in Proposition 4) by X U({αk}k) = −λe1 + (αk − λ)ek k>1

7 where λ = limk αk as in the previous proposition. U is clearly linear, and surjective since for P k αkek ∈ J we have X U((α2, α3,... ) − α1s∞) = αkek, k

with s∞ as in the previous proposition. To see that U is an isometry, let {αk}k ∈ Z and compute

2 kU({αk}k)kcyc ( ( l l−1 ) ) X X = sup max (α − α )2 + (α − α )2, α2 + α2 + (α − α )2 : l, p pl p1 pk+1 pk pl p1 pk+1 pk k k=1 k=1 with the “max” term coming from whether or not the initial −λ term is used to compute the cyclic quadratic variation. On the other hand we have

 n  2  X  k{αk}kkZ = sup αkek n  k=1 cyc ( ( l l−1 ) ) X X = sup max (α − α )2 + (α − α )2, α2 + α2 + (α − α )2 : l, p pl p1 pk+1 pk pl p1 pk+1 pk k k=1 k=1 with the “max” term coming from whether or not any of the “trailing zeros” were used in the computation of the cyclic quadratic variation. Therefore U is an isometry.

References

[1] Bernard Beauzamy, Introduction to Banach spaces and their geometry, Notas de Matematica [86], Mathematics Studies, vol. 68, North-Holland, Amsterdam, 1982.

[2] Robert C. James, A non-reflexive Banach space isometric with its second conjugate space, Pro. Nat. Acad. Sci. U.S.A. 37 (1951), 174-177.

[3] Robert Megginson, An introduction to Banach space theory, Graduate Texts in Mathemat- ics, vol. 183, Springer Verlag, New York, 1973

[4] Thomas Schlumprecht, Course Notes for I, Math 655-601, Fall 2011, http://www.math.tamu.edu/~schlump/course notes FA2012.pdf

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