ON COMPLEMENTED VERSIONS OF JAMES’S DISTORTION THEOREMS

WILLIAM B. JOHNSON^{†} AND NARCISSE RANDRIANANTOANINA^{‡}

Abstract. Examples are given to show that two natural questions asked in [5] about complemented versions of James’s distortion theorems have negative answers.

1. Introduction

The James’s distortion theorem for ℓ_{1} (respectively, for c_{0}) states that whenever a Banach space contains a subspace isomorphic to ℓ_{1} (respectively, c_{0}) then the Banach space contains subspaces that are almost isometric to ℓ_{1} (respectively, c_{0}). In [5], complemented versions of James’s distortion theorems were considered in the following senses:

Theorem 1. Let X be a Banach space whose dual unit ball is weak*-sequentially compact and ε > 0. If X contains a subspace isomorphic to c_{0}, then there exists a subspace Z of X and a projection P from X onto Z such that Z is (1 + ε)-isometric to c_{0} and ∥P ∥ ≤ 1+ε. Moreover, if X contains a subspace isometric to c_{0}, then there exists a subspace Z of X and a projection P from X onto Z such that Z is isometric to c_{0} and ∥P ∥ = 1.

Theorem 2. Let X be a Banach space which contains a complemented subspace isomorphic to ℓ_{1} and ε > 0. Then there exists a subspace Y of X and a projection P from X onto Y such that Y is (1 + ε)-isometric to ℓ_{1} and ∥P ∥ ≤ 1 + ε.

While Theorem 2 can be viewed as the exact analogue of the James’s distortion theorem for complemented copies of ℓ_{1}, Theorem 1 may be interpreted as combination of the James’s distortion theorem for c_{0} and the classical Sobczyk Theorem. These led to the following natural questions (see [5, Question 1, Question 2]):

Question 1. If a Banach space X contains a complemented copy of c_{0} and if ε > 0, does there exist a subspace Z of X and a projection P from X onto Z such that Z is (1 + ε)-isometric to c_{0} and ∥P ∥ ≤ 1 + ε?

1991 Mathematics Subject Classification. [2000]Primary: 46B20, 46B25. Secondary: 46B03, 46B45.

Key words and phrases. ℓ_{1}, c_{0}, renorming, James’s distortion theorems.

†Research partially supported by NSF DMS–0503688.

‡Research partially supported by NSF DMS–0456781.

January 31, 2007.

Question 2. If a Banach space X contains a complemented subspace isometric to ℓ_{1}, does there exists a subspace Z of X and a projection P from X onto Z such that Z is isometric to ℓ^{1} and ∥P ∥ = 1?

The aim of this note is to provide examples showing that, as expected, the answers to both questions are negative.

2. The examples

2.1. The c_{0}-case. In this subsection, we exhibit a Banach space X with the property that any complemented subspace of X that is almost isometric to c_{0} has large projection constant, thus answering Question 1 negatively. The space X is a renorming of ℓ_{∞} ⊕_{∞} c_{0}.

We denote by ∥ · ∥ the usual norm on ℓ_{∞} ⊕_{∞} c_{0}. In order to define the new norm on X, let J : ℓ_{1} → ℓ_{∞} be an isometric embedding of ℓ_{1} into ℓ_{∞} and Q : ℓ_{1} → c_{0} be a quotient map. For δ > 0, a norm ||| · ||| on X is defined by fixing its unit ball:

B_{(X,|||·|||)} := {(Jf, Qf); f ∈ ℓ_{1}, ∥f∥_{1} ≤ 1} + δB_{(X,∥·∥)}.

It is clear that ||| · ||| and ∥ · ∥ are equivalent norms on X and X contains a complemented subspace isomorphic to c_{0}.

Proposition 3. Let ε > 0 and Z be a subspace of (X, |||·|||) that is (1+ε)-isometric to c_{0} and is complemented in X. If P is a projection from X onto Z then

|||P ||| ≥ \frac{1 + δ}{4δ(1 + ε)^{3}} .

Proof. Throughout, we also denote by ||| · ||| the corresponding dual norm on X^{∗}. Let (V_{n})_{n≥1} be a basic sequence (1 + ε)-equivalent to the unit vector basis of c_{0} and whose closed linear span is Z. Let P be a projection from X onto Z. Then P is of the form

P = ∑_{n=1}^{∞} V_{n}^{∗} ⊗ V_{n}

where (V_{n}^{∗})_{n≥1} is a weak^{∗}-null sequence in X^{∗} and the sum can be taken with respect to the strong operator topology. Observe that X^{∗} = (ℓ_{∞})^{∗} ⊕_{1} ℓ_{1} isomorphically and thus for every n ≥ 1, V_{n}^{∗} = (x_{n}^{∗}, a_{n}^{∗}) where (x_{n}^{∗})_{n≥1} (respectively, (a_{n}^{∗})_{n≥1}) is a weak^{∗}- null sequence in (ℓ_{∞})^{∗} (respectively, ℓ_{1}). Since weak^{∗}-null sequences are weakly-null in (ℓ_{∞})^{∗} (see for instance [1, Theorem 15, p.103]), we have

weak − lim_{n→∞} x_{n}^{∗} = 0.

There exists a convex block (y_{n}^{∗})_{n≥1} of (x_{n}^{∗})_{n≥1} with

lim_{n→∞} ∥y_{n}^{∗}∥ = 0. (2.1.1)

There exists a strictly increasing sequence of integers (k_{n})_{n≥0} and positive scalars α_{j}^{(n)}, where k_{n−1} + 1 ≤ j ≤ k_{n}, ∑_{j=k_{n−1}+1}^{k_{n}} α_{j}^{(n)} = 1 for n ≥ 1, and

y_{n}^{∗} = ∑_{j=k_{n−1}+1}^{k_{n}} α_{j}^{(n)}x_{j}^{∗}.

For n ≥ 1, consider the corresponding block sequences:

\begin{aligned}W_{n} & = ∑_{j=k_{n−1}+1}^{k_{n}} V_{j} \\ W_{n}^{∗} & = ∑_{j=k_{n−1}+1}^{k_{n}} α_{j}^{(n)}V_{j}^{∗} \end{aligned} (2.1.2)

Then (W_{n})_{n≥1} is equivalent to the unit vector basis of c_{0}. Moreover, for every n, k ≥ 1, ⟨W_{n}^{∗}, W_{k}⟩ = δ_{n}^{k}, W_{n}^{∗} = (y_{n}^{∗}, b_{n}^{∗}) where b_{n}^{∗} = ∑_{j=k_{n−1}+1}^{k_{n}} α_{j}^{(n)}a_{j}^{∗}, and |||W_{n}||| ≤ 1 + ε. The latter implies that for n ≥ 1, W_{n} can be decomposed as

W_{n} = (Jf_{n}, Qf_{n}) + (x_{n}, a_{n}) (2.1.3)

with

(i) f_{n} ∈ ℓ_{1} satisfying ∥f_{n}∥_{1} ≤ 1 + ε; (ii) ∥(x_{n}, a_{n})∥ ≤ δ(1 + ε).

Since (f_{n})_{n≥1} is a bounded sequence in ℓ_{1}, we may assume (by passing to a subsequence if necessary) that for every n ≥ 1,

f_{n} = f_{0} + g_{n} + h_{n}

where

(a) lim_{n→∞} ∥f_{n}∥_{1} exists;

(b) weak^{∗} − lim_{n→∞} f_{n} = f_{0};

(c) (g_{n})_{n≥1} is a disjointly supported sequence in ℓ_{1};

(d) lim_{n→∞} ∥h_{n}∥_{1} = 0.

We claim that

lim_{n→∞} ∥f_{n}∥_{1} ≤ 3δ(1 + ε). (2.1.4)

To see this claim, let N ≥ 1, then

\frac{1}{N} ∑_{n=1}^{N} ∥f_{n}∥_{1} ≤ ∥f_{0}∥_{1} + \frac{1}{N} ∑_{n=1}^{N} ∥g_{n}∥_{1} + \frac{1}{N} ∑_{n=1}^{N} ∥h_{n}∥_{1}

= ∥f_{0}∥_{1} + \frac{1}{N} ∥ ∑_{n=1}^{N} g_{n}∥_{1} + \frac{1}{N} ∑_{n=1}^{N} ∥h_{n}∥_{1}

≤ 2∥f_{0}∥_{1} + \frac{1}{N} ∥ ∑_{n=1}^{N} f_{n}∥_{1} + \frac{2}{N} ∑_{n=1}^{N} ∥h_{n}∥_{1}.

Note that ∥f_{0}∥_{1} ≤ \underline{lim}_{N→∞}N^{−1}∥ ∑_{n=1}^{N} f_{n}∥_{1} and lim_{N→∞} N^{−1} ∑_{n=1}^{N} ∥h_{n}∥_{1} = 0. We deduce that

lim_{n→∞} ∥f_{n}∥_{1} ≤ 3 \overline{lim}_{N→∞} \frac{1}{N} ∥ ∑_{n=1}^{N} f_{n}∥_{1}.

We observe that

\frac{1}{N} ∥ ∑_{n=1}^{N} f_{n}∥_{1} = \frac{1}{N} ∥ ∑_{n=1}^{N} Jf_{n}∥_{∞}

≤ \frac{1}{N} ∥ ∑_{n=1}^{N} (Jf_{n}, Qf_{n})∥

≤ \frac{1}{N} ∥ ∑_{n=1}^{N} W_{n}∥ + \frac{1}{N} ∥ ∑_{n=1}^{N} (x_{n}, a_{n})∥

≤ \frac{1}{N} ∥ ∑_{n=1}^{N} W_{n}∥ + δ(1 + ε).

Since (W_{n}) is equivalent to the unit vector basis of c_{0}, we have lim_{N→∞} N^{−1}∥ ∑_{n=1}^{N} W_{n}∥ =

0. Combining all the above estimates, we get inequality (2.1.4).

We now show that if Π is the projection from X onto the closed linear span of (W_{n})_{n≥1} defined by Π = ∑_{n=1}^{∞} W_{n}^{∗} ⊗ W_{n}, then

|||Π||| ≥ \frac{1 + δ}{4δ(1 + ε)^{2}} . (2.1.5)

To see this, we first observe that for every y^{∗} ∈ ℓ_{1},

|||(0, y^{∗})||| = ∥y^{∗}∥_{1}(1 + δ). (2.1.6)

For every n ≥ 1, we have

1 = ⟨W_{n}^{∗}, W_{n}⟩
= ⟨(y_{n}^{∗}, b_{n}^{∗}), (Jf_{n} + x_{n}, Qf_{n} + a_{n})⟩
= ⟨y_{n}^{∗}, Jf_{n} + x_{n}⟩ + ⟨b_{n}^{∗}, Qf_{n} + a_{n}⟩
≤ (∥y_{n}^{∗}∥ + ∥b_{n}^{∗}∥)(∥f_{n}∥_{1} + δ(1 + ε)).

Taking limits as n → ∞, we deduce that

\underline{lim}_{n→∞}∥b_{n}^{∗}∥_{1} ≥ \frac{1}{4δ(1 + ε)}. (2.1.7)

We can estimate |||Π||| as follows:

|||Π||| ≥ sup_{n≥1} \frac{|||W_{n}^{∗}|||}{1 + ε}

= sup_{n≥1} \frac{|||(y_{n}^{∗}, b_{n}^{∗})|||}{1 + ε}

≥ \underline{lim}_{n→∞} \frac{|||(y_{n}^{∗}, b_{n}^{∗})|||}{1 + ε}

= \underline{lim}_{n→∞} \frac{|||(0, b_{n}^{∗})|||}{1 + ε}.

Thus (2.1.5) follows by combining (2.1.7) and (2.1.6). We conclude the proof by observing that |||Π||| ≤ (1 + ε)|||P |||.

2.2. The ℓ_{1}-case. Now we provide an example showing that Theorem 2 does not extend to the isometric case. In particular, the answer to Question 2 is negative.

First, recall that a norm ∥ · ∥ on a Banach space E is said to be strictly convex if Ext(B_{E}) = S_{E}. This is equivalent to the following property (see for instance [6, p. 246]):

(2.2.1) If x, y ∈ S_{E} satisfy ∥x + y∥ = 2, then x = y.

It is clear from (2.2.1) that if ∥ · ∥ is strictly convex then (E, ∥ · ∥) does not contain any ℓ_{∞}^{2} (the two dimensional ℓ_{∞}) isometrically. Indeed, if e_{1} = (1, 0) and e_{2} = (0, 1) are the unit vector basis of ℓ_{∞}^{2} then x = e_{1} and y = e_{1} + e_{2} fail to satisfy (2.2.1).

Define a norm | · | on ℓ_{1} that is equivalent to the usual norm and such that its dual norm | · |^{∗} is strictly convex. Such a dual norm on ℓ_{∞} can be taken by setting:

|(a_{i})|^{∗} := (∥(a_{i})∥_{∞}^{2} + ∑_{i=1}^{∞} 2^{−i}|a_{i}|^{2})^{1/2}.

Details on the existence of the norm | · | can be found in [6, pp. 241–254]. We define a Banach space Y by setting:

Y := (C[0, 1], ∥ · ∥_{∞}) ⊕_{∞} (ℓ_{1}, | · |).

Proposition 4. The Banach space Y contains a complemented subspace that is isometric to ℓ_{1} but ℓ_{1} is not isometric to a quotient of Y.

Let (e_{n}) be the unit vector basis of ℓ_{1} and for n ≥ 1, set v_{n} := e_{n}/|e_{n}|. Fix a sequence (f_{n})_{n≥1} in C[0, 1] that is isometrically equivalent to the unit vector basis of ℓ_{1} and for n ≥ 1, define U_{n} := (f_{n}, v_{n}) ∈ Y. We claim that (U_{n})_{n≥1} is isometrically equivalent to the unit vector basis of ℓ_{1} and its closed linear span is complemented. In fact, for any finite sequence (a_{n})_{n≥1} of scalars,

∑_{n≥1} |a_{i}| = ∥ ∑_{n≥1} a_{n}f_{n}∥_{∞}

≤ ∥ ∑_{n≥1} a_{n}U_{n}∥_{Y}

≤ ∑_{n≥1} |a_{i}|,

therefore, ∑_{n≥1} |a_{i}| = ∥ ∑_{n≥1} a_{n}U_{n}∥_{Y}. Moreover, if we denote by Z the closed linear span of (U_{n})_{n≥1} then Z is a complemented subspace of Y. Indeed, let T : (ℓ_{1}, |·|) → Y be defined by setting T (v_{n}) = U_{n} for all n ≥ 1 and Π be the second projection from Y onto (ℓ_{1}, | · |) then T ∘ Π is a projection from Y onto Z.

The fact that ℓ_{1} is not isometric to a quotient of Y follows from the next lemma, which we assume is well known.

Lemma 5. Let E and F be Banach spaces and T : c_{0} → E ⊕_{1} F be an isometry. Then there exists c_{j} ≥ 0, j = 1, 2 with:

(a) c_{1} + c_{2} = 1;

(b) if T = (T_{1}, T_{2}) then ∥T_{j}(e)∥ = c_{j}∥e∥ for j = 1, 2 and all e ∈ c_{0}.

In particular, if E ⊕_{1}F contains an isometric copy of c_{0} then either E or F contains an isometric copy of c_{0}.

Proof. Denote by c_{00} the space of finitely supported sequences of scalars and let (e_{n})_{n≥1} be the unit vector basis of c_{0}. Write T = (T_{1}, T_{2}) with T_{1} : c_{0} → E and T_{2} : c_{0} → F. We shall verify that for every x ∈ c_{00} with ∥x∥ = 1,

∥T_{j}(x)∥ = ∥T_{j}(e_{1})∥ for j = 1, 2.

To see this, we will show first that if x and y are disjointly supported unit vectors then

∥T_{j}(x)∥ = ∥T_{j}(y)∥ for j = 1, 2. (2.2.2)

Write 2x = (x − y) + (x + y). Then

2 = ∥T (2x)∥ = ∥T_{1}(2x)∥ + ∥T_{2}(2x)∥
≤ (∥T_{1}(x − y)∥ + ∥T_{1}(x + y)∥) + (∥T_{2}(x − y)∥ + ∥T_{2}(x + y)∥)
= ∥T (x − y)∥ + ∥T (x + y)∥ = 2.

For j = 1, 2, set a_{j} = ∥T_{j}(x−y)∥+∥T_{j}(x+y)∥−2∥T_{j}(x)∥. Then (a_{1}, a_{2}) is a positive element of ℓ_{1}^{2} whose norm is equal to zero so

2∥T_{j}(x)∥ = ∥T_{j}(x − y)∥ + ∥T_{j}(x + y)∥, j = 1, 2.

By reversing the role of x and y, we get (2.2.2).

Now, let x ∈ c_{00} with ∥x∥ = 1. Choose, n > 1 so that e_{n} and x are disjointly supported. From (2.2.2),

∥T_{j}(x)∥ = ∥T_{j}(e_{n})∥ = ∥T_{j}(e_{1})∥, j = 1, 2.

Setting c_{j} = ∥T_{j}(e_{1})∥ for j = 1, 2 proves the lemma.

End of the proof of Proposition 4. If ℓ_{1} is isometric to a quotient of Y then the dual space Y^{∗} = C[0, 1]^{∗} ⊕_{1} (ℓ_{∞}, | · |^{∗}) contains an isometric copy of ℓ_{∞} = ℓ_{1}^{∗} and hence of c_{0}. But since | · |^{∗} is strictly convex and C[0, 1]^{∗} is a L_{1}-space, this is in contradiction with Lemma 5 and thus completes the proof.

3. Concluding remarks

The notion of asymptotically isometric copies of ℓ_{1} (respectively, c_{0}) is closely related to James’s distortion theorems. We recall that a Banach space E is said to contain an asymptotically isometric copy of ℓ_{1} (respectively, c_{0}) if there exist a null sequence (ε_{n})_{n≥1} in (0, 1) and a sequence (x_{n})_{n≥1} in E such that for all finite sequence (t_{n})_{n} of scalars:

∑_{n}(1 − ε_{n})|t_{n}| ≤ ∥ ∑_{n} t_{n}x_{n}∥ ≤ ∑_{n} |t_{n}|,

respectively,

sup_{n}(1 − ε_{n})|t_{n}| ≤ ∥ ∑_{n} t_{n}x_{n}∥ ≤ sup_{n} |t_{n}|,

The norm introduced in the definition of the Banach space Y can be used to provide examples confirming the optimality of the James’s distortion theorems. We refer to [4] for earlier examples.

Proposition 6. (a) (ℓ_{1}, | · |) does not contain any subspace asymptotically isometric to ℓ_{1}.

(b) (ℓ_{∞}, | · |^{∗}) does not contain any subspace asymptotically isometric to c_{0}.

Proof. These statements follow from the norm | · |^{∗} being strictly convex and some known results. First, containing an asymptotically isometric copy of c_{0} and containing an isometric copy of c_{0} is equivalent in a dual space ([3]). Second, according to [2], a Banach space contains an asymptotically isometric copy of ℓ_{1} if and only if its dual contains an isometric copy of L_{1}[0, 1]. But since (ℓ_{∞}, | · |^{∗}) is strictly convex, it does not contain any isometric copy of ℓ_{1} and therefore it can not contain any isometric copy of L_{1}[0, 1]. Part (a) was already observed in [2] where an explicit formula for | · | was given (see [2, Corollary 12]).

The proof of Proposition 4 yields that the last part of the conclusion of Proposition 4 can be strengthened to “c_{0} is not isometric to a subspace of Y^{∗}”. However, in [3] Dowling proved that ℓ_{1} is a quotient of X if and only if c_{0} embeds isometrically into X^{∗}, so the more natural statement involving ℓ_{1} is only formally weaker.

References

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[6] M. Fabian, P. Habala, P. Hájek, V. Montesinos Santalucía, J. Pelant, and V. Zizler, Functional analysis and infinite-dimensional geometry, CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC, 8, Springer-Verlag, New York, 2001. MR 2002f:46001

Department of Mathematics, Texas A&M University, College Station, TX 77843

E-mail address: johnson@math.tamu.edu

Department of Mathematics and Statistics, Miami University, Oxford, OH 45056

E-mail address: randrin@muohio.edu