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An example regarding Kalton's paper ‘isomorphisms between spaces of vector-valued continuous functions’

Published online by Cambridge University Press:  02 August 2021

Félix Cabello Sánchez*
Affiliation:
Departamento de Matemáticas and IMUEx, Universidad de Extremadura, Avenida de Elvas, 06071Badajoz, Spain (fcabello@unex.es)
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Abstract

The paper alluded to in the title contains the following striking result: Let $I$ be the unit interval and $\Delta$ the Cantor set. If $X$ is a quasi Banach space containing no copy of $c_{0}$ which is isomorphic to a closed subspace of a space with a basis and $C(I,\,X)$ is linearly homeomorphic to $C(\Delta ,\, X)$, then $X$ is locally convex, i.e., a Banach space. We will show that Kalton result is sharp by exhibiting non-locally convex quasi Banach spaces $X$ with a basis for which $C(I,\,X)$ and $C(\Delta ,\, X)$ are isomorphic. Our examples are rather specific and actually, in all cases, $X$ is isomorphic to $C(K,\,X)$ if $K$ is a metric compactum of finite covering dimension.

Type
Research Article
Copyright
Copyright © The Author(s), 2021. Published by Cambridge University Press on Behalf of The Edinburgh Mathematical Society

The paper alluded to in the title contains the following striking result [Reference Kalton8, Theorem 6.4]: Let $I$ be the unit interval and $\Delta$ the Cantor set. If $X$ is a quasi Banach space containing no copy of $c_0$ which is isomorphic to a closed subspace of a space with a basis and $C(I,\,X)$ is linearly homeomorphic to $C(\Delta ,\, X)$, then $X$ is locally convex, i.e., a Banach space.

Here $C(K,\,X)$ denotes the space of continuous functions $F:K \longrightarrow X$. When $K$ is a compact space and $X$ a quasi Banach space $C(K,\,X)$ is also a quasi Banach space under the quasinorm $\|F\|=\sup \{\|F(t)\|: t\in K\}$.

When $X$ is a Banach space, the isomorphic theory of the spaces $C(K,\,X)$ is somehow oversimplified by Miljutin theorem (the spaces $C(K)=C(K,\,\mathbb {R})$ for $K$ uncountable and metrizable are all mutually isomorphic) and, above all, by Grothendieck's identity $C(K,\,X)=C(K)\check {\otimes }_{\varepsilon } X$ which implies that the isomorphic type of the Banach space $C(K,\,X)$ depends only on those of $C(K)$ and $X$. The situation for quasi Banach spaces is more thrilling and actually some seemingly innocent questions remain open: Is $C(I,\,\ell _p)$ isomorphic to $C(I^{2},\,\ell _p)$? Is $C(I,\,L_p)$ isomorphic to $C(\Delta ,\,L_p)$? These appear as Problems 7.2 and 7.3 at the end of [Reference Kalton8]. Problem 7.1, namely if $C(K)\otimes X$ (the subspace of functions whose range is contained in some finite-dimensional subspace of $X$) is always dense in $C(K,\,X)$, was posed by Klee and is connected with quite serious mathematics. While it seems to be widely open for quasi Banach spaces $X$, the answer is negative for $F$-spaces (complete linear metric spaces) as shown by Cauty's celebrated example [Reference Cauty4] (see also [Reference Kalton and Dobrowolski9]) and affirmative for locally convex spaces. See Waelbroeck [Reference Waelbroeck13, Section 8] for a discussion on Klee's density problem.

The aim of this short note is much more modest: we will show that Kalton's result is sharp by exhibiting non-locally convex quasi Banach spaces $X$ with a basis for which $C(I,\,X)$ and $C(\Delta ,\, X)$ are isomorphic. Our examples are rather specific and actually, in all cases, $X$ is isomorphic to $C(K,\,X)$ if $K$ is a metric compactum of finite covering dimension.

Recall that the (Lebesgue) covering dimension of a (not necessarily compact) topological space $K$ is the smallest number $n\geq 0$ such that every open cover admits a refinement in which every point of $K$ lies in the intersection of no more than $n+1$ sets of the refinement.

A quasi Banach space $X$ has the $\lambda$-approximation property ($\lambda$-AP) if for every $x_1,\,\ldots ,\, x_n\in X$ (or in some dense subset) there is a finite-rank operator $T$ on $X$ such that $\|T\|\leq \lambda$ and $\|x_i-Tx_i\|<\varepsilon$. We say that $X$ has the bounded approximation property (BAP) if it has the $\lambda$-AP for some $\lambda \geq 1$.

We end these preliminaries by recalling that a $p$-norm, where $0< p\leq 1$, is a quasinorm satisfying the inequality $\|x+y\|^{p}\leq \|x\|^{p}+\|y\|^{p}$ and that every quasinormed space has an equivalent $p$-norm for some $0< p\leq 1$, so says the Aoki–Rolewicz theorem.

Lemma If $K$ has finite covering dimension and $X$ has the BAP, then $C(K,\,X)$ has the BAP.

Proof. We first observe that if $K$ has finite covering dimension or $X$ has the BAP, then $C(K)\otimes X$ is dense in $C(K,\,X)$. The part concerning the BAP is obvious; the other part is a result by Shuchat [Reference Shuchat12, Theorem 1].

Given $g\in C(K)$ and $x\in X$, we denote by $g\otimes x$ the function $t\longmapsto g(t)x$. Since every function in $C(K)\otimes X$ can be written as a finite sum $\sum \nolimits _{i} g_i\otimes x_i$ with $g_i\in C(K),\, x_i\in X$ (which justifies our notation, see [Reference Shuchat12, Proposition 1]), it suffices to see that there is a constant $\Lambda$ such that, given $f_1,\,\ldots ,\,f_m\in C(K),\, y_1,\,\ldots ,\,y_m\in X$ and $\varepsilon >0$, there is a finite-rank operator $T$ on $C(K,\,X)$ such that $\|T\|\leq \Lambda$ and $\|f_i\otimes y_i - T(f_i\otimes y_i)\|<\varepsilon$. As $\varepsilon$ is arbitrary, there is no loss of generality in assuming that $\|f_i\|=\|y_i\|=1$ for $1\leq i\leq m$.

Take an open cover $U_1,\,\ldots ,\,U_r$ of $K$ such that for every $i,\,j$, one has $|f_i(s)-f_i(t)|<\varepsilon$ for all $s,\,t\in U_j$. Put $n=\dim (K)$ and take a refinement $V_1,\,\ldots ,\, V_k$ so that each point of $K$ lies in no more than $n+1$ of those sets. Finally, let $\phi _1,\, \ldots ,\, \phi _k$ be a partition of unity of $K$ subordinate to $V_1,\,\ldots ,\, V_k$.

For each $j$, pick $t_j\in V_j$ and define an operator $L$ on $C(K,\,X)$ by letting $L(F)=\sum \nolimits _{j\leq k} \phi _j\otimes F(t_j)$, that is, $(LF)(t)=\sum \nolimits _{j\leq k} \phi _j(t) F(t_j)$. Let us estimate $\|L\|$ assuming $X$ is $p$-normed: one has

\[ \|L(F)\|= \sup_{t\in K}\left\|\sum_{j\le k} \phi_j(t) F(t_j)\right\|, \]

but for each $t\in K$ the sum has no more than $n+1$ nonzero summands, so

\begin{align*} \left\|\sum_{j\le k} \phi_j(t) F(t_j)\right\| & \leq \left(\sum_{j\le k} \phi_j(t)^{p} \|F(t_j)\|^{p}\right)^{1/p}\leq \|F\| \left(\sum_{j\le k} \phi_j(t)^{p} \right)^{1/p}\\ & \leq \|F\|\|\textbf{I}: \ell_1^{n+1} \longrightarrow \ell_p^{n+1}\|=(n+1)^{1/p -1}\|F\| \end{align*}

We claim that $\|f_i\otimes y_i - L(f_i\otimes y_i)\|\le \varepsilon$ for all $i$. We have $L(f_i\otimes y_i)(t)=\sum \nolimits _{j\leq k}f_i(t_j)\phi _j(t) y_i$, hence

\[ \|f_i\otimes y_i - L(f_i\otimes y_i)\|= \left\|\sum_{j\leq k} f_i\phi_j -\sum_{j\leq k} f_i(t_j)\phi_j\right\|\|y_i\| = \left\|\sum_{j\leq k} f_i\phi_j -\sum_{j\leq k} f_i(t_j)\phi_j\right\|. \]

For each $j$ and each $t\in K$ one has $|f_i(t)\phi _j(t)-f_i(t_j)\phi _j(t)|\le \varepsilon \phi _j(t)$: this is obvious if $t\notin V_j$ since in this case $\phi _j(t)=0$, while for $t\in V_j$ we have $|f_i(t)-f_i(t_j)|\le \varepsilon$ by our choice of $V_1,\,\ldots ,\, V_k$ and thus

\[ \left|\sum_{j\leq k} f_i(t)\phi_j(t) -\sum_{j\leq k} f_i(t_j)\phi_j(t)\right|\le\varepsilon \sum_{j\leq k}\phi_j(t)=\varepsilon \]

holds for all $t\in K$; consequently, we have

\[ \left\|\sum_{j\leq k} f_i\phi_j -\sum_{j\leq k} f_i(t_j)\phi_j\right\|\leq \varepsilon. \]

Let $R$ be a finite-rank operator on $X$ such that $\|y_i-R(y_i)\|<\varepsilon$, with $\|R\|\leq \lambda$, where $\lambda$ is the ‘approximation constant’ of $X$, and define $T$ on $C(K,\,X)$ by $(TF)(t)=R((LF)(t))$. Clearly, $T$ has finite-rank since for an elementary tensor $f\otimes x$ one has

\[ T(f\otimes x)= \sum_{j\leq k} f(t_j) \phi_j\otimes R(x). \]

Finally, let us estimate $\|f_i\otimes y_i - T(f_i\otimes y_i)\|$. Write

\[ f_i\otimes y_i - T(f_i\otimes y_i) = f_i\otimes y_i - f_i\otimes R(y_i) + f_i\otimes R(y_i) - \sum_{j\leq k} f_i(t_j) \phi_j\otimes R(y_i) \]

and then

\begin{align*} \| f_i\otimes y_i - T(f_i\otimes y_i)\|^{p}& \leq \| f_i\otimes y_i - f_i\otimes R(y_i)\|^{p} + \left\|f_i\otimes R(y_i) - \sum_{j\leq k} f_i(t_j) \phi_j\otimes R(y_i) \right\|^{p}\\ & = \| f_i\|^{p}\|y_i - R(y_i)\|^{p} + \left\|f_i - \sum_{j\leq k} f_i(t_j) \right\|^{p} \| R(y_i) \|^{p} \leq \varepsilon^{p}+ \varepsilon^{p} \lambda^{p}, \end{align*}

so that $C(K,\,X)$ has the BAP with constant at most $\lambda (n+1)^{1/p -1}$.

The proof raises the question of whether the lemma is true for, say, the Hilbert cube $I^{\omega }$.

The other ingredient we need is a complementably universal space for the BAP. A separable $p$-Banach space is complementably universal for the BAP if it has the BAP and contains a complemented copy of each separable $p$-Banach space with the BAP. The existence of such spaces (one for each $0< p<1$) was first mentioned by Kalton himself in [Reference Kalton6, Theorem 4.1(b)]. A complete proof appears in the related issues of [Reference Cabello Sánchez, Castillo and Moreno1]. In any case, it easily follows from the Pełczyński decomposition method that any two separable $p$-Banach spaces complementably universal for the BAP are isomorphic, so let us denote by $\mathscr K_p$ the isomorphic type of such specimens and observe that since each separable $p$-Banach space with the BAP is complemented in one with a basis, it follows that $\mathscr K_p$ does have a basis. Needless to say, $\mathscr K_p$ is not locally convex since it contains a complemented copy of $\ell _p$.

Corollary If $K$ is a (non-empty) metrizable compactum of finite covering dimension, then $C(K,\,\mathscr K_p)$ is linearly homeomorphic to $\mathscr K_p$. In particular, $C(I,\,\mathscr K_p)$ and $C(\Delta ,\,\mathscr K_p)$ are linearly homeomorphic although $\mathscr K_p$ is not locally convex.

Proof. This clearly follows from the lemma since $C(K,\,\mathscr K_p)$ is separable, has the BAP and contains $\mathscr K_p$ complemented as the subspace of constant functions.

We do not know of any other non-locally convex quasi Banach space $X$ for which $C(I,\,X)$ and $C(\Delta ,\, X)$ are isomorphic, apart from the obvious ones arising as direct sums of $\mathscr K_p$ and Banach spaces lacking the BAP. An obvious candidate is the $p$-Gurariy space, introduced by Kalton in [Reference Kalton7, Theorem 4.3] and further studied in [Reference Cabello Sánchez, Garbulińska-Wȩgrzyn and Kubiś2]. Note that if $X$ is a quasi Banach space isomorphic to $X\oplus F$, with $F$ finite dimensional and $C(I,\,X)$ and $C(\Delta ,\, X)$ are not isomorphic then neither are $C(I,\,X\oplus c_0)$ and $C(\Delta ,\, X\oplus c_0)$.

It's time to leave. Perhaps the most important question regarding the general topological properties of quasi Banach spaces is to know whether every quotient operator $Q:Z \longrightarrow X$ (acting between quasi Banach spaces) admits a continuous section, namely a continuous $\sigma : X \longrightarrow Z$ such that $Q\circ \sigma =\textbf {I}_X$. More generally, let us say that $f\in C(K,\,X)$ lifts through $Q$ if there is $F\in C(K,\,Z)$ such that $f=F\circ Q$. Now, given $0< p<1$, a quotient operator between $p$-Banach spaces $Q:Z \longrightarrow X$ and a compactum $K$, consider the following statements:

  1. (1) $Q$ admits a continuous section.

  2. (2) Every continuous $f:K \longrightarrow X$ has a lifting to $Z$.

  3. (3) $C(K)\otimes X$ is dense in $C(K,\,X)$.

Clearly, (1)$\implies$(2): set $F=\sigma \circ f$, where $\sigma$ is the hypothesized section of $Q$. Besides, if (1) is true for some quotient map $\ell _p(J) \longrightarrow X$ then so it is for every $Q$. Similarly, if (2) is true for a given $K$ for some quotient map $\ell _p(J) \longrightarrow X$, then it is true for any quotient map onto $X$ and (3) holds.

Following (badly) Klee [Reference Klee10, Section 2], let us say that the pair $(K,\,X)$ is admissible if (3) holds, that $K$ is admissible if (3) holds for every quasi Banach space $X$ and that $X$ is admissible if (3) holds for every compact $K$. We do not know whether the $p$-Gurariy spaces are admissible or not.

We have mentioned Shuchat's result that every compactum of finite covering dimension is admissible. Actually one can prove that (2) holds for any $Q$ if $\dim (K)<\infty$. This indeed follows from Michael's [Reference Michael11, Theorem 1.2] but a simpler proof can be given using Shuchat's result, the argument of the proof of the lemma, and the open mapping theorem. Since every metrizable compactum is the continuous image of $\Delta$, this implies that for every compact subset $S\subset X$, there is a compact subset $T\subset Z$ such that $Q[T]=S$.

Long time ago, Riedrich proved that the spaces $L_p$ are admissible for $0\leq p<1$; see [Reference Caponetti and Lewicki3, Reference Ishii5] for more general results that cover all modular function spaces. We do not know if the quotient map $\ell _p \longrightarrow L_p$ has a continuous section or satisfies (2) for arbitrary compact $K$ and $0< p<1$.

Acknowledgements

Research supported in part by MICIN Project PID2019-103961GB-C21 and Junta de Extremadura Project IB-20038.

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