On subprojectivity of $C(K,X)$

TL;DR

This paper proves that the space of continuous functions from a scattered compact space to a subprojective Banach space retains the subprojectivity property.

Contribution

It establishes that $C(K,X)$ is subprojective when $K$ is scattered and $X$ is subprojective, extending known properties of Banach spaces.

Findings

$C(K,X)$ is subprojective if $K$ is scattered and $X$ is subprojective

Subprojectivity is preserved under these conditions

The result broadens understanding of structure in Banach space theory

Abstract

We show that the Banach space $C(K,X)$ is subprojective if $K$ is scattered and $X$ is subprojective.

Equations

$$Q(f)=\bigl{(}x^{*}_{n}\bigl{(}f(t_{n})-f(t_{\infty})\bigr{)}\bigr{)}_{n\in{\mathbb{N}}},$$

$$J\bigl{(}(\alpha_{n})_{n\in{\mathbb{N}}}\bigr{)}=\sum_{n=1}^{\infty}\alpha_{n}(P_{\mu_{n+1}}-P_{\mu_{n}})(f_{n})=\sum_{n=1}^{\infty}\alpha_{n}f_{n}\chi_{K_{\mu_{n+1}}\setminus K_{\mu_{n}}};$$

$$\sum _{n=1}^{\infty }\Vert (P_{{\mu }_{n+1}}-P_{{\mu }_n})(f_n)-f_n\Vert$$

Full text

A Banach space $X$ is called subprojective if every closed infinite-dimensional subspace of $X$ contains an infinite-dimensional subspace that is complemented in $X$. Subprojective spaces were introduced by Whitley [14] to give conditions for the conjugate of an operator to be strictly singular. They are relevant in the study of the perturbation classes problem for semi-Fredholm operators, which has a negative answer in general [5] but a positive answer when one of the spaces is subprojective [6]. Recently Oikhberg and Spinu made a systematic study of subprojective spaces [10], widely increasing the family of known examples of spaces in this class.

Here we prove that $C(K,X)$ is subprojective whenever $K$ is a scattered compact and $X$ is a subprojective Banach space; a compact space $K$ is said to be scattered or dispersed if every non-empty closed subset of $K$ contains an isolated point. In the case where $K=[0,\lambda]$ for some ordinal $\lambda$, this result was previously obtained for countable $\lambda$ [10, Theorem 4.1] and later for arbitrary $\lambda$ [3, Theorem 2.9].

This generalises the scalar case, where it was already known that $C(K)$ is subprojective if and only if $K$ is scattered [9, Theorem 11] [11, Main Theorem], and fully characterises the subprojectivity of $C(K,X)$, as the subprojectivity of $C(K,X)$ implies that of $C(K)$ and $X$ [10, Proposition 2.1].

We will use standard notation. If $(x_{n})_{n\in{\mathbb{N}}}$ is a sequence in $X$, then $[x_{n}:n\in{\mathbb{N}}]$ will denote the closed linear span of $(x_{n})_{n\in{\mathbb{N}}}$ in $X$. Given a (bounded, linear) operator $T\colon X\,{\longrightarrow}\,Y$, $N(T)$ and $R(T)$ denote the kernel and the range of $T$, respectively. An operator $T\colon X\,{\longrightarrow}\,Y$ is strictly singular if $T|_{M}$ is an isomorphism only if $M$ is finite-dimensional.

The injective tensor product of $X$ and $Y$ is denoted by $X\mathop{\hat{\otimes}_{\varepsilon}}Y$. Note that $C(K,X)$ can be identified with $C(K)\mathop{\hat{\otimes}_{\varepsilon}}X$ [12, Section 3.2], and they will be used interchangeably in the sequel. For countable ordinals $\lambda$, $\mu$ it was proved in [4] that the projective tensor product $C([0,\lambda])\mathop{\hat{\otimes}_{\pi}}C([0,\mu])$ is subprojective.

Let $K$, $L$ be compact spaces and let $\varphi\colon K\,{\longrightarrow}\,L$ be a continuous function. It is well known that $\varphi$ induces an operator $C(L)\longrightarrow C(K)$ that maps each $f\in C(L)$ to $f\circ\varphi\in C(K)$; we will denote this operator by $\tilde{\varphi}\colon C(L)\,{\longrightarrow}\,C(K)$.

Note that this is a stronger notion for $Z$ than merely being subprojective, as it requires the subspace to be complemented in $X$ and not just in $Z$. Also, a space $X$ is subprojective if and only if each of its subspaces is subprojective with respect to $X$.

Otherwise, if $M\cap N(P)$ is finite-dimensional, we can assume that $M\cap N(P)=0$ by passing to a further subspace of $M$ if necessary. If $P(M)$ is closed, then $P|_{M}$ is an isomorphism and, if $N$ is an infinite-dimensional subspace of $M$ such that $R(P)=P(N)\oplus H$ for some closed subspace $H$, then $X=N\oplus P^{-1}(H)$.

We are left with the case where $M\cap N(P)=0$ and $P(M)$ is not closed. Take a normalised sequence $(x_{n})_{n\in{\mathbb{N}}}$ in $M$ such that $\|P(x_{n})\|<2^{-n}$ for every $n\in{\mathbb{N}}$. Since any weak cluster point of $(x_{n})_{n\in{\mathbb{N}}}$ must be in $M\cap N(P)=0$, by passing to a subsequence [1, Theorem 1.5.6] we can assume that $(x_{n})_{n\in{\mathbb{N}}}$ is a basic sequence and that there exists a bounded sequence $(x^{*}_{n})_{n\in{\mathbb{N}}}$ in $X^{*}$ such that $x^{*}_{i}(x_{j})=\delta_{ij}$ for every $i$, $j\in{\mathbb{N}}$ and $\sum_{n=1}^{\infty}\|x^{*}_{n}\|\|P(x_{n})\|<1$. Then $K(x)=\sum_{n=1}^{\infty}x^{*}_{n}(x)P(x_{n})$ defines an operator $K\colon X\,{\longrightarrow}\,X$ with $\|K\|<1$ that maps $K(x_{n})=P(x_{n})$ for every $n\in{\mathbb{N}}$, and then $I-K$ is an automorphism on $X$ such that $[(I{-}K)(x_{n}):n\in{\mathbb{N}}]=[(I{-}P)(x_{n}):n\in{\mathbb{N}}]\subseteq Z\cap N(P)$ and, if $N$ is an infinite-dimensional subspace of $[(I{-}K)(x_{n}):n\in{\mathbb{N}}]$ complemented in $X$, then $(I{-}K)^{-1}(N)\subseteq M$ and is still complemented in $X$. ∎

Here we are using the identification $C(K,X)\equiv C(K)\mathop{\hat{\otimes}_{\varepsilon}}X$ and the fact that, if $X$ and $Y$ are Banach spaces and $M$ is a closed subspace of $X$ and $N$ is a closed subspace of $Y$, then $M\mathop{\hat{\otimes}_{\varepsilon}}N$ can be seen as a subspace of $X\mathop{\hat{\otimes}_{\varepsilon}}Y$ [12, Comments after Proposition 3.2].

Let us then assume that the result is true for every continuous surjection $K\longrightarrow[0,\mu]$ with $\mu<\lambda$. Consider, for each ordinal $\mu<\lambda$, the set $K_{\mu}=\varphi^{-1}([0,\mu])\subseteq K$, which is both open and closed, and the operator $P_{\mu}\colon C(K,X)\,{\longrightarrow}\,C(K,X)$ given by $P_{\mu}(f)=f\chi_{K_{\mu}}$, which is a projection with range isometric to $C(K_{\mu},X)$; also note that P_{\mu}(f)\mathrel{\textstyle\mathop{\vbox to14.0pt{\hbox{\textstyle\longrightarrow}\vss}}\limits_{\mu}}f for every $f\in\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$.

We will first prove that $\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$ is subprojective with respect to $C(K,X)$. Let $M$ be a closed infinite-dimensional subspace of $\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$. If there exists some $\mu<\lambda$ for which $P_{\mu}|_{M}$ is not strictly singular, we can assume that $P_{\mu}|_{M}$ is an isomorphism by passing to a further subspace of $M$ if necessary and then $P_{\mu}(M)$, seen as a subspace of $C(K_{\mu},X)$, contains an infinite-dimensional subspace complemented in $C(K_{\mu},X)$ by the induction hypothesis (with $\varphi|_{K_{\mu}}\colon K_{\mu}\,{\longrightarrow}\,[0,\mu]$) so $M$ contains an infinite-dimensional subspace complemented in $C(K,X)$ [10, Proposition 2.3]. This must necessarily be the case if $\lambda$ is not a limit ordinal, as then there exists some ordinal $\mu$ such that $\lambda=\mu+1$, for which $P_{\mu}$ is the identity on $M$ because functions in $M$ vanish at $\varphi^{-1}(\lambda)$.

Otherwise, if $M\subseteq\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$ but $P_{\mu}$ is strictly singular for every $\mu<\lambda$, then $\lambda$ must be a limit ordinal by the previous sentence. Also, for every $\mu<\lambda$ and $\varepsilon>0$, there exists $f\in M$ such that $\|f\|=1$ and $\|P_{\mu}(f)\|<\varepsilon$, and then there is $\nu>\mu$ such that $\|P_{\nu}(f)-f\|<\varepsilon$. By induction, starting with an arbitrary $\mu_{1}<\lambda$, there exists a strictly increasing sequence of ordinals $\mu_{1}<\mu_{2}<\cdots<\lambda$ and a sequence $(f_{n})_{n\in{\mathbb{N}}}$ of normalised functions in $M$ such that $\|P_{\mu_{n}}(f_{n})\|<2^{-n}/32$ and $\|P_{\mu_{n+1}}(f_{n})-f_{n}\|<2^{-n}/32$ for every $n\in{\mathbb{N}}$.

Now, for each $n\in{\mathbb{N}}$, there exist $t_{n}\in K$ such that $\|f_{n}(t_{n})\|=1$ and then a normalised $x^{*}_{n}\in X^{*}$ such that $x^{*}_{n}(f_{n}(t_{n}))=1$; note that $t_{n}\in K_{\mu_{n+1}}\setminus K_{\mu_{n}}$ because $f_{n}$ cannot attain its norm outside of $K_{\mu_{n+1}}\setminus K_{\mu_{n}}$. As $K$ is sequentially compact, by passing to a subsequence, we may assume that $(t_{n})_{n\in{\mathbb{N}}}$ converges to some $t_{\infty}\in K$. Let $Q\colon C(K,X)\,{\longrightarrow}\,c_{0}$ and $J\colon c_{0}\,{\longrightarrow}\,C(K,X)$ be the operators defined as

[TABLE]

[TABLE]

then $Q$ and $J$ are well defined, $\|Q\|=2$ and $J$ is an isometry into $C(K,X)$ (into $\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$, actually), and $QJ$ is the identity on $c_{0}$, so $JQ$ is a projection in $C(K,X)$ with range isometric to $c_{0}$. And, since

[TABLE]

it follows that $[f_{n}:n\in{\mathbb{N}}]$ is also isomorphic to $c_{0}$ and complemented in $C(K,X)$ [8, Proposition 1.a.9].

Finally, let $Z=\tilde{\varphi}(C([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$, which is a closed subspace of $C(K,X)$, fix $t_{0}\in\varphi^{-1}(\lambda)\subseteq K$ and consider the natural projection $P\colon C(K,X)\,{\longrightarrow}\,C(K,X)$ defined as $P(f)=1\otimes f(t_{0})\in C(K)\mathop{\hat{\otimes}_{\varepsilon}}X$; then $R(P)$ is isomorphic to $X$, which is subprojective by hypothesis, and $R(P)\subseteq Z$. Moreover, given $f\in Z\cap N(P)$, it holds that $f(t_{0})=0$ and $f\in\tilde{\varphi}(C([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$, so $f$ must be constant over $\varphi^{-1}(\lambda)$ and then $f|_{\varphi^{-1}(\lambda)}=0$. This means that $Z\cap N(P)=\tilde{\varphi}(C_{0}([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$, which is subprojective with respect to $C(K,X)$. Applying Proposition 1, $Z=\tilde{\varphi}(C([0,\lambda]))\mathop{\hat{\otimes}_{\varepsilon}}X$ is subprojective with respect to $C(K,X)$. ∎