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Let $Q$ be a locally compact Hausdorff space and $E$ be a Banach space. Let $C(Q)$ be the collection of all real-valued continuous functions on $Q$ and $C_b(Q,E)$ be the collection of all $E$-valued bounded continuous functions on $Q$.

Clearly $Q$ is completely regular. Let $\beta Q$ to be the Stone–Čech compactification, where $\beta Q = \overline{e(Q)}$ and $e:Q\to \prod_{f\in C_b(Q),[0,1])}[0,1]$ is an embedding.

Question: Is $C_b(Q,E)$ linearly isometrically isomorphic to $C(\beta Q,E)?$

It is known that $C_b(Q)$ is linearly isometrically isomorphic to $C(\beta Q)$, thanks to the following proposition. The following extension theorem is taken from Carothers's A Short Course in Banach Space Theory, Chapter 15 Theorem 15.1.

Theorem 15.1: Every $f\in C_b(Q)$ extends to a continuous function $F:\beta Q\to\mathbb{R}$ such that $F\circ e = f$.

The theorem provides a linear isometry from $C_b(X)$ onto $C(\beta Q)$ by the mapping $F\mapsto F\circ e$.

However, I do not know whether there exists a Stone—Čech compactification for ‘Banach space-valued’ mappings, that is,

Question: For every $f\in C_b(Q,E)$, does there exist a continuous extension $F:\beta Q\to E$ such that $F\circ e = f$?

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2 Answers 2

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You probably mean for the extension $F$ to be continuous. Since its image is then compact, this can only happen if the image of $f$ is relatively compact. And this condition is sufficient by the universal property of the compactification.

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  • $\begingroup$ Yes, you are right. Edited. $\endgroup$
    – Idonknow
    Commented May 8, 2018 at 13:49
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    $\begingroup$ Parschallen's answer means that the canonical map $\rho: C(\beta Q,E)\to C_b(Q,E)$, $f\mapsto f|_Q$ is hardly ever surjective: Continuous functions $\beta Q\to E$ have compact range but bounded continuous functions $f:Q\to E$ need not have compact range. $\endgroup$
    – Jochen Wengenroth
    Commented May 8, 2018 at 14:18
  • $\begingroup$ Then why would $C_b(Q)$ be linearly isometric isomorphic to $C(\beta Q),$ as the functions in latter set have compact range while the former may not. $\endgroup$
    – Idonknow
    Commented May 8, 2018 at 15:43
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    $\begingroup$ In the one dimensional case, boundedness and relative compactness are the same thing. $\endgroup$
    – Parschallen
    Commented May 8, 2018 at 16:13
  • $\begingroup$ @Parschallen Do you mean finite dimensional case? Because relative compactness always implies boundedness (as $A\subseteq \overline{A}$ and $\overline{A}$ is bounded). In finite dimensional case, bounded and totally bounded are equivalent notions and totally boundedness implies relative compact in a complete metric space. $\endgroup$
    – Idonknow
    Commented May 8, 2018 at 23:06
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Well, the point is that even if we know, after Jochen's remark, that the natural map $\rho:C(\beta Q,E)\to C_b(Q,E)$ fails to be onto it remains to see that these spaces are not isometric (under some "esoteric" isometry). This follows (in most cases) from classical stuff on tensor products and a result by Cembranos and Mendoza. Let us consider the case in which $Q=\mathbb N$ is infinite discrete and $E$ is an infinite dimensional, separable Hilbert space. The space $C_b(\mathbb N, E)$ consists of all bounded sequences in $E$ and, being isometric to the dual of $\ell_1(\mathbb N,E)$, cannot contain a complemented copy of $c_0$. The space $C(\beta\mathbb N, E)$ is isometric to the injective tensor product of $C(\beta N)=\ell_\infty$ and $E$ and (this is Cembranos–Mendoza's result) so it contains a complemented copy of $c_0$. Hence $C(\beta\mathbb N, E)$ and $C_b(\mathbb N, E)$ cannot be linearly homeomorphic, let alone isometric.

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