# Is $C_b(Q,E)$ linearly isometric isomorphic to $C(\beta Q,E)$ where $\beta Q$ is the Stone-Cech compactification of $Q?$

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 $Q$ while $C_b(Q,E)$ be the collection of all $E$-valued bounded continuous functions on $Q.$

Clearly $Q$ is completely regular. Let $\beta Q$ to be the Stone-Cech 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 isometric isomorphic to $C(\beta Q,E)?$

It is known that $C_b(Q)$ is linearly isometric 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-Cech compactification for 'Banach space-valued' mapping, that is,

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

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.
• 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. – Jochen Wengenroth May 8 '18 at 14:18
• 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. – Idonknow May 8 '18 at 15:43
• @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. – Idonknow May 8 '18 at 23:06