Is $C_b(Q,E)$ linearly isometrically isomorphic to $C(\beta Q,E)$ where $\beta Q$ is the Stone–Čech 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 $Q$ 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 $F\circ e = f$?

 A: 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.
A: 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.
