Assume that a Banach space of functions on $[0,1]$, $\mathcal{B}([0,1])$, with norm $||\cdot||_{B}$ can be embedded in the space of continuous functions, $\mathcal{C}([0,1])$, with the sup-norm $||\cdot||_{\infty}$. In other words, $\mathcal{B}([0,1]) \subset \mathcal{C}([0,1])$ and there exists a $C>0$ such that
$$ ||f||_{\infty} \leq C ||f||_{\mathcal{B}}, \quad f \in \mathcal{B}([0,1]). $$
My question is whether in that case the unit ball $\{f \in \mathcal{B}:||f||_{B} \leq 1 \}$ is necessarily compact in the topology of the sup-norm. For some of these spaces the Arzela-Ascoli theorem can be applied to show compactness but I am not sure if there is a more general result in that direction.
Thank you.
Best Answer
$C[0,1]$ is a "universal space" in the class of separble Banach spaces. This means any separable Banach space can be isometrically embedded in $C[0,1]$. (Ref. Geometric Functional Analysis by Holmes]. Take any infinite dimensional Banach space, embed it in $C[0,1]$ isometrically and you get an example where the ball is not compact in the sup norm.