I'm reading through C* algebras by Murphy, and I'm covering compact operators at the moment, and the following was stated:
"Let $X = C([0,1])$, and define $u \in B(X)$ by $u(f)(s) = \int_{0}^1 k(s,t)f(t) \: dt$ for $k \in C([0,1]^2)$. If we restrict $u$ to the closed unit ball of $X$, call it $S = \{f \in C \: | \: ||f||_{\infty} \leq 1\}$, then $u$ is pointwise bounded.
I.e, $\sup_{f \in S}|u(f)(s)| < \infty$ since $|u(f)(s)| \leq \int_{0}^1 |k(s,t)f(t)| \: dt \leq ||k||_{\infty} ||f||_{\infty} \leq ||k||_{\infty} < \infty$ since $f \in S$."
My question is, how is this not a uniform bound?
Since $f \in S$ and $s \in [0,1]$ was arbitrary, then from the above inequality, we should be able to conclude that $$||u(f)(s)||_{\infty} \leq ||k||_{\infty} \implies \sup_{f \in S} ||u(f)(s)||_{\infty} \leq ||k||_{\infty} < \infty$$ right? They go on to say that pointwise boundedness and equicontinuity implies total boundedess of the image $u(S)$ from the Arzela Ascoli theorem, which makes little sense to me (an explanation of this would also be helpful).
To me it seems that the image u(S) is uniformly equicontinuous and uniformly bounded, implying that $\overline{u(S)}$ is compact in $X$ from Arzela Ascoli, implying that $u$ is compact.
Murphy applies the following result, which can be found on the Wikipedia page with appropriate references:
Theorem: Let $X$ be compact Hausdorff. If $F \subseteq C(X)$ is pointwise bounded and equicontinuous, then $F$ is relatively compact.
Murphy doesn't claim that this isn't a uniform bound. To apply the version of Arzela-Ascoli you need, pointwise boundedness is enough, so why bother checking that something stronger holds?
[You are right that it is a uniform bound, but it doesn't matter here.]