Compact operator by proving Ascoli-Arzelà I need to prove that this operator satisfies Ascoli-Arzelà's hypothesis.

$T: C^0[0,1] \rightarrow C^0[0,1] $, defined $Tu(x)=\int_0^x a(x,t) u(t)dt$, where $a(x,t)=C^0 ([0,1] \times [0,1])$.

Equiboundedness is okay, I need to prove equicontinuity: taken $\{ u_n \} \subset C^0[0,1], ||u_n|| \leq M$
So I take $|Tu_n(x)-Tu_n(y)| \leq \ldots \leq L|x-y| + C$ , by bounding several times. Is it enouth to conclude?
 A: You need to prove that for every $\varepsilon > 0$ and $x \in [0,1]$ there is a $\delta > 0$ such that $|x-y| < \delta$ implies that $|Tu_n(x) - Tu_n(y)| < \varepsilon$. The kind of upper bound you exhibit is insufficient to do this (assuming from context that $C$ is a positive constant) since for $\varepsilon = C/2$ your upper bound is always bigger than $\varepsilon$.
Instead, assume without loss of generality that $y > x$ and bound
\begin{align*}
|Tu_n(x) - Tu_n(y)| &= \bigg| \int_0^x (a(x,t) - a(y,t)) u_n(t) dt + \int_x^y a(y,t) u_n(t) dt \bigg| \\
& \leq \int_0^x |a(x,t) - a(y,t)| |u_n(t)| dt + \int_x^y |a(y,t) u_n(t)| dt 
\\& \leq \int_0^x |a(x,t) - a(y,t)| M dt  + M \|a\|_\infty |y-x|
\end{align*}
Now $a$ is continuous on a compact set and hence uniformly continuous so there is a $\delta < (2M \|a\|_\infty)^{-1} \varepsilon$ such that $|x-y| < \delta$ implies that $|a(x,t) - a(y,t)| \leq M^{-1} \frac{\varepsilon}{2}$ for every $t$. So for $|x-y| < \delta$,
\begin{align}
|Tu_n(x) - Tu_n(y)| < \int_0^x \frac{\varepsilon}{2} dt + \frac{\varepsilon}{2} \leq \varepsilon
\end{align}
as desired.
A: If you have a nice strategy to induce (uniformly) bounded $L$ and sufficiently small $C>0$ (depending on $\epsilon$, of course), then it will work. But I suspect that it cannot be implemented in most cases. You don't have to get a Lipschitz-like, very concrete bound to show equicontinuity. Here's an approach.
Let $M$ be a constant such that $\sup_n \|u_n\|\le M$ and $|\alpha|\le M$, and define$$
m(\delta)=\sup\{|\alpha(t,x)-\alpha(t',x')|:|(t-t',x-x')|\le \delta\}.
$$ By continuity of $\alpha$, we have $\lim\limits_{\delta\to 0^+}m(\delta)=0.$ Let $\epsilon>0$ be given and we can find $\delta>0$ such that $m(\delta)\le \epsilon/M$. Note that we can choose $\delta$ so that $\delta<\epsilon/M^2$. We find that for all $0\le x- y\le \delta$ and $n\ge 1$,
$$\begin{eqnarray}
|Tu_n(x)-Tu_n(y)|&\le&\int_0^y |\alpha(x,t)-\alpha(y,t)||u_n(t)|\mathrm{d}t+\int_y^x |\alpha(x,t)||u_n(t)|\mathrm{d}t\\&\le& Mm(\delta)+M^2 \delta <2\epsilon.
\end{eqnarray}$$ This verifies equicontinuity of $(Tu_n)$ as desired.
