A question regarding an integral operator this was a question given in a list of many given by our teacher. It states:

If $k : [0, 1]  \times  [0, 1] \rightarrow K$ is continuous. Then we have to prove that the operator
  $T : L^1([0, 1]) \rightarrow C^0([0, 1])$ defined by
  $$Tf(x) :=\int_{ [0,1]}k(x, y)f(y) dy; $$
  for $f \in L^1([0, 1]), x \in [0, 1]$ is well-defined and compact.

Our teacher gave us a hint that we need to use the Arzela-Ascoli theorem and that we should observe that uniform continuity of $k$ is useful.
Could someone please help me out? Thank you.
 A: As $k$ is continuous there exist $\delta >0$ and $M>0$ such that 
\begin{equation}
|x-z| < \delta \Rightarrow |k(x,y)-k(z,y)| < \varepsilon
\end{equation}
and $|k(x,y)|\le M \ \forall (x,y) \in [0,1] \times [0,1]$.
Then
\begin{equation}
 1 |x-z|<\delta \Rightarrow |Tf(x) - Tf(z)| < \varepsilon \|f\|_{L^{1}}.
\end{equation} 
Hence $T$ is well-defined. Now add 
\begin{equation}
|Tf(x)|_{\infty} \le M \|f\|_{L^{1}}
\end{equation}
with 1 and we have by ascoli-Árzela theorem that $\left \{Tf_n\right \}_{f \in L^{1}}$ is equicontuinuous is $\|f_n\|_{L^{1}} \le C$. Hence we have $T(f_n)\rightarrow g$ in $C^0$ with $g \in C^0$.
A: To see that $T$ is well-defined, note that since $[0,1]\times [0,1]$ is compact and $k$ is continuous, $k$ must be uniformly continuous. Thus there exists some $\delta$ such that $k$ varies by less than $1$ over open balls of radius $\delta$. Since $[0,1]\times [0,1]$ is a finite union of open balls of radius $\delta$, this makes $k$ bounded, say by $C$. Clearly a bounded function times an $L^1$ function is $L^1$, hence the integral is well-defined.
To see that $T$ is compact, you need to show that the image of the unit ball in $L^1([0,1])$ has compact closure in $C^0([0,1])$. The unit ball in $L^1([0,1])$ is the set of functions $f\in L^1([0,1])$ such that $\int_{[0,1]} |f(y)|dy<1$. We want to verify that the image is uniformly bounded and equicontinuous. For uniformly bounded, note that 
$$\left|\int_{[0,1]}k(x,y)f(y)dy\right|\leq \int_{[0,1]}|k(x,y)||f(y)|dy\leq C\int_{[0,1]}|f(y)|dy<C.$$
For equicontinuity note that since $k$ is uniformly continuous for any $\epsilon>0$ we have some $\delta>0$ such that $|x-x'|<\delta$ implies $|k(x,y)-k(x',y)|<\epsilon$. Thus if $|x-x'|<\delta$ we have
$$\begin{align}
\left|\int_{[0,1]}k(x,y)f(y)dy-\int_{[0,1]}k(x',y)f(y)dy\right| &= \left|\int_{[0,1]}(k(x,y)-k(x',y))f(y)dy\right|\\
&\leq \int_{[0,1]}|k(x,y)-k(x',y)||f(y)|dy\\
&\leq \int_{[0,1]}\epsilon|f(y)|dy<\epsilon\\
\end{align}$$
hence the image of the unit ball is equicontinuous as well. Thus by Arzela-Ascoli it has compact closure, so $T$ is a compact operator.
