Show compactness with Arzelà–Ascoli

We have to show that for every sequence $f_n$ in the unit ball of $C([0,1])$, there exists a subsequence of $T(f_n)$ which converges.

Let $\|f\|$ denote the uniform norm on $C([0,1])$.

So let $f_n$ be such a sequence.

1) Let $M$ be an upper bound for the continuous function $|k|$ on the compact $[0,1]^2$. Now oObserve that $$ |T(f_n(x))|\leq\int_0^1|k(x,y)||f_n(y)|dy\leq M\|f_n\| \leq M $$ for all $n$ and all $x\in[0,1]$. So $$ \|T(f_n)\|\leq M $$ for all $n$. THis proves that the $T(f_n)$ are uniformly bounded.

2) Now $$ |T(f_n(x))-T(f_n(x_0))|\leq \int_0^1|k(x,y)-k(x_0,y)||f_n(y)|dy\leq \int_0^1|k(x,y)-k(x_0,y)|dy $$ for all $n$ and all $x,x_0$. Since $k$ is continuous on the compact domain of integration, it is uniformly continuous there. Thus the right-end side tends to $0$ when $x$ tends to $x_0$. This proves that the $T(f_n)$ are equicontinuous.

By Arzela-Ascoli, the exists a subsequence of $T(f_n)$ which converges.

A (linear) operator is called compact if and only if the image of the unit ball is relatively compact, i.e. $T(B(0,1))$ is relatively compact where

$$B(0,1) := \{f \in C[0,1]; \|f\|_{\infty} < 1\}$$

By Arzelà-Ascoli theorem a set $A \subseteq C[0,1]$ is relatively compact iff

1. $\forall x \in [0,1]: \sup_{f \in A} |f(x)|<\infty$
2. $A$ is equicontinuous, i.e. for all $x_0 \in [0,1]$, $\varepsilon>0$ there exists $\delta>0$ such that for all $x \in [0,1]$, $|x_0-x| \leq \delta$ and for all $g \in A$: $$|g(x)-g(x_0)| \leq \varepsilon$$

You have to prove these two properties for $A:=T(B(0,1))$, i.e.

1. For all $x \in [0,1]$: $$\sup_{\substack{f \in C[0,1] \\ \|f\|_{\infty} < 1}} \left| \int_0^1 k(x,y) \cdot f(y) \, dy \right| < \infty$$
2. for all $x_0 \in [0,1]$, $\varepsilon>0$ there exists $\delta>0$ such that for all $x \in [0,1]$, $|x_0-x| \leq \delta$ and for all $f \in B(0,1)$: $$\left| \int_0^1 (k(x,y)-k(x_0,y)) \cdot f(y) \, dy \right| < \varepsilon$$ Hint Use the continuity of $k$.