To say it directly: I never worked with the sentence of Arzelà Ascoli, therefore this thread.
I heard that one can often use it to show the compactness of operators.
For example let $T\colon C([0,1])\to C([0,1])$ be a linear operator, defined by
$$
(Tf)(x)=\int\limits_0^1 k(x,y)f(y)\, dy, k\colon [0,1]\times [0,1]\to\mathbb{R} \mbox{ continuous}.
$$
Can anybody explain me how I can here use Arzelà Ascoli to show the compactness of T?
It would be great if you could explain it slowly and not to short and complicated. Remember, that I want to learn how one can apply this sentence to show compactness and I never saw it before.
Thank you!
Best Answer
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.