First why the problem appeard.
I want to show that the linear and continuous operator $T:C([0,1])\rightarrow C([0,1])$ , $ (Tf)(t)=\int_{[0,1]}k(t,s)f(s)ds$ where $k:[0,1]^2\rightarrow\mathbb R$ is continuous, is compact. i.e that the set $T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\} \subset C([0,1])$ is relative compact.
To show this, I wanted to apply that $H\in C([0,1])$ is relative compact iff it is equicontinuous and bounded.
So I am trying to show that $T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\} \subset C([0,1])$ is equicontinuous. i.e $\forall \epsilon>0$ and $\forall y \in [0,1]$ $\exists(y-\epsilon,y+\epsilon)\subset [0,1] $ s.t $\forall c\in(y-\epsilon,y+\epsilon)$ and $ \forall f\in T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\}$ we have that $|f(c)-f(y)|\leq\epsilon$.
Fix some arbitrary $\epsilon$ and some $y\in[0,1]$
Then $\forall f\in T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\}$ and for some arbitrary $c\in(y-\epsilon,y+\epsilon)$ we have: $$ |\int_{[0,1]}k(c,s)f(s)-k(y,s)(f(s)ds|\leq\int_{[0,1]}|f(s)((k(c,s)-k(y,s))|ds\\\leq\max{_{x \in [0,1]}}|f(x)|\int_{[0,1]}|k(c,s)-k(y,s)|ds=\int_{[0,1]}|k(c,s)-k(y,s)|ds\leq....|c-y|$$
And now i dont know how to bound the last term to obtain that everything is smaller then some constant times epsilon.
Can someone help? Or is there maybe an easier way to prove that an operator is compact?
Thanks!
Use that $k$ is uniformly continuous. Given $\epsilon>0$ there exists $\delta>0$ such that $$ |c-y|<\delta\implies |k(c,s)-k(y,s)|<\epsilon\quad\forall s\in[\,0,1\,]. $$