emphsisingly, I just need a hint not a whole solution please.
Problem: Consider the operator $T:C([0,1])\to C([0,1])$defined by $$ (Tf)(t):=\int_{0}^{1}k(s,t)f(s)ds $$ where $k:[0,1]^{2}\to \mathbb{R}$ satisfies the following
- for all $t\in [0,1],$ the function $k_{t}(s)=k(t,s)$ is integrable in $s$: $$ \int_{0}^{1}k(s,t)ds<\infty, $$
- the function $ t\mapsto k_{t}\in L^{1}([0,1]) $ is continuous.
Show that $T$ is compact.
You should try to use the Arzela-Ascoli theorem to show that the image under $T$ of the closed unit ball of $C([0,1])$ is relatively compact. To this end, observe that for $f\in C([0,1])$ with $\|f\|\leq1$, we have $$|Tf(t)|\leq\int_0^1|k(s,t)|ds$$ for $t\in[0,1]$ (to help establish a uniform bound), and $$|Tf(t_1)-Tf(t_2)|\leq\int_0^1|k(s,t_1)-k(s,t_2)|ds$$ for $t_1,t_2\in[0,1]$ (to help establish equicontinuity).