# to prove an integral operator with a kernel is compact

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 might try constructing a proof which parallels that given inhttps://math.stackexchange.com/questions/2398606/cauchy-www-and-operator-theory-problem. Aug 21, 2017 at 0:19
• Yes, that's right. Maybe, It would be OK the same way proof and try it to this question. But I was in doubt. Or it maybe better to write the same method to this question and ask about its correctness here. Aug 21, 2017 at 0:25
• It's more or less what Aweygan proposed in his answer! Aug 21, 2017 at 0:33

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).