TITLE: to prove an integral operator with a kernel is compact
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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).