norm of integral operator in $C([0,1])$ If we define on $C([0,1])$ the operator $$ Tf(x) = \int_{0}^{1} K(t,s) f(s) ds$$ where $K$ is a continous function on two variables. I want to show that:
$1)$ $||T|| = \displaystyle\max_{t} {\int_{0}^{1} |K(t,s)| ds} $
$2)$ When $K(t,s) = \displaystyle\min(t,s)$, to prove that $T$ is compact
Thanks.
 A: Here is a sketch of the first part.
For each $f\in C[0,1]$, 
\begin{eqnarray*}
\|Tf\| &=& \max_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)f(s)ds\right|\\
       &\leq& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)f(s)\right|ds\\
       &\leq& \|f\|_{\infty}\max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\
\end{eqnarray*}
Therefore $\|T\|$ is at most $\max\limits_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds$.  
On the other hand, choose $t_{0}\in [0,1]$ such that $\left|K(t_{0},s)\right| = \max\limits_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds$.  Take $g = \text{sign}[K(t_{0},\cdot)]\in L^{1}[0,1]$, so that $K(t_{0},s)g(s) = \left|K(t_{0},s)\right|$ for every $s\in [0,1]$.  
Choose a sequence $(g_{n})_{n=1}^{\infty}$ in $C[0,1]$ which converges pointwise to $g$ almost everywhere.  Choose $f\in C[0,1]$ such that $\|f - g\|_{\infty} < \epsilon$ and then compute
\begin{eqnarray*}
\|Tg_{n}\|_{\infty} &=& \max\limits_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)g_{n}(s)ds\right|\\
&\to& \max\limits_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)g(s)ds\right|\\
&\geq& \left|\int_{0}^{1}K(t_{0},s)g(s)ds\right|\\
&=& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\
\end{eqnarray*}
I'll leave the details of the last few steps to you.
