norm of integral operator in $C([0,1])$

Solution 1:

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.