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



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.

  • $\begingroup$ We need to have $||g_{n}||_{\infty} = 1$, clearly $||g||_{\infty} = 1$ so I wanted to know if in the construction of the sequence, we can take it to satisfy this condition. $\endgroup$ – kagami Apr 19 '14 at 8:21
  • $\begingroup$ Good observation. You can choose $g_{n}$ to be an increasing sequence which converges pointwise to $g$ from below. This will ensure each $g_{n}$ is no larger than $1$. The reason this can be done is because $g$ is measurable. $\endgroup$ – roo Apr 19 '14 at 18:51
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    $\begingroup$ @roo How do we know that $f$ with the property that $\|f-g\|_\infty < \epsilon$ exists? $\endgroup$ – nan Apr 21 '17 at 4:38
  • $\begingroup$ Note: Years later, reading this, I realize that my answer makes the assumption that $K$ is defined on $[0,1]\times[0,1]$. I don't think this harms anything but I'll continue under that assumption. Since $sign(\cdot)$ is piecewise continuous and $K$ is continous, their composition $g$ is piecewise continuous and is thus measurable. Therefore there exists a sequence $(g_{n})_{n=1}^{\infty}$ in $C[0,1]$ which converges pointwise almost everywhere to $g$ on $[0,1]$. $\endgroup$ – roo Apr 21 '17 at 5:19
  • $\begingroup$ Do a little extra tweaking to ensure that $|g_{n}(t)|\leq 1 = |g(t)|$ for all $t\in[0,1]$, which is what I meant when I said from below. $\endgroup$ – roo Apr 21 '17 at 8:30

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