5
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.