1
$\begingroup$

Here is a structured question of my assignment:

Given the operator $K:C[0,1]\mapsto C[0,1]$ defined by$$(Kf)(t)=\int_{0}^{1}k(t,s)f(s)ds.$$

(1) Prove that $K$ is bounded linear operator and $$\left \| K \right \|\leq \max_{t\in [0,1]}\int_{0}^{1}\left | k(t,s) \right |ds.$$

(2) Denote $$M = \max_{t\in [0,1]}\int_{0}^{1}\left | k(t,s) \right |ds,$$ which is the RHS of the inequality (1).

Prove that there exists $t_0 \in [0,1]$ such that $$M=\int_{0}^{1} \left | k(t_0,s) \right |ds.$$

(3) Prove that $$F(x)=\int_{0}^{1}k(t_0,s)x(s)ds$$ is a bounded linear functional.

(4) Prove that $\left \| F \right \|=M.$

(5) Prove that $$\left \| K \right \|= \max_{t\in [0,1]}\int_{0}^{1}\left | k(t,s) \right |ds.$$


Here is my idea:

(1) I can prove that $K$ is linear but is it the way to use Cauchy Schwarz inequality to prove $K$ is bounded? If yes, would you please to show me the detail if sup-norm is used?

(2) I think the existence of $t_0$ is ensured by Extreme Value Theorem. Is it correct?

(3) Similar to (1), I manage to prove linear but I have no idea to prove that it is bounded. Show I would like to have some steps so that I can learn the trick behind.

(4) I can manage to prove=)

(5) I do not know how to combine the previous results to finish this part.

Thanks for reading and any helps would be appreciated.

$\endgroup$
2
$\begingroup$

(1) I would assume you are looking at $(C[0,1],\|\cdot\|_\infty)$, and $k\in C([0,1]\times[0,1])$, in which case we have:

$\|K(f)\|_\infty=\|\int_0^1k(t,s)f(s)ds\|_\infty\le\|f(s)\|_\infty\|\int_0^1k(t,s)ds\|_\infty=D\|f\|_\infty $

Thus $K$ is bounded and from above we have: $\|K\|=\sup_{0\ne f\in C[0,1]}\frac{\|K(f)\|_\infty}{\|f\|_\infty}\le\|\int_0^1k(t,s)ds\|_\infty=sup_{t\in[0,1]}\int_0^1k|(t,s)|ds$

(2) by assumption $k$ is continuous on a compact set, and thus must attain its minimum and maximum on $[0,1]\times[0,1]$, so there is such a $t_0\in[0,1]$ (after integrating, it becomes a one variable function).

(3) $\|F(x)\|_\infty=\|\int_0^1k(t_0,s)x(s)ds\|_\infty\le M\|x\|_\infty\int_0^11ds=M\|x\|_\infty$, so $F$ is bounded.

(5) we already know that $\|K\|\le\max_{t\in[0,1]}\int_0^1|k(t,s)|ds$ and $1\in C[0,1]$ where $\|1\|_\infty=1$, thus:

$\|K\|=\sup_{\|x\|_\infty=1}\|K(x)\|_\infty\ge\|K(1)\|_\infty=\max_{t\in[0,1]}\int_0^1|k(t,s)|ds.$

So we have $\|K\|\le\max_{t\in[0,1]}\int_0^1|k(t,s)|ds,$

and $\|K\|\ge\max_{t\in[0,1]}\int_0^1|k(t,s)|ds$

$\Rightarrow \|K\|=\max_{t\in[0,1]}\int_0^1|k(t,s)|ds$

$\endgroup$
  • $\begingroup$ Thank you so much. I understand your idea of the works. $\endgroup$ – nam May 14 '14 at 16:33

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.