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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.

Phira
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nam
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1 Answers1

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

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

Ellya
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