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First why the problem appeard.

I want to show that the linear and continuous operator $T:C([0,1])\rightarrow C([0,1])$ , $ (Tf)(t)=\int_{[0,1]}k(t,s)f(s)ds$ where $k:[0,1]^2\rightarrow\mathbb R$ is continuous, is compact. i.e that the set $T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\} \subset C([0,1])$ is relative compact.

To show this, I wanted to apply that $H\in C([0,1])$ is relative compact iff it is equicontinuous and bounded.

So I am trying to show that $T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\} \subset C([0,1])$ is equicontinuous. i.e $\forall \epsilon>0$ and $\forall y \in [0,1]$ $\exists(y-\epsilon,y+\epsilon)\subset [0,1] $ s.t $\forall c\in(y-\epsilon,y+\epsilon)$ and $ \forall f\in T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\}$ we have that $|f(c)-f(y)|\leq\epsilon$.

Fix some arbitrary $\epsilon$ and some $y\in[0,1]$

Then $\forall f\in T (\{f \in C([0,1]) : \max_{x\in[0,1]}|f(x)|\leq 1\}$ and for some arbitrary $c\in(y-\epsilon,y+\epsilon)$ we have: $$ |\int_{[0,1]}k(c,s)f(s)-k(y,s)(f(s)ds|\leq\int_{[0,1]}|f(s)((k(c,s)-k(y,s))|ds\\\leq\max{_{x \in [0,1]}}|f(x)|\int_{[0,1]}|k(c,s)-k(y,s)|ds=\int_{[0,1]}|k(c,s)-k(y,s)|ds\leq....|c-y|$$

And now i dont know how to bound the last term to obtain that everything is smaller then some constant times epsilon.

Can someone help? Or is there maybe an easier way to prove that an operator is compact?

Thanks!

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  • $\begingroup$ You lost an absolute value on $k(c, s)-k(y,s)$ in passing from the second to the third inequality. Anyway, to conclude you need to exploit the only assumption that is left, namely, continuity of $k$. $\endgroup$ – Giuseppe Negro Mar 15 '14 at 22:23
  • $\begingroup$ @GiuseppeNegro Thank you, I just corrected the post. The only way I know to bound $k$ would be by taking the maximum since by continuity of $k$ and compactness of $[0,1]$ it exist. But then $|c-y|$ will disappear and I can not use that it is smaller then epsilon..? $\endgroup$ – Thorben Mar 15 '14 at 22:32
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Use that $k$ is uniformly continuous. Given $\epsilon>0$ there exists $\delta>0$ such that $$ |c-y|<\delta\implies |k(c,s)-k(y,s)|<\epsilon\quad\forall s\in[\,0,1\,]. $$

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  • $\begingroup$ Thank you! That answer solves the probel. I forgot to use continuity on compact sets implies uniformly continuity. $\endgroup$ – Thorben Mar 15 '14 at 22:43

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