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The question is as follows:

Show that the linear operator $(Ax)(t)= \int_{0}^{1} \frac{e^{-\frac{t}{s}}}{\sqrt{s}} x(s) ds$ for $A : C[0,1] \to C[0,1] $ is compact.

$\textbf{About the question:}$

This question is the same as linked-to result.

I actualy do not know how to deal with such kind of questions!

Can you please let me now that how can I think about it? What are the conditions for them to be compact?

Thanks!

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Firstly, since $C[0,1]$ is infinite-dimensional, we need a norm to begin the study. Here I use $\|.\|_{\infty}$.

A linear operator is compact iff the image of the closed unit ball is relatively compact. We note $B$ the closed unit ball of $C[0,1]$. $B=\{f\in C[0,1],\ \|f\|_{\infty}\le1\}$.

From Arzelà–Ascoli theorem, $A(B)$ is relatively compact iff $A(B)$ is equicontinuous in $[0,1]$ and $\{f(t),\ f\in A(B)\}$ is relatively compact. Since $\{f(t),\ f\in A(B)\}\subseteq\Bbb R$, the set is relatively compact iff it's bounded.

  • Prove that $\{f(t),\ f\in A(B)\}$ is bounded:

Let $f \in B$, thus $\|f\|_{\infty}\le1$, and we have that \begin{align} \forall t\in[0,1],\ |A(f)(t)|&=|\int_{0}^{1} \frac{e^{-t/s}}{\sqrt{s}} f(s) ds|\\ &\le\int_{0}^{1} |\frac{e^{-t/s}}{\sqrt{s}} f(s)| ds\\ &\le\int_{0}^{1} \frac{|f(s)|}{\sqrt{s}} ds\\ &\le\int_{0}^{1} \frac{1}{\sqrt{s}}\ ds\cdot\|f\|_{\infty}\\ &\le2 \end{align}

  • Prove that $A(B)$ is equicontinuous:

Let $t \in [0,1]$, $\epsilon>0$. Since $g:\ (t,s)\mapsto\frac{e^{-t/s}}{\sqrt{s}}$ is a continuous function in $[0,1]^2$, there exists $\delta>0$ such that $\forall t'\in[t-\delta,t+\delta],\ \forall s\in[0,1],\ |g(t',s)-g(t,s)|\le\epsilon$.

Let $f \in B$, for all $t'\in[t-\delta,t+\delta]$, we have that \begin{align} |A(f)(t')-A(f)(t)|&=|\int_{0}^{1} \frac{e^{-t'/s}-e^{-t/s}}{\sqrt{s}} f(s) ds|\\ &\le\int_{0}^{1} |g(t',s)-g(t,s)|\cdot|f(s)| ds\\ &\le\epsilon\cdot\|f\|_{\infty}\\ &\le\epsilon \end{align}

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  • $\begingroup$ Many thanks! Very clever answer! $\endgroup$ – user510716 Dec 13 '17 at 5:03

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