0
$\begingroup$

Let $(f_{n})_{n}\subset C([0,1])$ such that $\vert\vert f_{n} \vert \vert_{\infty}\leq1$ for all $n \in \mathbb N$. Furthermore let $k \in C([0,1]^{2})$

define $T: (C([0,1]),\vert\vert \cdot \vert \vert_{\infty})\to (C([0,1]),\vert\vert \cdot \vert \vert_{\infty})$ where $Tf(x):=\int_{0}^{x} k(x,y)f(y)dy$

I have shown $T$ is a bounded linear operator, I now want to show that $(Tf_{n})_{n}$ is equicontinuous. I know that $k$ has to be uniformly continuous as it is continuous on the compact set $[0,1]^{2}$

But how do I use this fact?

$\endgroup$
1
$\begingroup$

Let $x,x' \in C[0,1]$ with $x <x'$ and $\|f\|_{\infty} \leq 1$. Then $|Tf(x)-Tf(x')|\leq \int_0^{x} |k(x,y)-k(x',y)| |f(y)|dy+\int_x^{x'} |k(x',y)||f(y)|dy$. In the first term use the fact that $k$ is uniformly continuous. The second term does not exceed $M\|f\|_{\infty}|x-x'|$. Can you complete the proof now?

$\endgroup$
  • $\begingroup$ Yes, but in the first term, there is a given $\delta$ (becuase of uniform continuity) and in the second term I need to select my $\delta:=\frac{\epsilon}{M \vert \vert f \vert \vert_{\infty}}$. Surely performing one action cancels the other out $\endgroup$ – SABOY Jun 5 '19 at 8:00
  • $\begingroup$ First note that we $\|f_n\| \leq 1$ so we are only interested in the case $\|f\| \leq 1$. So the first term does not exceed $\sup\ |[|k(x,y)-k(x',y)|: y\in [0,1]$. The second term does not exceed $M|x-x'|$. You just have to choose an appropriate $\delta_1$ for the first term and an appropriate $\delta_2$ for the second term and take $\delta$ to be the minimum of $\delta_1$ and $\delta_2$. $\endgroup$ – Kavi Rama Murthy Jun 5 '19 at 8:07

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.