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?


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?

  • $\begingroup$ 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

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