Show that $(Tf_{n})_{n}$ is equicontinuous

Question

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?

Answer 1

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?