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

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 $$x and $$\|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?
• 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 – SABOY Jun 5 '19 at 8:00
• 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 $\delta_2$. – Kavi Rama Murthy Jun 5 '19 at 8:07