Let $B$ be the unit ball in $C([a,b])$. Define for $f\in C([a,b])$, $$Tf(x)=\int_a^b (-x^2+e^{-x^2+y})f(y)dy.$$ Prove that $T(B)$ is relatively compact in $C([a,b])$.

My attempt:

If $|f(x)| \le 1$ on $[a,b]$. Then $T(f)$ is bounded. We only need to show that $T(f)$ is equi-continuous by Arzela-Ascoli theorem.$T(f)(x)-T(f)(z)=\int_a^b (x^2+z^2)f(y)dy+\int_a^b (e^{-x^2}-e^{-z^2})e^y f(y)dy$.

Then I don’t know how to proceed to get equi-continuous?

Could someone kindly help?

Thanks!

**Answer**

Let $k(x,y) = -x^2+e^{-x^2+y}$, then $k \in C([a,b]^2)$. It is uniformly continuous on $[a,b]^2$, so for any $\epsilon > 0, \exists \delta > 0$ such that

$$

|(s,t) – (x,y)| < \delta \Rightarrow |k(s,t) – k(x,y)| < \epsilon

$$

Thus,

$$

|T(f)(s) – T(f)(x)| = |\int_a^b k(s,t)f(t)dt – \int_a^b k(x,t)f(t)dt| \leq \int_a^b |k(s,t)-k(x,t)||f(t)|dt

$$

So if $|s-x| < \delta$, then

$$

|T(f)(s) – T(f)(x)| \leq \epsilon\|f\|_{\infty}(b-a)

$$

This gives equi-continuity.

**Attribution***Source : Link , Question Author : Tony , Answer Author : Prahlad Vaidyanathan*