Prove that $T(B)$ is relatively compact in $C([a,b])$. 
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!
 A: 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.
A: I think this might do it; note that the following argument makes no direct use of Arzela-Ascoli or related theorems:
$Tf(x) = \int_a^b (-x^2 + e^{-x^2 + y}) f(y)dy$
$= \int_a^b (-x^2)f(y)dy + \int_a^b (e^{-x^2 + y})f(y) dy = -x^2 \int_a^b f(y)dy + e^{-x^2} \int_a^b e^y f(y)dy; \tag{1}$
also, for $f(y) \in C[a, b]$, then
$\vert \int_a^b f(y)dy \vert \le \int_a^b \vert f(y) \vert dy \le \int_a^b \Vert f(y) \Vert dy = \Vert f(y) \Vert (b - a) \tag{2}$
and
$\vert \int_a^b e^y f(y) dy \vert \le \int_a^b \vert e^y f(y) \vert dy \le \int_a^b e^y \vert f(y) \vert dy$
$\le \int_a^b e^y \Vert f(y) \Vert dy = \Vert f(y) \Vert \int_a^b e^y dy = \Vert f(y) \Vert (e^b - e^a), \tag{3}$
where we have used $\vert f(y) \vert \le \Vert f(y) \Vert$ for all $y \in C[a, b]$; we see from (1)-(3) that
$\Vert Tf(x) \Vert = \Vert (-x^2) \int_a^b f(y)dy + e^{-x^2} \int e^y f(y) dy \Vert$
$\le \vert \int_a^b f(y) dy \vert \cdot \Vert -x^2 \Vert + \vert \int_a^b e^{-y} f(y) dy \vert \cdot \Vert e^{-x^2} \Vert$
$\le \Vert f(y) \Vert (b - a) \Vert -x^2 \Vert + \Vert f(y) \Vert (e^b - e^a) \Vert e^{-x^2} \Vert$
$=((b -a) \Vert -x^2 \Vert + (e^b - e^a) \Vert e^{-x^2} \Vert) \Vert f(y) \Vert. \tag{4}$
Setting
$M = (b - a)\Vert -x^2 \Vert + (e^b - e^a)\Vert e^{-x^2} \Vert, \tag{5}$
(4) shows that $M$ is a bound for $T$; that is,
$\Vert Tf(x) \Vert \le M\Vert f(x) \Vert \tag{6}$
for all $f(x) \in C[a, b]$ (note here that in fact $\Vert f(x) \Vert = \Vert f(y) \Vert$); furthermore, by (1), we see that the range of $T$ lies in $V = \text{span} \{x^2, e^{x^2} \}$, a two-dimensional subspace of $C[a, b]$.  $T$ thus maps the closed unit ball $\bar B_{C[a, b]}(1) \subset C[a, b]$ into the closed  ball $\bar B_V(M)$ of radius $M$ in $V$:
$T(\bar B_{C[a, b]}(1)) \subset \bar B_V(M); \tag{7}$
this implies
$\overline{T(\bar B_{C[a, b]}(1))} \subset \bar B_V(M) \tag{8}$
as well, since $\bar B_V(M)$ is closed; since $V$ is of finite dimension, $\bar B_V(M)$ is compact; by (8) the closure $\overline{T(\bar B_{C[a, b]}(1))}$ of $T(\bar B_{C[a, b]}(1))$ lies in the compact set $\bar B_V(M)$; thus it too must be compact, establishing the desired result.
