Prove that the integral operator $T:\mathcal{C}([0,1]) \to \mathcal{C}([0,1])$ is compact

Question

Let $T:\mathcal{C}([0,1]) \to \mathcal{C}([0,1])$ be the operator given by
$(Tf)(x) = \int \limits_{0}^{x} e^y \cdot f(y) dy, \: \forall f \in \mathcal{C}([0,1])$.
where $\mathcal{C}([0,1])$ is equipped with the $\lVert \cdot \rVert_{\infty}$-norm.
Prove that $T$ is compact.

My ideas:

1. I do not know how to approximate $T$ with compact operators in the operator-norm.
2. Take a sequence $(f_n)_{n \in \mathbb{N}}\subset \mathcal{C}([0,1])$ that is bounded w.r.t. $\lVert \cdot \rVert_{\infty}$. I would like to apply Arzela-Ascoli, but I am not sure if the sequence is equicontinuous.

Answer 1

As you said, we have to apply Ascoli-Arzelà theorem.

Of course the sequence $Tf_n$ is equally bounded, because $||f_n||\leq M$ for every $n$ and so

$$||Tf_n||\leq M sup_{x}(\int_0^xe^ydy)=M(e-1)$$

Moreover $Tf_n$ is uniformly equicontinuos because if you take an $\epsilon>0$ and $x_1\leq x_2\in [0,1]$ then

$|Tf_n(x_2)-Tf_n(x_1)|\leq M\int_{x_1}^{x_2}e^ydy=e^{x_2}-e^{x_1}\leq \epsilon$

for $|x_2-x_1|\leq \delta$ (here we are using that the function $e^x$ is uniformly continuos)

Remark The same statement holds if you take a generic uniformly continuos function $g$ instead of $e^x$.