Why is this integral operator NOT compact?

Question

Let $k \in C(\mathbb{R})$ be a function such that $k(t)>0$ for $|t|<c$, and $k(t)=0$ else. The operator $A: (C(\mathbb{R}),\|\cdot\|_{\infty}) \to (C(\mathbb{R}),\|\cdot\|_{\infty})$ given by \begin{align*} (A \varphi)(x) := \int_{\mathbb{R}} K(x,y) \varphi(y) dy \qquad x \in \mathbb{R} \end{align*} with $K(x,y):=k(x-y)$ is not compact? Why? (please prove this)

Answer 1

Since $K(x,y)$ depends only on $x-y$, $A$ commutes with translation. Moreover, if $\varphi$ is supported in an interval $[a,b]$, $A\varphi$ is supported in $[a-c,b+c]$. Take a function $\varphi$ supported on $[0,c]$ and positive in $(0,c)$, so that $A \varphi \ne 0$, then $A$ is an isomorphism on the closed linear span of the functions $t \mapsto \varphi(t + 3 n c)$, $n \in \mathbb Z$, which is infinite-dimensional. In particular it can't be compact.