Let $k$ be a bounded continuous function that is strictly positive and $\lim_{|x|\to\infty} k(x) \to 0$ and also $k \in L^1(\mathbb R)$. The question is, if the integral operator $T: C_b^0(\mathbb R) \to C_b^0(\mathbb R)$ with $$ Tf(x) := \int_{\mathbb R} f(y)k(x-y)dy $$ where $C_b^0(\mathbb R)$ is a space of continuous bounded functions, is compact.
I was able to prove that $T$ is actually a linear continuous map, but I don't know how to proceed to show that it is not compact. Any advice is appreciated.
Assume $\int_{\mathbb{R}}k(x)dx=1$ and $\int_{-L}^L k(x)\ge \frac{2}{3}.$ Take $\varphi(x)$ satisfied that $\operatorname{supp} \varphi\subset[-3L,3L],$ $\varphi\in[0,1],$ $\varphi|_{[-2L,2L]}=1.$ Then $T\varphi(x)|_{[-L,L]}\ge \frac{2}{3},$ and $T\varphi(x)\le \frac{1}{3}$ when $|x|\ge 4L$.
Now we can construct a counter-example. Let $f_n(x)=\varphi(x-6nL),$ then $\{f_n\}$ is a bounded set while $T\{f_n\}$ cannot have a sequence to converge, because the "main part" of $Tf_n(x)$ differs. $\|Tf_n-Tf_m\|\ge \frac{1}{3},$ $\forall m\neq n>0.$ So $T$ is not a compact operators.