integral operator with continuous integrable function
Let $k: [0,\infty) \mapsto \mathbb{R}$ be continuous and
$\int_{0}^{\infty} |k(x)|dx < + \infty$.
Set
$(Af)(x)=\int_{0}^{\infty} k(x+y) f(y) dy, f \in L^2(\mathbb{R})$.
Show that $A$ is bounded and compact. Arzela-Ascoli seems a reasonable guess with regards to the latter, but what about the former? Why is this integral even $L^{2}(\mathbb{R})$?
Solution 1:
Hint: $Af(-x)$ is the convolution of $f$ and $k_1(t)=k(-t)$. Use the inequality $\|f*g\|_2 \leq \|f\|_2 \|g\|_1$.
Ref. See the heading Integrable Functions in https://en.wikipedia.org/wiki/Convolution