Define the operator $Tf(x)=\int_0^\infty \frac{\sin(x-y)}{x-y} f(y) \text{d}y$ for $x > 0$. Is this bounded on $L^2(0,\infty)$? or maybe on some $L^p(0,\infty)$?
We can rewrite $Tf=g*Ef$ where $Ef$ is the extension of $f$ by $0$ outside $(0,\infty)$ and $g(x)=\frac{\sin(x)}{x}$. But we cannot apply Young's inequality because $g \notin L^1(\mathbb{R})$.