Bounded kernel operator

Question

I have to verify that this operator in $L^2(\Bbb R^3): (Gf)(x)=\int dy f(y) \frac{e^{-|x-y|}}{|x-y|} $ is bounded, using the following rule for the integral kernel: $ sup_x\int dy|a(x,y)|< \infty$ and $sup_y\int dx|a(x,y)|< \infty$, where $|a(x,y)|$ is the kernel. It seems to be unbounded and I don't understand why the space is $L^2(\Bbb R^3)$. Can you help me? Thank you.

Answer 1

It is a convolution operator with the positive function $g(x)={e^{-|x|}\over |x|}.$ Such operator is bounded if and only if $g\in L^1(\mathbb{R}^3).$ It coincides with the two conditions as $$a(x,y)=a(y,x)=g(x-y).$$ The function $g$ is radial, i.e. it depends on $ |x|.$ Applying spherical coordinates it can be shown that $g$ is integrable: $$\iiint g(x)\,dx=2\pi^2\int\limits_0^{\infty}re^{-r}\, dr=2\pi^2.$$ Similar operator on $L^2(\mathbb{R})$ is unbounded as $g(x)={e^{-|x|}\over |x|}$ is not integrable near $0.$ In general, if $g$ is not necessarily positive, the convolution operator on $L^2(\mathbb{R}^3)$ is bounded if and only if the Fourier transform of $g$ is bounded.