Riesz potential $I_{\alpha}\phi\notin L^2$ if $\frac{n}{2}<\alpha<n,$

Question

Riesz potential $I_{\alpha}(\phi)$ can be represented as the form $((4\pi^2|\xi|^2)^{-\frac{\alpha}{2}}\hat{\phi}(\xi))^{\vee}$. It is easy to see that $0<\alpha<\frac{n}{2}$, $I_{\alpha}(\phi)\in L^2$. I want to prove that if $\frac{n}{2}<\alpha<n,$ and $\int_{\mathbb{R}^n}\phi\neq 0$, then $I_{\alpha}\phi\notin L^2$

Answer 1

You don't say where $\phi$ lies. Let's say $\phi\in L^1$.

We have that $\hat\phi(0)=\int_{\mathbb{R}^n}\phi\neq 0$. Then, near $0$ $$ \bigl|(4\,\pi^2|\xi|^2)^{-\frac{\alpha}{2}}\,\hat{\phi}(\xi)\bigr|^2\sim|\xi|^{-2\alpha}, $$ which is integrable near $0$ if and only if $-2\,\alpha+n-1>-1$, that is, $\alpha<n/2$.