Q1) Let $f\in \mathcal C_c^\infty (\mathbb R^n)$ and $0<\alpha <\frac{n}{2}$. I want to show that $$\int_{\mathbb R^n}\frac{f(y)}{|x-y|^{n-\alpha }}dy$$ is integrable, i.e. that $$\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{|f(y)|}{|x-y|^{n-\alpha }}dxdy <\infty,$$ but unfortunately I didn't succeed. My idea was to use Hardy-Littlewood Sobolef inequality, i.e. $$\left|\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{f(x)f(y)}{|x-y|^\lambda }dydy\right|\leq C\|f\|_{L^p}\|f\|_{L^q},$$ whenever $\frac{1}{p}+\frac{\lambda }{n}+\frac{1}{q}=2,$ but since it's $\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{f(y)}{|x-y|^\lambda }dydx$ and not $\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{f(y)f(x)}{|x-y|^\lambda }dydx,$ my argument fail. Also I tried to majorate $$\int_{\mathbb R^n}\left|\int_{\mathbb R^n}\frac{f(y)}{|x-y|^\lambda }\right|dydx\quad \text{by}\quad \left|\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{f(y)f(x)}{|x-y|^\lambda }dydx\right|,$$ but unfortunately without success. Any idea ?
Q2) By the way, is there a way to proof that it's not integrable if $\alpha \geq \frac{n}{2}$.
Let the Riesz potential $$I_\alpha f(x) =\int_{\Bbb R^n}\frac{f(y)}{|x-y|^{n-\alpha }}dy$$
Note $$(I_\alpha f, g) = \int_{\Bbb R^n}\int_{\Bbb R^n}\frac{f(y)g(x)}{|x-y|^{n-\alpha }}dydx$$
and we know by Riesz representation theorem that
$$\|I_\alpha f\|_{q'} =\sup_{\|g\|_q = 1} |(I_\alpha f, g)|$$ Where $\frac{1}{q}+\frac{1}{q'}=1.$ Whereas by Hardy-Littlewood-Sobolev inequality we have
$$|(I_\alpha f, g)| \le C\| f\|_p \| g\|_{q}~~~~~\frac{1}{p}+\frac{\alpha}{n}+\frac{1}{q}=2.$$
Hence $$\|I_\alpha f\|_{q'} =\sup_{\|g\|_q = 1} |(I_\alpha f, g)|\le C\| f\|_p ~~~~\frac{1}{p}+\frac{\alpha}{n}+1-\frac{1}{q'}=2.$$
Taking $q'=1$,
$$\|I_\alpha f\|_{1} \le C\| f\|_p ~~~~\frac{1}{p}+\frac{\alpha}{n}=2.$$