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The problem was to prove the following that the operator

$$Tf(x)=\int_{\mathbb{R}^N}\frac{f(y)}{|x-y|^\alpha}dy$$

Is continuous from $$L^1 \to \ L_\mathrm{Weak}^{p}$$ where $0<\alpha<N$ and $p=(N-\alpha)^{-1}$.

I tried to use the Lorentz Space $$M^{q,\nu}(R^N)=L_\mathrm{Weak}^{\frac{p}{1-\nu}}(R^N)$$ but it didn't helped much.

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Consider $E$ a measurable set of finite measure then suppose WLOG that $f\geq0$ then by Tonelli's theorem we have

$$\int_{E}\int_{\mathbb{R}^N}\frac{f(y)}{|x-y|^\alpha}dydx=\int_{\mathbb{R}^N}f(y)\int_{E}\frac{1}{|x-y|^\alpha}dxdy.$$

Let us estimate the interior integral.

Nevertheless

Consider around the point $y$ a ball $B=B(y,r)$ so we have

$$\int_{E}\frac{1}{|x-y|^\alpha}dx\leq\int_{B}\frac{1}{|x-y|^\alpha}dx+\int_{E\setminus B }\frac{1}{|x-y|^\alpha}dx\leq N\omega_N\frac{r^{N-\alpha}}{N-\alpha}+r^{-\alpha}|E|$$

Minimizing the last term in $r$ we get

$$r=\left(\frac{\alpha |E|}{N\omega_N}\right)^{\frac{1}{N}}$$

So we have $$\int_{E}\frac{1}{|x-y|^\alpha}dx\leq C(N,\alpha)|E|^{1-\frac{\alpha}{N}}$$

It shows that $Tf$ is in $M^{1, 1-\frac{\alpha}{N}}=L^\frac{N} {\alpha}_w$. So try this method an adequate $p>1 $ and an intelligent use of Jensen's inequality.

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