The problem was to prove the following that the operator


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.


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


Let us estimate the interior integral.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.