Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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