I want to prove an inequality such that $$ \int_{B}|f(y)|dy\leq |B|^{1-\frac{1}{p}}\|f\|_{L^p(B)}, $$ where $B\subset\mathbb{R}^n$ is a ball, $p>1$ and $\|f\|_{L^p(B)}=(\int_{B}|f(y)|^pdy)^{\frac{1}{p}}$.

This inequality is a direct consequence of Hölder's inequality. But i want to prove this inequality with the help of maximal operator $M$, defined as $Mf(x)=\sup\limits_{B\ni x}\frac{1}{|B|}\int_{B}|f(y)|dy$, where supremum is taken over all balls containing $x$.

My proof is:

By the definition of maximal operator we can write $$ \frac{\chi_{_B}(x)}{|B|}\int_{B}|f(y)|dy\leq Mf(x),\qquad x\in \mathbb{R}^n $$ where $\chi_{_B}$ denotes the characteristic function of ball $B$. Then if we take Lebesgue norm over ball $B$ we get $$ \frac{|B|^{1/p}}{|B|}\int_{B}|f(y)|dy\leq \|Mf\|_{L^p(B)}. $$ We know that $M$ is bounded on $L^p(\mathbb{R}^n )$. Using this fact can we write $$\|Mf\|_{L^p(B)}\leq C \|f\|_{L^p(B)},$$ where $C$ is a positive constant.

  • $\begingroup$ You get a weaker inequality in a result because of constant $C$. $\endgroup$ – Virtuoz Dec 20 '15 at 6:55

Your Answer

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

Browse other questions tagged or ask your own question.