# Hardy-Littlewood maximal function not integrable in B(0,1)

It is well known that if $f\in L^1(\mathbb{R}^n)$, then the Hardy-Littlewood maximal function: $$Mf=\sup_{r>0}\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)|dy$$ is not in $L^1(\mathbb{R}^n)$. Does there exist a $f$ such that $Mf$ is not integrable in $B(0,1)$?

The answer to your question is no by theorem of Stein given on the screen shot for the full proof see here

Yes. In fact such an example must exist, because

if $\lambda>0$ and $g(x)=f(\lambda x)$ then $Mg(x)=Mf(\lambda x)$.

In detail: Let $F=\chi_{B(0,1)}$, and check that $$\int MF=\infty.$$

So there exists $R_k\in(0,\infty)$ such that $$\int_{|x|<R_k}MF(x)>k^3.$$

Define $$f_k(x)=R_k^{n}F(x R_k),$$and let $$f=\sum\frac1{k^2}f_k.$$ Then $$\int_{|x|<1}Mf\ge\frac1{k^2}\int_{|x|<1}Mf_k=\frac1{k^2}\int_{|x|<R_k}MF>k.$$

• see hte theorem below – Guy Fsone Jan 25 '18 at 9:13

This may help Let $x\in B$ that is $|x|\le 1$ and let fix $r>0$ then we have $B(x,r)\subset B(0,r+1)$ indeed for $y\in B(x,r)$ we have $$|y|\le |x|+|x-y|\le r+1$$

Therefore

$$\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)|dy \le \frac{1}{\color{blue}{|B(0,1)|r^n}}\int_{B(0,r+1)}|f(y)|dy\\=\left(\frac{r+1}{r}\right)^n\frac{1}{\color{blue}{|B(0,r+1)|}}\int_{B(0,r+1)}|f(y)|dy\le \color{red}{\left(\frac{r+1}{r}\right)^nMf(0) }$$

• Of course there's no reason to think that $Mf(0)<\infty$. – David C. Ullrich Jan 23 '18 at 16:11
• @DavidC.Ullrich yes it is true . even if it is finite taking the sup over $r$ leads to a heavy blow up – Guy Fsone Jan 23 '18 at 16:21
• Right. So I really don't see how this helps... – David C. Ullrich Jan 23 '18 at 18:15