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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 $B(0,1)$?

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

enter image description here

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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 $$\int_{|x|<1}Mf\ge\frac1{k^2}\int_{|x|<1}Mf_k=\frac1{k^2}\int_{|x|<R_k}MF>k.$$

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

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

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

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