The Hardy–Littlewood maximal inequality states that for $d ≥ 1$, $1 < p ≤ ∞$, and $f ∈ L^p(\mathbb{R}^d)$, there is a constant $C_{p,d} > 0$ such that

$$||Mf||_{L^p(\mathbb{R}^d)}\leq C_{p,d}||f||_{L^p(\mathbb{R}^d)},$$ where

$$Mf(x)=\sup_{r>0}{\dfrac{1}{|B(x,r)|}}\int_{B(x,r)}|f(y)|dy.$$ Is it true that $$\Big(\int_A |Mf(x)|^pdx\Big)^{1/p}\leq C_{p,d}\Big(\int_A |f(x)|^pdx\Big)^{1/p} $$

where A is a measurable set?

  • $\begingroup$ I think that you should write out precisely what you want to be true. If you only replace the restriction to $A$ in the first inequality, then it seems false. $\endgroup$ – H. H. Rugh Sep 18 '16 at 9:33
  • $\begingroup$ @H.H.Rugh I wrote what I want to be true. $\endgroup$ – Michael Sep 18 '16 at 9:48

No if e.g. $f$ has support outside of $A$ the RHS is zero but the LHS is non-zero if $f$ has non-vanishing integral in ${\Bbb R}^d$ and $A$ has positive measure.

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  • $\begingroup$ If $f\in L^p(A)$, if you set $g(x)=f(x)\cdot \boldsymbol 1_A(x)$, then $g\in L^p(\mathbb R^n)$, and has all the hypothesis of the theorem. Therefore, it must be true... $\endgroup$ – Surb Sep 18 '16 at 9:54
  • $\begingroup$ That's why I asked to make the statement more precise. $f$ was not supposed to have support in $A$. $\endgroup$ – H. H. Rugh Sep 18 '16 at 9:55
  • $\begingroup$ What if $|f|>0$ a.e and $A$ has positive measure? $\endgroup$ – Michael Sep 18 '16 at 10:11
  • $\begingroup$ Not with a uniform bound without further conditions, since you may get arbitrarily close to the situation I described. If you happen to know something about the ratio of the $L^p(A)$ and $L^p(A^c)$ norm then you are in business (with an obvious estimate) but I can't think of a 'natural' general condition. $\endgroup$ – H. H. Rugh Sep 18 '16 at 10:38

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