Is is true that for a locally integrable function, we always have $Mf(x)\geq |f(x)|$ a.e.? I think that is true but I can not find any reference for that.
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Are you asking if there is a specific $M$ where the bound holds for all $x \in \mathbb{R}$? – Dionel Jaime Jul 16 '18 at 03:25
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I think you may be asking about the third basic property listed here but I am not quite sure I understand your question yet. https://en.wikipedia.org/wiki/Maximal_function#Basic_properties – Mason Jul 16 '18 at 03:26
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@Mason Well that's extremely different ... – Dionel Jaime Jul 16 '18 at 03:29
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@Mason I am familiar with that properties but I did wonder if the OP is true. – Tongou Yang Jul 16 '18 at 03:44
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1@Mason Yes it is the centred maximal function. – Tongou Yang Jul 16 '18 at 03:55
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If $x$ is a Lebesgue point of $f$, then $Mf(x) \geq |f(x)| $. Indeed, we have $$ \tag{1} Mf(x) = \sup_{x\in B} \frac{1}{|B|} \int_B |f(y)| dy \geq \lim\limits_{r\to 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| dy = |f(x)|, $$ where the equality follows by the definition of Lebesgue point. Since for integrable $f$ almost every point is a Lebesgue point, then $(1)$ holds true almost everywhere (where $f$ is defined and is integrable).
Hayk
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