Problem of the Week #220 - Aug 16, 2016

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Euge

MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $f : \Bbb R^n \to \Bbb R$ be a function such that $f$ and its maximal function $\mathcal{M}f$ belong to $\mathcal{L}^1(\Bbb R^n)$. Show that $f(x) = 0$ for almost every $x\in \Bbb R^n$.

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Euge

MHB Global Moderator
Staff member
This week's problem was correctly solved by Opalg . You can read his solution below.

If $f$ is not almost everywhere zero then there exist $r, \varepsilon > 0$ such that $$\displaystyle \int_{B(0,r)}|f(y)|\,dy > \varepsilon.$$ For $x \in\Bbb{R}^n,$ $B(0,r) \subseteq B(x,|x|+r).$ It follows that $$\displaystyle \int_{B(x,|x|+r)}|f(y)|\,dy > \varepsilon$$ and therefore $$(\mathcal{M}f)(x) \geqslant \frac1{V(|x|+r)}\int_{B(x,|x|+r)}|f(y)|\,dy > \frac{\varepsilon}{V(|x|+r)}.$$ But $$\displaystyle \lim_{|x|\to\infty}\frac{V(|x|+r)}{V(|x|)} = 1,$$ and $V(|x|)$ is a constant times $|x|^n$. So whenever $|x|$ is sufficiently large, $(\mathcal{M}f)(x) \geqslant C|x|^{-n}$ for some positive constant $C$. But $|x|^{-n}$ is not an $L^1$-function as $|x|\to\infty,$ and hence neither is $\mathcal{M}f.$

In conclusion, if $\mathcal{M}f$ is an $L^1$-function then $f$ must be almost everywhere zero.

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