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Let $f: \mathbb{R}^d \to \mathbb{R}^*$ be measurable. Let $f^*$ denote the Hardy-Littlewood maximal function.

1) If $f$ is nonzero on a set of positive measure prove that there exists $0 \lt A \lt \infty$ such that $f^*(x) \geq A ||x||^{-d}$ whenever $||x||$ is sufficiently large.

2) If $f \in L^1(\mathbb{R}^d)$ is compactly supported, prove that there exists $0 \lt B \lt \infty$ such that $f^*(x) \leq B ||x||^{-d}$ holds whenever $||x||$ is sufficiently large.

Both of these parts make sense intuitively: for the first one, I want to show that if $f$ is nonzero on a set of positive measure, then I can bound the "supremum" below by a nonzero function that depends on $x$. For the second part, I have that $f$ is integrable, so it is finite almost everywhere, so $f^*$ should be finite almost everywhere. Hence $f^*$ can be bounded above by some finite function which depends on $x$. I'm not sure where the compact support comes in though. Is my intuition correct? I have no idea how to prove this rigorously. How do I even start with this kind of problem?

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1) Without loss of generality we may assume $A\subset B_R(0)$, where $B_R(z)$ is the ball of radius $R>0$ centered at $z$. If $\|x\|\ge2\,R$ then $A\subset B_{\|x\|+R}(x)$ and $$ f^*(x)\ge\frac{1}{|B_{\|x\|+R}(x)|}\int_A|f|=C_d(\|x\|+R)^{-d}\int_A|f|\ge\Bigl(C_d\,(2/3)^d\int_A|f|\Bigr)\,\|x\|^{-d}. $$

2) Suppose that the support of $f$ is contained in $B_R(0)$. Take $x$ with $\|x\|>2\,R$. Then $B_R(0)\cap B_r(x)=\emptyset$ if $r\le\|x\|-R$. If $r>\|x\|-R$ then $$ \frac{1}{|B_{r}(x)|}\int_{B_r(x)}|f|\le C_d\,r^{-d}\|f\|_1\le C_d\,(\|x\|-R)^{-d}\|f\|_1\le\Bigl(C_d\,2^d\,\|f\|_1\Bigr)\,\|x\|^{-d}. $$

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  • $\begingroup$ What happens to your proof of 1) if the set $A$ where $f$ in nonzero has infinite measure? Wouldn't that make $R$ and $||x||$ infinite as well? Is the proof still valid if that happens? $\endgroup$ – Johanna Nov 20 '14 at 15:53
  • $\begingroup$ There exists $A=\cup_{n=1}^\infty A\cap B_n$. Since $|A|>0$, there is $n$ such that $|A\cap B_n|>0$. $\endgroup$ – Julián Aguirre Nov 20 '14 at 15:56
  • $\begingroup$ I don't understand how that helps if $|A| = \infty$? It seems to me that we need to contain all of $A$ in a ball of finite radius, so we need $|A \cap B_n| = |A|$, right? $\endgroup$ – Johanna Nov 20 '14 at 15:59
  • $\begingroup$ No. $f^*(x)\ge(f\,\chi_{A\cap B_n})^*(x)$ for all $x$. $\endgroup$ – Julián Aguirre Nov 20 '14 at 16:02

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