Analysis Problem of the Day 70

Today’s problem appeared as Problem 5 on the UCLA Fall 2023 Analysis Qual:

Problem 5. Let $\omega: \mathbb{R} \to (0,\infty)$ be a locally integrable function, and define the corresponding Borel measure $\omega(E) = \int_E \omega(x)dx.$ Let $M$ be the Hardy-Littlewood maximal function, i.e.

$Mf(x) = \sup_{r>0} \frac{1}{2r} \int_{x-r}^{x+r}|f(y)|dy.$

Suppose that $\frac{1}{\omega}$ is locally integrable and

$\omega(\{x:|Mf(x)|>\lambda}) \leq \frac{C}{\lambda^2} \int |f(x)|^2 \omega(x) dx$

for some $C>0.$ Show that

$\sup_{x \in \mathbb{R},r>0} \left(\frac{1}{2r} \int_{x-r}^{x+r} \omega(y) dy\right) \left(\frac{1}{2r} \int_{x-r}^{x+r} \frac{1}{\omega(y)}dy \right) < \infty.$

Solution: Note that one has to somehow obtain an integrand of $\frac{1}{\omega}.$ Since one has the freedom of choosing $f,$ it would make sense to set $f=\frac{1}{\omega}.$ But $\frac{1}{\omega}$ might not be integrable over the entire real line, so one must also restrict $f$ appropriately, i.e. $f(x):=\frac{\chi_{[x_0-r,x_0+r]}(x)}{\omega(x)}.$ With such an $f,$ we attempt to use the given inequality. But so far, it is going in the wrong direction. One might then note that if we set $\lambda := \frac{1}{4r} \int_{x_0-r}^{x_0+r} \frac{1}{\omega(x)} dx,$ that the given inequality becomes

$\left(\frac{1}{2r} \int_{|Mf|>\lambda} \omega(y) dy\right) \left(\frac{1}{2r} \int_{x_0-r}^{x_0+r} \frac{1}{\omega(y)}dy \right) \leq C$

uniformly in $x_0$ and $r.$ Since $\omega$ is always positive, it would thus suffice to show that $\{x:|Mf(x)|>\lambda\} \supset \{x_0-r,x_0+r\}.$ Now, note that $Mf$ at a given point $x$ is the supremum over averages of $f$ centered at $x.$ In particular, since $f$ is positive and is supported on $[x_0-r,x_0+r] \subseteq [x-2r,x+2r]$ for $x \in [x_0-r,x_0+r],$ then

$4r Mf(x) \geq \int_{x-2r}^{x+2r}|f(y)|dy = \int_{x_0-r}^{x_0+r}|f(y)|dy,$

i.e.

$|Mf(x)| \geq \lambda$

on $[x_0-r,x_0+r].$ But this completes the proof.

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