Exercise 1.6.11 (Two-sided Hardy-Littlewood inequality)

Question

Let $f:\mathbb{R}	o\mathbb{C}$ be an absolutely integrable function, and let $\lambda > 0$. Show that
$$m \left(\left\{x \in \mathbb{R}: \sup_{	ext{interval $I$}\ \wedge\ x \in I} \frac{1}{|I|} \int_{I} |f| \ \mathrm{d}m \geq \lambda ight\} ight) \leq \frac{2}{\lambda} \int_{\mathbb{R}}{|f|\,\mathrm{d}m}$$
(i.e., supremum in question ranges over all intervals $I$ of positive length that contain $x$).

Elu Thingol · Accepted Answer

Notice that 
$$\left\{x \in \mathbb{R}: \sup_{	ext{interval $I$}\ \wedge\ x \in I} \frac{1}{|I|} \int_{I} |f| \ \mathrm{d}m \geq \lambda ight\}
=
\bigcup_{n\in\mathbb{N}} \left\{x \in [-n,n]: \sup_{h>0\ \wedge\ [x-h,x+h]\subseteq [-n,n]} \frac{1}{2h} \int_{I} |f| \ \mathrm{d}m \geq \lambda ight\}.$$

By \href{./exercise_1-2-11}{upwards monotonicity of the Lebesgue measure}, it will thus suffice to show that for any compact interval, we have
$$\forall [a,b] \subseteq \mathbb{R}: \quad m \left(\left\{x \in [a,b]: \sup_{h>0\ \wedge\ [x-h,x+h]\subseteq [a,b]} \frac{1}{2h} \int_{[x-h,x+h]} |f| \ \mathrm{d}m \geq \lambda ight\} ight) \leq \frac{2}{\lambda} \int_{\mathbb{R}}{|f|\,\mathrm{d}m}.$$

By modifying $\lambda$ by an epsilon, we may replace the non-strict inequality here with strict inequality:
$$\forall [a,b] \subseteq \mathbb{R}: \quad m \left(\left\{x \in [a,b]: \sup_{h>0\ \wedge\ [x-h,x+h]\subseteq [a,b]} \frac{1}{2h} \int_{[x-h,x+h]} |f| \ \mathrm{d}m > \lambda ight\} ight) \leq \frac{2}{\lambda} \int_{\mathbb{R}}{|f|\,\mathrm{d}m}.$$

Fix $[a,b]$. We apply the \href{./lemma_1-6-17}{rising sun lemma} to the function
$$F:[a,b]	o\mathbb{R}, \qquad F(x):= \int_{[a,x]} |f|\ \mathrm{d}m - (x-a)\lambda.$$
By \href{./exercise_1-6-5}{Exercise 1.6.5}, $F$ is continuous, and so we can find at most countable sequence of intervals $(I_n)_{n\in\mathbb{N}}$ with the properties given by the rising sun lemma. By equivalently expressing the below property
\begin{align*}
&\frac{1}{2h}\int_{[x-h,x+h]} |f|\ \mathrm{d}m > \lambda\
&\iff \frac{1}{2h}\int_{[a,x+h]} |f|\ \mathrm{d}m - (x+h-a)\lambda > \frac{1}{2h}\int_{[a,x+h]} |f|\ \mathrm{d}m - (x-a)\lambda\
&\iff F(x+h) > F(x-h),
\end{align*}
we observe from the second condition of the rising sun lemma that
$$\left\{x \in [a,b]: \sup_{h>0\ \wedge\ [x-h,x+h]\subseteq [a,b]} \frac{1}{2h} \int_{[x-h,x+h]} |f| \ \mathrm{d}m \geq \lambda ight\} \subseteq \bigcup_{n \in \mathbb{N}} I_n.$$
By \href{./lemma_1-2-15}{countable additivity}, we may, on one hand, upper bound the measure of the left-hand side by $\sum (b_n - a_n)$
$$m \left(\left\{x \in [a,b]: \sup_{h>0\ \wedge\ [x-h,x+h]\subseteq [a,b]} \frac{1}{2h} \int_{[x-h,x+h]} |f| \ \mathrm{d}m > \lambda ight\} ight) \leq \sum_{n=1}^{\infty} (b_n - a_n).$$
On the other hand, since $F(b_n) - F(a_n) \geq 0$, we have
$$\int_{I_n} |f|\ \mathrm{d}m \geq \lambda(b_n - a_n)$$
and thus,
$$\sum_{n\in\mathbb{N}}(b_n-a_n) \leq \frac{1}{\lambda}\sum_{n\in\mathbb{N}} \int_{I_n} |f|\ \mathrm{d}m.$$
As the $(I_n)_{n\in\mathbb{N}}$ are disjoint intervals in $I$, we map apply monotone convergence and monotonicity to conclude that
$$\sum_{n\in\mathbb{N}} \int_{I_n} |f|\ \mathrm{d}m \leq \int_{[a,b]} |f|\ \mathrm{d}m.$$
and the claim follows.

Exercise 1.6.11 (Two-sided Hardy-Littlewood inequality)

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Exercise 1.6.11 (Two-sided Hardy-Littlewood inequality)

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