Rising Sun Inequality (Dunford-Schwartz maximal inequality)

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be an absolutely integrable function, and let $f^*:\mathbb{R} \rightarrow \mathbb{R}$ be the one-sided signed Hardy-Littlewood maximal function $$f^*(x) := \sup_{h>o} \frac{1}{h}\int_{[x,x+h]} f(t) dt.$$ Establish the rising sun inequality $$\lambda \mu (\{f^* > \lambda\})\leq \int_{\{f^*>\lambda\}} f(t)dt,$$ furthermore, the above is in fact equal when $\lambda > 0$.

I have been stuck on this for the past couple of days. I am still trying to show the case when $\lambda = 0$.

Here is the hint from the problem: when $f$ is compactly supported on the compact interval $[a,b]$, then $F(x) = \int_a^x f(t)dt-(x-a)\lambda$ is continuous on the compact interval; apply the Rising Sun lemma to $F(x)$.

Side note: for $\lambda >0$ $$\mu (\{x\in \mathbb{R} : \sup_{h>o} \frac{1}{h}\int_{[x,x+h]} |f(t)| dt > \lambda\})\leq \frac{1}{\lambda}\int_\mathbb{R}|f(t)|dt$$ is called One-sided Hardy-Littlewood maximal inequality. This can be proven with the hint from above instead of the standard method using Vitali cover lemma.

Answer 1

I'll show the inequality part.

First note that it suffices to do the case when $\lambda = 0$. Since then we can apply the case to the absolutely integrable function $g = f|_{[a, b]} - \lambda 1_[a, b]$, for some compact interval $[a, b]$, to obtain $0 \leq \int_{x \in [a, b]: g^*(x) > 0}g(x)\ dx$, which after unpacking $g$ again becomes

$m(\{x \in [a, b]: f^*(x) > \lambda\}) \leq \int_{x \in [a, b]: f^*(x) > \lambda}f(x)\ dx$, $\forall \lambda \in \mathbb{R}$.

The result then follows from upwards monotonicity and the dominated convergence theorem.

Define the restricted maximal function $\displaystyle f^*_{[a, b]}: [a, b] \rightarrow \mathbb{R}$ by $f^*_{[a, b]}(x) := \sup_{h>0; [x, x+h] \subset [a, b]} \frac{1}{h} \int_{[x,x+h]} f(t)\ dt$. By the dominated convergence theorem, it'll suffice to show that $\displaystyle 0 \leq \int_{x \in [a, b]: f^*_{[a, b]}(x) > 0} f(x)\ dx\ \ \ \ \ (1)$.

Apply the rising sun lemma to $F:[a, b] \rightarrow \mathbb{R}$ defined as: $\displaystyle F(x) := \int_{[a,x]} f(t)\ dt.$ $F$ is continuous, and so we can find an at most countable sequence of intervals ${I_n = (a_n,b_n)}$ with the properties given by the rising sun lemma. Futhermore, by the proof of the rising sun lemma we have:

$\bigcup_n I_n = \{x \in (a, b): \exists y \ \text{such that}\ x < y < b\ \text{and}\ F(y) > F(x) \} = \{x \in (a, b): \sup_{h>0, [x, x+h] \subset [a, b]}F(x + h) > F(x)\} = \{x \in (a, b): f^*_{[a, b]}(x) > 0\}$.

Since $F(b_n) - F(a_n) \geq 0$, we have $\displaystyle \int_{I_n} f(t)\ dt \geq 0$, so $\sum_{n}\int_{I_n} f(t)\ dt = \int_{\cup_{n}I_n} f(t)\ dt = \int_{x \in [a, b]: f^*_{[a, b]}(x) > 0} f(x)\ dx \geq 0$ (again by DCT), as desired.

Answer 2

Assume that $f$ is compactly supported. To prove that both sides are equal when $\lambda > 0$, try to answer the following questions in the given sequence:

1. How is open set $S$ that is mentioned in the Wikipedia article that you provide connected to the set $\mu\{f^{*}(x) > \lambda\}$?
2. If you have figured out the equivalence between them, how can you reformulate $S$ in terms of the function $F(x):= \int_{[a,x]} |f|\ \mathrm{d}m - (x-a)\lambda$ that is mentioned in the hint that you provide?
3. Finally, applying the rising sun lemma to the function $F$, how can you make use of the condition $F(a_i) = F(b_i)$?

Did you figure it out? Can you generalise the proof now to non-compactly supported functions? If you are still struggling, you can look up the solution either here or here.