Let ${f: {\bf R} \rightarrow {\bf C}}$ be an absolutely integrable function, and let ${\lambda > 0}$. Given the One-sided Hardy-Littlewood maximal inequality:
$\displaystyle m( \{ x \in {\bf R}: \sup_{h>0} \frac{1}{h} \int_{[x,x+h]} |f(t)|\ dt \geq \lambda \} ) \leq \frac{1}{\lambda} \int_{\bf R} |f(t)|\ dt$.
Show the Two-sided Hardy-Littlewood maximal inequality:
$\displaystyle m( \{ x \in {\bf R}: \sup_{x \in I} \frac{1}{|I|} \int_{I} |f(t)|\ dt \geq \lambda \} ) \leq \frac{2}{\lambda} \int_{\bf R} |f(t)|\ dt$,
where the supremum ranges over all intervals ${I}$ of positive length that contain ${x}$. Can we do this by simply using the one-sided result on the reflected function $x \rightarrow f(-x)$?
Best Answer
Another way of doing this more in line with what you were thinking: Consider the maximal operator $$M_- f(x):= \sup_{h>0} \dfrac{1}{h}\int_{x-h}^{x} |f|\, dx,$$ and with $M_+$ denoting your one-sided maximal operator. Then we have $$ M_- f(x)= (M_+ f_1) (-x), \qquad \text{where} f_1(y):= f(-y). $$ In particular, $M_-$ satisfies the same estimate as $M_+$. Now use (and prove!) that if $M$ denotes your second (uncentered) maximal function, then $$\{x: Mf(x)>\lambda\}\subset \{ x: M_+f(x)>\lambda\} \cup \{ x: M_-f(x)>\lambda\}.$$ Note that this inclusion is not "trivial", as using the easy estimate $Mf(x)\leq M_+f(x)+M_-f(x)$ yields a factor of $\lambda/2$ instead of $\lambda$ in both terms on the right of the display (this gives a factor of 4 for the uncentered bound, instead of 2). The point is that the easy inequality can be improved by controlling $1/|I|\int_I |f|$ by an average of $M_\pm f(x)$ for any $x\in I$.