# Left and Right Discrete Maximal Functions

Define the uncentered maximal function $$\widetilde{M}f(n)=\sup_{s,r\in\mathbb{Z}^{+}}\dfrac{1}{s+r+1}\sum_{k=-r}^{k=s}\left|f(n+k)\right|,$$ where $\mathbb{Z}^{+}=\left\{0,1,2,\ldots\right\}$. Define the left and right maximal functions $M_{L}f$ and $M_{R}f$, respectively by $$M_{L}f(n)=\sup_{r\in\mathbb{Z}^{+}}\dfrac{1}{r+1/2}\left\{\dfrac{\left|f(n)\right|}{2}+\sum_{k=-r}^{k=-1}\left|f(n+k)\right|\right\}$$ $$M_{R}f(n)=\sup_{s\in\mathbb{Z}^{+}}\dfrac{1}{s+1/2}\left\{\dfrac{\left|f(n)\right|}{2}+\sum_{k=1}^{k=s}\left|f(n+k)\right|\right\}$$

In the paper "Discrete Tanaka's Theorem", the authors claim that $$\widetilde{M}f(n)=\max\left\{M_{L}f(n),M_{R}f(n)\right\}$$ It is easy to see the direction $\widetilde{M}f(n)\leq\max\left\{M_{L}f(n),M_{R}f(n)\right\}$, but I am at a bit of a loss for the reverse inequality.

Indeed, suppose $f$ is the compactly supported function $$f(n)=\begin{cases} 1 & {n=0} \\ 2 & {n=1} \\ 0 & {\text{otherwise}}\end{cases}$$ Then $M_{L}f(0)=1$ and $M_{R}f(0)=5/3$, but $\widetilde{M}f(0)=3/2$. Am I missing something?

• It's worth mentioning that the authors of the referenced paper have since submitted a revised version taking care of this error. – Matt Rosenzweig Apr 9 '15 at 14:02

It holds in the continuous case (and was used by Tanaka), but in the discrete case the central point with its weight $1/2$ throws a monkey wrench into the argument.
By the way, it seems that the conjecture that motivated the paper has been settled in the original, continuous formulation by Ondřej Kurka in On the variation of the Hardy-Littlewood maximal function. Although the sharp constant $C=1$ remains conjectural.