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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?

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  • $\begingroup$ It's worth mentioning that the authors of the referenced paper have since submitted a revised version taking care of this error. $\endgroup$ – Matt Rosenzweig Apr 9 '15 at 14:02
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and since the reverse inequality is obvious, we conclude that...

Yes, the word obvious and mistaken claims are often in close proximity of each other. Your counterexample is valid; the stated identity is false.

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.

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  • $\begingroup$ Thank you. So the correctness of this claim would be academic if it were not used later in the proof of Theorem 1. But as currently stated in the paper, the proof of Lemma 3 is incorrect since it relies on the above claim. $\endgroup$ – Matt Rosenzweig Dec 20 '14 at 15:01
  • $\begingroup$ It does seems to have implications for the rest of the paper. You may want to contact the authors. Although the paper is already published in PAMS 2012. $\endgroup$ – user147263 Dec 20 '14 at 18:54

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