I am searching for a reference for the (sharp) Hardy maximal function on the torus $\mathbb{T}^2:=\mathbb{R^2}/\mathbb{Z}^2$, for instance I would need result result of the following type : if $g\in H^1(\mathbb{T}^2)$ then for a.e. $x,y\in \mathbb{T}^2$ satisfying $|x-y|\leq \lambda$, one has \begin{align*} |g(x)-g(y)|\lesssim |x-y|(M_\lambda |\nabla g|(x) + M_\lambda |\nabla g|(y)), \end{align*} where for any $f\in L^1(\mathbb{T}^2)$ and $x\in \mathbb{T}^2$ we have introduced \begin{align*} M_\lambda f(x) := \sup_{0<r<\lambda} \frac{1}{|B_r(x)|} \int_{B_r(x)} |f|. \end{align*} The corresponding result is true on the whole space and I am quite sure that it should also be true on $\mathbb{T}^2$ but I lack a reference. Thanks for any help.
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$\begingroup$ I do not have a reference. However, for $\lambda < 1/2$ this is a local result: define a function $f$ on $\mathbb{R}^2$ so that $f(x_1,x_2) = g(x_1+\mathbb{Z},x_2+\mathbb{Z})$ for $x_1,x_2 \in [-1,2]$ and re-use the result in $\mathbb{R}^2$. For $\lambda > 1/2$ a minor modification is needed. $\endgroup$– Mateusz KwaśnickiSep 17, 2017 at 20:06
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$\begingroup$ Thanks Mateusz. What about the continuity of $M_\lambda$ from $L^2(\mathbb{T}^2)$ to itself ? Do you see any other way to justify the transference ? $\endgroup$– Ayman MoussaSep 18, 2017 at 5:56
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$\begingroup$ Is it not enough to note that $M_\lambda^{(\mathbb{R}^2)} f(x_1,x_2) = M_\lambda^{(\mathbb{T}^2)} g(x_1+\mathbb{Z},x_2+\mathbb{Z})$ for $x_1,x_2 \in [0,1]$ and $\lambda < 1/2$ (with the definitions from my previous comment)? $\endgroup$– Mateusz KwaśnickiSep 18, 2017 at 8:10
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