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Let $f \in W^{1, 1} (\mathbb R^d)$. For every $\varepsilon > 0$, we consider the local maximal function $M_\varepsilon f: \mathbb R^d \to \mathbb R$, defined by

$$M f_{\varepsilon} (x) = \sup_{r \leq \varepsilon} \frac{1}{|B_r (x)|} \int_{B_r (x)} |f(y) - f(x)| \, dy.$$

Question: Is it true that there exists some $C > 0$ such that for every $K > 0$, there exists some $\delta> 0$ such that for all $f \in W^{1, 1}$ with $\|f\|_{W^{1, 1}} \leq K$ and $\varepsilon < \delta$, we have that $M_\varepsilon f \in L^1$, and

$$\|M_\varepsilon f\|_{L^1} \leq C\|f\|_{W^{1,1}}?$$

Remark: It is known (see, eg here) that the usual global maximal function is bounded from $W^{1, p}$ to itself for all $p > 1$, but not for $p = 1$. I was unable to find results on whether it is even bounded from $W^{1, 1}$ to $L^1$.

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  • $\begingroup$ rmk: by homogeneity, you can remove $K$ and the condition $\|f\|_{W^{1,1}}\le K$ $\endgroup$ Commented Oct 15, 2023 at 15:38
  • $\begingroup$ The global maximal function is unbounded. If $f \in L^1_{loc}$ and one do not subtract $f(x)$ in the definition, then $Mf(x) \geq C|x|^{-d}$ for large $x$ (take a large $R$ where the integral of $|f|$ is positive, $|x|$ large and evaluate the mean over a radius $2|x|$). If you subtract the mean, the same happens if $f$ has a compact support and take $x$ large and outside the support. $\endgroup$ Commented Oct 16, 2023 at 6:55

1 Answer 1

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Yes I think this is true.

Let $\mathcal{G} := \{ \times_{i=1}^n[k_i,k_i+1] : k_i\in \mathbb{Z} \} $ be the grid of cubes of side length $1$ and vertices in $ \mathbb{Z}^n$. For $Q\in \mathcal{G} $ let $2Q$ be the cube with same center but double the side length. Notice that for $\varepsilon < \frac 12$ if $f$ is supported in $2Q$ then $M_\varepsilon f$ is supported in $3Q$. Let also $\phi \in C^\infty(\mathbb{R}^n), 0\leq \phi \leq 1$ supported in $ [-1,1]^n $ and $ \phi(x) = 1, \forall x \in [-\frac 12, \frac 12]^n $ and $ \phi_Q(x) = \phi(x-c_Q)$, where $c_Q$ is the center of the cube $Q$. Then for $f\in W^{1,1}(\mathbb{R^n})$ we have \begin{align*} \Vert M_\varepsilon f \Vert_{L^1(\mathbb{R}^n)} & \leq \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^1(\mathbb{R}^n)} \\ & = \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^1(3Q)} \\&\leq C_n \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^{\frac{n}{n-1}}(3Q)} \\ & \leq C_n \sum_{Q\in \mathcal{G}} \Vert f \phi_Q \Vert_{L^{\frac{n}{n-1}}(\mathbb{R}^n)} \\ & \leq C_n \sum_{Q\in \mathcal{G}} \Vert f \phi_Q \Vert_{W^{1,1}(\mathbb{R}^n)} \leq C_n \Vert f \Vert_{W^{1,1}(\mathbb{R}^n).} \end{align*} The constant $C_n$ is a dimensional constant which is different in each occurance. Passing from the third to the fourth line you use the boundedness of the maximal function and the second to last inequality is Sobolev's embedding theorem.

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  • $\begingroup$ @GiorgioMetafune I think it works now. Thanks for the observation :) $\endgroup$ Commented Oct 21, 2023 at 17:17
  • $\begingroup$ Wow, very nice proof. $\endgroup$
    – Nate River
    Commented Oct 22, 2023 at 7:08

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