When reading the characterization of Besov space with $L_p$-modulus of continuity in the 7th chapter “Fractional Order Space” of Sobolev space written by Adams(Page 243), I encounter some small specific problem. We define $$\Delta_hw(x):=w(x)-w(x-h)$$ We want to prove \begin{equation}||\Delta_hw||_p\leq |h||w|_{1,p}\qquad(*)\end{equation} The original words in the book are “We majorize $|w(x-h)-w(x)|$ by the integral of $|{\rm{grad}} w|$ along the line segment joining $x-h$ to $x$, and use Holder's inequality to majorize that by $|h|^{1/p’}$ times the one-dimensional $L_p$ norm of the restriction of $|{\rm{grad}} w|$ to that segment. Finally, we take $p$-th powers, integrate with respect to $x$, and take a $p$-th root to get that $||\Delta_hw||_p\leq |h||w|_{1,p}$”. I know through Holder’s inequality, $$|w(x-h)-w(x)|=\int_{x-h}^x|grad w|\cdot1\leq(\int_{x-h}^x|grad w|^pds)^{1/p}\cdot|h|^{1/p’}$$ Then $$||\Delta_h w||_p=(\int_\mathbb{R}|w(x-h)-w(x)|^p)^{1/p}\leq|h|^{1/p’}(\int_\mathbb{R}\int_{x-h}^x|grad w|^pdsdx)^{1/p}$$ It seems that through integral mean value theorem, $$(\int_\mathbb{R}\int_{x-h}^x|grad w|^pdsdx)^{1/p}=(\int_\mathbb{R}\int_{x-h/2}^{x+h/2}|grad w|^pdsdx)^{1/p}=|h|^{1/p}(\int_\mathbb{R}|grad w(\delta_x)|^pdx)^{1/p}$$ Where $\delta_x\in(x-h/2,x+h/2)$. But I can not convince myself it is bounded above by $|h|^{1/p}||gradw||_p$. Another idea is to use maximal function? Since \begin{align*} (\int_\mathbb{R}\int_{x-h/2}^{x+h/2}|grad w|^pdsdx)^{1/p}&=|h|^{1/p}(\int_\mathbb{R}\frac{1}{|h|}\int_{x-h/2}^{x+h/2}|grad w|^pdsdx)^{1/p}\\ &\leq |h|^{1/p}(\int_\mathbb{R}\mathrm{M}(|gradw|^p)(x)dx)^{1/p} \end{align*} And I want to use $||\mathrm{M}(f)||_p\leq C||f||_p$ when $p\geq2$. But I do not know how to use it to solve this problem. So any idea to prove equation $(*)$? Thanks in advance!
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$\begingroup$ Why you cannot use maximal function inequality (the one that you wrote at the end)? $\endgroup$– Paata IvanishviliCommented Jul 19, 2019 at 14:27
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$\begingroup$ I think I can use it. But is $\mathrm{M}(|gradw|^p)(x)\leq\mathrm{M}(|gradw|)^p(x)$ right? If it holds, I think the proof is complete. $\endgroup$– JohnLeeCommented Jul 19, 2019 at 14:54
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$\begingroup$ Yes. For $p>1$ you can use the inequality $\|M(|\nabla f|)\|_{p}| \leq C_{p} \| \nabla f\|_{p}$. $\endgroup$– Paata IvanishviliCommented Jul 19, 2019 at 15:30
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$\begingroup$ If $|h|^{1/p}(\int_\mathbb{R}\mathrm{M}(|gradw|^p)(x)dx)^{1/p}$ is $|h|^{1/p}(\int_\mathbb{R}(\mathrm{M}(|gradw|)(x))^pdx)^{1/p}$, everything is okay, But it is not... $\endgroup$– JohnLeeCommented Jul 20, 2019 at 0:46
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$\begingroup$ You can put maximal function at the very beginning (before you apply holder). Eventually you will not need to use holder at all. $\endgroup$– Paata IvanishviliCommented Jul 20, 2019 at 1:52
1 Answer
1) One way would be to apply Tonelli's theorem
\begin{align*} &\int_{\mathbb{R}} \int_{x-h}^{x}|w'(s)|^{p}ds dx = \int_{\mathbb{R}} \int_{\mathbb{R}} |w'(s)|^{p} \chi_{[x-h,x]}(s) dsdx =\\ & \int_{\mathbb{R}} \int_{\mathbb{R}}|w'(s)|^{p}\chi_{[x-h,x]}(s) dxds= |h| \int_{\mathbb{R}}|w'(x)|^{p}dx & \end{align*} You can notice that when $p=1$ you are not loosing too much (in this case there is no Holder's inequality).
2) Another more sophisticated way (which only works for $p>1$) would be to apply maximal function \begin{align*} |w(x-h)-w(x)|\leq \int_{x-h}^{h} |w'(s)|ds \leq |h| \tilde{M}(w') (x) \end{align*} where $\tilde{M}$ is the uncentered maximal function operator. Next, raise to the power $p$ ($p>1$) and apply Hardy--Littlewood, you will gain the constant $C_{p}$ which behaves as $\frac{p}{p-1}$.
