Skip to main content
MathOverflow

Return to Question

added 10 characters in body
Source Link
JohnLee
JohnLee
  • 53
  • 4

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!

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, 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!

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!

Source Link
JohnLee
JohnLee
  • 53
  • 4

Characterization of Besov space with Lp-modulus of continuity

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, 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!