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Let $H^s(\mathbb{R}^n)$ be the Sobolev space of order $s>1$.

I want to show that

the norms

$\|f\|_s^2:=\int\limits_{\mathbb{R}^d} (1+|\xi|^2)^s |\mathscr{F}f(\xi)|^2d\xi$

and

$[f]_s^2:= \|f\|_{H^{m}(\mathbb{R}^n)}^2+\int\limits_{\mathbb{R}^d\times \mathbb{R}^d} \frac{|f(x)-f(y)|^2}{|x-y|^{n+\sigma 2}}d(x,y)$

are equivalent on that space, where $m$ is the smallest integer such that $\sigma:=s-m\in (0,1)$

I know that for smooth functions $f$ one has $\mathscr{F}(\partial^m f)=i^\alpha x^\alpha \mathscr{F}f$ but I still stuck somehow, when doing the calculations.

Yimin
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Furkan
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  • related https://math.stackexchange.com/questions/4757253/how-do-i-show-that-the-definition-of-sobolev-slobodeckij-spaces-ws-p-omega?rq=1 ā€“ Yimin Dec 30 '23 at 02:09
  • Not sure if this is entirely related. They compare fractional sobolev spaces defined in a certain way with the regular sobolev spaces when the order of derivatives is a integer. Im asking whether these norms for $s>1$ are equivalent. In Hitchhikers guide for $s<1$. they provide a proof. But nothing is mentioned about $s>1$. ā€“ Furkan Dec 30 '23 at 02:57
  • It should rather be $\nabla^mf(x)-\nabla^mf(y)$ in your last integral. These kind of equivalences can be found in Triebel's books about function spaces. ā€“ LL 3.14 Dec 30 '23 at 09:55

1 Answers1

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So, first, there is an error in your formula, it should be $$ [f]_s^2 = \|f\|_{H^m}^2 + \iint \frac{|\nabla^mf(x)-\nabla^mf(y)|^2}{|x-y|^{d+2\sigma}}\,\mathrm d x \,\mathrm d y. $$ Define the Fourier transform $$ \widehat{u}(y) = \mathcal F(u)(y) = \int_{\mathbb R} e^{-2iĻ€ \,x\cdot y} \,u(x)\,\mathrm d x. $$ Recall that for $\sigma\in(0,1)$, the fractional Laplacian verifies $\mathcal{F}((-\Delta)^{\sigma/2} u) = |2\pi y|^\sigma \,\widehat{u}(y)$ and also $$ (-\Delta)^{\sigma/2} u(x) = c_{d,2\sigma} \int\frac{u(x)-u(y)}{|x-y|^{d+\sigma}} \,\mathrm d y $$ which is just that fact that the Fourier transform of a product is the convolution of the Fourier transforms (see e.g. the last part of the answer here). From the integral formula and by symmetry in $x$ and $y$ it follows that $$ \int u\,(-\Delta)^{\sigma} u = \frac{c_{d,s}}{2} \iint \frac{|u(x)-u(y)|^2}{|x-y|^{d+2\sigma}}\,\mathrm d x \,\mathrm d y =: |u|_{\dot{H}^\sigma}^2. $$ On the other hand, by Plancherel formula $$ \int u\,(-\Delta)^{\sigma} u = \int |2\pi y|^{2\sigma}\, |\widehat{u}(y)|^2\,\mathrm d y. $$ Therefore, taking $u = \nabla^mf$, $$ |\nabla^mf|_{\dot{H}^\sigma}^2 = (2\pi)^{2s} \int |y|^{2\sigma}\, |y^{\otimes m}\widehat{u}(y)|^2\,\mathrm d y \\ = (2\pi)^{2s} \int |y|^{2s}\, |\widehat{u}(y)|^2\,\mathrm d y. $$ On the other hand, by Plancherel again $\|f\|_{H^m}^2 = \|f\|_{L^2}^2 + \|\nabla^m f\|_{L^2}^2 = \int (1+|2\pi y|^{2m})\, |\widehat{f}|^2\,\mathrm d y$, hence finally $$ [f]_s^2 = \int (1+|2\pi y|^{2m}+\tfrac{2}{c_{d,2\sigma}}|2\pi y|^{2s})\, |\widehat{f}(y)|^2\,\mathrm d y. $$ The equivalence between your two norms now just follow from the fact that the function $$ y \mapsto \frac{1+|2\pi y|^{2m}+\tfrac{2}{c_{d,2\sigma}}|2\pi y|^{2s}}{(1+|2\pi y|^2)^s} $$ is bounded above and below by positive constants.

LL 3.14
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