1

By definition of the Sobolev space $W^{m,p}$ we have : $$W^{m,p}(\Omega)=\{u\in L^p(\Omega)\ |\ \forall \alpha \text{ such that } |\alpha|\le m, D^{\alpha}u\in L^p(\Omega)\}$$

Can someone give me a reference where it is explained how we find the others definitions given by: $$H^m(\mathbb{R}^n)=\{u\in L^2(\mathbb{R}^n)\ ∣\ \int_{\mathbb{R^n}}|\hat{u}(\xi)|^2 (1+|\xi|^2)^m d\xi < \infty\}$$ and $$u \in H^s(\mathbb{R}^n) \quad\text{ iff }\quad (1-\Delta)^{s/2}\in L^2(\mathbb{R}^n)$$

Thanks in advance

2 Answers2

0

One reference is Stein's Singular integrals. Bessel and Riesz potentials are covered there, and you need those to establish the relation with the Laplacian.

detnvvp
  • 8,237
0

I advise you "Real Analysis Modern Techniques And Their Applications" by G.Folland. Formally to answer your question, you have

$\widehat{\Delta u}(\xi)=\sum_{i=1}^n \widehat{\partial_{xi} \partial_{xi} u}(\xi)=-4\pi^2 |\xi|^2 \widehat{u}(\xi)$

Now, $H^s(\mathbb{R}^n):=\lbrace u \in L^2(\mathbb{R}^n) : \Lambda^s u \in L^2(\mathbb{R}^n) \rbrace$ with $\omega_s(\xi):=(1+|\xi|^2)^{s/2}$, and $\Lambda^s u := \mathcal{F}^{-1}(\omega_s \widehat{u})$, and by Plancherel theorem we have the $H^s$-norm

$\displaystyle \left \| u \right \|_{H^s}= \left \| \Lambda^s u \right \|_{L^2}= \left\| \mathcal{F}(\Lambda^s u) \right\|_{L^2}=\left \| \omega_s \widehat{u} \right \|_{L^2}=\left( \int_{\mathbb{R}^n} |\widehat{u}(\xi)|^2 (1+|\xi|^2)^s d\xi \right)^{1/2}$

Therefore we can write $\Lambda^s = (Id - (1/4\pi) \Delta^{s/2})$. Note that the constants depend on how you define the Fourier transform.

user288972
  • 2,360
  • here is a link of thebook:http://www.olumcamp.ir/Math/Books/Analysis/data/Real%20Analysis%20folland.pdf but i didn't find the answer here –  Mar 17 '16 at 12:21
  • Link-only and reference-only links are not very helpful here. Could you include the major points in your referred work? – Rory Daulton Mar 17 '16 at 12:24
  • The fact that you say is an immediate consequence of the definition, and the Fourier transform of the Laplace operator. @user317150 – user288972 Mar 17 '16 at 12:28