Construction of Sobolev space

Question

I am reading about the construction of Sobolev spaces from $L^2$. In the book I read (Introduction to Partial Differential Equations, by Gerald B. Folland) that the operator used to construct those spaces for any $s \in \mathbb{R}$, which is defined to be the operator $\Lambda^s$ such that:

$\widehat{\Lambda^{s}u}(\xi) = (1 + |\xi|^2)^{s/2}\widehat{u}(\xi)$

Where $\widehat{u}$ is the Fourier transform of $u$, is equivalent to the operator $\Lambda^s = (I - (2\pi)^{-2}\Delta)^{s/2}$.

How can you show that? I supposed $\Delta$ is the Laplacian because it is a book about PDE, but I'm not sure either, because I don't get the link between the Laplacian and the Fourier transformation!

Answer 1

In general, if $u \in C^2(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, we 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 $\Lambda^s = (Id - (1/4\pi) \Delta^{s/2})$.

Answer 2

The link is that if $f\in C^2 \cap L^2$, $$\mathcal{F}[\Delta f](\xi) = -|\xi|^2 \mathcal{F}[f](\xi)$$

So taking the laplacian is like multiplying by $-|\xi|^2$ in Fourier

So, by extension, you can define an operator $\Delta$ by

$\Delta f (x) = \mathcal{F^{-1}} \left[ -|\xi|^2 \mathcal{F}[f](\xi) \right](x)$$

Now, about your operator $\Lambda^s$, the trick is to consider $s$ as an even integer, and see what it means for a $C^{s}$ function.

$$(1+|\xi|^2)^{n} = \sum_{k=0}^n {n\choose k} |\xi|^{2k}$$

And if you take the Fourier inverse, it become a sum of derivative.

And with $s$ a real, it's like taking fractionnal derivatives