Definition of Sobolev spaces: Fourier transform of tempered distribution

Question

I consider in "McLean - Strongly Elliptic Systems and Boundary Integral Equations" the definition of the Sobolev space for $s \in \mathbb R$ $$ H^s(\mathbb R^n) := \{u \in \mathcal S^*(\mathbb R^n) \colon \mathcal J^s u \in L^2(\mathbb R^n) \} $$ with the inner product $$ (u,v)_{H^s(\mathbb R^n)} := (\mathcal J^s u, \mathcal J^s v)_{L^2(\mathbb R^n)} $$ and the induced norm $$ \|u\|_{H^s(\mathbb R^n)} := \|\mathcal J^s u\|_{L^2(\mathbb R^n)}. $$

$\mathcal J^s \colon \, \mathcal S(\mathbb R^n) \to \mathcal S(\mathbb R^n)$ is the Bessel potential of order $s \in \mathbb R$ defined by $$ \mathcal J^s u(x) := \int_{\mathbb R^n}(1 + |\xi|^2)^{\frac s 2} \hat u(\xi) \mathrm e^{\mathrm i 2 \pi \xi \cdot x} \mathrm d\xi \quad \text{ for } x \in \mathbb R^n $$ where $\hat u$ is the Fourier transform of $u,\,$ $\mathcal S(\mathbb R^n)$ is the Schwartz space and $\mathcal S^*(\mathbb R^n)$ is the space of tempered distributions. The Bessel potential has a natural extension to $\mathcal J^s \colon \, \mathcal S^*(\mathbb R^n) \to \mathcal S^*(\mathbb R^n)$ by $$ \langle \mathcal J^s u, \varphi \rangle := \langle u,\mathcal J^s \varphi \rangle. $$ My question is: If $u \in H^s(\mathbb R^n)$ and so $J^s u \in L^2 (\mathbb R^n),$ is then $\hat u$ a function for all $s \in \mathbb R?$ In McLean (and many other books) the Sobolev norm has the integral representation $$ \|u\|^2_{H^s(\mathbb R^n)} = \int_{\mathbb R^n}(1 + |\xi|^2)^{s} | \hat u(\xi)|^2 \mathrm d\xi $$ for all $s \in \mathbb R$ and so $\hat u$ must be function. But I haven't a proof for this. Is there anybody who can help me?

Thanks.

Answer 1

HINT: One has $ (1+\lvert\xi\rvert^2)^\frac{s}{2}\ge 1$. This, and Plancherel's theorem.

EDIT: This actually addresses another question, namely, "When $s\ge 0$, is it true that $u\in H^s$ implies $u\in L^2$?". Indeed, when $s<0$, it might well happen that $u\in H^s$ is not a function. (The Dirac delta $\delta$ is such that $\delta\in H^{-s}$ for $s> \frac{n}{2}$).

What is always true, however, is that the Fourier transform $\hat{u}$ is a function if $u\in H^s$. This can be seen as follows. By definition, the pointwise product $$(1+\lvert \xi\rvert^2)^{\frac{s}{2}}\hat{u}=\hat{v}$$ is a function, that is, it belongs to $L^1_\mathrm{loc}$. Now we observe that any pointwise product $f\cdot g$ with $f\in L^1_\mathrm{loc}$ and $g\in L^\infty_{\mathrm{loc}}$ is an element of $L^1_\mathrm{loc}$. Apply this observation to $f=\hat{v}$ and $g=(1+\lvert \xi\rvert^2)^{\frac{-s}{2}}$.

Answer 2

We know from the comments above for $h:= w_s \hat u \in L^2(\mathbb R^n)$ that it holds $$ \langle w_s \hat u, \varphi \rangle = \langle \hat u, w_s \varphi \rangle = \langle h, \varphi \rangle_{L^2(\mathbb R^n)} $$ for all $\varphi \in S(\mathbb R^n)$ and if we choose $\varphi := w_{-s} \psi \in \mathcal S (\mathbb R^n)$ with $\psi \in \mathcal S (\mathbb R^n)$ it follows $$ \langle w_s \hat u, w_{-s} \psi \rangle = \langle \hat u, w_s w_{-s} \psi \rangle = \langle \hat u, \psi \rangle = \langle h, w_{-s} \psi \rangle_{L^2(\mathbb R^n)} = \langle w_{-s} h, \psi \rangle_{L^2(\mathbb R^n)}. $$ That means we get for all $\psi \in \mathcal S (\mathbb R^n)$ the relation $$ \langle \hat u, \psi \rangle = \langle w_{-s} h, \psi \rangle_{L^2(\mathbb R^n)}. $$ So $\hat u$ has the representation by the function $\xi \mapsto w_{-s}(\xi) h(\xi) = (1+|\xi|^2)^{-\frac s 2} h(\xi)$ with $h \in L^2(\mathbb R^n).$ I think that is the proof I wanted. Any improvements?