$L^2$ and Sobolev space

Question

In Raymond's book on Pseudodifferential Operator page 18, he says , where $S'$ is the tempered distributions, we define sobolev space of exponent $s$ as

$u \in S'$ with $\lambda^s \hat{u} \in L^2$. This is equivalent to $\hat{u}$ is a function satisfying $$ ||u||_s^2 =(2 \pi)^{-n} \int (1+|\xi|^2 )^s |\hat{u}(\xi)|^2 \,d \xi < \infty $$ $\lambda^s (\xi) = (1+|\xi|^2 )^\frac{s}{2}$.

My concern is if $u \in S'$, then $u:S \rightarrow \Bbb R$ is a continuous semi linear norm. How does this guarantee $\hat{u}$ is a "function"?

Let me make this more precise.

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Let us denote $u \in S'$ by $L_u$ as it is a linear functional. So $L_u: S \rightarrow \Bbb C$ is a tempered dsitribution. We define $\hat{u}$ by $$L_{\hat{u}} (\varphi) = L_{u}(\hat{\varphi}(-x)), \quad \forall \varphi \in S$$

So this is our $\hat{u}:=L_{\hat{u}}$. This is an element in $S'$.

So what exactly does it mean to say that $\lambda^s \hat{u} \in L^2$? So I guess we embed $L^2$ into $S'$ I.e. there exists $\varphi \in L^2$ such that

$$\lambda^s \hat{u} (f) = \int f \bar{\varphi} $$ for all $f \in S$?

Answer 1

Your confusion seems to be on understanding the assumption $\lambda^s \hat u \in L^2$, since $\hat u$ is only an element of $S'$.

1. Important to keep in mind is the identification of $f\in L^1_{loc}$ with the distribution $T_f\in S'$ defined by $(T_f,\phi) = \int f\bar{\phi}$ for every $\phi\in S$. The map $ T : L^1_{loc} \to S'$ that sends $f\mapsto T_f$ is a continuous embedding, and one usually abusively writes $f=T_f$, but I'll avoid using it below.

2. On the one hand, as $\lambda^s\in C^\infty$ with polynomial growth of derivatives, $\lambda^s \hat u$ is well-defined as an element of $S'$. On the other hand, the assumption $\lambda^s \hat u\in L^2$ is read as (to be understood as) there exists $f\in L^2$ such that $T_f = \lambda^s \hat u$. [By the above identification $f=\lambda^s\hat u$, we treat $\lambda^s \hat u$ as a function.]

3. This implies that $\hat u$ is $T_g$ for some $g\in L^2$. Indeed, set $g = f/\lambda^s$, a well defined quotient of measurable functions. As $\lambda^s$ is bounded a.e. away from $0$, and $f\in L^1_{loc}$, $g$ is in $L^1_{loc}$. Then $$ (\hat u,\phi) =( \hat u,\lambda^s\lambda^{-s} \phi ) = (\lambda^s \hat u,\lambda^{-s} \phi ) = (T_f,\lambda^{-s} \phi)=(T_{f\lambda^{-s}}, \phi) = (T_g,\phi)$$ so $\hat u = T_g$.

PS for $s\ge 0$, $u$ is also a function, since $\hat u \in L^2$ by $L^\infty-L^1$ Holder's inequality. There are distributions that are not functions in $H^s$ for $s<0$. For instance, the Dirac delta which is in $S'(\mathbb R^d)$, is also in $H^{-d/2-\epsilon}(\mathbb R^d)$ for every $\epsilon>0$.

Answer 2

It i,s just the definition of $L^{2}$ norm. Just square $\lambda^{s}\hat {u}$ and integrate. I thing the book defines $L^{2}$ space by taking the reference measure as Lebesgue measure divided by $(2\pi)^{n}$.