$L^2$ and Sobolev space 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. 

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 $f \in S$? 

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 $(2\pi)^{n}$. 
