Let $s$ be any real number. $D$ is dense in $H^s$.
$D$ is the space of $C^\infty$ functions with compact support. $u\in H^s$ means, by definition of this space of functions, $\hat{u}\in L^2(\mathbb{R};(1+|\xi|^2)^sd\xi)$ and $\hat{u}\in L^2_{loc}(\mathbb{R})$. And $\hat{u}=Fu$, the Fourier Transform.
Question: I have always proved density, given $f\in H^s$, by trying to find $f_n\in D$ which $\lim ||f_n-f||_{H^s}=0$. BUT I have found the following proof. I don't understand why does it prove it.
Proof He considers a distribution $u\in H^s$ such that, for any test function $\phi$ (I think he intends to say "for any $\phi\in D"$) we have: $\int_\mathbb{R^d} \hat{\phi}(\xi)(1+|\xi|^2)^s\bar{\hat{u}}(\xi)d\xi=0$ Which means that $\hat{u}$ is a tempered distribution. As the multiplication by $(1+|\xi|^2)^{-s}$ is continuous in $S'$ (Space of tempered distributions), we have that $\bar{\hat{u}}=0$ as a tempered distribution. Thus $u\equiv{}0$.
Why does this conclude the density of one space into the other? I can understand each step of the proof, it's the strategy what I don't see, its relation with density (With proving $H^s\subset \bar{D}$).
Thank you in advance
First note that the integral you've written (that the author assumes is zero) is precisely the inner product $\langle \phi, u\rangle_{H^s}$.
Now pick an element $u$ in the orthogonal complement of $D$ in $H^s$. Our goal is to prove that $u$ is zero, and hence $\overline D = H^s$. (This implies the result by the standard fact that for a Hilbert space, $(D^\perp)^\perp = \overline D$.) By assumption, for any $\phi \in D$, $\langle u, \phi\rangle_{H^s} = 0$, and thus by the proof you just gave $u = 0$.