I have already shown that the $\delta$-distribution is in $H^{-s}(\mathbb{R}), s>n/2$:
Every function $\phi \in H^s(\mathbb{R}^n)$ is continuous hence it can be evaluated at a point so using Sobolev embedding theorem we can say:
$|\delta(\phi)|=|\phi(0)| \leq ||\phi||_{\infty}=||\phi||_{C_b(\mathbb{R}^n)} \leq C||\phi||_{H^s(\mathbb{R}^n)}$
Hence $\delta$ is a continuous linear functional on $H^s(\mathbb{R})$.
Now I need to show using Fourier transformation that $\delta \in H^{-1}(\mathbb{R})$. So what I want to show is that $||\phi||_{\infty} \leq C||\phi||_{H^{1}(\mathbb{R)}}$. Now if we denote $F^{-1}$ as the inverse Fourier transformation we get:
$||\phi||_{\infty}=||F^{-1} \overline{\phi}||_\infty=\sup_{x \in \mathbb{R}} \frac{1}{(2\pi)^{1/2}}|\int_{\mathbb{R}}e^{ix.\xi}\overline{\phi}(\xi)d\xi| \leq \frac{1}{(2\pi)^{1/2}}\sup_{x \in \mathbb{R}}\int_{\mathbb{R}}|e^{ix.\xi}||\overline{\phi}(\xi)|d\xi \leq \frac{1}{(2\pi)^{1/2}}\int_{\mathbb{R}}|\overline{\phi}(\xi)|d\xi $
(Why is the supremum attained at $x=0$? Or why does the $e$ vanish?)
Now $C||\phi||_{H^1(\mathbb{R})}=C(\int_\mathbb{R}(1+|\rho|^2)|\overline{\phi}(\rho)|^2d\rho)^{1/2}$
Now I'm unsure how to compare these two integrals. I'm especially unsure about the $^{1/2}$ in the norm.
Is it correct what I have shown up to now and how could I continue?