DEFINITION: Let $s\in \mathbb{R}$. Then $$ H^{s}(ℝ)=\left\{f\in S'(\mathbb{R}):(1+|\xi|^2)^\frac{s}{2}, \hat{f} \in L^{2}(\mathbb{R}) \right\} $$
OBSERVATION: $f \in H^{s}(\mathbb{R})$ means that $\hat f$ is a measurable function. We introduce $||f||^{2}_{s}=\int_{\mathbb{R}}(1+|\xi|^2)^{s}|\hat{f}(\xi)|^{2}d\xi = ||(1+|\xi|^2)^\frac{s}{2}\hat{f}||^{2}_{L^{2}}$, which comes from the scalar product $(f, g)_s=\int_\mathbb{R} (1+|\xi|^2)^s\hat{f}(\xi)\overline{\hat{g}(\xi)}d\xi$.
PROPOSITION: $(H^s(\mathbb{R}), (.,.)_s)$ is a Hilbert Space
Proof. $(.,.)_s$ is well-defined because if $f,g\in H^s(\mathbb{R})$, then $$ (f, g)_s=\int_{\mathbb{R}}(1+|\xi|^2)^\frac{s}{2}\hat{f}(\xi)\overline{(1+|\xi|^2)^\frac{s}{2}\hat{g}(\xi))}d\xi \le||(1+|\xi|^2)^\frac{s}{2}\hat{f}||_{L^2}||(1+|\xi|^2)^\frac{s}{2}\hat{g}||_{L^2}\lt+\infty $$ due to the Cauchy-Schwartz inequality.
So, what I'm trying to do is I want to prove the above proposition. And now I am stuck with it. Can anyone help me get this done? :-D
