$\Vert . \Vert_{H^1}$ is induced by an inner product and $(H_0^1, \Vert . \Vert_{H^1})$ is a Hilbert space

Question

Any function $f \in L^2 := L^2_{\mathbb C}[0,1]$ is uniquely determined by its Fourier series and $f=\sum_{n \in \mathbb Z}f'_ne_n$ where

$f'_n=<f,e_n>$ and $e_n(x) := e^{i2\pi n x}$.

Denote by $L_0^2$ the subspace with $f'_0=0$ and define $H_0^1:=\{f \in L_0^2 : \Vert f \Vert _{H^1} < \infty\}$ and $\Vert f \Vert_{H^1}^2:=\sum_{n \neq 0}(2n\pi)^2\vert f'_n\vert^2$.

I have to show that $\Vert . \Vert_{H^1}$ is induced by an inner product and that $(H_0^1, \Vert . \Vert_{H^1})$ is a Hilbert space. But I'm confused by all those notations and definitions and I'm not sure where to start. Could someone help me? Thanks in advance!

Answer 1

Consider the map $U : H_0^1\mapsto \ell^2$ defined by $$ f \in H_0^1 \mapsto (2\pi \hat{f_1},4\pi\hat{f_2},6\pi\hat{f_3},\cdots)\in \ell^2 $$ Show that $\|Uf\|_{\ell^2}=\|f\|_{H_0^1}$, and argue that $U$ is surjective. Because $\ell^2$ is a Hilbert space, then $H_0^1$ must also be a Hilbert space because it is isometrically isomorphic to a Hilbert space.