If $u = \sum_{k=1}^\infty d_k w_k$ where $d_k = (u,w_k)_{L^2(U)}$, why is $\sum_{k=1}^\infty d_k^2 = \|u\|^2_{L^2(U)}$. I'm going through a proof from Evans, and I'm stuck on what I think must be an easy linear algebra question.
Let $\{w_k\}_{k=1}^\infty$ be an orthonormal basis of $L^2(U)$, where $w_k\in H_0^1(U)$ is an eigenfunction with eigenvector $\lambda_k$.
If $u\in H_0^1(U)$, we can write
$u = \sum_{k=1}^\infty d_k w_k$
where $d_k = (u,w_k)_{L^2(U)}$. Additionally,
$\sum_{k=1}^\infty d_k^2 = \|u\|^2_{L^2(U)}$.
Why does this last equality hold? Is there some simple way to determine $||u||^2_2$ in terms of the orthonormal basis?
Any help appreciated!
 A: This is just the Parseval Identity:
Let $H$ be an inner product space, and $(e_n)_{n=1}^\infty$ an orthonormal basis for $H$. Then for any $x \in H$ we have:
$$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2$$
This follows from linearity and continuity of the inner product: 
Since $(e_n)_{n=1}^\infty$ is an orthonormal basis we have:
$$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n$$
So:
\begin{align}
\|x\|^2 = \langle x, x \rangle = \left\langle \sum_{n=1}^\infty \langle x, e_n \rangle e_n, \sum_{n=1}^\infty \langle x, e_n \rangle e_n\right\rangle = \sum_{n=1}^\infty \langle x, e_n \rangle \overline{\langle x, e_n \rangle} = \sum_{n=1}^\infty \left|\langle x, e_n\rangle\right|^2
\end{align}
A: We compute:
\begin{align*}
\|u\|_{L^{2}(U)}&=\langle u,u\rangle_{L^{2}(U)}\\
&=\left\langle\sum_{k=1}^{\infty}d_{k}w_{k},\sum_{\ell=1}^{\infty}d_{\ell}w_{\ell}\right\rangle_{L^{2}(U)}\\
&=\sum_{k,\ell=1}^{\infty}d_{k}d_{\ell}\langle w_{k},w_{\ell}\rangle\\
&=\sum_{k=1}^{\infty}d_{k}^{2},
\end{align*}
since $\langle w_{k},w_{\ell}\rangle=\delta_{k,\ell}$ by definition of an orthonormal basis.
