If $u = \sum_{k=1}^\infty d_k w_k$ where $d_k = (u,w_k)_{L^2(U)}$, why is $\sum_{k=1}^\infty d_k^2 = \|u\|^2_{L^2(U)}$.

Question

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!

Answer 1

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}

Answer 2

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.