I was not able to solve the following homework problem:
Let $U$ be a bounded open set with $C^\infty$ boundary in $\mathbb{R}^n$. Let $L$ be a linear, symmetric uniformly elliptic operator. Let $w_k \in H^1_0(U)$ be an orthonormal basis of eigenfunctions of $L$ with respect to the $L^2$ inner product. Thus, $Lw_k = \lambda_k w_k$. Let $u \in H^1_0(U) \cap H^2(U)$ and let $u = \sum_{k=1}^\infty d_k w_k$. We know that, this series converges in $H_0^1(U)$. Prove that under the additional assumption $u \in H^2(U)$, the series also converges in $H^2(U)$.
Can someone tell me how to solve this problem?
My approach was to define an appropriate inner product on $H^1_0(U) \cap H^2(U)$ using the bilinear form that defines the weak solutions for the corresponding BVP. But I couldn't get anywhere in showing that bilinear form generates a norm that is equivalent to the standard norm on $H^1_0(U) \cap H^2(U)$?
Suggestions?
Throughout I'm assuming that you're working with the Dirichlet laplacian, i.e. $w_k=0$ on the boundary.
Consider the case $L=-\Delta$. Define the inner product $\langle u,v\rangle_X:= \langle u,v\rangle_{L^2}+ \langle \Delta u,\Delta v\rangle_{L^2}$ in $X:=H^1_0\cap H^2(\Omega)$. By global regularity of solutions, we know that in $X$ we have that this inner product (or rather the induced norm) is equivalent to the one in $H^2$ (use $\| D^2 u\|_{L^2}\lesssim \| \Delta u\|_{L^2}$ and that $\| Du\|_{L^2}^2\lesssim \| u\|_{L^2}\| \Delta u\|_{L^2}$).
Note that the sequence $(w_k)_k$ is orthogonal in $X$ with this inner product, and $\| w_k\|_X^2= 1+\lambda_k^2$.
Fix $u\in X$ and $d_k=\langle u,w_k\rangle_{L^2}$, so that $u=\sum\limits_k d_kw_k$ in $L^2$, and call $u_m$ the $m$-th partial sum. For any $k>l$ $$ \| u_m-u_l\|_{X}^2 = \left\| \sum_{k=l+1}^m d_kw_k\right\|_{X}^2 = \sum_{k=l+1}^m d_k^2\| w_k\|^2_X = \sum_{k=l+1}^m d_k^2(1+\lambda_k^2). $$ Now note, by definition of $d_k$, we have $$ d_k^2=\langle u,w_k\rangle_{L^2}^2 = \lambda_k^{-2} \langle u, \Delta w_k\rangle_{L^2}^2 = \lambda_k^{-2} \langle \Delta u, w_k\rangle_{L^2}^2. $$ Plugging this into the above estimate we have $$ \| u_m-u_l\|_X \leq \sum_{k=l+1}^m (\lambda_k^{-2}+1)\langle \Delta u, w_k\rangle_{L^2}^2 \leq (1+\lambda_l^{-2})\sum_{k=l+1}^m \langle \Delta u, w_k\rangle_{L^2}^2. $$ Since $\lambda_k\to \infty$ as $k\to \infty$, the right hand side in the last display tends to $0$ as $l\to \infty$.
Now I'll leave it to you to prove that the above still works for $L=-\partial_i(a_{ij}\partial_j)$ with $a_{ij}$ symmetric and smooth (I'm not sure exactly what class of $L$ you're working with, but it should at least contain this one). Consider the inner product $\langle u,v\rangle_X:= \langle u,v\rangle_{L^2}+ \langle Lu, Lv\rangle_{L^2}$.