Fix a domain $\Omega \subset \mathbb{R^2}$. Define $H_0^1(\Omega)$ to be the closure of test functions $D(\Omega)$ with respect to the dirichlet norm $$ \|\varphi\|_\nabla^2 = \int_\Omega |\nabla \varphi|^2\,. $$ I want to show that this is equivalent to defining $H_0^1(\Omega)$ to be the space $$ \mathscr{H}_0^1(D) = \{ f \in D^\star(\Omega) : \|f\|_\lambda^2 := \sum_n \langle f,e_n \rangle^2 \lambda_n < \infty\}\,, $$ where $(e_n)$ and $(\lambda_n)$ are the sequences of eigenfunctions and eigenvalues, respectively, of the Laplace operator $- \Delta$ on $\Omega$. Note that $(e_n)$ is an orthonormal sequence w.r.t. the $L^2$-norm and that both $H_0^1$ and $\mathscr{H}_0^1$ are Hilbert spaces.
It is easy to see that $\|\varphi\|_\nabla = \|\varphi\|_\lambda$ for test functions and thus $H_0^1$ is a closed subspace of $\mathscr{H}_0^1$.
So the question is: how to show that for an arbitrary $f \in \mathscr{H}_0^1$ we have $f \in H_0^1$ with $\|f\|_\lambda = \|f\|_\nabla$?
Edit: I guess I have to assume that the eigenfunctions $(e_n)$ vanish on the boundary $\partial \Omega$. Now the terms $(f,e_n)$ make more sense since by regularity theory the eigenfunctions are test functions.
Edit2: One heuristic idea is that $$ \sum_n \langle f,e_n \rangle^2 \lambda_n = \sum_n \langle (-\Delta)f,e_n \rangle \langle f, e_n \rangle \simeq (-\Delta f, f)_{L^2} = \|f\|_\nabla^2 $$ and this quantity being finite would imply some extra regularity on the distribution $f$, allowing us to deduce that it belongs to $H_0^1$.
In fact, let $f=\sum_{n}e_nf_n$ where $f_n=\left<f,e_n\right>$. Then \begin{eqnarray} \|f\|_\nabla^2&=&\int_{\Omega}\|\nabla f\|^2dx=\int_{\Omega}(-\Delta f)fdx\\ &=&\int_{\Omega}\sum_n(-\Delta e_n f_n)\sum_{n}e_nf_ndx\\ &=&\int_{\Omega}\sum_n(\lambda_n e_n f_n)\sum_{n}e_nf_ndx\\ &=&\sum_{m,n}f_mf_n\int_{\Omega}\lambda_me_me_ndx\\ &=&\sum_{n}\lambda_n|f_n|^2\\ &=&\sum_{n}\lambda_n\left<f,e_n\right>^2\\ &=&\|f\|_\lambda^2. \end{eqnarray}