Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$ be bounded and open. Assume $\overline\Omega$ is a $d$-dimensional properly embedded $C^\infty$-submanifold of $\mathbb R^d$.
Let$^1$ $(e_n)_{n\in\mathbb N}\subseteq H_0^1(\Omega)$ such that $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $L^2(\Omega)$ and an orthogonal basis of $H_0^1(\Omega)$ and satisfies $$\mathfrak a(u,e_n)=\lambda_n\langle u,e_n\rangle_{L^2(\Omega)}\;\;\;\text{for all }u\in H^1(\Omega),$$ where$^1$ $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for }u,v\in H^1(\Omega)$$ and $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ is nondecreasing with $\lambda_n\xrightarrow{n\to\infty}\infty$.$^2$
Assume $u\in L^\infty(\Omega)$. We can write $u=\sum_{n\in\mathbb N}\langle u,e_n\rangle_{L^2(\Omega)}e_n$, where the sum exists in $L^2(\Omega)$. Does it necessarily hold that already $u\in H^1(\Omega)$?
In order to show that $u\in H^1(\Omega)$, we can observe that $$\int_\Omega e_n\nabla\varphi=-\int\varphi\nabla e_n\tag1$$ for all $\varphi\in C_c^\infty(\Omega)$ and $n\in\mathbb N$. If $u_N:=\sum_{n=1}^N\langle u,e_n\rangle_{L^2(\Omega)}e_n$, then it clearly holds $$\int_\Omega u\nabla\varphi=\lim_{N\to\infty}\int_\Omega u_N\nabla\varphi=\lim_{N\to\infty}\sum_{n=1}^N\langle u,e_n\rangle_{L^2(\Omega)}\int_\Omega e_n\nabla\varphi\tag2$$ for all $\varphi\in C_c^\infty(\Omega)$ and $N\in\mathbb N$.
Does the right-hand side converge to an expression of the form $-\int_\Omega\varphi g$ for some $g\in L^2(\Omega,\mathbb R^d)$ s.t. we can at least conclude that $u\in H^1(\Omega)$? If so, how do we show that even $u\in H^1(\Omega)$?
$^1$ By the Poincaré inequality, $\mathfrak a$ is an inner product on $H_0^1(\Omega)$ which is equivalent to the usual inner product on $H_0^1(\Omega)$.
$^2$ i.e. $((e_n,\lambda_n))_{n\in\mathbb N}$ is the eigenpair sequence corresponding to the weak Laplacian on $\Omega$.