Let $u_k$ be eigenfunctions for operator $-\Delta$ over domain $\Omega\subset \mathbb R^2$, open bounded, smooth boundary. Then, we know that $E:=\{u_k\}_{k=1}^\infty$ forms a basis for $L^2$ and we have $-\Delta u_k=\lambda_k u_k$ (with $0$ boundary condition)
Now, let $u\in W_0^{1,1}(\Omega)\setminus W_0^{1,2}(\Omega)$ be given, and $u_\epsilon:=u\ast \eta_\epsilon$ be the mollification of $u$. (We assume $u\equiv 0$ in a nbhd of $\partial \Omega$ so that $u_\epsilon\in C_c^\infty(\Omega)$ for $\epsilon>0$ small enough. )
Hence, we may write $$ u_\epsilon=\sum_{k=1}^\infty d^\epsilon_k u_k,\,\,\,u=\sum_{k=1}^\infty d_k u_k $$ where $d_k^\epsilon:=\left<u_\epsilon,u_k\right>_{L^2}$ and $d_k:=\left<u,u_k\right>_{L^2}$.
By the definition of $u$, we have $$\|\nabla u_\epsilon\|_{L^2}=\sum_{k=1}^\infty (d_k^\epsilon)^2\lambda_k <\infty \text{ but } \|\nabla u\|_{L^2}=\sum_{k=1}^\infty (d_k)^2\lambda_k=\infty$$
Let $K:=\{k\in\mathbb N,\,d_k\neq 0\}$ and $K_\epsilon:=\{k\in\mathbb N,\,d_k^\epsilon\neq 0\}$ and we assume that $$ \sup_{k\in K}\{\lambda_k\}=M<\infty. $$ Then, can we prove that $$ \sup_{\epsilon>0}\sup_{k\in K_\epsilon}\{\lambda_k\}<\infty, $$ or even $$ \lim_{\epsilon\to0}\sup_{k\in K_\epsilon}\{\lambda_k\}=M? $$