by a slight abuse of notation, denote this extension again by $u_n$. The $\gamma$-convergence of $(\Omega_n)$, together with the compact embedding of $H^1_0(B_1)$ in $L^1(B_1)$, imply that (i) ${\lambda}_1({\Omega}_n) \rightarrow {\lambda}_{\varepsilon}$, and (ii) $u_n\rightharpoonup v_{\varepsilon}$ weakly in $H^1_0(B_1)$ and strongly in $L^1(B_1)$. Assuming that $$\label{L infty}
\lim_{n \to \infty }\|u_n\|_\infty = \|v_{\varepsilon}\|_\infty.$$ we get $$\liminf_{n \rightarrow +\infty} \frac{\int_{{\Omega}_n} u_n dx}{\|u_n\|_\infty |{\Omega}_n|} \ge  \frac{\int_{B_1} v_{\varepsilon}dx}{\|v_{\varepsilon}\|_\infty |B_1|}.$$ Since the right-hand side is arbitrarily close to $1$ when ${\varepsilon}\downarrow 0$, the proof of Theorem \[the7\] is then completed by a diagonal selection procedure.

It remains to show . This kind of assertion is known to be true in a general setup. In essence it is a consequence of the subharmonicity of the eigenfunctions $u_n$, $n \ge 1$. (For a similar result for the torsion function see [@HLP Theorem 2.2].) For the sake of completeness, we give below a proof. It slightly differs from the one in [@HLP Theorem 2.2].

First note that since $u_n \to v_{\varepsilon}$ strongly in $L^1$, it follows that $$\liminf _{n \to +\infty} \|u_n\|_\infty  \ge  \|v_{\varepsilon}\|_\infty .$$ To prove that $ \limsup _{n \to +\infty} \|u_n\|_\infty  \le  \|v_{\varepsilon}\|_\infty $ we argue as follows. Being convergent, the sequence $({\lambda}_1(\Omega_n))$ is bounded, 