The first eigenvalue of Laplacian It's well known that for any bounded domain $\Sigma$, there exist unique $\lambda_\Sigma >0$
  and nonnegative $\varphi\in H_0^1(\Sigma)$ such that
  \begin{equation}\label{Eq-La-Bel-op}
    \left\{
      \begin{aligned}
        &-\Delta\varphi=\lambda_\Sigma\varphi
        \quad\mbox{in }\Sigma, \\
        &\varphi=0 \quad\quad\quad\quad~\mbox{on } \partial\Sigma,\\
      \end{aligned}
     \right.
  \end{equation}
  where $\Delta$ is the Laplace operator and $\lambda_\Sigma$ is the first eigenvalue of it.
Question: For any $\varepsilon>0$ small, does there eixist some smooth domain  $\Omega$ such that $\Sigma\subsetneq\Omega$ and 
$\lambda_{\Sigma}\leq\lambda_{\Omega}+\varepsilon$?
 A: One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)


*

*if $\Sigma$ is convex and $\Omega_n$ are convex sets tending to $\Sigma$ in Hausdorff distance, then $\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$. 

*If $\Omega_n$ are uniformly Lipschitz and converging to $\Sigma$ in Hausdorff distance, then $\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$

*If $\Omega_n\subset \mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $\Sigma$ in Hausdorff distance, then $\lambda_k(\Omega_n)\to \lambda_k(\Sigma)$.


Depending on the regularity and connectedness properties of your set $\Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
