Eigenvalues of the Laplacian in $L^1$ space Let $\Omega \subset \mathbb{R}^N$ be bounded domain with smooth boundary. The first eigenvalue of $-\Delta \colon W_0^{1,2}(\Omega) \cap W^{2,2}(\Omega) \rightarrow L^2(\Omega)$ is strictly positive. This is well known and proved by using the energy method. 
But I learned that $-\Delta$ can be extended up to $L^1(\Omega)$.
In this case, is the first eigenvalue still strictly positive? 
 A: Throughout this answer, let $\Omega \subseteq \mathbb{R}^d$ be non-empty, open and bounded. To answer the question, I think some clearifications are needed first.


*

*The Sobolev space $W^{2,2}_0(\Omega)$ (which is by definition the closure of all test functions on $\Omega$ with respect to the $W^{2,2}$-norm) is probably not the right domain for the Laplace operator for this question, since the Laplace operator defined on this domain does not need to have any eigenvalues at all. As an example, let $\Omega = (0,1) \subseteq \mathbb{R}$. Then every function in $f \in W^{2,2}_0(\Omega)$ is an element of $C^1([0,1])$ and both the function $f$ and its derivative $f'$ vanish at $0$ and $1$. Hence, the equation $f'' = \lambda f$ does not have a solution in $W^{2,2}_0(\Omega)$ for any $\lambda \in \mathbb{C}$.

*The operator $\Delta: L^2(\Omega) \supseteq W^{2,2}_0(\Omega) \to L^2(\Omega)$ is not even self-adjoint (though it is certainly symmetric): It follows from form methods that the operator 
\begin{equation}
    \Delta: L^2(\Omega) \supseteq \{f \in W^{1,2}_0(\Omega): \; \Delta f \in L^2(\Omega)\} \to L^2(\Omega)
\end{equation}
is a self-adjoint operator on $L^2(\Omega)$, and the domain of this operator is strictly larger then $W^{2,2}_0(\Omega)$ (loosely speaking, since for a function $f \in W^{2,2}_0(\Omega)$ not only $f$ itself but also the derivatives of $f$ have to vanish at the boundary). Since 
\begin{equation}
    \lambda - \Delta: \{f \in W^{1,2}_0(\Omega): \; \Delta f \in L^2(\Omega)\} \to L^2(\Omega) 
\end{equation}
is bijective for every $\lambda \in \mathbb{C} \setminus \mathbb{R}$, the operator 
\begin{equation}
     \lambda - \Delta: W^{2,2}_0(\Omega) \to L^2(\Omega)
\end{equation}
cannot be surjective for any such $\lambda$. As the spectrum of every linear operator is closed, it follows that the operator considered in the question has the entire complex plane as its spectrum. In particular, the operator is not self-adjoint.

*As pointed out in 2. the appropriate domain to realise the Dirichlet Laplace operator on general bounded open sets $\Omega$ is
\begin{equation}
    \tag{*} \{f \in W^{1,2}_0(\Omega): \; \Delta f \in L^2(\Omega)\}.
\end{equation}
When defined on this domain, the Laplace operator is self-adjoint and has compact resolvent. In particular, the spectrum of the operator $\Delta$ with this domain is discrete and consists of eigenvalues only. It follows from Poincaré's inequality (which holds for arbitrary open sets that are contained in strip - and therefore, in particular, for bounded open sets) that the first eigenvalue of $-\Delta$ is strictly positive. 

*In case that $\Omega$ is sufficiently regular - say, simply connected and with $C^2$-boundary - it follows from elliptic regularity theory that the domain $(*)$ coincides with $W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$. That's probably why the Laplace operator in the question is supposed to be defined on a Sobolev space. 

*To finally answer the question: Let $\Delta$ be the Laplace operator on $L^2(\Omega)$ defined on the domain $(*)$. Then $\Delta$ can indeed be extended to an operator on $L^1(\Omega)$. This can, for instance, be done by $C_0$-semigroup methods: The $C_0$-semigroup on $L^2(\Omega)$ generated by $\Delta$ is positive and contractive with respect to the $L^1$-Norm. Hence, the semigroup can be extended to $L^1(\Omega)$ and it is again a $C_0$-semigroup on $L^1(\Omega)$. Let us denote the generator of this $C_0$-semigroup by $\Delta_1$. Then the domain of $\Delta_1$ containes the domain $(*)$ and it acts as the Laplace operator on $(*)$; hence, $\Delta_1$ can be seen as a natural extension of $\Delta$ to $L^1(\Omega)$. It follows from ultra contractivity theory of $C_0$-semigroups and from Gaussian estimates that the semigroup on $L^1(\Omega)$ is analytic and consists of compact operators (so the spectrum of $\Delta_1$ consists of eigenvalues, again), and that the spectrum of $\Delta_1$ coincides with the spectrum of $\Delta$. In particular, the first eigenvalue of $-\Delta_1$ is strictly positive. 

*I'm not quite sure, but I don't think that one can find a "good" description of the exact domain of $\Delta_1$. Even if $\Omega$ is very regular, this is probably not possible since elliptic regularity theory does not work on $L^1(\Omega)$.
The details for the arguments sketched in 5. can for example be found in Section 7 of [Wolfgang Arendt, Semigroups and Evolution Equations: Functional Calculus, Regularity and Kernel Estimates, in Evolutionary equations. Vol. I, 1–85 (2004)]
