Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and let $L$ be a second order elliptic operator on $\Omega$, whose associated bilinear form $a(\cdot, \cdot)$ is bounded and coercive on $H^1_0(\Omega)$. Let $G$ be the Green's function of $L$, and let $w$ be an eigenfunction of $L$ with eigenvalue $\lambda$, which we assume is in $ L^{\infty}(\Omega)$.
Assuming for the moment that $w$ is sufficiently smooth, the property $L G(\cdot, y) = \delta_y$ of the Green's function and self-adjointness of $L$ implies that $$ w(y) = \int_{\Omega} w(x) \delta_y(x) \, d x = \int_{\Omega} w(x) L_x G(x, y) \, d x = \int_{\Omega} G(x,y) L_x w(x) \, d x = \int_{\Omega} \lambda G(x,y) w(x) \, d x. $$ In general, $w$ is only in $H^1_0(\Omega)$ and may not be continuous, so the expression $\int_{\Omega} w(x) \delta_y(x) d x$ may not be meaningful. However, the identity $$w = \int_{\Omega} \lambda G(x,\cdot) w(x) \, dx$$ is still meaningful in the sense of equality of functions in $H^1_0(\Omega)$. I suspect it is true, but I am not sure if my proof is correct. Any alternative proof is also welcome.
My attempt
By assumption, $w \in L^{\infty}(\Omega)$. Thus we also have $\lambda w \in L^{\infty}(\Omega)$, so property (iii) of Green's functions below implies that the function $u$ defined by $$ u(y) = \int_{\Omega} G(x,y) \lambda w(x) \, dx $$ is in $H^1_0(\Omega)$ and satisfies $L u = \lambda w$ in the sense of distributions. But we also have $L w = \lambda w$, so by uniqueness of solutions, $u = w$. Thus we have shown that $$ w(y) = \int_{\Omega} G(x,y) \lambda w(x) \, dx $$ Is this correct?
More details
$L$ is a second order elliptic operator associated to a symmetric positive bilinear form $a$. That is, for $u \in H^1_0(\Omega)$ and $f \in H^{-1}(\Omega) \colon=(H^1_0(\Omega))^{*}$, the equation $L u = f$ is understood in the weak sense $a(u, v) = (f,v)$ for all $v \in H_0^1(\Omega)$ (where $H^1_0(\Omega)$ is the closure of $C^{\infty}_{0}(\Omega)$ in the Sobolev space $H^1(\Omega)$). We assume that $a$ is bounded and coercive on $H^1_0(\Omega)$, so by the Lax-Milgram theorem the above equation has a unique solution for any $f \in H^{-1}(\Omega)$.
Let $G$ be the Dirichlet Green's function of $L$, characterized by the following:
- For any $y \in \Omega$, $G(\cdot, y)$ is locally integrable and $L G(\cdot, y) = \delta_y$ in the sense of distributions, i.e. $\int_{\Omega} L_x G(x, y) \varphi(x) dx \colon = a(G(\cdot, y), \varphi) = \varphi(y)$ for all $\varphi \in C^{\infty}_c(\Omega)$.
- For any $y \in \Omega$ and radius $r > 0$, $G(\cdot, y) \in H^1(\Omega \setminus B_r(y))$, and $G(\cdot, y)$ vanishes on $\partial \Omega$.
- For any $f \in L^{\infty}(\Omega)$, the function $u$ given by $u(y) = \int_{\Omega} G(x,y) f(x) \, dx$ belongs to $H^1_0(\Omega)$ and satisfies $L u = f$ in the sense of distributions, i.e. $a(u, \varphi) = (f, \varphi)$ for all $\varphi \in C^{\infty}_c(\Omega)$.