Let $\Omega\in \mathbb{R}^n$ be a bounded domain with a fixed $x\in \Omega$. Let $u$ be a function subject to $$\begin{cases} \Delta u=\delta (y-x), & \forall y \in \Omega\\ u=0, & \forall y \in \partial\Omega\end{cases} $$
Prove that $u$ has a fixed sign and determine this sign.
My approach: Intuitively speaking, since $\delta$ is a nonnegative "function" inside $\Omega$, so by the maximum principle we have $u(y)<\max_{\partial\Omega} u=0$ which implies the sign is negative. However, this argument is not rigorous as $\delta (y-x)$ is a distribution rather than a function. Could anyone make my argument rigorous? I think the main part of the argument is to deal with the singularity at $y=x.$