Symmetry of the Green function

Remarks

Proof idea for a smooth boundary: fix $x \ne y$ and apply Green's second identity to $v(z) = G(z, x)$ and $w(z) = G(z, y)$ on $\Omega$ with small balls around $x$ and $y$ removed. Both functions are harmonic there and vanish on $\partial\Omega$, so only the small spheres contribute. As the radii tend to $0$, the sphere around $x$ gives $G(x, y)$ and the sphere around $y$ gives $G(y, x)$, up to the same constant, and the two must be equal. For a general domain one approximates it from inside by smooth domains, whose Green's functions increase to $G$. Example: on the unit disc, $G(z, w) = \log\abs{\frac{1 - \bar w z}{z - w}}$, and this is symmetric because $\abs{1 - \bar w z} = \abs{1 - \bar z w}$.

Going deeper

For a student

Symmetry means that the Green operator $f \mapsto \int_\Omega G(\cdot, y) f(y) \, dy$ is self-adjoint on $L^2(\Omega)$, which matches the self-adjointness of the Dirichlet Laplacian. It lets one treat $G(x, \cdot)$ as a function of the second variable, for example to see that $G(x, y)$ is harmonic in $y$ away from $x$. For operators that are not self-adjoint, the Green's function of the adjoint operator appears instead, and symmetry fails.