 Solution

For a $C^2$ function, let $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(y+r\omega )d\omega$ be its spherical mean. Differentiation and the divergence theorem giveIf $u$ is harmonic, this derivative is zero; since $M(0)=u(y)$ by continuity, every spherical mean equals the value at the centre. Integrating $4\pi r^{2}M(r)$ over $0<r<R$ gives the same statement for the volume mean on a ball. This proves the mean value property for harmonic functions.

If $u$ is subharmonic, $\Delta u\ge 0$, so $M^{\prime }(r)\ge 0$. ThusThese are the corresponding mean value inequalities. A subharmonic function has the maximum principle on a ball: apply the second-derivative test to $u+ϵ|x-y{|}^2$, whose Laplacian is strictly positive and hence cannot have an interior maximum, then let $ϵ\downarrow 0$.

For $w=|\varphi {|}^2$, the given equation says $\Delta \varphi =iV\varphi$. Since $V$ is real,The subharmonic maximum principle gives $w(x)\le {\max }_{\partial B}w$ throughout the ball. Taking square roots yieldsNo sign condition on the real potential $V$ is needed.