= 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> give
$$
M'(r)=\frac1{4\pi r^2}\int_{\partial B(y,r)}\partial_nu\,dS
=\frac1{4\pi r^2}\int_{B(y,r)}\Delta u\,dx.
$$
If $u$ is harmonic, this derivative is zero; since $M(0)=u(y)$ by continuity, \b[every spherical mean equals the value at the centre]. Integrating $4\pi r^2M(r)$ 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, $\Delta u\geq0$, so $M'(r)\geq0$. Thus
$$
\boxed{u(y)\leq\frac1{|\partial B(y,R)|}\int_{\partial B(y,R)}u\,dS,
\qquad u(y)\leq\frac1{|B(y,R)|}\int_{B(y,R)}u\,dx.}
$$
These are the corresponding mean value inequalities. A <subharmonic function> has the maximum principle on a ball: apply the second-derivative test to $u+\epsilon|x-y|^2$, whose Laplacian is strictly positive and hence cannot have an interior maximum, then let $\epsilon\downarrow0$.
For $w=|\phi|^2$, the given equation says $\Delta\phi=iV\phi$. Since $V$ is real,
$$
\Delta w=2|\nabla\phi|^2+2\operatorname{Re}(\overline\phi\Delta\phi)
=2|\nabla\phi|^2\geq0.
$$
The subharmonic maximum principle gives $w(x)\leq\max_{\partial B}w$ throughout the ball. Taking square roots yields
$$
\boxed{\sup_{B(y,R)}|\phi|\leq\sup_{\partial B(y,R)}|\phi|.}
$$
No sign condition on the real potential $V$ is needed.
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