= Solution
For the usual <Dirichlet problem> on a bounded regular volume, suppose $\phi_1,\phi_2$ both satisfy the stated data. Their difference $u=\phi_1-\phi_2$ is a <harmonic function>, with $\nabla^2u=0$ in $V$ and $u=0$ on $\partial V$. In <Green's first identity>, choose both functions equal to $u$:
$$
\int_V|\nabla u|^2\,dV=\int_{\partial V}u\frac{\partial u}{\partial n}\,dS=0.
$$
The nonnegative continuous integrand forces $\nabla u=0$. Thus $u$ is constant on each connected component, and its zero boundary values force each constant to vanish. This proves \b[uniqueness] for the <Dirichlet problem>.
If the boundary value is a constant $c$, the constant function $c$ is itself a <harmonic function> with those data. Uniqueness therefore yields
$$
\boxed{\phi\equiv c\text{ throughout }V.}
$$
Boundedness, or an appropriate condition at infinity replacing it, is essential. On the upper half-space both $u=0$ and $u=z$ are <harmonic functions> with zero data on its boundary plane. They show why unrestricted uniqueness must not be asserted for an unbounded volume.
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