= Solution
With the <Minkowski metric> $(+---)$, the momentum conjugate to the real field vector is $\boldsymbol\pi=\dot{\boldsymbol\phi}$. The <Hamiltonian density> is
$$
\mathcal H=\tfrac12\boldsymbol\pi^2+\tfrac12\sum_i(\partial_i\boldsymbol\phi)^2+V(\boldsymbol\phi).
$$
\b[Assume the potential has a <global minimum>] $V_{\min}$ and the boundary conditions permit a constant minimizing field. Then
$$
E-\int d^dx\,V_{\min}
=\int d^dx\left[\tfrac12\boldsymbol\pi^2+\tfrac12(\nabla\boldsymbol\phi)^2+V(\boldsymbol\phi)-V_{\min}\right]\geq0.
$$
The bound is attained by
$$
\boxed{\boldsymbol\phi(t,\mathbf x)=\boldsymbol\phi_*,\qquad
V(\boldsymbol\phi_*)=V_{\min},\qquad\dot{\boldsymbol\phi}_*=0.}
$$
The <homogeneous vacuum minimum criterion> proves existence of a homogeneous lowest-energy configuration under those assumptions. On infinite space one subtracts the constant <vacuum energy> before comparing finite energy differences. This identifies the classical vacuum configuration; it does not assert that the quantum field has no fluctuations.
The claim does not follow from “no derivative dependence in $V$” alone. For example, the allowed functional form $V(\phi)=-\phi^4$ is unbounded below, so there is no state of lowest energy to identify. Boundary conditions forcing different vacua at opposite ends can instead require an inhomogeneous minimum, such as a domain wall. The positive quartic potential used in the following parts does have homogeneous minima, so these caveats do not obstruct its vacuum calculation.
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