= Solution
For the usual <scalar field theory> with a nonnegative potential $U$, shifted to zero in the vacuum, write its static energy in $d$ dimensions as
$$
E[\phi]=T+P,\qquad T=\frac12\int|\nabla\phi|^2d^dx,\qquad P=\int U(\phi)d^dx.
$$
Apply <Derrick scaling> $\phi_\lambda(x)=\phi(\lambda x)$. The gradient acquires a factor $\lambda$ and the volume a factor $\lambda^{-d}$, giving
$$
E(\lambda)=\lambda^{2-d}T+\lambda^{-d}P.
$$
A smooth finite-energy static solution is a stationary point of this energy, so its <Derrick virial identity> is
$$
\boxed{(d-2)T+dP=0.}
$$
For $d\geq3$, both terms are nonnegative, forcing $T=P=0$. The field is a constant vacuum: \b[there is no nontrivial finite-energy static soliton in this scalar-gradient-plus-nonnegative-potential model]. The assumptions matter: extra derivative terms, gauge fields, time dependence or a potential not bounded below by its vacuum value can change the argument.
For $d=2$, the relation becomes $P=0$ and says nothing further about $T$. A gradient-only energy is scale invariant, so <Derrick theorem> alone does not rule out nonconstant solutions; <sigma-model lumps> are examples. For a scalar field with isolated potential minima, $P=0$ constrains the continuous field to one vacuum and hence makes it constant. If the vacuum values form a nontrivial manifold, the scaling relation alone leaves open a nonconstant map into it. In neither case does the two-dimensional scaling argument itself prove existence.
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