The inequality need not be strict. Start with the nearest-neighbour <weighted graph> $\mathbb Z^3$, attach a new <leaf of a graph>[leaf] $b$ only to the origin $o$, and give its edge positive weight. Take $A=\{o\}$ and $B=\{o,b\}$. A walk starting from $b$ must move to $o$ at its first non-killed step, while any walk reaching $b$ from elsewhere must first pass through $o$. Hence the <equilibrium potential of a finite set> is the same for the two sets:
$$
h_A=h_B.
$$
Using $\operatorname{cap}(C)=\mathcal E(h_C,h_C)$ for a finite set $C$ gives
$$
\operatorname{cap}(A)=\operatorname{cap}(B)
$$
despite $A\ne B$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 / 1 / b / iii / Solution

{title2=$h_A$}

For a finite vertex set $A$, its equilibrium potential is $h_A(x)=P_x(H_A<\infty)$. It equals one on $A$, is harmonic off $A$, and belongs to the <Dirichlet energy space with zero boundary at infinity>.

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 Incoming links: Equilibrium potential of a finite set