Assume $A\subset B$ and use the corrected strict identity from part (i). If a walk starts at $y\in A$ and never returns to $A$, then transience makes $L_B<\infty=\widetilde H_A$. Thus
$$
\begin{aligned}
\operatorname{cap}(A)
&=\sum_{y\in A}\mu_yP_y(\widetilde H_A=\infty)\\
&\leq\sum_{y\in A}\mu_yP_y(L_B<\widetilde H_A,L_B\geq0)\\
&=P_{e_B}(H_A<\infty)\\
&\leq\sum_{z\in B}e_B(z)
=\operatorname{cap}(B).
\end{aligned}
$$
This proves monotonicity of the <capacity of a finite set for a transient random walk>.

Solution

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

For the normalized <Green function of a transient weighted graph>, the Green identity is
$$
\mathcal E(g(\mathord\cdot,0),f)=f(0),
\qquad f\in H_0.
$$
Taking $f=g(\mathord\cdot,0)$ and using $h=g(\mathord\cdot,0)/g(0,0)$ therefore gives
$$
\mathcal E(h,h)
=\frac{\mathcal E(g(\mathord\cdot,0),g(\mathord\cdot,0))}{g(0,0)^2}
=\frac1{g(0,0)}.
$$
This is also the <capacity of a finite set for a transient random walk> $\operatorname{cap}(\{0\})$.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 / 2 / c / Solution

{title2=$\operatorname{cap}(A)$}

The capacity of a finite set is the total mass of its <equilibrium measure of a finite set>:
$$
\operatorname{cap}(A)=\sum_xe_A(x)=\mathcal E(h_A,h_A).
$$
Capacity is monotone under inclusion, though strict inclusion need not give strict inequality.

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 Incoming links: Capacity of a finite set for a transient random walk