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> $\operatorname{cap}(\{0\})$.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-226/2/c/solution