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Let $v$ be the voltage with $v(a)=1$, $v(b)=0$, and harmonic values at every other vertex. For any other admissible $f$, write $f=v+g$, where $g(a)=g(b)=0$. Its discrete Dirichlet energy expands asDiscrete summation by parts turns the cross term intobecause $v$ is harmonic in the interior and $g$ vanishes at the boundary. Thus $v$ minimizes the energy. Under a unit voltage drop, its energy equals the total current from $a$ to $b$, namely the effective conductance $1/R_{\mathrm{eff}}(a,b)$. Therefore the Dirichlet principle gives

Let the square contain the Euclidean ball of radius $R$, set $r_x=\max (1,\Vert x\Vert )$, and define the logarithmic cutoffThen ${\phi }_0=1$ and $\phi =0$ on the boundary. On an edge at radius comparable to $r$, the mean value theorem gives $|{\phi }_x-{\phi }_y|\le C/(r\log R)$. There are $O(r)$ edges in the annulus of radius $r$, hence the discrete Dirichlet energy satisfiesThis logarithmic cutoff is the discrete manifestation of recurrence in two dimensions.