To show that $R(r)$, the effective resistance between two given vertices of a finite network, is a concave function, we need to prove the inequality:
$$ \frac{1}{2}\left[R\left(r_1\right)+R\left(r_2\right)\right] \leq R\left(\frac{1}{2}\left(r_1+r_2\right)\right) $$
Step 1: Interpret Effective Resistance via Rayleigh's Monotonicity and Convexity
The effective resistance $R(r)$ can be characterized using the Dirichlet energy associated with a given resistance configuration.
Given a network with edge resistances $r(e)$, let $L(r)$ be the associated Laplacian matrix. The effective resistance between two vertices $u$ and $v$ is given by:
$$ R(r)=\left(\delta_u-\delta_v\right)^T L(r)^{\dagger}\left(\delta_u-\delta_v\right) $$
where $L(r)^{\dagger}$ is the Moore-Penrose pseudoinverse of $L(r)$, and $\delta_u$ and $\delta_v$ are the indicator vectors of the vertices.
The key property used here is that $L(r)$ is a function of $r$, and its pseudoinverse preserves concavity.
Step 2: Effective Resistance as a Convex Function of Conductances Define the conductance $c(e)=\frac{1}{r(e)}$. The effective resistance can be expressed in terms of conductances:
$$ R(c)=\left(\delta_u-\delta_v\right)^T L(c)^{\dagger}\left(\delta_u-\delta_v\right) $$
A well-known result in electrical network theory states that $R(c)$ is a convex function of $c$. That is, for two sets of conductances $c_1$ and $c_2$,
$$ R\left(\frac{1}{2}\left(c_1+c_2\right)\right) \leq \frac{1}{2} R\left(c_1\right)+\frac{1}{2} R\left(c_2\right) $$
Since resistance is the reciprocal of conductance $(r=1 / c)$, the transformation from conductance to resistance reverses the inequality direction, making $R(r)$ concave in $r$.
Step 3: Conclusion By applying the convexity of $R(c)$ and the transformation $r=1 / c$, we obtain:
$$ \frac{1}{2} R\left(r_1\right)+\frac{1}{2} R\left(r_2\right) \leq R\left(\frac{1}{2}\left(r_1+r_2\right)\right) $$
Thus, $R(r)$ is concave in $r$, as required.
https://gist.github.com/viadean/7588d6d0252267c08d8e4b25dde7a153
Explanation of the Code: