Laplacian on the Sierpiński gasket

Example

Let $p_1, p_2, p_3$ be the vertices of an equilateral triangle in $\R^2$ and let $F_i(x) = (x + p_i)/2$ for $i = 1, 2, 3$. The Sierpiński gasket $K$ is the unique nonempty compact set with $K = F_1(K) \cup F_2(K) \cup F_3(K)$, the attractor of this iterated function system. Let $V_0 = \set{p_1, p_2, p_3}$ and $V_{m+1} = F_1(V_m) \cup F_2(V_m) \cup F_3(V_m)$. Let $\Gamma_m$ be the graph on $V_m$ in which $x \sim_m y$ when $x$ and $y$ are two vertices of one triangle $F_{w_1} \circ \cdots \circ F_{w_m}(\text{triangle } p_1 p_2 p_3)$. For $u \colon V_m \to \R$, set $$\mathcal{E}_m(u) = \left(\frac{5}{3}\right)^m \sum_{x \sim_m y} (u(x) - u(y))^2,$$ the sum over edges of $\Gamma_m$.

The factor $5/3$ is the one that makes these energies compatible. Take $u(p_1) = 1$ and $u(p_2) = u(p_3) = 0$, so $\mathcal{E}_0(u) = 2$. Among all extensions of $u$ to $V_1$, the one of least energy $\mathcal{E}_1$ follows the "$1/5$–$2/5$ rule". At the midpoint of $p_j p_k$ its value is $(2 u(p_j) + 2 u(p_k) + u(p_i))/5$, where $i$ is the third index. Here this gives $2/5$ at the midpoints of $p_1 p_2$ and $p_1 p_3$, and $1/5$ at the midpoint of $p_2 p_3$. The three small triangles contribute $18/25$, $6/25$ and $6/25$ to the sum, in total $6/5$, and $(5/3) \cdot (6/5) = 2 = \mathcal{E}_0(u)$.

Because of this compatibility, $\mathcal{E}_m(u|_{V_m})$ is nondecreasing in $m$ for every $u$, and its limit $\mathcal{E}(u)$ is a resistance form on $K$. Let $\mu$ be the self-similar probability measure that gives mass $3^{-m}$ to each level-$m$ cell. For a suitable function $u$ and a point $x \in V_m \setminus V_0$, the Laplacian of $u$ with respect to $\mu$ is $$\Delta_\mu u(x) = \frac{3}{2} \lim_{m \to \infty} 5^m \sum_{y \sim_m x} (u(y) - u(x)).$$ The factor $5^m = (5/3)^m \cdot 3^m$ combines the energy scaling with the measure scaling. The function on $V_1$ found above satisfies $\sum_{y \sim_1 x} (u(y) - u(x)) = 0$ at each of the three midpoints. For example, at the midpoint of $p_2 p_3$ the neighbours have values $0, 0, 2/5, 2/5$, and $4/5 - 4 \cdot (1/5) = 0$. Its continuation by the same rule at all levels is a harmonic function on $K$, that is, $\Delta_\mu u = 0$ off $V_0$.