Discrete Cauchy integral formula : The interior values are always convex combinations of exterior values for harmonic functions? Let $T={\mathbb Z}^2$. For $t=(x,y)\in T$, the neighborhood 
$N(t)$ of $t$ is the four-point set $\lbrace x\pm 1;y\pm 1\rbrace$. A map
$f:T \to {\mathbb R}$ is harmonic iff $4f(t)=\sum_{s\in N(t)}f(s)$ for
any $s\in t$. 
Let $A\subseteq {\mathbb Z}^2$. For $a\in A$, say that $a$ is interior if $N(a)\subseteq A$. Denoting by $I(A)$ the set of interior points in $A$, $a\in A$ is a limit point if $a\not\in I(A)$ but $N(a)\cap I(A)\neq \emptyset$. Note that some points in $A$ may be neither interior nor limit points (e.g. when $A$ is a singleton).
My question is, is it always true that given a finite $A\subseteq {\mathbb Z}^2$ and given an interior point $a\in A$, if we denote by $L_1,L_2,L_3,\ldots,L_t$
the limit points of $A$, then there are nonnegative coefficients 
$c_1,c_2,\ldots,c_t$, depending only on $A$ and $a$, with
$\sum_{k}c_k=1$, such that $f(a)=\sum_{k}c_kf(L_k)$ for any harmonic $f$.
 A: I'll say that a function is harmonic on a set $A$ if $4f(t)=\sum_{s\in N(t)}f(s)$ holds for all $t\in I(A)$. 
Claim 1. If a harmonic function $h$ on $A$ is nonnegative at all limit points of $A$, then it is nonnegative at all interior points of $A$.
Proof. Suppose to the contrary that $h(c)<0$ for some $c\in I(A)$. Let $M$ be the minimum of $h$ on $I(A)$, and let $B=\{x\in I(A): h(x)=M\}$. Since $B$ is not all of $\mathbb Z^2$, there is a point $b\in B$ that neighbors a point $b'\notin B$. The harmonic property implies $h(b')=M$. Therefore, $b'\notin I(A)$. Since $b'$ neighbors $b$, it is a limit point of $A$. But then $h(b')\ge 0$, a contradiction. $\qquad \Box$
Corollary: if $h=0$ at all limit points of $A$, then it is zero at all interior points of $A$.
Claim 2. For each $i=1,\dots,t$ there exists a  harmonic function $\omega_i$ on  $A$ such that $\omega_i(L_j)=\delta_{ij}$ (Kronecker delta). This function $\omega_i$ satisfies $0\le \omega_i \le 1$ in $I(A)$. Its values in $I(A)$ are uniquely determined, even if $\omega_i$ itself is not.
Proof. The uniqueness part follows from Corollary. To show existence, define the energy of function $f$ as $$E(f) = \sum_{\{a,b\}\cap I(A)\ne\varnothing, \ b\in N(a) } |f(a)-f(b)|^2$$
Let $\omega_i:A\to \mathbb R$ be a function of smallest energy with prescribed values at limit points (such a function exists by compactness). Then $\omega_i$ is harmonic on $A$, for otherwise we could decrease its energy by redefining it at some interior point. The inequality $0\le \omega_i\le 1$ holds by  Claim 1. $\qquad \Box$
Now we can let $c_i=\omega_i(a)$ and conclude.
