Given a weighted graph, where the nodes add and subtract from each other. Is there convergence? Informally: Assume you have a set of weighted nodes (each node has a value between 0 and 1). Further, the nodes may influence each other by either adding to connected nodes or subtracting from them. Is there a convergent distribution of values?  
Formally: 
Assume a graph $G =\langle N, w, R_+, R_-\rangle$, where


*

*$N$ is a set of nodes, 

*the weighting $w$ is a function from $N$ to $[0,1]$

*the support relationship $R_+$ is a binary relation on $N$

*the attack relationship $R_-$ is a binary relation on $N$


Let the damping factor $d$ be such that $0 < d \leq 1$. 
Let $f^i$ ($i \in \mathbb{N} $) be a function from N to 
$\mathbb{R}$ such that for any  $v\in N$, 


*

*if $i = 0$, then $f^i(v) = w(v)$, 

*otherwise,
$$  
   f^i(v) = w(v) + d \ 
                   (\sum_{x \in \{y | y R_+ v \} } f^{i-1}(x) -
                    \sum_{x \in \{y | y R_- v \} } f^{i-1}(x) )   
$$


Does $f^i$ converge? 
 A: If we interpret $f$ as vector indexed by $N$ and $R$'s as matrices representing the characteristic function of the relation, i.e., that have $1$ in appropriate row and column if and only if $u\ R\ v$, then we can set $R = R_+ - R_-$ and observe that
\begin{align}
f^i
 &= w + d(R_+ f^{i-1} - R_- f^{i-1}) \\
 &= w+d(R_+-R_-)f^{i-1} \\
 &= w+dRf^{i-1} \\
 &= w+dR(w+dRf^{i-2}) \\
 &= w+dR(w+dR(w+dR(w+dRf^{i-3}))) \\
 &\hspace{6pt}\vdots \\
 &= \sum_{k=0}^i (dR)^kw = \left(\sum_{k=0}^i (dR)^k\right)w
\end{align}
The matrix power series converge iff $\rho(dR) < 1$ (see convergent matrix and spectral radius), in particular it does when $\|dR\|<1$.
In such case $I-(dR)$ is invertible and $\sum_{k=0}^{\infty} (dR)^k = (I-dR)^{-1}$.
To give you some concrete examples, for $d = 1$ we can use
$$N = \{1,2\}, w(1) = 1, w(2) = 1, R_+=N^2, R_- = \varnothing$$
and then $f$ grows infinitely, by $1$ every turn. 
You can also create a cycle:
$$N = \{1,2\}, w(1) = 1, w(2) = 0, R_+ = \{1 \to 2, 2\to 1\}, R_- = \{1 \to 1, 2\to 2\},$$
where $f^i(x)$ flips between $w(x)$ and $1-w(x)$.
You can also adjust that example so that $R_-$ is non-reflexive and perhaps also include a cost for attacking/supporting.
For $d = \frac{1}{2}$ you can take 
$$N = \{1,2,3,4,5\}, w(n) = 1, R_+ = N^2, R_- = \varnothing$$
so $f^i(n) = 1 + \frac{1}{2}(5\cdot f^{i-1}(n)-0)$ which is again unbounded.
I hope this helps $\ddot\smile$
A: In terminology of the first answer, the infinite sum $\sum_{k=0}^{\infty} (dR)^k$ can be generalised to $\sum_{k=0}^{\infty} c_k R^k$, where $c_k$ are suitable coefficients, i.e. those of a Taylor series. Then it should be possible to obtain convergence for all matrices $R$. 
More specifically, consider $\sum_{k=0}^{\infty} \frac{1}{k!} R^k$ (which formally is the Maclaurin series for the exponential function). This converges, because $\frac{\rho(R)^k}{k!}$ converges to zero.
