Stability of numerical scheme using von Neumann condition I'm trying to investigate the numerical scheme
$$\sum_{j=-m_{1}}^{m_{1}} \alpha_{j} U_{i+j}^{n+1}=\sum_{j=-m_{2}}^{m_{2}} \beta_{j} U_{i+j}^{n}$$
and I am trying to find a set of conditions for the defining coefficients, i.e., $\{\alpha_j\}$ and $\{\beta_{j}\}$ such that the scheme is stable in $l^{\infty}.$
So far the only approach that came in my mind was the von Neumann theorem, however I've no idea of how to apply it to this problem. Any help or hint is appreciated.
 A: EDIT: von Neumann stability analysis is based on Fourier series and applies to the case of $l^2$, rather than $l^{\infty}$, so it is not a natural first choice unless there is extra information available. Regardless, there is at least one alternative as demonstrated below.

ORIGINAL: Let us analyze the implicit method for solving the initial value problem for the heat equation on the real line and see if we can generalize the underlying principles. 
The initial value problem takes the form $$u_{t}(x,t) = u_{xx}(x,t), \quad u(x,0) = g(x)$$ where $x\in \mathbb{R}$ and $t \ge 0$. Euler's implicit method takes the form $$ \frac{v^{n+1}_j-v^{n}_j}{k} = \frac{v^{n+1}_{j+1}-2v^{n+1}_j+v^{n+1}_{j-1}}{h^2}$$
where $0<k$ and $0<h$ are the step sizes in time and space. We use $v^{n}_j$ to approximate $u(jh,nk)$. This equation is equivalent to
$$ -\lambda v^{n+1}_j + (1 + 2\lambda) v_j^{n+1} - \lambda v^{n+1}_{j-1} = v^{n}_j$$ where $\lambda = k/h^2$ characterizes the grid. It is worth emphazising that this is an equation involving elements in $l^{\infty}$. We have $$Av^{n+1} = v^{n}$$ where $A : l^{\infty} \rightarrow l^{\infty}$ is given by $$ (Ax)_j = -\lambda x_{j+1} + (1+2 \lambda) x_j - \lambda x_{j-1}.$$
It is worth repeating how we measure size in $l^{\infty}$. By definition we have $$\|x\|_{\infty} = \underset{j \in \mathbb{Z}}{\sup} \{|x_j|\},$$ and $$\|A\|_\infty = \underset{\|x\|_\infty = 1}{\sup} \|Ax\|_\infty.$$ It follows immediately that $$\|A\|_\infty \leq (1+ 4\lambda)$$ so that not only have we correctly identified the co-domain of $A$, but $A$ is a bounded operator. At this point, it is clear that our equation has a unique solution. To that end we write $A$ as $A = I - E$ where $E : l^{\infty} \rightarrow l^{\infty}$ is the operator given by $$(Ex)_j = \lambda x_{j+1} - 2\lambda x_j + x_{j-1}.$$ We observe that $$\|E\|_\infty \leq 4 \lambda.$$ In particular, we have $$\|E\|_\infty < 1$$ when $\lambda<\frac{1}{4}$. It follows that $A$ is invertible and $$A^{-1} = \sum_{j=0}^\infty E^j$$ when $\lambda<\frac{1}{4}$. I suspect that $A$ is invertible in general, but the analysis presented here does not cover the case of $\lambda \ge \frac{1}{4}$. Regardless, it is now sensible to speak of the solution of the solution of Euler's equation whenever $\lambda<\frac{1}{4}$. We will rewrite it as
$$ (1 + 2 \lambda) v^{n+1}_j = v^n_j + \lambda v^{n+1}_j + \lambda v^{n+1}_{j-1}.$$
At first glance this accomplishes nothing, but the application of the triangle inequality  allows us to conclude that
$$ (1 + 2 \lambda) |v^{n+1}_j| \leq \|v^n\|_\infty + 2 \lambda \|v^{n+1}\|_\infty$$
from which it follows that
$$  \|v^{n+1}_j\|_\infty \leq \|v^n\|_\infty,$$
This shows that Euler's implicit method is stable for the initial value problem for the heat equation on the real line provided $\lambda<\frac{1}{4}$.

What are the key ingredients in this analysis? What carries to your more general case? We need something akin to diagonal dominance, i.e., the ability to write our operator as $I - E$ where $\|E\|_\infty < 1$. In your more general case you have an equation of the form $$(I-E)v^{n+1} = B v^n$$ and you will need to select assumptions such that $$\|E\|_\infty < 1, \quad\text{and} \quad \|(I - E)^{-1} B\|_\infty \leq 1.$$ It is conceivable that your application can accept exponential growth as a function of $t=nk$, but that depends on the differential equation of which I know nothing.
