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Let $f$ be a function that maps $\mathbb{Z}^2$ to $\mathbb{R}$ and consider the operator $T$ which replaces the value of $f$ at $(i,j)$ by the average of the values of $f$ at its four neighbors (left, right, down, up): $$ Tf(i,j) = \frac{f(i-1,j) + f(i+1,j) + f(i,j-1) + f(i,j+1)}{4}.$$ The discrete-version of Liouville's theorem say that the equation $$ Tf = f$$ does not have any solutions $f$ which are bounded.

My question is whether its possible to prove this by demonstrating that $T$ is a contraction mapping in some sense.

Specifically, let $\mathbb{S}$ be the set of bounded $f: \mathbb{Z}^2 \rightarrow \mathbb{R}$ with the equivalence relation $f = g$ whenever $f-g$ is a constant. Is there a metric on $\mathbb{S}$ with respect to which $T$ is a strict contraction?

Note that is related, but not identical with, my other question. In particular, a positive answer to this question would likely imply an answer to that question as well.

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No. If such a metric existed then $T^2$ would also be a contraction, hence would also have a unique fixed point (namely the equivalence class of constant functions), which it doesn't: $T^2$ fixes every function such that $f(i, j) = g(i + j \bmod 2)$ for some $g$, and there are infinitely many equivalence classes of such functions.

(More generally, in order for such a metric to exist for a general set map $T : X \to X$ it is necessary that $T^n$ has a unique fixed point for all $n \in \mathbb{N}$. As it turns out, this is sufficient.)

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  • $\begingroup$ Very nice. Actually, I had meant $Tf(i,j)$ to include $f(i,j)$ besides the four neighboring points, so that $T^2$ (and higher powers) all have a unique fixed point. The inverse fixed point theorem you cite then implies such a metric exists. $\endgroup$ – morgan Jun 7 '12 at 5:30
  • $\begingroup$ I hope you don't mind a follow up question: for the case I mentioned in the previous comment (i.e., $Tf(i,j) = (1/5) (f(i,j) + f(i+1,j) + f(i-1,j) + f(i,j-1) + f(i,j+1)$) is there an explicit expression for the metric that turns $T$ into a contraction? $\endgroup$ – morgan Jun 7 '12 at 5:31

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