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It is well known that $\mathbb{Z}^d$ has Liouville property, i. e. every bounded harmonic function on this graph is constant.

(harmonic means that the value of $f$ in a point $x$ is equal to the average of $f$ over neighbours of $x$ in the lattice $\mathbb{Z}^d$).

What are the nicest/shortest/most ingenious proofs of this fact that you know?

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closed as too localized by Gjergji Zaimi, Wadim Zudilin, Steve Huntsman, Mark Sapir, Willie Wong Jan 17 '11 at 1:41

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This is almost a contest problem. Actually it has appeared in contests :) I think artofproblemsolving would be a better forum for your question. – Gjergji Zaimi Jan 13 '11 at 12:52
I disagree - although there are completely elementary and short proofs of this fact, there are also more sophisticated ones that illustrate important mathematical techniques or results (e. g. Krein-Milman theorem, Markov chains). – Michal Kotowski Jan 13 '11 at 13:13
There are at least 5 proofs in the link. The only additional thing requested in your question is which one is one's favorite proof. You haven't really explained how that is of any use to you or anyone else.… – Gjergji Zaimi Jan 13 '11 at 13:18
So, if you know several (elementary) proofs, what do you ask?! And what's your motivation to ask? (This could be curiosity, of course.) – Wadim Zudilin Jan 13 '11 at 13:19
I wonder wether there is an elementary proof for functions bounded from belov (a priori not from above). – Fedor Petrov Jan 13 '11 at 15:40

Yuval Peres gave a fairly short proof of this result in his lectures at the 2009 Cornell Probability Summer School (

If one wants to use more machinery from probability theory, the Hewitt-Savage 0-1 law implies that the tail $\sigma$-field $\mathcal{T}$ associated with the SRW on $\mathbb{Z}^d$ is trivial. But the tail $\sigma$-field is a superset of the 'invariant' sigma field $\mathcal{I} := \{F\in\mathcal{F}:F\circ \theta_n = F \text{ for all }n\}$, and the invariant sigma field being $\mathbb{P}^x$-trivial for all $x$ is equivalent to the graph having the Liouville property.

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