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- Path: sparky!uunet!zaphod.mps.ohio-state.edu!not-for-mail
- From: edgar@function.mps.ohio-state.edu (Gerald Edgar)
- Newsgroups: sci.math
- Subject: Re: Elementary solution needed: harmonic bi-sequences
- Date: 27 Aug 1992 14:29:13 -0400
- Organization: The Ohio State University, Dept. of Math.
- Lines: 29
- Message-ID: <17j6tpINNo77@function.mps.ohio-state.edu>
- References: <1992Aug27.170048.26459@ugle.unit.no>
- NNTP-Posting-Host: function.mps.ohio-state.edu
-
- In article <1992Aug27.170048.26459@ugle.unit.no> ap@levangerhs.no (Andrei Prasolov) writes:
- > There is a problem which seems to have a very simple solution.
- >I remember that I heard one in my youth. That solution used
- >the epsilon-delta language only. However, I cannot recall that.
- >
- >DEF. Let`s call a bi-sequence (a(i,j), i,j from Z) HARMONIC if for all i,j
- >
- >(*) a(i,j) = 1/4 (a(i,j+1) + a(i,j-1) + a(i+1,j) + a(i-1,j)).
- >
- >PROBLEM. Prove that any bounded (|a(i,j)| < C) harmonic bi-sequence is constant.
- >
-
- Here is a proof that I like. Whether you consider it "elementary" depends
- on your background.
-
- The symmetric random walk on the two-dimensional integer lattice Z^2
- is recurrent: it visits each node repeatedly (with probability one).
- Apply the function a to that random walk. The result is (by (*)) a
- martingale. But a bounded martingale converges (almost surely),
- while _this_ bounded martingale runs repeatedly through all the values of
- the function a. Therefore a is constant.
-
- So maybe it's not elementary, but it is short and cute.
-
- --
- Gerald A. Edgar Internet: edgar@mps.ohio-state.edu
- Department of Mathematics Bitnet: EDGAR@OHSTPY
- The Ohio State University telephone: 614-292-0395 (Office)
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