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- Path: sparky!uunet!mcsun!uknet!acorn!eoe!jrickard
- From: jrickard@eoe.co.uk (John Rickard)
- Newsgroups: sci.math
- Subject: Re: Elementary solution needed: harmonic bi-sequences
- Message-ID: <1388@eouk8.eoe.co.uk>
- Date: 31 Aug 92 15:57:25 GMT
- References: <1992Aug27.170048.26459@ugle.unit.no>
- Organization: EO Europe Limited, Cambridge, UK
- Lines: 71
- X-Newsreader: Tin 1.1 PL3
-
- 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.
-
- Rewrite (*) as:
-
- a(i,j) = (1/8) a(i,j+1) + (1/8) a(i,j-1) + (1/8) a(i+1,j)
- + (1/8) a(i-1,j) + (1/2) a(i,j) (1)
-
- By substituting (1) in itself and repeating a number of times we get,
- for each N >= 0:
- (N)
- a(i,j) = sum B (m,n) a(i+m,j+n) (2)
- m,n
-
- where
- (0)
- B (m,n) = ( 1, if m=n=0
- ( 0, otherwise
-
- (N+1) (N) (N)
- B (m,n) = (1/8) B (m,n+1) + (1/8) B (m,n-1)
-
- (N) (N)
- + (1/8) B (m+1,n) + (1/8) B (m-1,n)
-
- (N)
- + (1/2) B (m,n) (3)
-
- We have (proofs left as exercise for the reader):
-
- (N)
- For fixed N, n, B (m,n) is monotonically increasing for m < 0,
- reaches its maximum at m = 0, and is monotonically decreasing
- for m > 0. It is 0 for m sufficiently far from 0. (4)
-
- (N)
- sum B (0,n) -> 0 as N -> infinity. (5)
- n
-
- From (2):
- (N) (N)
- | a(0,0) - a(1,0) | = | sum ( B (m,n) - B (m-1,n) ) a(m,n) |
- m,n
- (N) (N)
- < C sum | B (m,n) - B (m-1,n) | (6)
- m,n
-
- From (4) we have:
-
- (N) (N) (N)
- sum | B (m,n) - B (m-1,n) | = 2 B (0,n) (7)
- m
-
- Combining (6) and (7):
- (N)
- | a(0,0) - a(1,0) | < 2C sum B (0,n) (8)
- n
-
- By (5), the RHS of (8) tends to 0 as N -> infinity, so a(0,0) and
- a(1,0) must be equal. Similarly any two adjacent values of a(i,j)
- must be equal, so a(i,j) is constant.
-