From - Sun May 25 18:25:49 1997
From: abian@iastate.edu (Alexander Abian)
Newsgroups: sci.math,sci.astro
Subject: BROUWER FIXED POINT THEOREM (was: Abian's Fixed Point Theorem)
Date: 24 May 1997 17:33:24 GMT
Organization: Iowa State University, Ames, Iowa, USA
Lines: 120
Message-ID: <5m78p4$1vl$1@news.iastate.edu>
Xref: ar4dec01 sci.math:2181 sci.astro:1539
In article <33775EEC.5278@cogentex.com>,
Daryl McCullough wrote:
>Professor Abian,
>
>This is on another topic, (luckily, a topic with the same title!),
>but is there any relationship between the fixed point theorem for
>partial orders and *topological* fixed point theorems? (--for example,
>the theorem from complex analysis that every continuous function
>from the unit disc to itself must have a fixed point--)
Abian answers:
Yes there is a relationship between the fixed point theorems for partial
orders and "topological" ones. However, the relationship, very often,
is hidden and buried under 20 tons of cold lava.
Let me remark about the Theorem that you inquired which is the Brouwer
Fixed Point Theorem stating that:
(2) Any continuous map f from a closed 2-dimensional disk D
into itself has a fixed point.
(*) (Incidently, in one of my papers it is shown that all one needs
for the proof of the above Theorem is the hypothesis that the
boundary of D is mapped by f into D. Thus, a part of D
could be mapped in the plane of D but outside of D)
The one-dimensional case of the Brouwer Fixed Point Theorem is the
familiar:
(1) Any continuous map f from a closed real unit interval [0, 1]
into itself has a fixed point.
The remark (*) also applies to (1).
The proof of (2) as well as (1), of course, dramatically uses the
fact that
Under a continuous map points close to each other are mapped
into points which are close to each other.
More precisely:
(3) Under a continuous map f every point sufficiently close
to x is mapped within any preassigned closeness to f(x).
There are several proofs of (1). I give below one which will motivate
a proof of (2).
PROOF of (1).
Assume on the contrary that f has no fixed point.
Using f let us define a new mapping g from [0, 1] onto the boundary B
of [0, 1] by the following recipe (note that B has only two points
0 and 1) :
(i) join f(x) to x if the movement is toward 0, define g(x) = 0
as shown below:
0....<---x------f(x)...........1 thus g(x) = 0
(ii) if the movement is toward 1, define g(x) = 1 as shown below:
0........f(x)--------x--->.....1 thus g(x) = 1
(Since we assumed that f has no fixed point, g is well defined)
Observe that
(4) if f(x) is to the right of x then g(x) = 0 and
if f(x) is to the left of x then g(x) = 1
Clearly,
(5) g(0) = 0 and g(1) = 1 thus the boundary B of [0, 1]
remains pointwise fixed under g.
Based on the fact that f is continuous, we show that g is also
continuous.
Since f is continuous
(6) for any m which is sufficiently close to x it is the case that
f(m) is within a preassigned closeness to f(x).
Since we assumed that f has no fixed point, f(x) is either to the right
of x or f(x) is to the left of x. There are points of both kinds.
Clearly, the boundary point 0 is such that f(0) is to the right of 0,
whereas the boundary point 1 is such that f(1) is to the left of 1.
If f(x) is to the right of x by (6) so is f(m) to the right of m and
therefore by (4), we have g(x) = g(m) = 0. Thus g(m) is identical to
g(x) and therefore g(m) is within any preassigned closeness to g(x).
Again, if f(x) is to the left of x, by (6), so is f(m) to the left of
m and therefore by (4), we have g(x) = g(m) = 1 and again g(m) is
within any preassigned closeness to g(x). Consequently g as defined
by (4) is a continuous mapping from the interval [O, 1] ONTO a set of
two points 0 and 1 (the ontoness also follows from (5)).
But this contradicts that the continuous function g must take every
intermediate value between 0 and 1. In other words, this contradicts
that the continuous function g must map a CONNECTED set such as
[0, 1] onto a connected set and not onto a disjoint set of two points.
In other words we arrived at a contradiction to the fact that:
(6) a continuous function cannot tear apart a connected set.
Thus, our assumption is false and hence f has a fixed point.
P.S. I deliberately went through a very lengthy prove above, because
the above proof will serve as a motivation of Brouwer's Fixed Point
Theorem for continuous mappings from a closed disc D into itself,
which I will post shortly.
--
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