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### Fixed-Point Theorem

```Date: 05/02/2002 at 08:06:39
From: Leslie
Subject: Real Analysis, Fixed-Point Theorem

If f:[a,b]->[a,b] is continuous then there is a point c in [a,b] such
that f(c)=c.

This doesn't seem intuitive to me the way some other problems do.  It
seems to me I should be able to find a number of counterexamples.  Am
I wrong?  If so, how can I get started on the construction of this
proof.
```

```
Date: 05/02/2002 at 10:06:02
From: Doctor Paul
Subject: Re: Real Analysis, Fixed-Point Theorem

Have you tried to find a counter example?  I think you'll find
yourself quite unable to do so...

Here is an intuitive reason why the theorem is true:

First, without loss of generality, assume a < b.

Now draw the square whose corners are the points:

(a,a)
(a,b)
(b,a)
(b,b)

Now draw the diagonal connecting the point (a,a) to the point (b,b).
Note that you are drawing a portion of the line y = x.

Now draw a random continuous (this is important) function that begins
at some point f(a) on the y axis [note that a <= f(a) <= b] and ends
at some point f(b) on the y axis (note that a <= f(b) <= b]. Does the
function you drew intersect the diagonal you drew? My guess is that it
does. Don't spend too much time trying to draw a function that doesn't
touch the diagonal - you'll find it impossible.

This is what the theorem is saying - that anytime you draw such a
function, it must intersect the line y = x at least once. It will
either intersect the line at x = a, at x = b, or at some point in the
interval (a,b).

For a proof, consider the following:

Let a, b be real numbers with a < b.

Let f: [a,b] --> [a,b] be continuous and let g: [a,b] --> [a,b] be
defined by g(x) = x.

Now define h(x) = f(x) - g(x) = f(x) - x.

g(x) is clearly continuous on [a,b] and since f(x) is also continuous
on [a,b], we know that h(x), which is the difference of two continuous
functions, will also be continuous on [a,b].

Case 1: if f(a) = a, we're done.

Case 2: assume f(a) > a.  If f(b) = b, we're done.

Case 3: assume f(a) > a and f(b) < b.

then f(a) - a > 0 which implies

h(a) > 0.

we also have:

f(b) - b < 0 which implies

h(b) < 0.

Let's summarize what we know:

h(x) is continuous on [a,b], h(a) > 0, and h(b) < 0.

The Intermediate Value Theorem will now give us that there exists a
point c in the interval (a,b) such that h(c) = 0.  But this means
f(c) - c = 0 which implies f(c) = c which is exactly what we wanted
to show.

I hope this helps.  Please write back if you'd like to talk about
this some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Analysis
College Calculus
High School Analysis
High School Calculus
High School Functions

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