# Unique Fixed Point

Bashyboy

## Homework Statement

Let ##(X,d)## be some metric space, and let ##f : X \to X## be such that ##d(f(x),f(y)) \le a d(x,y)## for every ##x,y \in X## for some ##a \in (0,1)## (such a map is a called a contraction map)

If ##f## is a contraction and ##X## is compact, show that ##f## has a unique fixed point.

## The Attempt at a Solution

I will first show that ##g : X \to \Bbb{R}## defined by ##g(x) = d(f(x),x)## is continuous. First note that ##d(f(x),x) - d(x,y) \le d(f(x),y)## and ##d(x,y) - d(f(y),y) \le d(f(y),x)##, as well as ##d(f(x),y) \le d(f(x),x) + d(x,y)## and ##d(f(y),x) \le d(f(y),y) + d(x,y)##. Therefore

\begin{align*}
|d(f(x),x) - d(f(y),y)| & = \left|\bigg(d(f(x),x) - d(x,y)\bigg) + \bigg(d(x,y) - d(f(y),y) \bigg) \right| \\
& \le |d(f(x),x) - d(x,y)| + |d(x,y) - d(f(y),y)| \\
& \le d(f(x),y) + d(f(y),x) \\
& \le d(f(x),x) + d(x,y) + d(f(y),y) + d(x,y) \\
& \le 2(\alpha +1)d(x,y), \\
\end{align*}

from which we can conclude ##g## is continuous. This seems right, albeit a bit roundabout. Is there a more clever way of adding in ##0##? I think there is a way of viewing it as a composition of continuous functions. Let ##e : X \to X \times X## defined by ##e(x) = (f(x),id(x)) = (f(x),x)##, which is continuous by the maps-to-products theorem. Then ##g = d \circ e##, a composition of continuous functions (this obviously answers the original question in my OP). I am still interested in a less roundabout way of using the triangle inequality.

My goal is to try and use the solution found https://math.stackexchange.com/questions/118536/prove-the-map-has-a-fixed-point I tried redefining ##g## to be ##g(x) = a d(f(x),x)## and ##g(x) = \frac{1}{a} d(f(x),x)##, but I couldn't get it to work. I think part of the problem lies in the fact that I am working with ##\le## and not strict inequality. Does anyone see a way of modifying the solution found in the link above.

Homework Helper
Gold Member
If ≤ is your problem, can you consider b, where a<b<1 and get the strict < that will work for you?
EDIT: Get the strict < as long as d(x,y) >0.

Last edited:
Bashyboy
Ah! Blast! If ##a \in (0,1)##, then ##a d(x,y) < d(x,y)## so that ##d(f(x),f(y)) \le a d(x,y) < d(x,y)##, and so we can easily use the method in the given link. Therefore, since ##g## is continuous and ##X## is compact, it achieves a minimum over ##X## at, say, the point ##x \in X##; hence, ##g(x) \le g(y)## for every ##y \in X##. Now, if ##0 = g(x) = d(f(x),x)##, then we are finished. However, if ##g(x) > 0##, then ##d(f(f(x),f(x)) \le ad(f(x),x) < d(f(x),x)## which contradicts the fact that the minimum is attained at ##x##. Therefore ##d(f(x),x)=0## and so ##f(x) = x##.

The only thing that confuses me is, isn't ##d(f(f(x),f(x)) \le ad(f(x),x) < d(f(x),x)## a contradiction especially when ##d(f(x),x)=0## or ##f(x)=x##, since this would say ##d(f(f(x)),f(x)) < 0##? Also. I just realized that I didn't prove uniqueness; I'll have to think about that one some more.

Bashyboy
D'oh! Uniqueness is trivial. Suppose that ##x## and ##y## are both fixed points. Then ##f(x) = x## and ##f(y) = y##. Therefore ##d(f(x),f(y)) = d(x,y)## by mere substitution; but this contradicts the fact that ##d(f(x),f(y)) < d(x,y)## for all ##x, y \in X##.

I still have that worry I expressed in my last post. Namely, if ##d(f(x),x) > 0##, then ##d(f(f(x)),f(x) < d(f(x),x)## contradicts minimality. This forces ##d(f(x),x) = 0##. However, we still get a contradiction even in this case; the contradiction that ##d(f(f(x)),f(x) < d(f(x),x) = 0## which says that ##d(f(f(x)),f(x)## is negative. Where am I going wrong? I basically using Davide Giraudo's proof, found in the link I gave above; his answer is five years old and no else has seen this issue, so I must be making a mistake somewhere...

Bashyboy
Bump!

Bashyboy
I figured it out: ##d(f(x),f(y)) < d(x,y)## if and only if \$x \neq y##.

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