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.

Homework Equations

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.

 

[FactChecker]

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.

 

[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##.