- #1

Bashyboy

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- 5

## 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.