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Suppose $(X,\ d)$ is a compact metric space and $f:X\to X$ is a continuous function with the property that $$d(f(x),\ f(y))\le\frac{1}{2}d(x,\ y)$$ $\forall x,\ y\in X$.

Show that $f$ has a unique fixed point.

Hint: Define $f^1=f$ and $f^{n+1}=f\circ f^n$. Consider the intersection of the sets $A_n=f^n(X)$.

Siddhartha
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    While this would be using a jackhammer to crack open a walnut, the Banach fixed-point theorem does cover this case. Y. Forman has given a good direct proof. – bitesizebo Apr 22 '18 at 19:23
  • @bitesizebo What's the easier way to do this, then? – Siddhartha Apr 22 '18 at 19:24
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    There isn't (to my knowledge) an easier proof than Y. Forman's. I just thought it might be interesting to know that this is essentially a subcase of a more powerful theorem known as the Banach fixed point theorem. It deals with when $X$ is just complete rather than compact. – bitesizebo Apr 22 '18 at 19:29
  • @bitesizebo if one ignores the hint, there is a very quick proof using compactness – Will Jagy Apr 22 '18 at 20:30

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The key point here is that a compact metric space has finite diameter (see, e.g., Compact metric space: proof $\text{diam}(K)$). $A_n$ is a nested sequence of sets (because the range of $f^n$ is contained in the range of $f^{n+1}$) with shrinking diameter approaching zero (because the diameter of $f(S)$ is $\leq \frac12$ the diameter of $S$ for any set $S \subset X$). So its intersection is a single point $p$; since $f(p)$ must also be in the intersection, $p$ is a fixed point.

BallBoy
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ignoring the hint, here is the easiest proof: define $$ g(x) = d(x,f(x)). $$ This is continuous, the space is compact, so $g$ assumes its minimum at some $x_0.$ Assume that $g(x_0) = \delta > 0.$ Then $g(f(x_0)) = d(f(x_0), f(f(x_0)) \leq \frac{1}{2} d(x_0, f(x_0)) = \frac{1}{2} \delta < \delta.$ This contradicts the assumption that $\delta \neq 0.$ That is, $g(x_0) = 0$ and $x_0$ is a fixpoint

Will Jagy
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