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The following question comes from Chapter 6.4, Exercise 4 on page 156 in the set of notes Topology Without Tears.

Using Exercise 2 and 3, show that while $f: \mathbb R \to \mathbb R$ given by $f(x)=\cos(x)$ does not satisfy the condition of the Contraction Mapping Theorem, it nevertheless has a unique fixed point.

The "Exercise 2 and 3" mentioned in the question (which I've already done) were

  1. Extend the Contraction Mapping Theorem by showing that if $f$ is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point.

  2. Use the Mean Value Theorem to prove the following:

    Let $f: [a,b] \to [a,b]$ be differentiable. Then $f$ is a contraction if and only of there exists an $r \in (0,1)$ such that $|f'(x)|\le r$, for all $x \in [a,b]$.

Can anybody please show me how to use those to show that $f(x)=\cos(x)$ has a unique fixed point?

I know, from a Numerical Analysis course, that this fixed point can be found using a method such as Newton's Method, but I am not sure how to prove that it actually exists (and is unique).

user229672
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    Try $g(x)=f^2(x)=f(f(x))$ with exercise 3, then apply exercise 2. You could also show this by showing directly that there are no fixed points on $\mathbb R\setminus [0,1]$, whereas restricting $f$ to $[0,1]\to[0,1]$ allows you to apply the Contraction mapping theorem. – Jonas Meyer Jul 07 '16 at 21:35
  • Many previous posts are closely related, such as: http://math.stackexchange.com/questions/201770/convergence-of-x-n-cos-x-n-1 – hardmath Jul 07 '16 at 22:04

2 Answers2

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This isn't the suggested proof at all, but it's a lot simpler, and I hope it's of some value. @Jonas Meyer's comment seems to give a good route to solving the problem as suggested, so there's no reason to duplicate.

A fixed point of $x \mapsto \cos x$ must be between $-1$ and $1$, because at a fixed point $\cos x = x$, and the codomain of $\cos$ is $[-1, 1]$.

For $-1 < x < 0$, cosine is positive but $x$ is negative, so we need only consider $0 \le x \le 1$.

At $0$, $g(x) = \cos x - x$ is positive. At $1$, $g(1)$ is negative (by direct evaluation). Throughout the interval, $g'(x) < 0$, so $g$ has at most one zero, say $b$, in the interval, and at that point $\cos b - b = 0$, so $\cos b = b$.

John Hughes
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fixed point of a function taking the real numbers to the reals is a place where the graph crosses the line $y=x.$ Existence and uniqueness can be shown with ordinary calculus theorems

Will Jagy
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