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Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous such that for any real $x$,

$|f(x) - f(f(x))| \leq \frac{1}{2} |f(x) -x|$.

Must $f$ have a fixed point?

The question seems to invite an eventual application of the standard contraction mapping theorem. But this approach has not led me to showing there is a fixed point.

Is it true that there is a fixed point, and if so why?

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Let $x_i$ be the sequence defined by $x_0=0$ and $x_{i+1}=f(x_i)$. Then the inequality implies that $$|x_{i+1}-x_{i+2}|<\frac{1}{2}|x_{i+1}-x_i|.$$ Therefore $|x_{i+1}-x_i|<2^{-i}|x_1-x_0| \rightarrow 0$. For all $m>n>N$, one has $$|x_m-x_n|\leq |x_m-{x_{m-1}}|+\ldots +|x_{n+1}-x_n|<(2^{-m+1}+\ldots+2^{-n})|x_1-x_0|<2^{-N+1}|x_1-x_0|$$

It follows that $\{x_i\}_{i=1}^\infty$ is a Cauchy sequence, hence convergent to some $x\in \mathbb{R}$. Since $x_{i+1}=f(x_i)$, by taking the limit, we get $x=f(x)$.

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  • $\begingroup$ Why is it Cauchy? $\endgroup$ – CuriousKid7 Aug 27 '18 at 0:03
  • $\begingroup$ A Cauchy sequence $x_n$ has the property that for every $\epsilon>0$ there exists $N$ such that $m,n>N \Rightarrow |x_m-x_n|<\epsilon$. In $\mathbb{R}$, every Cauchy sequence is convergent. $\endgroup$ – Marco Aug 27 '18 at 0:07
  • $\begingroup$ I understand, but I only see that $d(x_{i}, x_{i+1}) \to 0$, which is not enough to show it's Cauchy. $\endgroup$ – CuriousKid7 Aug 27 '18 at 0:30
  • $\begingroup$ @CuriousKid7 a sequence whose subsequent terms go to $0$ exponentially fast is Cauchy. $\endgroup$ – Robert Wolfe Aug 27 '18 at 3:28
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To further improve on Marcos answer:

the sequence is Cauchy, since for all $n<m$, we have $$ |x_m - x_n| \leq \Sigma_{i=n}^{m-1}|x_{i+1}-x_n| \leq 2^{-n+1} |x_1-x_0| $$

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