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So, I have a continuous function $f$ from $\left[0,1\right]$ to $\left[0,1\right]$ such that : $$\forall \, x_1,x_2\in\left[0,1\right]:\; \mid x_1-x_2\mid\;\leq \;\mid f(x_1)-f(x_2)\mid$$ It's easy enough to prove that $f$ is both surjective and injective so $f$ is a bijection, and also $\{f(1),f(0)\}=\{0,1\}$. Now I'm trying to prove that $f$ must have a fixed point. When drawing an example it seems obvious, but I'm not sure how to go about the proof, setting aside the trivial case when $f(1)=1$ and $f(0)=0$. Since $f$ is a bijection and if we assume we're in the case $f(1)=0, f(0)=1$ then $f$ is strictly decreasing but not sure how that would help. So could anyone give some kind of hint or an observation I might have missed ? Thanks.

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Assume that we are in the case $f(1)=0$, $f(0)=1$. Consider the function $g:[0,1]\to\mathbb{R},\; x\mapsto f(x)-x$. Then, $g$ is also continuous. Note $g(0)=f(0)=1$ and $g(1)=f(1)-1=-1$. By the intermediate value theorem (Bolzano's theorem), there exists a $x\in[0,1]$ such that $g(x)=f(x)-x=0$, which completes your proof.

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Think of of the link between a fix-point and the function $g: [0,1] \rightarrow [0,1], x\mapsto x$. If you were to find a fix-point what kind of equality would hold? Try to link that to $g$ and $f$.

In the end you should use the intermediate value theorem.

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You have already shown that $f(x)$ is bijective. Take a look at $g(x)=f(x)-x$ for the case $f(1)=0$, $f(0)=1$. If $g(x)$ has a zero, we are done by definition of a fixpoint. Now $g(1)=-1$, $g(0)=1$, so by continuity, there exists a $t\in[0,1]$, for which $g(t)=0$.

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  • $\begingroup$ Ok now I wish I had thought of that, it was really not that hard and I had done it before. But I don't see why the fact that $f$ is bijective matters ? Even if it's not bijective, a continuous mapping from $\left[0,1\right]$ to $\left[0,1\right]$ will still have a fixed point. Am I missing something again ? $\endgroup$ Mar 30 '20 at 18:00

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