If $f$ is Lipschitz continuous on a closed interval $[a,b]$ such that $f([a,b])\subseteq [a,b]$ then it has a unique fixed-point

Question

I am stucked at this problem:

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Prove or give a counter-example for the following sentence:

If $f:[a,b]\to\Bbb{R}$ is Lipschitz continuous on a closed interval $[a,b]$ and $f([a,b])\subseteq [a,b]$ then $f$ has a unique fixed-point in $[a,b]$.

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I am having hard time trying to prove it and to find a counter-example.

(Note: since $f([a,b])\subseteq [a,b]$ and since $f$ is continuous on $[a,b]$ we know that $f$ has a fixed-point in $[a,b]$ but we do not know wether it is unique)

Thanks for any hint/help.

Answer 1

The simplest counterexample I can conceive is

$$f(x)=x$$

Then every point is a fixed point.

Answer 2

Clearly false. Just take $f$ to be the identity function and $a<b$. Clearly $f$ is Lipschitz continuous and $f([a,b])=[a,b]\subseteq[a,b]$ and all points are fixed points.