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Let the function $f:[0,1]\rightarrow \mathbb{R}$ be continuous on $[0,1]$ and differentiable on $(0,1)$ and $|f'(x)|<1$ for $ \forall x \in (0,1)$

I want to prove this statement: there exist at most one $c\in[0,1]$ such that $f(c)=c$.

Proof 1: Suppose on the contrary that there are two distinct points such that $f(c)=c$ and $f(d)=d$ then by MVT there exist a point $e$ such that $f'(e)=1$ which is a contradiction so we have at most one fixed point.

Also note that we can find a function like $f(x)=x+{1\over 2}$ which satisfy all the property but does not have any fixed point.

Ignore the example. It is incorrect. We can instead take constant function.


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  • $\begingroup$ In your "counterexample" $|f'(x)|=1$. $\endgroup$ – Botond Feb 18 at 9:11
  • $\begingroup$ @Botond you are right. Thanks for pointing out the silly mistake. $\endgroup$ – StammeringMathematician Feb 18 at 9:22
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Your proof is incomplete, since you did not explain how to deduce from the mean value theorem that such a point $e$ must exist. It is simple, though:$$1=\frac{f(c)-f(d)}{c-d}=f'(e),$$for some $e$.

And your example is not an example since it is not a map from $[0,1]$ into itself.

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  • $\begingroup$ I want to find an example of a real valued function $f:[0,1] \to R$ such that it satisfies the above property but have no fixed point. Does such a function exist. Thanks. $\endgroup$ – StammeringMathematician Feb 18 at 9:15
  • $\begingroup$ No, it doesn't exist. Every continuous map from $[0,1]$ into itself (with no extra conditions) has some fixed point. Just consider the map $x\mapsto f(x)-x$ and apply the intermediate value theorem. $\endgroup$ – José Carlos Santos Feb 18 at 9:17
  • $\begingroup$ I know the result for function from $[0,1]$ to itself. I am asking for $[0,1]$ to $\mathbb{R}$. $\endgroup$ – StammeringMathematician Feb 18 at 9:19
  • $\begingroup$ @JoséCarlosSantos he wants it mapping into R, so one could take the constant function 2 $\endgroup$ – Calvin Khor Feb 18 at 9:20
  • $\begingroup$ @CalvinKhor You are right. $\endgroup$ – José Carlos Santos Feb 18 at 10:09

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