# Proof verification there exist at most one $c\in[0,1]$ such that $f(c)=c$.

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.

• In your "counterexample" $|f'(x)|=1$. – Botond Feb 18 at 9:11
• @Botond you are right. Thanks for pointing out the silly mistake. – StammeringMathematician Feb 18 at 9:22

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.
• 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. – StammeringMathematician Feb 18 at 9:15
• 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. – José Carlos Santos Feb 18 at 9:17
• I know the result for function from $[0,1]$ to itself. I am asking for $[0,1]$ to $\mathbb{R}$. – StammeringMathematician Feb 18 at 9:19