Prove that there exists at most one $c\in[0,1]$ such that $f(c)=c$ if $|f'(x)|<1$ for $ \forall x \in (0,1)$ Let the function $f:[0,1]\rightarrow \mathbb{R}$ be continuous on $[0,1]$ and differentiable on $(0,1)$.
And also: $|f'(x)|<1$ for $ \forall x \in (0,1)$
And I want to prove this statement: there exist at most one $c\in[0,1]$ 
such that $f(c)=c$.
So I'm thinking of using Lagrange's (I think that is as well called mean value theorem), or Rolle's theorem.
But I don't know how to apply it in this situation.
Any help would be appreciated.
 A: The mean value theorem says that if you have two distinct $c,d$ such that $f(c)=c$ and $f(d)=d$, then there is an $a\in (c,d)$ such that $f'(a)=1$. Therefore there can be at most one such $c$.
A: Hint:
If indeed the range is contained in $[0,1]$, you can show the existence of a root applying the intermediate value theorem to the function $g(x)=f(x)-x$, and unicity proving $g$ is monotonic.
A: The statement of the above theorem, which is a specific use of the Banach contraction principle, should be stated as follows:

Given $f : [0,1] \to [0,1]$ such that $f'(x) < 1$ for all $x \in [0,1]$, then there is a unique point $x \in [0,1]$ such that $f(x) = x$.

Proof: We'll first debate the uniqueness, then we'll debate the existence.
Suppose $f(x)=x$ and $f(y)=y$. Note that if $g(z)=f(z)-z$, then $g'(z)<0$ for all $z$, hence $g$ is a strictly decreasing function. Therefore, $g(x)=g(y)$ implies $x=y$.
About existence, note that $g(0) \geq 0$ and $g(1) \leq 0$, hence by the intermediate value theorem, there is an $x$ such that $g(x) = 0$, or that $f(x)=x$.
A: Proof of existence: Assume that $f(x) \neq x$ for all $x \in [0,1]$. The function $g(x) = \frac{f(x)-x}{|f(x)-x|}$ is continuous and takes only values $\pm 1$. Since $[0,1]$ is connected, it follows that $g(x) = 1$ for all $x$ or $g(x) = -1$ for all $x$. In the first case, we have $f(x) > x$ and a contradiction at $x=1$ and in the second case, $f(x) < x$ and a contradiction at $x = 0$.
