# Fixed point and MVT problem.

I've been asked to use the mean value theorem prove the following. I think that I have done all the steps correctly but I would appreciate anyone looking it over.

Question: Given $f$ is differentiable everywhere and $\forall x,f\prime(x)<1$ , show that there can be at most one fixed point for $f$.

Suppose, aiming for contradiction, $\exists a,b \in \mathbb{R}:f(a)=a \wedge f(b)=b \wedge a<b$

Since $f$ is differentiable everywhere, $f$ is continuous on $[a,b]$ thus by the mean value theorem $\exists c \in (a,b): f \prime (c) = \frac{f(b)-f(a)}{b-a}$

Since $f\prime(c)<1 \Rightarrow \frac{f(b)-f(a)}{b-a}<1$

$\Rightarrow f(b)-f(a)<b-a$

$\Rightarrow b-a<b-a$

However this is a contradiction, thus there can be at most one fixed point for $f$.

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That looks fine. –  in_wolframAlpha_we_trust Oct 14 '13 at 7:58

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