Conditions yielding a unique fixed point of a continuous differentiable function.

Question

Let $f$ be a function defined on $[0,1]$ which is continuous for each point in $[0,1]$ and differentiable for each point in $(0,1)$. Suppose that $f^\prime (x) \neq 1$ for every $x \in (0,1)$. Prove that there exists exactly one point $x \in [0,1]$ such that $f(x) = x$ [Hint: You might find both the intermediate value theorem and the mean value theorem useful.]

Could someone kindly assist me with this?

Answer 1

Hint:

Consider the function $g(x) = x - f(x) $. Notice

$g(0) = - f(0) < 0 $

$g(1) = 1 - f(1) > 0 $

Now, can you finish the problem?

To show uniqueness, another hint: suppose $x,y$ are both fixed points, then by Mean value theorem, we can find $\xi \in (x,y) $ such that

$$ |x-y| = |f(x) - f(y)| = |f'(\xi) |x-y| $$

Then, .....