Let $f$ be a differentiable function on $[0, 1]$ with $0$ and $1$ be its fixed points. Also, it is given that $$f'(0)>1, f'(1)>1.$$ Then prove that
(i) $f$ has at least one fixed point in $(0, 1)$.
(ii) $f$ has a fixed point $\xi\in (0, 1)$ such that $f'(\xi)\leq 1$.
My effort: I started by considering $g(x)=f(x)-x$. Clearly, $g$ is a differentiable function which means $g'(x)=f'(x)-1$. This means $g'(0)>0, g'(1)>0$. But I am not able to conclude anything. Please help.
Best Answer
You also have $g(0) = 0$ and $g(1) = 0$ because $0, 1$ are fixed points of $f$. Now, because $g'(0) > 0$, you must have $g(x_1) > 0$ for some $x_1 \in (0, 1)$. Otherwise, for a contradiction, if $g(x) \leq 0$ were instead true for all $x \in (0, 1)$, then $g(x) - g(0) = g(x) - 0 \leq 0$ always. Hence, the fraction $$ \frac{g(x) - g(0)}{x - 0} \leq 0, \ \forall x \in (0, 1) \implies g'(0) = \lim_{x \to 0, x \in (0, 1)}\frac{g(x) - g(0)}{x - 0} \leq 0 $$ contrary to $g'(0) > 0$.
For a similar reason, there must also be some $x_2 \in (0, 1)$ such that $g(x_2) < 0$. Obviously $x_1 \neq x_2$. So now, you can conclude via IVT.