f is a differentiable function with $f(0)>0$. f has only one fixed point on [0,1] Prove that $f′(x_0)<1$ Question: Let $f:[0,1]\rightarrow [0,1]$ be a differentiable function with $f(0) > 0$. Asuming $x_0\in(0,1)$ is the only fixed point of $f$, prove that $f'(x_0) < 1$
My Attempt: 
Let $g(x) := f(x)-x$ for $x\in[0,1]$ 
By given condition $g(0) = f(0)-0 > 0$ 
As $f$ has unique fixed point on [0,1] so $g$ will have only one solution on [0,1].
So if $g(0)>0$ then $\forall x<x_0,\; g(x)>0\; and \; \forall x>x_0,\; g(x)<0$
Now $g'(x)=f'(x)-1$
$f$ is differentiable on [0,1] so is $g$.
Consider Right-hand derivative of $g$ at $x_0$ 
$A=\lim_{h \rightarrow 0}\frac{g(x_0+h)-g(x_0)}{h}$ for $h>0$
Using fact that $g(x_0)=0\;and\;\forall x>x_0,\; g(x)<0$ we have $A<0$
So we can say that $g'(x_0)<0$ Therefore $f'(x_0)<1$
I want to know if my proof is correct or not. Is there any simple proof for this question.
 A: Your approach is fine, but your argument (and the conclusion) needs a minor fix . . .

By hypothesis, you have$\;\,g(x)=0\;\;$if and only if$\;\,x=x_0\;\,$for some$\;x_0\in (0,1)$.

Along the lines of your attempt, we first show
$$
\begin{cases}
g(x) > 0&\text{if}\;\,0\le x < x_0\\[4pt]
g(x) < 0&\text{if}\;\,x_0 < x\le 1\\
\end{cases}
$$
We can argue as follows . . .


*

*Since$\;0\;$ is not a fixed point, it follows that$\;f(0) > 0$,$\;$hence$\;g(0) > 0$.$\;$Then since$\;g(x)\ne 0\;\,$for$\;0\le x < x_0$,$\;$it follows by the Intermediate Value Theorem that$\;g(x) > 0\;\,$for$\;0\le x < x_0$.$\\[6pt]$

*Since$\;1\;$ is not a fixed point, it follows that$\;f(1) < 1$,$\;$hence$\;g(1) < 0$.$\;$Then since$\;g(x)\ne 0\;\,$for$\;x_0 < x \le 1$,$\;$it follows by the Intermediate Value Theorem that$\;g(x) < 0\;\,$for$\;x_0 < x \le 1$.


From there, your argument works as is (and you can even use a two-sided derivative at$\;x=x_0$) to show $\;g'(x_0)\le 0$,$\;$hence$\;f'(x_0)\le 1$.

However, to see that it's not possible to show$\;f'(x_0) < 1$,$\;$you can verify that the function
$$f(x)=-8x^3+12x^2-5x+1$$
restricted to the domain$\;[0,1]\;$satisfies the hypothesis and has a unique fixed point at$\;x_0={\large{\frac{1}{2}}}$,$\;$but$\;f'(x_0)=1$.

As regards the flaw in your attempt . . .

You have that$\;g\;$is differentiable, but the fact that
$$\frac{g(x)-g(x_0)}{x-x_0} < 0$$
for all$\;x\in[0,1]\;$with$\;x\ne x_0\;$only implies
$$\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}\le 0$$
hence you can claim$\;g'(x_0)\le 0$,$\;$but as the counterexample shows, you can't claim$\;g'(x_0) < 0$.
