I see there is a post almost related to this, but I think the conditions are slightly different.
Suppose $f$ is differentiable on (a,b) and continuous on [a,b], with $a\leq f(x) \leq b, \forall x \in [a,b]$ and $\mid f'(x)\mid<1,\forall x \in (a,b)$. Prove that $f(x) = x$ has a unique solution in $[a,b]$
So, I let, $g(x) = f(x) – x \Rightarrow g(a) = f(a) – a \geq 0$ and $g(b) = f(b) – b \leq 0$. So, from the Intermediate value theorem, we know, $\exists c \in [a,b]$, s.t.$g(c ) = 0 \Rightarrow g(c ) = f(c ) – c = 0 \Rightarrow f(c ) = c$, but then again, I don't understand how is that going to help us prove uniqueness. Any help? Thanks.
Best Answer
Suppose that we have two solutions $a,b$, where $a\neq b$ then by MVT:
$$\frac{f(a)-f(b)}{a-b}=f'(\xi)$$
where $\xi \in (a,b)$. But $a,b$ are solutions, so:
$$\frac{f(a)-f(b)}{a-b}=\frac{a-b}{a-b}=1$$
Thus $f'(\xi)=1$. But $f'(\xi)<1$.