TITLE: $f'(c) \ge 0 , \forall c \in (a,b)$ then $f$ is increasing in $[a,b]$ , proof of this without Mean Value theorem
QUESTION [5 upvotes]: Let $f: [a,b] \to \mathbb R$ be a function differentiable in $(a,b)$ , it is known that if $f'(c) \ge 0 , \forall c \in (a,b)$ then $f$ is increasing in $[a,b]$ and this can be proved by Lagrange Mean value theorem ; I would like to know , is there any other proof of this ?

REPLY [7 votes]: Assume that there are points $c$, $d$ with $a<c<d<b$ such that
$${f(d)-f(c)\over d-c}=-p<0\ .$$
I claim that there is a point $\xi\in[c,d]$ with $f'(\xi)\leq-p$.
Proof. Using binary division we can find an increasing  sequence $(c_n)_{n\geq0}$ and a decreasing sequence $(d_n)_{n\geq0}$ with $$c\leq c_n<d_n\leq d,\qquad d_n-c_n={d-c\over2^n}\qquad(n\geq0)\ ,$$
such that
$${f(d_n)-f(c_n)\over d_n-c_n}\leq -p\qquad(n\geq0)\ .\tag{1}$$
The $c_n$ and the $d_n$ have a common limit point $\xi\in[c,d]$. If
$$\xi=c_n\qquad(n\geq n_0)$$ one immediately concludes from $(1)$ that $f'(\xi)\leq-p$. Otherwise we may assume $c_n<\xi<d_n$ for all $n\geq0$. Rewriting the left hand side of $(1)$ we then can say say that
$${\xi-c_n\over d_n-c_n}{f(\xi)-f(c_n)\over \xi-c_n}+{d_n-\xi\over d_n-c_n}{f(d_n)-f(\xi)\over d_n-\xi}\leq -p\qquad(n\geq0)\ .\tag{2}$$
Now let an $\epsilon>0$ be given. Then for some sufficiently large $n$ the difference quotients of $f$ in $(2)$ are $\geq f'(\xi)-\epsilon$, and this implies
$$f'(\xi)-\epsilon\leq -p\ .$$
Since this is true for all $\epsilon>0$ the claim follows.