One implication in $f'(x) \ge 0 \iff f \ \text{is monotonically increasing}$

Question

I am trying to understand why $f'(x) \ge 0 \iff f \ \text{is monotonically increasing}$ with the usual set of assumptions. To do this I am trying to prove the two implications. It is relatively easy to get why $\impliedby$ holds since an increasing $f$ implies$${f(x_0+h)-f(x_0) \over h} \ge 0$$ no matter what the $h$ is.

The second implication $\implies$ proves to be more tricky. I am able to show it rewriting the mean value theorem as

$$f(b)=f'(\xi)(b-a)+f(a)$$ and concluding that for $a,b$ satisfying $a<b$ we do get $f(b)\ge f(a)$.

Is there an easier way to see $\implies$ without using the mean value theorem?

I think this is the usual way. You can go look at how the mean value theorem is proven and use this instead of the theorem as black box. — M. Winter, Mar 02 '17 at 08:59
The mean value theorem is the proper tool here (as I see it). The MVT itself is a consequence of Rolle's theorem. — Martin R, Mar 02 '17 at 09:00
Related (duplicate?): $f'(c) \ge 0 , \forall c \in (a,b)$ then $f$ is increasing in $[a,b]$ , proof of this without Mean Value theorem — Martin R, Mar 02 '17 at 09:02
@BrandonSweeting FCT, in its usual formulations, talks about the primitive of a continuous function, so you'd technically need $f$ to be $C^1$ for it to be usable. However, I gather the "usual assumptions" are "$f$ differentiable". — , Mar 02 '17 at 09:07
Some clarification on what the "usual set of assumptions" are would be helpful, as possible answers depend on it. — Martin R, Mar 02 '17 at 09:08

Ronald J. Zallman · Answer 1 · 2017-03-02T09:20:18.533

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Yes, by the fundamental theorem of calculus. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be differentiable with $f'(x) \geq 0 \forall x \in \mathbb{R}$. Taking $x \in \mathbb{R}$ and $h \in \mathbb{R}_{\geq 0}$. \begin{align*} f(x+h) - f(x) = \int_{x}^{x+h} f'(t) dt \geq 0 \end{align*} Rearranging, $f(x + h) \geq f(x)$.

Edit: this assumes integrability of $f'(x)$.

edited Mar 02 '17 at 09:20

answered Mar 02 '17 at 09:06

Ronald J. Zallman

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This is what I had in mind. To clarify, the second part of the FTC doesn't assume continuity of the derivative in being able to compute its definite integrals via its primitive? – Brandon Sweeting Mar 02 '17 at 09:08
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@BrandonSweeting: But is requires integrability (Lebesgue or Riemann) of the derivative, as far as I know. – Martin R Mar 02 '17 at 09:11
And btw, the proof of the second part of FTC (as given here https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part) uses the MVT. – Martin R Mar 02 '17 at 09:12
That would definitely make sense. There seems to be something subtle (or not so subtle) that I'm missing. – Brandon Sweeting Mar 02 '17 at 09:16
You can prove the FTC without the MVT. – Ronald J. Zallman Mar 02 '17 at 09:17
Sorry about this, yeah. Under minimal assumptions on f this would not work then. – Ronald J. Zallman Mar 02 '17 at 09:19
How is that so? – Brandon Sweeting Mar 02 '17 at 09:19
The derivative of a function is not necessarily integrable. See http://math.stackexchange.com/questions/257069/what-is-an-example-that-a-function-is-differentiable-but-derivative-is-not-riema – Ronald J. Zallman Mar 02 '17 at 09:26
Does this mean I should delete the post? I'm new here so I don't know. – Ronald J. Zallman Mar 02 '17 at 09:27

One implication in $f'(x) \ge 0 \iff f \ \text{is monotonically increasing}$

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