# Show that if f is differentiable and f'(x) ≥ 0 on (a, b), then f is strictly increasing

Show that if f is differentiable and f'(x) ≥ $$0$$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.

So so far I'm trying to do this by contradiction:

Suppose not, that is suppose we have function $$f$$ where f(x)$$\geq0$$ on (a,b) where f ' is not identically $$0$$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $$x_1$$ and $$x_2$$ where a <$$x_1$$ < $$x_2$$ < b and f($$x_1$$) = f($$x_2$$). Then for all y $$\in$$ [$$x_1$$, $$x_2$$], f(y) = f($$x_2$$) which means that f is constant and f '(y) = 0.

Since f'(y)=$$0$$ for all y $$\in$$ [$$x_1$$, $$x_2$$] this means f ' is identically $$0$$ which is a contradiction. Thus f is strictly increasing. $$\square$$

I'm not sure if there is a better way to do this but any help or comments would be appreciated!

• I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all). – freakish Dec 3 '18 at 11:00

Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $$x_1 such that $$f(x_1)=f(x_2)$$ is ok, but it is not clear how to conclude from that the fact that $$f$$ is constant of $$[x_1,x_2]$$.

My advice is to either use Rolle's theorem on $$x_1,x_2$$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.

• I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way? – Nolando Dec 3 '18 at 12:02
• @Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$\forall x_1, x_2: x_1<x_2\implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$\exists x_1,x_2:x_1<x_2\land f(x_1)\geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)\geq f(x_2)$. – 5xum Dec 3 '18 at 12:41
• Thanks! I was thinking wondering if i should go along the lines of $\dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $\geq$ 0 then f(x2)-f(x1) $\geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $\leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either. – Nolando Dec 3 '18 at 13:27

To prove that by contradiction, firstly we can prove that $$f'(x)>0 \implies f(x)$$ strictly increasing.

Suppose indeed that exist $$x_1 such that $$f(x_1)>f(x_2)$$ then by MVT

$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$

Form here we can now extend the result to $$f'(x)\ge 0$$, with $$f'(x)=0$$ only for a set of isolated points, $$\implies f(x)$$ strictly increasing, indeed assuming again by contradiction that $$f(x)$$ is not strictly increasing then it would exist an interval with $$f'(x)=0$$ which is against the hypotesis.