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)$\geq$$0$ 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!

  • $\begingroup$ 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). $\endgroup$ – 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 $x_1<x_2$ 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.

  • $\begingroup$ 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? $\endgroup$ – Nolando Dec 3 '18 at 12:02
  • $\begingroup$ @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 $f(x_1)\geq f(x_2)$. $\endgroup$ – 5xum Dec 3 '18 at 12:41
  • $\begingroup$ 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. $\endgroup$ – 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 $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT


which is a contradiction.

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.


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