# Will'differentiable Convexity' of a function ensure the positive of a strictly increasing function's first derivative?

I Know that strict monotonic function can have a zero first derivative in quite. And not just the cantor function.

Would the pseudo-convexity or convexity of said function, ensure that its always this first derivative is positive?. (so long as f twice differen-tiable)

Otherwise, a Strictly Increasing function, which to my knowledge cannot have local minima, would indeed have one. As as a differentiable convex function is pseudo convex and whenever a pseudo convex function has a first derivative/gradient= zero, the function has a local mini-ma.

This would be a contradiction in terms for a strictly increasing function wouldn't it?

Or would it only ensure that at worst, the function has a zero positive derivative at its minimum value; if the function is only defined relative to a restricted range and domain?.

Would pseudo-concavity in addition help stop this from happening.

Ie, would this mean that if the derivative dropped to zero that such a point would be both a local minim-a and a local maxim-a, and in the convex case, global maxima.

So that it would either be contradiction in terms, or a constant function which, again would be ruled out, if f is strictly increasing? .

• That function only has negative derivative where it is not strictly increasing and its not technically a strictly increasing function.Ie when we consider negative elements of the domain. There its strictly the function value is strictly decreasing. I am talking about a function that strictly increases completely; or when restricted to that interval that is strictly increasing. – William Balthes Apr 29 '17 at 13:17
• oh, my bad, I oversaw the part about monoton. – user251257 Apr 29 '17 at 13:21

If I understand right, you're asking:

Let $I$ be an open interval of real numbers, and let $f$ be a strictly increasing, twice-differentiable, real-valued function on $I$. If $f$ is convex, is $f'$ non-vanishing (i.e., is $f'$ strictly positive on $I$)?

Assuming that interpretation is correct, "Yes".

Contrapositively, if there exists a point $b$ such that $f'(b) = 0$, then there exists a point $a < b$ such that $f'(a) > 0$. (Since $f$ is differentiable and strictly increasing, $0 \leq f'$ on $I$. If no such $a$ exists, then $f' = 0$ on some interval, contrary to the fact that $f$ is strictly increasing.) By the Mean Value Theorem applied to $f'$ on $[a, b]$, there exists a $c$ in $(a, b)$ such that $$f''(c) = \frac{f'(b) - f'(a)}{b - a} = -\frac{f'(a)}{b - a} < 0,$$ so $f$ is not convex.

If instead $I$ has a left-hand endpoint $a$, you can have $f'(a) = 0$, e.g., $f(x) = x^{2}$ on $[0, 1]$.

• Thanks for this. Your interpretation is correct; although I just have to careful about the terminology with regard to open interval. But in any case, yes you have it correct, Any convex function would likewise be at local or global minimum if the derivative dropped to zero; (except at the end points) and it would not be strictly monotonic by any stretch. What about if its a closed interval F:[0,1] to [0,1];F(1)=1 I – William Balthes Apr 29 '17 at 13:23
• I presume one cannot guarantee, that if F:[0,1] to [0,1] ; F(1)=1 F(0)=0, Could F'(0)>0 be guaranteed,(it is a minimum value in that case, but it may not be a maxima, or minima as such) Would some kind of pseudo concavity condition help ie some condition that implies that ifthe first derivative is zero, its f is a local or global maximum contradicting the fact that F'(0) is a minimum – William Balthes Apr 29 '17 at 13:34
• I think that because it only has the local implication; that this may not work, although I might be wrong, as the F'(0) could be a local maxima, if one were to extend the domain, (to the negative numbers, F may not be strictly increasing there, and so it might be a saddle point) ; So perhaps outright concavity would be required, to ensure it for that the end, point. Otherwise I suppose one would have to manually fix; F'(0)>0 maximum compatible condition help here, ie a function. I am not sure if increasing the it to ' would help, although in my case, there are other conditons as well. – William Balthes Apr 29 '17 at 13:38
• I think that is the issue with convex function even with regard to their continuity; they are generally lipschitz continuous but often have issues along the end points when defined on closed intervals in certain cases – William Balthes Apr 29 '17 at 14:01