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? .
If I understand right, you're asking:
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 $[0, 1]$.