# Convex functions

What is a condition in terms of first derivatives for a function $f$ to be strictly convex? I am thinking that convex functions satisfy $f(a)\ge f(b)+(a-b)f'(b)$ the derivation is available here. However I am not sure how to impose the strictness condition since there exist strictly convex functions with $f''=0$ at certain points.

Or is there another condition in terms of first derivatives ?

Thank you.

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@Egbert, $x^4$ is a strictly convex function with vanishing 2nd derivative at $0$. – plm May 24 '12 at 9:34
Whoops... I'm sincerely sorry. Time for a break :( – Egbert May 24 '12 at 9:40

There was a discussion of the size of the set $N$ of nonstrictly positiveness of the hessian of a strictly convex function on MO.
Assume $f$ is $C^1$ on $\mathbb R$. Egbert's condition (that $f'$ is increasing) is necessary and sufficient, and it is equivalent to strict inequality in your formula. This is not incompatible with vanishing, since you take the limit as $a$ tends to $b$. In the limit you may have equality even though $f(a)>f(b)+(a-b)f'(b)$.
The condition is clearly sufficient because $f(a)=f(b)+\int_b^af'(x)dx>f(b)+(a-b)f'(b)$.
It is necessary because if the first derivative did not increase in some interval then integrating (i.e. using the fundamental theorem of calculus) you would have $f$ linear on that open interval, therefore not strictly convex.
For a $C^2$ function on $\mathbb R$, a criterion (at least a necessary condition) for strict convexity, in terms of $f''$, is that $N$ is nowhere dense. I think (as the MO OP) this must hold also for functions in all (open sets of) euclidian spaces.