Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
@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
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.