MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a formal way to characterise strong convexity about the optimum value of a strictly convex function? I have an objective that looks something like this: $J(p,q) = \sum_{i=1}^{n}d(\pi_{i},q_{i}) + \alpha\sum_{i,j}w_{ij}d(p_{i},q_{j})$ where d is a jointly convex Bregman divergence (like KL or squared Euclidean) and $\pi_{i}$'s and $w_{ij}$'s and $\alpha$ are given parameter values. At the optimal value $(\bar{p}, \bar{q})$, neither the first term, nor the second term becomes zero (this comes from the practical usage of this objective). Is it possible to show strong convexity of this function around $(\bar{p}, \bar{q})$. My objective is to show positive definiteness of the Hessian around $(\bar{p}, \bar{q})$ and I think it is easier to prove strong convexity compared to proving positive definiteness of Hessian.

share|cite|improve this question
Do you wish to do this because strong convexity implies strict convexity? Or are you looking to see to what extent does the reverse direction hold? – Suvrit Mar 8 '12 at 17:31
Suvrit, I want to see to what extent strict convexity implies strong convexity around the minimizer. My final objective is to show positive definiteness of Hessian around the minimizer. Strong convexity $\Leftrightarrow$ positive definite Hessian, and positive definite Hessian around minimizer is going to help me in convergence rate analysis. – Ayan Mar 8 '12 at 18:35

Well, not a full answer, but in general a strictly convex function does not need to be strongly convex around its minimizer. An obvious example is $f(x) = x^4$ in the real axis. While this is "locally strongly convex" away from $x=0$, its "local modulus of strong convexity" decreases to zero for $x\to 0$.

However, I am not sure if one can produce this situation in is your case of Bregman divergences but I would guess it could be possible.

share|cite|improve this answer
Actually, I worked out the details and it turned out the showing positive definiteness of Hessian is easier for some special cases. However, I would really like to have a more "general" solution. Is anyone aware of any such theorem in convex analysis? – Ayan Mar 15 '12 at 0:08

Your Answer


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.