most recent 30 from http://mathoverflow.net 2013-05-24T08:44:39Z http://mathoverflow.net/feeds/question/90597 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90597/local-strong-convexity-of-a-strictly-convex-function Ayan 2012-03-08T17:22:47Z 2012-03-09T09:15:08Z

<p>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. </p>

http://mathoverflow.net/questions/90597/local-strong-convexity-of-a-strictly-convex-function/90675#90675 Dirk 2012-03-09T09:15:08Z 2012-03-09T09:15:08Z

<p>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 $x\to 0$.</p> <p>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.</p>