Local strong convexity of a strictly convex function

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.

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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 $x\to 0$.