$f(x,y)$ has a Hessian matrix is positive definite for all $(x,y)$, prove that $(a,b)$ is the unique absolute minimum for $f$?

Question

suppose that $f(x,y) \in C^2$ with one critical point $(a,b)$ and that $f$ has the property that its Hessian matrix is positive definite for all $(x,y)$ except possibly at $(a,b)$ how can you prove that $(a,b)$ is the unique absolute minimum for $f$?

I know that since the Hessian matrix is positive definite that all the critical points will be local minimums, and that $f(x,y)$ is a convex function. However does the fact that the hessian matrix is positive definite except possibly at $(a,b)$ impact the result at all?

Answer 1

Hint.

Near the minimum

$$ f(x) \approx \frac 12x^{\dagger}H x $$

now due to epigraph convexity

$$ f(x_1) + x_1^{\dagger}H(x_2-x_1) \le f(x_2)\Rightarrow f(x_2)-f(x_1) \ge x_1^{\dagger}H(x_2-x_1) $$