I have a function $G(Q) : \mathbb{R}^n \rightarrow \mathbb{R}$ that is known to be convex. I also know that $Q^*$ is a minimum of $G(D)$. If I apply Taylor's theorem to $G(Q)$ at $Q^*$, I get:
$$ G(Q) = G(Q^*) + \nabla G(Q^*) (Q - Q^*) + \frac{1}{2} (Q - Q^*)^T \nabla^2 G(\hat{Q}) (Q - Q^*) $$
for some $\hat{Q}$ in the neighbourhood of $Q^*$. Since $Q^*$ is a minimum then the gradient is zero, leaving
$$ G(Q) - G(Q^*) = \frac{1}{2} (Q - Q^*)^T \nabla^2 G(\hat{Q}) (Q - Q^*) $$
I want to show that the right-hand side is strictly greater than zero. To do that, I need the Hessian $\nabla^2 G(\hat{Q})$ to be positive definite. But I only know from the convexity of $G(Q)$ that the Hessian is positive semi-definite. My intuition tells me that if $\hat{Q}$ is a non-stationary point (i.e., $G(Q)$ is strictly convex) then the Hessian at $\hat{Q}$ is positive definite. But I don't know how to show this or if it's even correct. My questions is then, is $\nabla^2 G(\hat{Q})$ positive definite if $G(Q)$ is convex and $\hat{Q}$ is in the neighbourhood of $Q^*$? What if $G(Q)$ is strictly convex instead?
This is not true. A counterexample in the convex case is $f(x) = 0$. And in the strictly convex case you can take $f(x) = x^{42}$ on $\mathbb{R}$.
Edit: Take $g(x) = x \, (1-\cos(1/x))$. Then, $g$ is continuous and, thus, there exists a function $G : \mathbb{R}\to\mathbb{R}$ with $G(0) = G'(0) = 0$ and $G''(x) = g(x)$. Then, $G$ is strictly convex, $0$ is the global minimizer, but $G''(1/(2\,\pi\,n)) = 0$ for all $n \in \mathbb{N}$.