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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?

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It seems you have answered your own question? As long as $Q^*$ is a strict minimum, then the left-hand side will be strictly positive: $$0<G(Q)-G(Q^*)=\frac{1}{2}(Q-Q^*)^T\nabla^2G(\hat{Q})(Q-Q^*),$$ which also suggests a counterexample might be constructed if $Q^*$ is not a strict minimum, for example take $f(x,y)=x^4$ in $\mathbb{R}^2$.

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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 $n \in \mathbb{N}$.

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  • $\begingroup$ If $f(x) = x^4$ (I assume you made a typo but the same thing holds for $x^{42}$) then $f''(x) = 12 x^2$, which is only zero for $x=0$, which happens to be the optimum point of $f$. I'm asking about $f''$ evaluated at a point that's not an optimum. $\endgroup$ – Jeff E Dec 2 '15 at 21:43
  • $\begingroup$ @JeffE: See the edit. $\endgroup$ – gerw Dec 3 '15 at 7:36

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