Callen claims in his book (chapter 5 in my copy at least) that the condition of minimum energy for fixed entropy is exactly equivalent to the condition of maximum entropy for fixed energy. I have seen this claim restated multiple times, and there are a bunch of answers in this site, like this one, which directly cite Callen's claim and derviation.
I have some problems with the way he derives this equivalency. Callen starts his derivation by stating that in the absence of internal constraints, $dS = 0$ and $d^2S < 0$ since entropy is maximum at this point. He then states that this implies that $$ \frac{\partial^2 S}{\partial X^2} < 0 $$ for all $X$. I understand he is roughly referring to the hessian of $S$, however, if this is the case, what he should probably say is that the Hessian matrix of the entropy $\mathcal{H}(S)$ is negative definite at that point, which means that $v^{T}\mathcal{H}(S) v < 0~\forall v$.
Callen finishes his derivation by stating that, since he can prove from this that
$$ \frac{\partial^2 U}{\partial X^2} > 0 $$ energy is a minimum at this point.
I don't think this implies the existence of a minimum, which, similarly to what was stated before requires the Hessian to be positive definite, which is stronger than all $\partial^2_X U$ being positive.
My question is twofold:
1) Does the partials $\partial^2_X U$ being positive imply that there is a minimum as Callen states?
2) If it doesn't then his derivation that the internal energy has a minimum at that point doesn't hold (as shown here, in wikipedia for instance), what would be a correct derivation of this fact?