Let $A\subseteq\Bbb R^n$ be an open set which contains the closed unit ball $\overline{B(0,1)}$ and let $f\in C^2(A,\Bbb R).$ Suppose the Hessian $H_f(c)$ is positive definite $\forall c\in B(0,1).$ Prove $\exists x\in \Bbb R^n,\|x\|=1$ s. t. $f(x)>f(y),\forall x\in B(0,1).$
This is my reasoning that I would like to improve/correct if possible:
Since $f$ is continuous,$H_f(c)$ positive definite on the open unit ball $B(0,1),$ and the closed unit ball $\overline{B(0,1)}$ is compact, $f$ must attain its global extrema on the boundary $\Bbb S^{n-1}$ of the ball. If the statement were false, that is $\forall x\in\Bbb S^{n-1},\exists y_0\in B(0,1), f(x)\le f(y)$. In particular, it would also mean $\max\limits_{x\in\Bbb S^{n-1}}f\le f(y_0),$ so $y\in B(0,1)$ would be the global maximum, which is a contradiction to the prior conclusion as that would imply $H_f(y_0)$ is negative semidefinite.
Is there anything to add?