In the second order optimality condition we have to prove that if $x^*$ is a stationary point of a function $f(x)$ and $\triangledown^2f(x^*)\succ0$, then $x^*$ is a strict local minimum point of $f(x)$.
But in the middle of the theorem, I don't understand the meaning of continuity of the Hessian and why it remains positive definite around a ball $x \in B(x^*,r)$.
On
I think I cracked it by myself.
Given a vector $d \in \mathbb{R}^n$, let's define the function:
$$f(x) = d^TH(x)d,$$ in which $H(x)$ is the hessian of $x$. Since the second derivatives are continuous, $f(x)$ is also continuous. Let's take any sequence inside of the ball $B(x_0,r), r>0$, by definition of continuity:
$$\lim_{x \rightarrow x_0} f(x) = f(x_0),$$ so, if that sequence converges to $x_0$ such that $f(x) \leq 0$ for all $x$ in the sequence, it will not converge to $f(x_0)$, because $f(x_0) > 0$, contradicting the definition.
Let $H$ be a symmetric and positive definite matrix, that is, there exists some $c>0$ with $$ d^T H d \ge c $$ for every $d$ on the unit sphere.
Now, let $H_n$ be a sequence of symmetric matrices, not positive definite, converging to $H$. Then, there exists a sequence $d_n$ on the unit sphere with $$ d_n^T H_n d_n \le 0. $$ As the unit sphere is compact, by going to a subsequence, we may assume $d_n$ converges to say $d$ without loss of generality. And we obtain the contradiction $$ 0 \ge d_n^T H_n d_n \to d^T H d \ge c > 0.$$
Aside: Many just argue that eigenvalues depend continuously on the components of the matrix... I find it a little bit too technical.