Positive definite and continuous function I'm  reading a proof with the following statement.
Let $\nabla^{2} f(x) $ be the hessian matrix and continous. Assume it is positive definite in $x'$. Now there exists an open ball around $x'$ such that $\nabla^{2} f(x) $ remains positive definite for all $x$ in this ball.
Can someone explain why this is true? 
 A: A symmetric matrix is positive definite if and only if there is $C>0$ so that 
$$v^T A v \ge C$$
for all $v$ so that $\|v\| =1$. So as the space $\{ v: \|v\| =1\}$ is the sphere which is compact, if $A$ change continuously, then the above inequality still holds (with $C$ changed $C/2$ for example). 
A: By Sylvester's criterion, a symmetric matrix will be positive definite iff all of its leading principal minors (upper-left subdeterminants) are positive.  
We note, however, that each of these leading principal minors varies continuously with the entries of the matrix.  That is: suppose that $A$ is a positive definite matrix. If we change the entries of $A$ by a small enough amount, these principal minors will remain positive.  In other words, we can find an open ball of matrices around $A$ that all have positive leading principal minors.
So, there is an $\epsilon > 0$ such that if $\|A - \nabla^2 f(x')\| < \epsilon$ (here $\|\cdot\|$ is a "matrix norm"), then $A$ has positive minors.
By the continuity of $\nabla^2 f$, there is a $\delta$ such that if $\|x - x'\| < \delta$, then $\|\nabla^2f(x) - \nabla^2f(x')\| < \epsilon$.  That is, if $\|x - x'\| < \delta$, then $\nabla^2f(x)$ is positive definite.  So, the ball of radius $\delta$ around $x'$ is the ball that we're looking for.
