Necessary condition for local minima; non-negative Hessian matrix The problem I have is the following. Any results on Taylor expansions etc. can be assumed:
Let F : R^n -> R  be a C2 function. Let x_0 be a local minimum of F. Prove that the Hessian matrix of F is non-negative.
Thanks for your help.
 A: Let $\def\R{\mathbb R}h \in \R^n$. Define $g \colon \R\to \R$ by $g(t) = F(x_0 + th)$. Then $g$ is a $C^2$-function, having a local minimum at $0$. Hence $g'(0) = 0$, $g''(0) \ge 0$. By the chain rule, we have for $t\in \R$:
\begin{align*}
  g'(t) &= F'(x_0 + th)h\\
  g''(t) &= F''(x_0 + th)[h,h]
\end{align*}
For $t = 0$, we get 
$$ g'(0) = F'(x_0)h, g''(0) = F''(x_0)[h,h] $$
So we have $0 \le F''(x_0)[h,h]$. As $h$ was arbitrary, the bilinear mapping $F''(x_0)$ is non-negative, and hence it is representing matrix, the Hessian.
A: In elementary calculus for a function to have a local minimum the following condition must hold around $x_0$
$$f(x_0+\epsilon )\gt f(x_0)\Rightarrow f(x_0+\epsilon )- f(x_0)\gt 0$$
By Taylor's approximation we have
$$f(x_0+\epsilon)\approx f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2+O(\epsilon^3)$$
We can substitute the approximation of $f(x_0+\epsilon)$ to the definition of minimum
$$f(x_0+\epsilon )- f(x_0)=f(x_0)+\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2-f(x_0)\gt 0$$
$$\frac{df}{dx}|_{x_0}\epsilon+\frac 12\frac{d^2f}{dx^2}|_{x_0}\epsilon^2\gt 0$$
We can generalize it to n variables such as
$$\nabla f(x_0)\vec \epsilon+ \vec \epsilon^T \frac 12\nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since it is given that $\nabla f(x_0)=0$ the condition reduces to
$$\vec \epsilon^T \nabla^2 f(x_0)\vec \epsilon\gt 0$$
Since the term $\vec \epsilon^T (I)\vec \epsilon\gt 0$ is always positive definite for every $\epsilon$ the condition for local minimum reduces to
$$\nabla^2 f(x_0)\gt 0$$
