Does being a local minimum imply a positive definite hessian? If $p\in R^{m}$ is a local minimum of  $F:R^{m}\rightarrow R$, then can we conclude that $\dfrac{\partial ^2F}{\partial x \partial x'}[p]$ is positive definite?
I guess you guys answers have concluded that $\dfrac{\partial ^2F}{\partial x \partial x'}[p]$ can be only positive semi-definite.
How can I prove this result then?
 A: A nontrivial counterexample is the function $f(x,y) = x^4+y^4$.  
Clearly, $f$ has a local minimum at $(x,y) = (0,0)$. 
However, the Hessian at $(0,0)$ is the zero-matrix, which is not positive definite. 
A: No. You can only conclude that it is positive semidefinite. Take
$f(x,y) = x^2$ for a simple example.
To show that the Hessian must be positive semidefinite, first consider
a scalar function has a local minimum at $x_0$ and is twice differentiable at $x=x_0$.
Then $f'(x_0)=0$ and so the second order Taylor expansion gives
$f(x) = f(x_0) + {1 \over 2} f''(x_0) (x-x_0)^2+ r(x-x_0)$, where
$\lim_{\delta \to 0} { r(\delta) \over \delta^2} = 0$. Since $x_0$ is a local
minimizer, we have ${1 \over 2} f''(x_0) (x-x_0)^2+ r(x-x_0)  \ge 0$ for
$x$ in a neighbourhood of $x_0$. Dividing by $(x-x_0)^2$ and taking the limit
as $x \to x_0$ gives $f''(x_0) \ge 0$.
Now consider a general $F$ which has a local minimum at $x_0$. Let $h$ be some direction and consider
$f(t) = F(x_0+th)$. We have $f'(0) = DF(x_0)h = 0$ and
$f''(0) = \langle h , D^2F(x_0)h \rangle $. From the
previous paragraph we conclude that $\langle h , D^2F(x_0)h \rangle \ge 0$ for
all $h$ and so $D^2F(x_0)$ is positive semidefinite.
