Unique critical point and psd implies pd and hence strict relative maximum

Question

Let $f(x)$ be of class $C^{(2)}$ on an open set A, $x_0\in A\subseteq R^n$ a critical point. In addition, the hessian matrix of f(x) at $x_0$, $H(x_0)=\{f_{ij}\}|_{x=x_0}$, is negative semi-definite. Can we show that if $x_0$ is the unique critical point in A, then $H(x_0)$ is negative definite?

Answer 1

No, take $A=(-1,1)$, $f(x)=x^4$. Then $x_0=0$ is the unique critical point of $f$, $f''$ is positive semi-definite, but $f''(x_0)=0$.