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

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?

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 $f''(x_0)=0$.
• How about I add some conditions? Let $f(x)$, $0< f(x)\leq M<\infty$ be of class $C^{(2)}$ on an open set A, $x=(x'_1,x'_2)'$, and $x^*\in A\subseteq R^n$ is a critical point. Also, for any fixed $x_1(x_2)$, $\frac{\partial^2 f(x)}{\partial x_2 \partial x_2'}(\frac{\partial^1 f(x)}{\partial x_1 \partial x_1'})$ is negative definite. In addition, the hessian matrix of $f(x)$ at $x^*$, $H(x^*)=\frac{\partial^2 f(x)}{\partial x \partial x'}|_{x=x^*}$, is negative semi-definite. Can we show that if $x^*$ is the unique critical point in A, then $H(x^*)$ is negative definite? – Charles Su Jun 28 '16 at 19:29