# Negative Definite vs Semi-definite Hessian - Sufficient vs Necessary conditions?

When a Hessian matrix is negative definite at a critical point then that critical point is a local maximum (Sufficient Condition).

As per the calculus wiki: Link, when the Hessian is negative semi-definite then, we can only conclude that it is not a local minimum. This seems to suggest that negative semi-definiteness is a necessary condition, not a sufficient one.

Can anyone provide an example of a multiple variable function where we have a negative semi-definite Hessian but not a local maximum? As per my thinking, if we evaluate the hessian to be negative semi-definite at the critical point it must also be a local maximum, but clearly calculus wiki disagrees.

The simplest example is $$-x^3$$ in the single variable case, or $$-x_1^3-x_2^3$$ in the case of two variables. The Hessian matrix is negative semi-definite at $$(0,0)$$, but there is no maximum at this point.
• @Kinno: Yes, the mere finding of a negative semi-definite Hessian does not imply that there is no maximum at this point. Consider $-x_1^4-x_2^4$ for instance, whose Hessian is nsd at $(0,0)$. Hmm, I am not sure that there is a general method allowing you to conclude in all cases. I would recommend to go back to the definition of a maximum and try to study whether $f(x_1,x_2) \leq f(x_1^*,x_2^*)$ for any $(x_1,x_2)$. In our example $-x_1^4-x_2^4 \leq 0$ and so $(0,0)$ corresponds to a global maximum of $f$, even though the Hessian is not negative definite at this point. Oct 18 at 12:02