2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I see. So it is indeed a necessary condition not a sufficient one. Could you refer me to a method of further inspection to verify that the critical point is indeed a maximum if we have negative semi-definiteness (a non-graphical one). $\endgroup$
    – Kinno
    Oct 18 at 11:48
  • $\begingroup$ @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. $\endgroup$
    – Bertrand
    Oct 18 at 12:02
  • $\begingroup$ Great thank you. I have a feeling this might have to do something with establishing concavity at the intervals on the side of our critical point. Regardless, I will look further into it. $\endgroup$
    – Kinno
    Oct 18 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.