<p>The simplest example is <span class="math-container">$-x^3$</span> in the single variable case, or <span class="math-container">$-x_1^3-x_2^3$</span> in the case of two variables. The Hessian matrix is negative semi-definite at <span class="math-container">$(0,0)$</span>, but there is no maximum at this point.</p>

CC BY-SA 4.0
2021-10-18T11:48:13.107Z
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).
Kinno
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2021-10-18T12:02:52.507Z
@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.
Bertrand
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2021-10-18T12:10:37.177Z
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.
Kinno
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2021-10-18T11:38:40.817Z
2021-10-18T11:38:40.817Z
Bertrand
<p>When a Hessian matrix is <strong>negative definite</strong> at a critical point then that critical point is a local maximum (Sufficient Condition). <br />
<br />
As per the calculus wiki:
<a href="https://calculus.subwiki.org/wiki/Second_derivative_test_for_a_function_of_two_variables" rel="nofollow noreferrer">Link</a>, when the Hessian is <strong>negative semi-definite</strong> 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. <br />
<br />
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 <strong>negative semi-definite</strong> at the critical point it must also be a local maximum, but clearly calculus wiki disagrees.</p>

<mathematical-economics><optimization><linear-algebra>
Negative Definite vs Semi-definite Hessian - Sufficient vs Necessary conditions?
Kinno
# Negative Definite vs Semi-definite Hessian - Sufficient vs Necessary conditions?

(Asked by: username_0 on 2021-10-18)

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](https://calculus.subwiki.org/wiki/Second_derivative_test_for_a_function_of_two_variables), 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.

**Tags:** mathematical-economics, optimization, linear-algebra

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## 1 answers

### Answer by username_1 on 2021-10-18. Score: 4 | Accepted answer

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.

> **Comments:**
> **username_0**: 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).  
> **username_1**: @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.  
> **username_0**: 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.  

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