<ol>
<li>My question is under which contexts Negative Definitness (ND) vs Negative Semi-Definitness (ND) is required for classifying a global maximiser. And also Global vs Local.</li>
<li>I also want to understand what seems like contradiction in different sources, but I'm sure is just something I'm missing.</li>
</ol>
<p><strong>Current Understanding:</strong></p>
<ol>
<li>Necessary FOC for local max: <span class="math-container">$\nabla f(\mathbb{x}) = \mathbb{0}$</span></li>
<li>Necessary SOC for Local Max: Hessian <span class="math-container">$H$</span> is NSD.</li>
<li>Sufficient SOC for Local Max: <span class="math-container">$H$</span> is ND.</li>
<li>Sufficient SOC for Global Maximum: The function is concave over it's (whole) domain.</li>
</ol>
<ul>
<li>To ensure our local maximum point is a Global Max, we need to check that the function is Concave at all points. (As unlike quasi-concavity/convexity  Concavity and Convexity are not global properties, they are defined at points).</li>
</ul>
<p><strong>&quot;Conflicting&quot; Sources.</strong></p>
<ol>
<li>First perspective: Discovering that the Hessian is NSD, is not enough to establish that the function is Concave. This would also explain why are first SOC above is  a Necessary Condition and not a Sufficient Condition. Sources for this, and here's a quote:</li>
</ol>
<blockquote>
<p><em>Theorem: (multivariable second derivative test) At a critical point, if the Hessian function is positive (negative) definite, then the function has a minimum (maximum). If the Hessian is indefinite, the critical point is a saddle—you go up in some directions and down in others. If the Hessian is semidefinite, you cannot tell what is happening without further analysis, though if it is positive semidefinite you cannot have a maximum and negative semidefinite you cannot have a maximum.</em> <a href="https://staff.imsa.edu/%7Efogel/MVC/PDF/Notes%2031--The%20Second%20Derivative%20Test.pdf" rel="nofollow noreferrer">Source</a>. <a href="https://youtu.be/iPuMrI8wAl4?t=346" rel="nofollow noreferrer">And another source</a></p>
</blockquote>
<ol start="2">
<li>The proposition that is put forward by my course pack seems to be that NSD of the hessian is enough to establish concavity. I have attached an exact screenshot below.</li>
</ol>
<ul>
<li>This screenshot is in the wider context of a chapter on constrained optimisation and KKT, but it certainly reads like a stand alone statement.</li>
<li>I'm also unsure on the practical difference between <strong>concave and strict concave</strong>?</li>
</ul>
<p>For example they give a function <span class="math-container">$u(x_1, x_2) = x^a_1x_2^b$</span> where they take the hessian and show that <span class="math-container">$h_{11} &lt; 0$</span>, <span class="math-container">$|H| = 0$</span> therefore the function <span class="math-container">$u(x_1, x_2) = x^a_1x_2^b$</span> is concave. So any stationary points are global maximisers.</p>
<p><strong>My Theory/Remaining questions</strong></p>
<ol>
<li>My course pack is talking about testing the general function in the Hessian and hence all at once seeing if it's ND or NSD, and i suppose if if the Hessian of the while function is NSD it's concave? I'm not sure how ND vs NSD would materially effect the shape of the function?.</li>
<li>The other example sources cited and the first three conditions I mention are referring to testing specific critical points (I think), evaluating the hessian at this points to establish shape within a neighbourhood around the critical point? Hence local max min or saddle?</li>
<li>I'm unclear why NSD would NOT establish local concavity around a point. But across the entire function NSD does establish concavity.</li>
<li>Are there other ways of assessing the concavity of a function apart frrom the hessian, and what is concavity vs strict concavity?</li>
</ol>
<p><a href="https://i.sstatic.net/dzxBf.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dzxBf.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-04-08T03:26:47.907Z
For other ways of checking concavity/convexity, the following playlist covers many interesting results: https://youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u
Amit
CC BY-SA 4.0
2023-04-08T07:30:41.203Z
$u(x_1, x_2)=x_1^ax_2^b$ is concave for $a>0, b>0, a+b\leq 1$. When $a>0, b>0, a+b>  1$, $u(x_1, x_2)=x_1^ax_2^b$ is not concave.
Amit
CC BY-SA 4.0
2023-04-23T10:42:43.470Z
Thanks again for this @Amit. for the last example $a >0, b>0$ and $a+b > 1$ would I be correct in saying the utility function would be Quasi-Concave and monotonically increasing?
CormJack
CC BY-SA 4.0
2023-04-23T11:15:15.657Z
Yes it is quasi-concave and it also satisfy $(\forall (x_1, x_2), (x_1', x_2')\in\mathbb{R}^2_+)((x_1\geq x_1' \wedge x_2\geq x_2')\rightarrow u(x_1,x_2)\geq u(x_1',x_2'))$
Amit
<mathematical-economics><consumer-theory><optimization>
Testing for Concavity - Local Maximum & Global Maximum
CormJack
# Testing for Concavity - Local Maximum & Global Maximum

(Asked by: username_0 on 2023-04-07)

1. My question is under which contexts Negative Definitness (ND) vs Negative Semi-Definitness (ND) is required for classifying a global maximiser. And also Global vs Local.
2. I also want to understand what seems like contradiction in different sources, but I'm sure is just something I'm missing.

**Current Understanding:**

1. Necessary FOC for local max: $\nabla f(\mathbb{x}) = \mathbb{0}$
2. Necessary SOC for Local Max: Hessian $H$ is NSD.
3. Sufficient SOC for Local Max: $H$ is ND.
4. Sufficient SOC for Global Maximum: The function is concave over it's (whole) domain.

* To ensure our local maximum point is a Global Max, we need to check that the function is Concave at all points. (As unlike quasi-concavity/convexity Concavity and Convexity are not global properties, they are defined at points).

**"Conflicting" Sources.**

1. First perspective: Discovering that the Hessian is NSD, is not enough to establish that the function is Concave. This would also explain why are first SOC above is a Necessary Condition and not a Sufficient Condition. Sources for this, and here's a quote:

> *Theorem: (multivariable second derivative test) At a critical point, if the Hessian function is positive (negative) definite, then the function has a minimum (maximum). If the Hessian is indefinite, the critical point is a saddle—you go up in some directions and down in others. If the Hessian is semidefinite, you cannot tell what is happening without further analysis, though if it is positive semidefinite you cannot have a maximum and negative semidefinite you cannot have a maximum.* [Source](https://staff.imsa.edu/%7Efogel/MVC/PDF/Notes%2031--The%20Second%20Derivative%20Test.pdf). [And another source](https://youtu.be/iPuMrI8wAl4?t=346)

2. The proposition that is put forward by my course pack seems to be that NSD of the hessian is enough to establish concavity. I have attached an exact screenshot below.

* This screenshot is in the wider context of a chapter on constrained optimisation and KKT, but it certainly reads like a stand alone statement.
* I'm also unsure on the practical difference between **concave and strict concave**?

For example they give a function $u(x\_1, x\_2) = x^a\_1x\_2^b$ where they take the hessian and show that $h\_{11} < 0$, $|H| = 0$ therefore the function $u(x\_1, x\_2) = x^a\_1x\_2^b$ is concave. So any stationary points are global maximisers.

**My Theory/Remaining questions**

1. My course pack is talking about testing the general function in the Hessian and hence all at once seeing if it's ND or NSD, and i suppose if if the Hessian of the while function is NSD it's concave? I'm not sure how ND vs NSD would materially effect the shape of the function?.
2. The other example sources cited and the first three conditions I mention are referring to testing specific critical points (I think), evaluating the hessian at this points to establish shape within a neighbourhood around the critical point? Hence local max min or saddle?
3. I'm unclear why NSD would NOT establish local concavity around a point. But across the entire function NSD does establish concavity.
4. Are there other ways of assessing the concavity of a function apart frrom the hessian, and what is concavity vs strict concavity?

[![enter image description here](https://i.sstatic.net/dzxBf.jpg)](https://i.sstatic.net/dzxBf.jpg)

**Tags:** mathematical-economics, consumer-theory, optimization

> **Comments:**
> **username_1**: For other ways of checking concavity/convexity, the following playlist covers many interesting results: https://youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u  
> **username_1**: $u(x_1, x_2)=x_1^ax_2^b$ is concave for $a>0, b>0, a+b\leq 1$. When $a>0, b>0, a+b>  1$, $u(x_1, x_2)=x_1^ax_2^b$ is not concave.  
> **username_0**: Thanks again for this @username_1. for the last example $a >0, b>0$ and $a+b > 1$ would I be correct in saying the utility function would be Quasi-Concave and monotonically increasing?  
> **username_1**: Yes it is quasi-concave and it also satisfy $(\forall (x_1, x_2), (x_1', x_2')\in\mathbb{R}^2_+)((x_1\geq x_1' \wedge x_2\geq x_2')\rightarrow u(x_1,x_2)\geq u(x_1',x_2'))$  

---

## 0 answers

