<p>I assume the notation <span class="math-container">$\mathbb R^2_+$</span> refers to <span class="math-container">$[0,\infty)^2$</span>.</p>

<p>Note that the set on which a function is <strong><em>defined</em></strong> need not be the same set on which a function is <strong><em>differentiable</em></strong>. In particular, it's typical that differentiability requirements are imposed on open sets (see e.g. the <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part" rel="nofollow noreferrer">fundamental theorem of calculus</a>). This is because defining differentiability on boundary points requires <a href="https://en.wikipedia.org/wiki/Semi-differentiability" rel="nofollow noreferrer">extra care</a> that's usually unnecessary for the issue at hand (such as defining concavity of a function via the Hessian matrix). </p>

<p>The Cobb-Douglas function is defined on <span class="math-container">$\mathbb R^2_+=[0,\infty)^2$</span>, (continuous also on that domain), and differentiable on <span class="math-container">$(0,\infty)^2$</span>, which is open and convex.</p>

CC BY-SA 4.0
2020-04-10T08:23:38.473Z
Thanks so much for the clarification. If so, the Hessian only allows us to conclude that the open set of (0,infinity)^2 is concave, not R2+?
Jacob Bak
CC BY-SA 4.0
2020-04-11T00:22:21.537Z
@JacobBak: Strictly speaking you're right, if we go by Osborne's definitions. Two remarks though. First, some authors (e.g. MWG) require only convexity (and not openness) in stating the Hessian-concavity/convexity proposition. Second, the Cobb Douglas function [can be shown](https://econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/concavity.pdf) to be concave on $\mathbb R^2_+$ without the Hessian approach.
Herr K.
CC BY-SA 4.0
2020-04-11T08:19:59.083Z
Your second remark helped me figure out that it is concave. However, with regards to MWG, footnote number 4 in Section M.C of the appendix states that the Hessian method applies to open sets. It would be very helpful if you could give me some reference that can teach me regarding methods for non-open sets as well.
Jacob Bak
CC BY-SA 4.0
2020-04-11T22:12:21.483Z
@JacobBak: Sorry, I must have missed that footnote in MWG. I take back what I said in the first remark above.
Herr K.
CC BY-SA 4.0
2020-04-10T07:49:04.783Z
2020-04-10T07:49:04.783Z
Herr K.
<p>My textbook argues that the Cobb-Douglass utility function <span class="math-container">$u=(x1)^a(x2)^b$</span> with <span class="math-container">$a,b&gt;0$</span> and <span class="math-container">$a+b&lt;1$</span> is concave on <span class="math-container">$R2+$</span> by computing the Hessian and showing it to be negative semidefinite for all points in <span class="math-container">$R2+$</span>.</p>

<p>However, I feel this method is flawed because <span class="math-container">$R2+$</span> is not an open set. A function is  concave on the set <span class="math-container">$A$</span> if and only if its Hessian is negative semidefinite for all <span class="math-container">$x$</span> in <span class="math-container">$A$</span>, but the assumption is that <span class="math-container">$A$</span> is an open and convex set. This does not hold, so the above methodology seems flawed. I am getting confused about this, so I would really appreciate some help please!</p>

<p>For reference, the textbook I am using is this: <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t" rel="nofollow noreferrer">https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t</a></p>

<microeconomics><mathematical-economics><utility><optimization><real-analysis>
Concavity of Cobb-Douglass Utility Function on Non-Open set
Jacob Bak
# Concavity of Cobb-Douglass Utility Function on Non-Open set

(Asked by: username_0 on 2020-04-10)

My textbook argues that the Cobb-Douglass utility function $u=(x1)^a(x2)^b$ with $a,b>0$ and $a+b<1$ is concave on $R2+$ by computing the Hessian and showing it to be negative semidefinite for all points in $R2+$.

However, I feel this method is flawed because $R2+$ is not an open set. A function is concave on the set $A$ if and only if its Hessian is negative semidefinite for all $x$ in $A$, but the assumption is that $A$ is an open and convex set. This does not hold, so the above methodology seems flawed. I am getting confused about this, so I would really appreciate some help please!

For reference, the textbook I am using is this: <https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t>

**Tags:** microeconomics, mathematical-economics, utility, optimization, real-analysis

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## 1 answers

### Answer by username_1 on 2020-04-10. Score: 4

I assume the notation $\mathbb R^2\_+$ refers to $[0,\infty)^2$.

Note that the set on which a function is ***defined*** need not be the same set on which a function is ***differentiable***. In particular, it's typical that differentiability requirements are imposed on open sets (see e.g. the [fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part)). This is because defining differentiability on boundary points requires [extra care](https://en.wikipedia.org/wiki/Semi-differentiability) that's usually unnecessary for the issue at hand (such as defining concavity of a function via the Hessian matrix).

The Cobb-Douglas function is defined on $\mathbb R^2\_+=[0,\infty)^2$, (continuous also on that domain), and differentiable on $(0,\infty)^2$, which is open and convex.

> **Comments:**
> **username_0**: Thanks so much for the clarification. If so, the Hessian only allows us to conclude that the open set of (0,infinity)^2 is concave, not R2+?  
> **username_1**: @JacobBak: Strictly speaking you're right, if we go by Osborne's definitions. Two remarks though. First, some authors (e.g. MWG) require only convexity (and not openness) in stating the Hessian-concavity/convexity proposition. Second, the Cobb Douglas function [can be shown](https://econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/concavity.pdf) to be concave on $\mathbb R^2_+$ without the Hessian approach.  
> **username_0**: Your second remark helped me figure out that it is concave. However, with regards to MWG, footnote number 4 in Section M.C of the appendix states that the Hessian method applies to open sets. It would be very helpful if you could give me some reference that can teach me regarding methods for non-open sets as well.  
> **username_1**: @JacobBak: Sorry, I must have missed that footnote in MWG. I take back what I said in the first remark above.  

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