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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

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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). This is because defining differentiability on boundary points requires extra care 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.

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  • $\begingroup$ 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+? $\endgroup$
    – Jacob Bak
    Apr 10, 2020 at 8:23
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    $\begingroup$ @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 to be concave on $\mathbb R^2_+$ without the Hessian approach. $\endgroup$
    – Herr K.
    Apr 11, 2020 at 0:22
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    $\begingroup$ 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. $\endgroup$
    – Jacob Bak
    Apr 11, 2020 at 8:19
  • $\begingroup$ @JacobBak: Sorry, I must have missed that footnote in MWG. I take back what I said in the first remark above. $\endgroup$
    – Herr K.
    Apr 11, 2020 at 22:12

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