# Concavity of Cobb-Douglass Utility Function on Non-Open set

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

I assume the notation $$\mathbb R^2_+$$ refers to $$[0,\infty)^2$$.
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.
• @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. Apr 11 '20 at 0:22