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 $a,b&gt;0$ and $a+b&lt;1$ 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 $[0,\infty)^2$.

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.