Derivative of a function involving a characteristic function.

Question

Consider the function $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ defined by

$$f(x,y,z) = y^2z\chi_{(0,\infty)^3}.$$

I would like to find

$$\frac{\partial^3f}{\partial x\,\partial y\,\partial z}.$$

I would have thought that if $x,y,z >0$ then

$$\frac{\partial^3f}{\partial x\,\partial y\,\partial z} = \frac{\partial^2}{\partial x\,\partial y}y^2 = \frac{\partial}{\partial x}2y = 0.$$

In the other case the function would be identically $0$ and hence it will have $0$ derivative. Is this correct?

Answer 1

Yes, this is entirely correct. You may want to make sure that the derivative is actually defined at the 'edges' where $x=y=0$, $x=z=0$ or $y=z=0$.

Answer 2

Your analysis is correct. Note that partial derivatives commute, so an alternative calculation is $$\frac{\partial^3 f}{\partial x\partial y\partial z}=\frac{\partial^2}{\partial y\partial z}0=0.$$