Finding the weak derivative of order $3$ of $f(x)=\operatorname{sgn} \sin(x)$ where $\operatorname{sgn}$ is the sign function

Question

Let $$f(x)=\operatorname{sgn} \sin(x)$$ where $\operatorname{sgn}$ is sign function. I need to find the weak derivative of order 3 for $f(x)$?

Answer 1

I'm not convinced that approximation of $\mathrm{sgn}\,x$ by $x/\sqrt{x^2+\epsilon}$ makes life easier. Some points to make:

• The function $f$ is locally constant on $\mathbb R\setminus (\pi \mathbb Z)$. Therefore, its derivative on this open set is zero.

• The distributional derivative is local: its restriction to an open interval such as $(\pi n-\pi/2, \pi n+\pi/2)$ is determined only by the values of $f$ on that interval.

• You should know the distributional derivative of the Heaviside (step) function.

• But if you don't, it's the Dirac delta, which acts on test functions by evaluating them at a point.

• When you take derivatives of a distribution $f$, you pass them to the test functions $\varphi$ via $\langle f',\varphi\rangle =\langle f,-\varphi'\rangle$. This is why, for example, the derivative of the Dirac delta acts on test functions by evaluation of $-\varphi'$ at a point. And so on.