Why is the derivative of |x| not equal to sgn(x) + 2xδ(x)?

[IniquiTrance]

The wikipedia article on $\sgn (x)$ (http://en.wikipedia.org/wiki/Sign_function) states that,

$$\frac{d}{dx}\vert x\vert = \sgn(x)$$

and $\frac{d}{dx}\sgn(x) = 2\delta(x)$. I'm wondering why the following is not true:

$$\begin{align*}<br /> \vert x\vert &= x\sgn(x)\\<br /> \Longrightarrow \frac{d}{dx}\vert x \vert &= \sgn(x) + 2x\delta(x) <br /> \end{align*}$$

by the product rule for derivatives. Is it because this derivative is indeterminate at $x=0$, and $2x\delta(x) \equiv 0$ for all $x \neq 0$?

 

[DavideGenoa]

The stated derivative of $\text{sgn}$ is the derivative of $\text{sgn}$ as a distribution, but for distributions that are not identifiable with differentiable functions the chain rule don't apply in general.

 

[jostpuur]

It is quite common that formal calculations still give correct results in some sense, even if they are not fully justified.

In this case I don't see anything wrong with the formula

$$\frac{d}{dx}|x| = \textrm{sgn}(x) + 2x\delta(x)$$

Simply subsitute $2x\delta(x)=0$ and you get the previous formula

$$\frac{d}{dx}|x| = \textrm{sgn}(x)$$

For test function $f$ we have

$$\int\limits_{-\infty}^{\infty} f(x) 2x\delta(x)dx = 0$$

so in this sense $2x\delta(x)=0$ holds "for all $x$", not only for $x\neq 0$.

 

[Xiuh]

The formula is true, because $x\cdot \delta$ is actually the zero distribution.