A controversy regarding the generalization of the Sign function to dual numbers

Here is a link to a long discussion regarding generalization of $\operatorname{sign}z$ function to dual numbers.

There are basically two proposed versions:

1. $\operatorname{sign}(a+\varepsilon b) = \operatorname{sign}(a) + 2 b \delta(a) \varepsilon$ - proposed by user M.G. in their answer.

2. $\operatorname{sign}(a+\varepsilon b) = \operatorname{sign}(a) + 2 \operatorname{sign}(b) \delta(a) \varepsilon$ - this version is mine

The test results with matrices in Mathematica are inconclusive as they give different results depending on whether $b$ is variable or a numerical constant (link to the question on Mathematica.SE).

The version (1) obviously does not keep the very important property of the Sign function: $\operatorname{sign} (u v)=\operatorname{sign} u\cdot \operatorname{sign} v$.

Another my point is that for $b>0$, $\operatorname{sign} x=\operatorname{sign} bx$, so the right-hand side should not depend on $b$ except for its sign.

I do not understand the arguments raised against my version (that it breaks with scaling and change of basis). Can anyone please outline in a simple language the possible arguments against my generalization or better explain the existing ones? Why it would not work?

So first remark: the sign function is not differentiable in usual terms so one has to use derivatives in the distributions sense.

Second remark: Notice that the derivative of the function $x↦ u(λ x)$ evaluated at the point $x$ (which I write $(u(\lambda x))'$) is different from $u'(λ x)$ (the derivative of $u$ evaluated at the point $λ x$).

Let $u$ be a $0$-homogeneous distribution (what you call invariance by stretching, i.e. $u(λx)=u(x)$). Then its derivative (in the sense of distributions, or in the classical sense if this derivative exists in the classical sense) is $-1$-homogeneous since

• on the one side $(u(λx))' = (u(x))'=u'(x)$ as the function in $0$-homogeneous
• on the other side $(u(λ x))' = λ u'(λ x)$

The last identity is valid for distributions since for any smooth test function $\varphi$ $$ \langle (u(λ x))',\varphi\rangle = -\langle u(λ x),\varphi'\rangle = -\frac{1}{|\lambda|}\langle u,\varphi'(x/\lambda)\rangle \\ = -\langle u,\frac{1}{|\lambda|}\varphi'(x/\lambda)\rangle = -\frac{\lambda}{|\lambda|}\langle u,(\varphi(x/\lambda))'\rangle \\ = \frac{\lambda}{|\lambda|}\langle u',\varphi(x/\lambda)\rangle = \lambda\,\langle u'(\lambda x),\varphi\rangle $$ where I used the definition of the derivative of distribtions to get the 1st and 5th identities, the chain rule for smooth functions to get the 4th identity, and the definition of compositions of distributions to get identities 2 and 6.

Hence, in the sense of distributions and in the case of the function $\mathrm{sign}$, $\mathrm{sign}'(\lambda x) = \mathrm{sign}'(x)/\lambda$. More precisely, since the derivative of $\mathrm{sign}$ is $\delta_0$: $$ \delta_0(bx) = \mathrm{sign}'(bx) = (\mathrm{sign}(bx))'/b = (\mathrm{sign}(x))'/b = \delta_0(x)/b $$ which is why $\mathrm{sign}'= \delta_0$ is not "invariant by stretching", and this is why formula 1 makes more sense than formula 2 in your question (I think it was the only point you were missing?)