Do these statements about analysis with dual numbers make any sense?

Question

I am reading Color for the Sciences by Jan Koenderink, and in Ch. 3 he introduces the dual number system to define the space of possible power spectra for a beam of light. However, his statements do not seem mathematically rigorous and I cannot see a way they could be made sensible.

He appears to be assuming there is a natural way to extend functions defined on the reals to the duals, but this post suggests this is impossible.

I see how the "blip function" he defines is ruled out. However, this is not the only way to extend the blip function on $\mathbb R$ to the duals, and it seems one could simply define a blip function on the duals that agrees with the one on $\mathbb R$ and also with his criterion for acceptable functions.

Finally, his claim at the end that the space is Hausdorff implies that there exists a topology, which he has not specified. Perhaps there is a natural topology on this space that is too obvious to state?

I suggest not using images for large bodies of text because they are not friendly for people with disabilities. — Cameron L. Williams, Jun 14 '21 at 16:06
Not sure if this helps, but just realized the two pages I included are available in the Google books preview, p 67–68: https://www.google.com/books/edition/Color_for_the_Sciences/VrofEAAAQBAJ?hl=en&gbpv=1&printsec=frontcover — WillG, Jun 14 '21 at 16:39
Some of his statements are mathematically rigorous but that is not the point of his exposition as is clear from the text. They could be made sensible (for some definition of sensible) but the effort is not worth it. — Somos, Jun 14 '21 at 17:03
@Somos Fair but I would like to have some sense of how this could be made sensible. For example, how is the "blip" function ruled out? I could define $f(x_0+r\epsilon)=1+r\epsilon$ and this does not seem to conflict with his restriction on acceptable functions. With $f(x)=0$ elsewhere, this reduces to the same blip function when restricted to $\mathbb R.$ — WillG, Jun 14 '21 at 17:06
@Somos Since this question is about mathematical rigor I have to ask: What is the definition of "physical function"? — WillG, Jun 14 '21 at 17:13

score 1 · Answer 1 · answered Jun 14 '21 at 17:14

1

Formally, one can see the set of dual numbers as the quotient $\Bbb R[t]/(t^2)$ and write $\epsilon$ for the class of the indeterminate $t$ in the quotient. One equips the set of dual numbers with the Euclidean topology. No extension of the blip function to the set of dual numbers is even continuous, because the restriction to the real axis is already not continuous. His space $\Bbb S$ is an infinite dimensional subspace of $\mathcal{C}^\infty(\Bbb R)$ --- which is a Frechét space equipped with the family of semi-norms $\|f\|_{k,n} = \sup_{x \in [-n,n]}|f^{(k)}(x)|$ --- because this is his assumption: he just restricts the discussion to smooth functions.

From what I see, the author is being mathematically sound and not really sloppy here.

answered Jun 14 '21 at 17:14

Ivo Terek

80,301

But he does not restrict to continuous functions, he only restricts to functions with $f(x+\epsilon y)=f(x)+\epsilon Hy$. I could extend the blip function to the dual numbers in a way that satisfies this: Let $f(x_0+\epsilon y)=1+\epsilon y,$ and $f(x)=0$ for all $x\neq x_0+\epsilon y$. – WillG Jun 14 '21 at 17:20
This function restricted to $\mathbb R$ is the same blip function, but it satisfies his criterion. So I don't think he successfully rules out discontinuous functions with his criterion, as he claims. – WillG Jun 14 '21 at 17:23
But those functions are smooth, he says it on the text. The extension you're writing is not completely defined (what is $f(x+\epsilon y)$ for $x \neq x_0$ and $y\neq 0$?), and you can't make it continuous. – Ivo Terek Jun 14 '21 at 17:24
Define it however you want, but the limit along the real axis as $x \to x_0$ will be zero while the value at $x_0$ is $1$. You can't have a continuous extension for a discontinuous function. – Ivo Terek Jun 14 '21 at 17:26
Just let $f(x)=0$ whenever $x$ cannot be written as $x_0+\epsilon y$ for some $y\in\mathbb R$. Of course it is not continuous or smooth when restricted to $\mathbb R,$ but that is precisely my point. He claims that restricting to $f(x+\epsilon y)=f(x)+\epsilon Hy$ only allows for smooth functions, which is clearly false. – WillG Jun 14 '21 at 17:26
Perhaps his sentence "Such functions are smooth..." should be "In addition, I demand we only consider smooth functions..." ? – WillG Jun 14 '21 at 17:29
You can't just say that $f(x+\epsilon y)=0$ if $x \neq x_0$, this is not what he calls "locally affine". You would necessarily need $f(x+\epsilon y)= \epsilon y$ for $x \neq x_0$ to get the extension. But I see your point, even this corrected extension is not smooth, and I agree that the author should have made clear that smoothness is an assumption, not a consequence. – Ivo Terek Jun 14 '21 at 17:37
As a final question, is this a proper subspace of $C^\infty(\mathbb R)?$ If not, it seems like his presentation is a needlessly elaborate way of saying "Let's restrict to $C^\infty(\mathbb R)$." – WillG Jun 14 '21 at 18:21
Absolutely, there are smooth functions which are not locally affine in his sense. Take any smooth $f$ on $\Bbb R$ and extend with $f(x+\epsilon y)=f(x)+\epsilon p(y)$ where $p(y)$ is any function such that $p(0)=0$. This is not in $\Bbb S$ if $p$ is, say, quadratic. – Ivo Terek Jun 14 '21 at 23:41

Do these statements about analysis with dual numbers make any sense?

1 Answers1