Hey, I was inspired by your little notice, so I thought I'd give an example of your black-box convergence at work. It's not perfect, wrote it in about three hours, but I think it's a good example of what you're driving at. It nearly constructs a set,
\(
\mathcal{B} \subset \mathcal{C}^1(\mathbb{R},\mathbb{R})\\
\)
That satisfies the conjugacy property,
\(
\forall f,g \in \mathcal{B}\,\,\exists \phi \in \mathcal{B}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\\
\)
I couldn't achieve this, unfortunately. But I got very very close. If \( \overline{\mathcal{B}} \) is the closure of \( \mathcal{B} \) as a monoid, then,
\(
\forall f,g \in \mathcal{B}:\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\,\,\text{or}\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,f(\phi(x)) = \phi(g(x))\\
\)
Which is so close to finding a set of functions satisfying the conjugacy property. The problem being when,
\(
\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1\\
\)
Which kind of throws a wrench in the construction; but only a small one--we just need to let \( \phi \) be in the closure of our monoid. I'm very confident we can use the closure of \( \mathcal{B} \) as our desired set, but I'm a little wonky on how to construct superfunctions for every element in the closure in a nice clean manner as I did for all of \( \mathcal{B} \) as a whole. In this space we'll find that the successor operation \( s \in \overline{\mathcal{B}} \) but not in \( \mathcal{B} \). This is great progress though, I think everything is coming together on trying to solve conjugacy equations in a differentiable manner. I'm going to take a closer look at \( C^k \) solutions now.
Regards, James
PS
I also think it's important to add, if we call \( \mathcal{B}^{-1} \) the set of functional inverses of \( \mathcal{B} \) then I believe that,
\(
\mathbb{B} = \overline{\mathcal{B}} \cup \overline{\mathcal{B}^{-1}}
\)
Will be our desired group with a conjugacy property \( \forall f,g \in \mathbb{B}\,\,\exists \phi \in \mathbb{B}\,\,\text{s.t.}\,\,f\phi = \phi g \). But, I'm wary on how to show this precisely. I think I'm 90% there though.
EDIT: I did notice a couple of typos in the PDF, please ignore them. I wrote this pretty quick. Especially when constructing the super function \( F(x) \) and when I use the Lipschitz condition. Recall that,
\(
F_n(x) = f^{\circ -n} (\Phi(x+n)) = \Phi(x) + \tau_n(x)\\
\)
Which is a small expression I forgot to add, upon which,
\(
|F_{n+1}(x) - F_n(x)| \le \lambda^n|f^{-1}(\Phi(x+n)) - \Phi(x+n)|\\
\)
and not what I wrote, I wrote it a tad too fast. But this function does look like \( x+n \), minus some error in growth, but it doesn't really matter because \( \lambda^n \) will still do the job it's supposed to do. Also, this is still going to need a good amount of finesse, I'm trying to think of alternative convergence factor functions than \( e^x \) (of which using \( e^{\mu x} \) is necessary for \( \mu > 0 \) in the general case, which I didn't add as of yet), as I think this might produce problems with simple functions like \( f(x) = \lambda\cdot(x+1) \) for \( \lambda > 1 \) unless we choose an appropriately small enough \( \mu \) dependent on \( \lambda \) (\( \frac{e^{\mu}}{\lambda} < 1 \)).
EDIT: Argh, this is becoming more and more frustrating, I'm having trouble making the construction work for \( f'(x) \to A >1 \) as \( x\to\infty \); which in turn means that \( \phi \) may NOT belong in the closure of \( \mathcal{B} \). God damn this is hard. The exponential convergents seem to only work when \( f'(x) \to \infty \).
EDIT: Woohoo! I fixed the problem I was facing. The PDF I sent is slightly incorrect, but I believe I have the right result now. I made a bit too fast of a jump when considering \( f'(x) \to A > 1 \) as \( x \to \infty \); we have to handle this case in a more special way, without using exponential convergents. So take the construction of the superfunction in this paper as only for the case \( f'(x) \to \infty \) and everything is still pretty much the same. I'll post an update in a bit.
\(
\mathcal{B} \subset \mathcal{C}^1(\mathbb{R},\mathbb{R})\\
\)
That satisfies the conjugacy property,
\(
\forall f,g \in \mathcal{B}\,\,\exists \phi \in \mathcal{B}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\\
\)
I couldn't achieve this, unfortunately. But I got very very close. If \( \overline{\mathcal{B}} \) is the closure of \( \mathcal{B} \) as a monoid, then,
\(
\forall f,g \in \mathcal{B}:\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,g(\phi(x)) = \phi(f(x))\,\,\text{or}\,\,\exists \phi \in \overline{\mathcal{B}}\,\,\text{s.t}\,\,f(\phi(x)) = \phi(g(x))\\
\)
Which is so close to finding a set of functions satisfying the conjugacy property. The problem being when,
\(
\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1\\
\)
Which kind of throws a wrench in the construction; but only a small one--we just need to let \( \phi \) be in the closure of our monoid. I'm very confident we can use the closure of \( \mathcal{B} \) as our desired set, but I'm a little wonky on how to construct superfunctions for every element in the closure in a nice clean manner as I did for all of \( \mathcal{B} \) as a whole. In this space we'll find that the successor operation \( s \in \overline{\mathcal{B}} \) but not in \( \mathcal{B} \). This is great progress though, I think everything is coming together on trying to solve conjugacy equations in a differentiable manner. I'm going to take a closer look at \( C^k \) solutions now.
Regards, James
PS
I also think it's important to add, if we call \( \mathcal{B}^{-1} \) the set of functional inverses of \( \mathcal{B} \) then I believe that,
\(
\mathbb{B} = \overline{\mathcal{B}} \cup \overline{\mathcal{B}^{-1}}
\)
Will be our desired group with a conjugacy property \( \forall f,g \in \mathbb{B}\,\,\exists \phi \in \mathbb{B}\,\,\text{s.t.}\,\,f\phi = \phi g \). But, I'm wary on how to show this precisely. I think I'm 90% there though.
EDIT: I did notice a couple of typos in the PDF, please ignore them. I wrote this pretty quick. Especially when constructing the super function \( F(x) \) and when I use the Lipschitz condition. Recall that,
\(
F_n(x) = f^{\circ -n} (\Phi(x+n)) = \Phi(x) + \tau_n(x)\\
\)
Which is a small expression I forgot to add, upon which,
\(
|F_{n+1}(x) - F_n(x)| \le \lambda^n|f^{-1}(\Phi(x+n)) - \Phi(x+n)|\\
\)
and not what I wrote, I wrote it a tad too fast. But this function does look like \( x+n \), minus some error in growth, but it doesn't really matter because \( \lambda^n \) will still do the job it's supposed to do. Also, this is still going to need a good amount of finesse, I'm trying to think of alternative convergence factor functions than \( e^x \) (of which using \( e^{\mu x} \) is necessary for \( \mu > 0 \) in the general case, which I didn't add as of yet), as I think this might produce problems with simple functions like \( f(x) = \lambda\cdot(x+1) \) for \( \lambda > 1 \) unless we choose an appropriately small enough \( \mu \) dependent on \( \lambda \) (\( \frac{e^{\mu}}{\lambda} < 1 \)).
EDIT: Argh, this is becoming more and more frustrating, I'm having trouble making the construction work for \( f'(x) \to A >1 \) as \( x\to\infty \); which in turn means that \( \phi \) may NOT belong in the closure of \( \mathcal{B} \). God damn this is hard. The exponential convergents seem to only work when \( f'(x) \to \infty \).
EDIT: Woohoo! I fixed the problem I was facing. The PDF I sent is slightly incorrect, but I believe I have the right result now. I made a bit too fast of a jump when considering \( f'(x) \to A > 1 \) as \( x \to \infty \); we have to handle this case in a more special way, without using exponential convergents. So take the construction of the superfunction in this paper as only for the case \( f'(x) \to \infty \) and everything is still pretty much the same. I'll post an update in a bit.