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 Generalized Kneser superfunction trick (the iterated limit definition) JmsNxn Long Time Fellow Posts: 419 Threads: 81 Joined: Dec 2010 03/24/2021, 10:20 PM (This post was last modified: 03/24/2021, 10:21 PM by JmsNxn.) It's back to the drawing board! So I thought I'd post how close I came in the third iteration of this paper. Call $ \mathcal{B} \subset \mathcal{C}^1(\mathbb{R},\mathbb{R})\\ f\,\,\text{is a bijection}\,\,\\ f'(x) > 0\\ \lim_{x\to\infty}f'(x) \ge 1\\ f^{\circ n}(x) \to \infty\\ f^{\circ -n}(x) \to -\infty\\$ Note in this space that the transfer operator $s : x \mapsto x+1$ exists. Then there exists a super function operator on $\mathcal{B}$, call it $\uparrow$, $ \uparrow : \mathcal{B} \to \mathcal{C}^1(\mathbb{R},\mathbb{R})\\$ And if $ \lim_{x\to\infty} g(x)/f(x) = I_{g,f} \ge 1\\$ There exists a $\phi \in \mathcal{C}^1(\mathbb{R},\mathbb{R})$ in which, $ g(\phi(x)) = \phi(f(x))\\ \phi\,\,\text{is a bijection}\\ \phi'(x) > 0\\ \lim_{x\to\infty} \phi'(x) \ge 1\\ \phi^{\circ n}(x) \to \infty\\$ BUT I CAN'T GET $\phi^{\circ -n}(x) \to -\infty$. I think this might be off base even trying to prove it with this much information. So I nearly have a set $\mathcal{B}$ which has a conjugate property, but this pesky condition at negative infinity has me stumped. Largely because the behaviour at positive infinity is used to construct $\phi$ and consequently it encodes nothing about the behaviour at negative infinity. Damn it, so close! As I said, back to the drawing board trying to find a set of functions which satisfies the conjugate property. I was so close, too! Regards, James JmsNxn Long Time Fellow Posts: 419 Threads: 81 Joined: Dec 2010 03/29/2021, 02:54 AM (This post was last modified: 03/29/2021, 03:22 AM by JmsNxn.) Honestly, at this point in my evaluation of my ideas (my approach), I don't think constructing a monoid globally on $\mathbb{R}$ is hopeful. I definitely got ahead of myself in thinking I could get a monoid $\mathcal{B} \subset \mathcal{C}^1(\mathbb{R}, \mathbb{R})$. I think, doing this in a local setting is easier--and far more probable. Even if we focus on the trivial case. That would be when we can just multiply our Schroder functions. By this I mean, if $f(0)=0=g(0)$ and $f'(0) = \lambda,\, g'(0) = \mu$ and $f,g$ are holomorphic in a neighborhood of $0$ and $|\mu|,|\lambda| \neq 0,1$, then, $ \phi(z) = \Psi^{-1}(\lambda \mu^{-1}\Phi(z))\\$ Is holomorphic in a neighborhood of zero. Here, $ \Phi(g(z)) = \mu\Phi(z)\\ \Psi^{-1}(\lambda z)= f(\Psi^{-1}(z))\\$ And this is of course a group under composition; in which $\phi$ belongs if $\lambda \neq \mu$... I think this may be a more tractable approach to constructing a general categorical theory.  This set of sheaves $\mathcal{B}$ do satisfy the conjugate property. But it's a little useless globally (it just means there's a taylor series in a neighborhood of zero). I do think its doable in the global sense, but probably in a local setting, is the correct way to approach the larger theory of conjugation. I definitely can't show it on the real line, but I may be able to do it locally in the complex plane (not necessarily about a fixed point). I'll have to entirely alter my approach though. Nonetheless the group, under composition, of sheaves $\mathcal{B}$, $ f(0) = 0\\ f\,\,\text{is holomorphic at}\,0\\ |f'(0)| \neq 0,1\\$ Is a very good place to start. Of which the conjugate property is almost satisfied here. And furthermore, this is a GROUP which almost satisfies the conjugate property. We'd just have to allow for $|f'(0)| =1$ and somehow massage this case to allow for the conjugate property--while still staying in the group. I forgot how useless I am at deep questions in real-analysis, so I'll stick to holomorphy. I'm so angry I can't get it on the real-line ):< Regards, James « Next Oldest | Next Newest »

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