Generalized Kneser superfunction trick (the iterated limit definition) JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 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 Ultimate Fellow Posts: 1,176 Threads: 123 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 MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/26/2021, 12:56 AM (This post was last modified: 05/27/2021, 03:58 PM by MphLee. Edit Reason: ERRONEOUS CLAIM ) PROGRESS UPDATE (May 25, 2021): I'll keep this as main hub for updates regarding superfunctions trick and superfunction-closed spaces. I'm making progress on the work on the Jabotinsky-Nixon theory link seems not completely contained in this so I'll post those updates in the Jabotinky-Nixon thread. I found some interesting trival results about groups that are extremely relevant to invertible superfunction-theory. I save them here for the future. Now about superfunction closed space I'd like to restrict attention on the special case where the superfunction space is a group, i.e. every function in the space is invertible. No further assumption on the space. We say that a group is superfunction-closed if $\forall g,f\in G$ that are different from the identity do exists an element $x\in G$ such that $xf=gx$. Remember: in a group, if an element is conjugated with the identity it IS the identity. In fact $1_Gx=xg$ implies $1_G=x^{-1}1_Gx=g$ We can conclude that a non-trivial group $G$ is supefunction-closed iff it has only two conjugacy classes: $[1]_\sim =\{1\}$ and $[g]_\sim =G/\{1\}$ $G=[1]_\sim \cup [g]_\sim$  (g is a nontrivial element) G being non-trivial means that exists $g\in G$ s.t. $g\neq 1_G$, i.e. exists a non-trivial element g. In other words this means that G has not less than 2 element. What I found is that if G is finite and is supefunction closed then it only contains cyclic elements of order two. i.e.  $g^2=1_G$. That means that the theory of ranks inside finite groups, when restricted to finite groups could finally provide examples for periodic sequences of subfunctions * (e.g. solution $x,y$ to the Nixon's system of equations $xf=yx \\yf=xy$ that dates back to the diamond operation thread). Lets call an element a torsion element if  $t^n=1_G$ for some non-zero n. For exemple, every involution is a torsion. The function i(x)=-x is a torsion in the group of bijections ${\mathbb R}\to{\mathbb R}$; the function j(x)=1/x is a torsion in the group of bijections ${\mathbb R}^+\to{\mathbb R}^+$. So torsion elements have only a finite number of different iterates. Torsion free elements have infinite integer iterates. Also if $G$ is taken infinite and torsion-free, torsion free means that it has not torsion elements except the identity, then we can extend  $G$ to a new group  $\mathfrak G$ that is superfunction-complete using the HNN extension, an advanced group theoretic technique (actually Exercise 11.78 in Rotman's An Introduction to the Theory of Groups) that reminds somehow the procedure of adding layers after layers of new superfuntions like adding ranks going and taking the limit of this process. Here some bookmarks for future studies: Embedding Theorems for Groups [url=https://math.stackexchange.com/questions/773437/infinite-groups-with-only-2-conjugacy-classes?rq=1][/url] * That statement is not accurate. When G is finite and super-closed then IT HAS ONLY TWO ELEMENTS. So the only periodic chain of subfunctions has lenght 1, i.e. it is a fixed point. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/26/2021, 10:31 AM PROGRESS UPDATE (May 26, 2021): Found some more bookmarks at MO Is there an infinite group with exactly two conjugacy classes?. The first is an "higly non trivial" result by Osin (2010) Theorem 1.1 (pag. 2) Any countable group G can be embedded into a 2-generated group C such that any two elements of the same order are conjugate in C Strengthening Higman-Neumann-Neumann previous embedding theorem Quote:HNN Embedding thm  any countable group G can be embedded into a countable group B such that every two elements of the same order are conjugate in B that is, modulo some details, given a countable group G of function we can add new function (extend the group) to obtain a new group where we can always solve the superfunction equation. Not only that you could think that all those groups are pretty boring and there are not many of them. Instead we have Quote:Corollary 1.3 (pag. 2) There exists an uncountable set of pairwise nonisomorphictorsion-free2-generated groups with exactly2conjugacy classes. In the MO answer by Dan Sălăjan there is an interesting note that could be interesting if we look at superfunction-closed groups as groups where, in some sense, there exist infinite chains of hyperoperations. Quote:As a psychological curiosity, Per Enflo writes in his Autobiography that the existence of groups with two conjugacy classes was a key insight behind his many solutions to outstanding problems in Functional Analysis. "I made important progress in mathematics in 1966, but it was more on the level of new insights, than actual results. When thinking about topological groups in the spirit of Hilbert's fifth problem (I had gradually modified the Hibert problem to some very general program: To decide whether different classes of topological groups shared properties with Lie groups) I was wondering whether there exist "very non-commutative" groups i.e. groups, where all elements except e, are conjugate to each other. I constructed such groups, by finding the right finite phenomenon and then make an induction. I understood, that this is a very general construction scheme (or "philosophy"), that probably could be applied to various infinite or infinite-dimensional problems*. And actually – this philosophy is behind several of my best papers - the solution of the basis and approximation problem, the solution of the invariant subspace problem for Banach spaces, the solution of Smirnov's problem on uniform embeddings into Hilbert space and more." (Dan Sălăjan, MO, 2013) *Boldface is mine. Correction to the previous post: if G is finite and is super-closed then it has to be the unique group with two element. https://proofwiki.org/wiki/Finite_Group_...2_Elements MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 05/26/2021, 11:36 PM Sounds interesting, Mphlee. Just please don't tell me we're going to have to start using Lie Groups MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/26/2021, 11:55 PM (05/26/2021, 11:36 PM)JmsNxn Wrote: Sounds interesting, Mphlee. Just please don't tell me we're going to have to start using Lie Groups Good news: I'm not going to.. Bad news: ..yet... I fear that it will be needed when I'll be ready to convert all o this from discrete to continuous. Lie groups=Groups+Topology... literally. Good news: I won't be able to start using those anytime soon... 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