Doubts on the domains of Nixon's method. MphLee Long Time Fellow    Posts: 375 Threads: 30 Joined: May 2013 02/27/2021, 11:03 AM (This post was last modified: 02/28/2021, 11:16 PM by MphLee. Edit Reason: Fixed many errors and typos. ) [Edited massively to fix MANY errors and typos: I hope it is more readable now. I apologize.] This follows from the discussion held at: MphLee, Generalized Kneser superfunction trick (the iterated limit definition), (January 21, 2021), Tetration Forum Quote:Can we write, $\mathcal{F} = \{f \in \mathcal{C}(\mathbb{R}^+,\mathbb{R}^+),\,f\,\text{is an isomorphism},\,f' \neq 0\}$ So that $f$ is say, a diffeomorphism (I believe that's the word, if not; it's something like that) of $\mathbb{R}^+$. Just so my shallow brain can think of a representative of the category; and it's not all up in the air. Let's additionally assume that: $|f(x)| \le Ae^{Bx}$ For some constants $A,B$. Which will make the exponential convergents behave well.  And it would imply it's inverse at worse grows like $\log$ somethin' somethin'. This would be a perfectly good algebraic space where we could derive, $\forall f,g \in \mathcal{F} \exists \phi \in \mathcal{F} f\phi=\phi g$ Now I haven't proven that, not entirely sure how to, but it's manageable--I could probably prove something close enough to continue the discussion.  ( JmsNxn , 28 January 2021) What you describe is "just" a group, i.e. a very poor category that has only one object (${\mathbb R}^+$) and s.t. every morphism is invertible. [note 1] Working on that we may try to extend the reasoning to the category of differential manifolds... this is good and bad. Bad because it is extremely complex. Good because it is an overdeveloped field of research, old and very relevant physics... ergo if we set up the right starting point/dictionary we might be able to recover everything in the existing literature [note 2]. Observations and doubts: you are taking the group of diffeomorphisms of ${\mathbb R}^+$: I'm very shaky on this, but I don't know if ${\mathbb R}^+=[0,+\infty)$ is a differential manifold. Let's admit it is. It is an object of the category "$\rm Diff$" of differential manifolds and what you are describing is the group of automorphisms of ${\mathbb R}^+$ in that category: $\mathcal F\overset{?}{=}{\rm Diff}({\mathbb R}^+):={\rm Aut}_{\rm Diff}({\mathbb R}^+)\subseteq \mathcal C^1({\mathbb R}^+,{\mathbb R}^+)$ Quick review. An endomorphism is a morphism whose domain and codomain coincide. An automorphism is an endomorphism that is an isomorphism. By definition, being an isomorphism (e.g. in the cat. $\rm Top$) means that there exists an inverse that is still in the category (e.g. continuous). In our case let $f:X\to Y$ be a continuous function. Being an isomorphism in $\rm Top$ (cat. of top. spaces) is a stronger condition than just being bijective: $f^{-1}$ MUST be continuous too. If it is, they (the mathematicians) call $f$ an homeomorphism. There are categories $C$ where a homomorphism in $f\in {\rm Ob}_C$ can be bijective as a set-theoretic map but its inverse do not respect the structure failing therefore to be an iso in that particular category $C$. An example I heard of is the map $[0,1)\to{\mathbb S}^1$ that maps $r\mapsto (\cos r2\pi ,\sin r2\pi)$: it is bijective (iso in $\rm Set$) but not a homeomorphism (not iso) in $\rm Top$. To be a $\mathcal C^k$-Diffeomorphims for a function $f$ means that it is differentiable ($\mathcal C^k$), it is invertible, and its inverse $f^{-1}$ is also differentiable ($\mathcal C^k$): it is an iso in the category $\rm Diff$. What are you doing? You consider the set/monoid $\mathcal C^1({\mathbb R}^+,{\mathbb R}^+)$. It is not a group because it contains non invertible $\mathcal C^1$ functions. Subsequently you take the subset of bijective ones. Not all of them but just the ones that have a $\mathcal C^1$ inverse: briefly, you consider the group of $\mathcal C^1$-diffeomorphisms. You continue by restricting to the subset of diffeomorphisms s.t. $f'\neq 0$. On this point I have four, very ignorant, questions for you. Is ${\mathbb R}^+$ really a manifold? How do we define an atlas of charts on ${\mathbb R}^+$? The problem I see is that $[0, x)$ is open in ${\mathbb R}^+$ with subspace topology... but how we would define a homeomorphims from that open to an open of some Euclidean space ${\mathbb E}^n$ ($n=1$, ${\mathbb E}^1={\mathbb R}$)? Should we consider instead $\rm Diff(0,+\infty)$?    why is the non-zero derivative condition necessary? I mean that if $f:{\mathbb R}^+\to {\mathbb R}^+$ is a diffeomorphism then $f'\neq 0$. I don't know how to prove it but, in my mind, if the derivative $f'(x)$ becomes zero at $x$ then the inverse $f^{-1}$ at the point $y:=f(x)$ should have a vertical slope, making $f^{-1}$... not good (not continuous/differentiable!?);    if $f:{\mathbb R}^+\to {\mathbb R}^+$ is a diffemorphism then is $0$ necessarily a fixed point? If $f\in \mathcal F$ is $\mathcal C^1$ and the derivative is nowhere zero then it has to be strictly increasing or decreasing right? If it is so then all functions in $\mathcal F$ have a fixed point at $0$: let $f(0)=b$ and $f$ increasing, where is the pre-image $f^{-1}[0,b)$ supposed to be located? The function has a zero somewhere, say in the point $z_0$. The function must be increasing in the interval $[z_0,+\infty)$ and bounded by $b$ so it has to have an horizontal asymptote $y=b$ and a vertical asymptote at $x=z_0$. Contradiction: $f$ cannot be $\mathcal C^1$ on ${\mathbb R}^+$. The same goes for the decreasing case. I know this is not a proof. I hope you can make it formal or explain to me where my intuition is off. Anyways this would produce fatal consequences for our needs. If all the $f\in \mathcal F$ fix $0$ then the successor or $\exp$ are not in our space and we can not consider the superfunction equations $\chi s=\chi f$ in it;    let $\mathcal F_{eb}$ be the subset of $f\in \mathcal F$ that are exponentially bounded. $f\in{\mathcal F}_{eb}:{\Leftrightarrow} \exists A,B\in{\mathbb R}^+, \forall x\in{\mathbb R}^+:f(x)\leq Ae^{Bx}$ Is ${\mathcal F}_{eb}$ still closed under composition? It doesn't seem so. Exponentiation is in the space but its iterates seem to be outside. Assume for the sake of argument that it is closed: scaling by $\lambda$ is in our space. By you closure proposition a Schroeder function of $\exp$ is in the space, i. e. a solution to $\phi\exp=\lambda\cdot \phi$ It could in principle because it grows slowly enough (?). In that case, since the logarithm must be in our space, an Abel function of exp will be in there as well ($\log_\lambda\circ{\rm Schroeder}_\lambda={\rm Abel}$). Contradiction! Its inverse tetration can't. We conclude that ${\mathcal F}_{eb}$ can't be a group. Sorry for the naive questions. I'm very limited in this field but probably these problems constitute partially the reason that induced you to add that Quote:    [t]he trouble I see with this space is that the super function of $f\in Q$ will not exist here $F\not\in Q$. I don't envy you if you're trying to create a general structure to where the superfunction sits. (JmsNxn, 06 February 2021) Could you try to make more precise your closure statement grounded on the stable knowledge of your papers? For example it seems to me that your statement should be something more like this. $\forall f,g \in \mathcal{F}_{eb}, \exists \phi \in \mathcal{C}^k:f\phi=\phi g$ I don't know if $k=\infty$ (smooth) or $k=\omega$ (analytic). Can you confirm or make the details precise? [note 1] Even without extending it to categories I believe that we can find a lot of work on those groups. To express this in a very inaccuarate way: the kind of "closure theorem" you propose is equivalent, if we ignore the exp bound condition, to this one "The group $\mathcal F$ has only one conjugacy class." I'll explain better this conceptual link in the post I'm prepairing but this is a specific case of "congugacy problem": when we study Real $\mathcal C^1$-flows we are indirectly studying the structure of conjugacy class of a group of diffeomorphisms. [note 2] To somehow prove that I'm not having hallucinations here two promising references that are taking it from this point of view. O'Farrell, Roginskaya, Conjugacy of real diffeomorphisms, A suvery, 2010 Farinelli, Conjugacy classes of diffeomorphisms of the interval in C1-regularity, 2012/18 This last one approaches the problem of approximating the solution by a sequence (or a path) of diffeomorphisms $\phi_n$ ($\phi_x$) converging to the (not exactly) superfunction. 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: 1,214 Threads: 126 Joined: Dec 2010 03/02/2021, 10:43 PM Hey MPHlee, Honestly I was just probing what might happen. I wasn't standing by what I wrote absolutely. I suggest reading the construction of $C^\infty$ hyper-operations, which I just posted. I was trying to visualize how we might generalize this to more exotic scenarios, and I was largely hypothesizing (guessing?) what might happen. That is, guessing what type of structure we may be able to develop--I think you're right though. What I wrote is definitely not correct, but it was more of an exercise in thought. After working through hyper-operators, I think I can more clearly say what I can say. But still, I honestly do not know.  I chose the space rather arbitrarily as, let's say, "I think it might look something like this." Perhaps a better way to say it is, to list the requirements of what we need in the process of the proof. The first thing we need is good control over our function $f$ such that, $\Phi(s) = \Omega_{j=1}^\infty e^{s-j}f(z) \bullet z\\$ Converges.  This will be $C^k$ if $f$ is. Further, this should be pretty simple to derive convergence. All we really need from this is that $f(\mathbb{R}^+)\subset \mathbb{R}^+$ is defined. So that the infinite composition converges. Second thing we need is for $f^{-1}$ to be well behaved, which is why I said diffeomorphism. I said that loosely, and you are correct. diffeomorphism is probably too strong, I simply meant it as a differentiable isomorphism. I chose $\mathbb{R}^+$ because it's convenient (and yes that would require it fixes zero). This is by no means necessary. However, $f'(x) > 0$ is needed, or at least, where $f'(x) = 0$ is easily controlled, not too sure. I'd just stick to non-zero derivative tbh. All we really need is for iterates $f^{-1}$ to be defined well enough, and for $f^{-n} \Phi(s+n)$ to be a meaningful thing. Again, this is very malleable. In the construction of the hyper-operators I use either a function $f^{-1} : (\alpha,\infty) \to \mathbb{R}$ bijectively, or $f^{-1} : \mathbb{R} \to (\alpha,\infty)$ bijectively. So we have a good amount of freedom here. If I were being honest, we just want $f^{-n} \Phi(s+n)$ to be a meaningful thing. The next thing we need is good control as a lipschitz condition of $f^{-1}$. We want, $|f^{-1}(x) - f^{-1}(y)| \le g(min(x,y)) |x-y|\\$ Where as $x,y \to \infty$ the value $g(min(x,y)) \le \lambda < 1$. This allows us to apply Banach's fixed point theorem very cleanly and efficiently. Again, this may not be necessary, but for a quick simple proof, yes this is good. This is why I said that exponential growth, or something like it, would be good. It would mean that $f^{-1}$ looks something like the logarithm. Again, I did this in the construction of the hyper-operators, where $\text{slog}$ and its higher order equivalents, have a lipschitz condition AT LEAST as good as the logarithm. Which is when $g(x) = \frac{1}{x}$. Which makes things pretty quick and not very messy. But again, not really necessary. As long as $f^{-1}$ is decently behaved and has some kind of lipschitz condition, we are all green. Now all of this is enough for us to show that, $F = \lim_{n\to\infty} f^{-n} \Phi(s+n)\\$ Converges to a CONTINUOUS function such that $F(x+1) = f(F(x))$. We will know that $\Phi$ is $C^k$ if $f$ is, but deriving that $F$ is $C^k$ is something I am not sure how to do yet. I managed to do it for hyper-operations, but it required a lot of nice things to happen. Mainly that, $\Lambda_n(s) = f^{-n}\Phi(s+n) - \Phi(s)$ Had some pretty convenient differential properties. Namely that $\Lambda^{(j)}_n(s)$ looks like $\Lambda_1^{(j)}(s)$ for large values of $s$. This is a very very crucial step in the proof to derive $C^\infty$, and I'm not sure how to guarantee this as  a condition on $f$. It happened conveniently with $\phi$ and $\tau$ and its converging sequence $\tau_n$; which look like $s$ for large $s$ and similarly its derivatives looked like $s^{(j)}$ for large $s$. Luckily, something similar happens with hyper-operators. Although I am not certain, I do believe that the missing ingredient will be something of the form $g(x) \le \frac{1}{x}$. Or that, the lipschitz constant of the inverse function $f^{-1}$ looks like the logarithmic lipschitz constant. This would mean, we want $f$ to grow faster, or like the exponential $e^x$ and that its inverse $f^{-1}$ is slow like the logarithm.  Again though, there's probably more to it. I think something like $f^{(j)}(x) > 0$ for all $j$ may come into play at some point. I cannot say for certain. But this isn't necessary, because I never showed $e \uparrow^n x$ satisfied this--but for large $x$ this will be true (again though, I don't use this.) Again, don't take everything I'm saying to heart. I'm still trying to get a feel for the general case. I also have to clean up the hyper-operators a bit--and in doing so, hopefully be better at identifying what actually allowed for them to be $C^\infty$. The proof seems to be pretty hardwired to the super-fast growth of hyper-operators, and the excessively slow growth of their inverses. When I get the time, I might run through a more elementary example, and see if we can construct a super function to something simple like, $f(x) = x^3 +1\\$ My suspicions are the method should still work in this case, but I may have to modify the approach; or at least fine-tune it. Especially when it comes to trying to make the super-function $C^\infty$. I don't foresee any problems with constructing a continuous super-function--it's the $C^\infty$ part that has me scratching my head. Regards, James. PS: I'll leave the spaces and category; or what sets these things live in, and how the conjugacy classes are structured to you, lol. I'll just focus on trying to create a reasonable set of criteria on $f$ such that we can get a super function $F$ which is as differentiable as $f$. My guess is that it's going to be really tricky to nail down a single criteria. However, what I just laid out above is the gist of how the hyper-operators were constructed. « Next Oldest | Next Newest »

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