(05/17/2022, 08:38 PM)MphLee Wrote: Oh! Finally something I can fully grasp. This makes really lot of sense. What I take home from all of this is that complex functions are really well behaved gadgets. They possess a kind of elastic/rigid quality that reduces the degrees of freedom and stores a lot of global information of the objects in the local behavior, much like an hologram or a fractal, where every local part contains the information on the whole (holos). Unlike real functions where is like we play with rabber, with some plastic material: even if we have infinitesimal data of all at all degrees (smoothness) we have still lot of freedom (eg. not every smooth is analytic).
What you say in the first paragraph seems incredible to me but also reinforces my previous statement about the rigidity of the holomorphic functions. Basically, you are saying that the global behavior of a function, i.e. its values at an uncountable number of points \(0\in U\subseteq \mathbb C\) is determined uniquely by its behavior at a countable subset of points*, because \({\rm im}x=\{x_n\,:\, n\in\mathbb N\}\) is at most countable....
*sure that set \({\rm im}x\) needs to "accumulate" around \(0\), i.e. for every open containing zero, the sequence must fall infinite times in it.
This means that, if we see from another direction, that... functions defined only on a finite/countable set, sure that set must be special, uniquely extends to an uncountable set of complex numbers.. THIS is magic.
Let \(x:\mathbb N\to \mathbb U\) and \(f,g\in{\mathcal H}(U,\mathbb C)\), where \(U\in {\mathfrak U}(0)\). Then your statement is literally the following
Quote:Injectivity of precomposition if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
\[x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)\]
is injective. This translates as: \(x^*f=x^*g\), i.e. \(\forall n.\, f(x_n)=g(x_n)\), implies \(f=g\)
I read this as: holomorphic functions are so rigid, that if they agree infinitely time inside a small region, then they agree on the entirety of that small region.
Corollary 0 The first shocking consequence is that the cardinality of \({\mathcal H}(U,\mathbb C)\) is equal or smaller then the cardinality of \(\mathbb C\).
\[|{\mathcal H}(U,\mathbb C)|\leq |{\rm Hom}(\mathbb N,\mathbb C)|=|\mathbb C|^{|\mathbb N|}=(2^{\aleph_0})^{\aleph_0}={\mathfrak c}\]
This translates as "there are not many holomorphic functions" and as we can associate to every holomorphic function \(f\) on \(U\) an unique complex number \(i(f)\in\mathbb C\).
This is so good that it seems suspicious. There should be some technicality that you omitted stating the conditions probably.
Corollary 1 Note that as an easy corollary we have that there exists (at least one) surjection
\[{\mathcal H}(U,\mathbb C)\leftarrow {\rm Hom}(\mathbb N,\mathbb C):e\] going in the opposite way sending each \(y:\mathbb N\to \mathbb C\) to an holomorphic \(e(y):U\to \mathbb C \) such that
\[e(x^*(f))=f\]
Question
We have injectivity of \(x^*\) for \(x\) well behaved, it decays in a good manner at zero. But is \(x^*\) also surjective? In other words, do every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) come as a restriction at \(x\) of some holomorphic function \(f\in {\mathcal H}(U,\mathbb C)\)?
Yes, you say. You define it to be \(y\mapsto P\mapsto {\iota\circ P \circ \iota^{-1}}\). Call this construction \[{\mathfrak e}_x(y)\]
Quote:Surjectivity of \(x^*\) if \(x\in {\mathcal O}(1/n^{1+\delta})\subseteq {\rm Hom}(\mathbb N,U)\)
\[x^*:{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C)\]
is injective. That reads as: for every \(y\in {\rm Hom}(\mathbb N,\mathbb C)\) exists a \(f\in {\mathcal H}(U,\mathbb C)\) such that \(x^*f=y\).
Construction Let \(\iota(z)=1/z\), \(W_{x}\) the holomorphic function constructed by Weirstrass factorization theorem, it satisfies \(x^*W_{x}=W_x\circ x=0\). Define \(P_{x,y}(z)=W_x(z)\sum_{k=0}^\infty y_k\frac{a_{x,k}}{z-x_k}\) the function you defined in your post.
\[{\mathcal H}(U,\mathbb C)\to {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x\quad{\rm defined\,\, as}\quad {\mathfrak e}_x(y)=\iota \circ P_{\iota\circ x,\iota\circ y}\circ \iota\]
This construction satisfies \[x^*{\mathfrak e}_x(y)={\mathfrak e}_x(y)\circ x=y\]
Corollary 3 \({\mathfrak e}_x\) must be injective. This means there is a bijection (\((x^*)^{-1}={\mathfrak e}_x\)).
\[x^*:{\mathcal H}(U,\mathbb C)\simeq {\rm Hom}(\mathbb N,\mathbb C):{\mathfrak e}_x\]
If this holds exactly as I stated, then it is like an upgraded version of the recursion theorem. The analogy is a bit blurry but \(x^*\) is like evaluation at \(0\) and \({\mathfrak e}_x\) is like recursion (of what?).
ENIGMA This is all cool and exciting... but then why haven't we solved the existence and uniqueness problem for tetration and for complex iteration in general?
I mean... just define \(y_k=1/m[k]n \) or \(w_k=1/m[i]k\). We find a function \(P\) that satisfies \(P(1/k^{1+\delta})=1/y_k\) and an \(f\) is an holomorphic interpolation with \(f|_{\mathbb N}(k^{1+\delta})=m[k]n\).
At this point we have only two possibilities. Since extensions are interpolations, every holomorphic extension must be an holomorphic interpolation. But it must be unique and obtained by this interpolation procedure.
So if the interpolation obtained by Weierstrass is an extension good; if it isn't we have proved that there is not an holomorphic extension.
Hey, Mphlee
I can clarify why you think this is too good to be true. Everything you are saying upto that point is exactly true. But when you get into interpolating functions things get difficult. You also kind of missed one of the points I was making.
Let's say we take \(x_k = \frac{1}{k}\), and lets try to interpolate as \(x_k \to 0\) where each value equals \(y_k = \frac{1}{e\uparrow \uparrow k}\). Well, you're forgetting that these are point pairs. So, as per the rigidity of holomorphic functions, the function \(1/e \uparrow \uparrow z\) CANNOT BE HOLOMORPHIC as \(|z| \to \infty\).
So, if we were to make an interpolation, we'd have some weird function \(f(z)\), such that \(f(1/k) = 1/e\uparrow \uparrow k\), but it will also be holomorphic at zero. That instantly disallows it from being tetration. This doesn't help too much with interpolating, as it will not satisfy the functional equation properly. This is largely because we are interpolating as points go to infinity, and there's no unique way to do that (there's an essential singularity at \(z=\infty\)).
The only way you have uniqueness is if... \((x_k,y_k) \to 0\) and \(f(z) \) is holomorphic about zero, and \(f(x_k) = y_k\). THATS where you have a uniqueness criterion.
On the flip side, if we take points which go to infinity... we have no such uniqueness. As an example, consider \(g(z) = A(z)\sin(\pi z)\). You can make \(A\) whatever you want, and this function will still satisfy \(g(k) = 0\). This is because this function will not be holomorphic at \(z = \infty\), there's an essential singularity there.
So this idea extends really only for \(|z| < \delta\), and accumulation points at zero. We have to have holomorphy at the accumulated point. So trying to cheat the math and interpolating tetration's data points won't work unfortunately, though it'd be nice if it did.
This actually touches on the fractional calculus approach. Yes we are interpolating \(f(k) = \alpha \uparrow \uparrow k\), but we are also restricting that \(|f(z)| \le e^{\rho \Re(z) + \tau |\Im(z)|}\) for \(\rho \in \mathbb{R}^+\) and \(\tau < \pi /2\). It turns out, this produces a uniqueness condition at infinity. Where, if you have a function \(g(k) = 0\) and satisfies this bound, then \(g=0\). Notice, that the \(\sin(\pi z)\) function I wrote above, doesn't satisfy these bounds, so it doesn't matter to us.
All in all, what I wrote is only useful for local scenarios. It will not help us at interpolating sequences to infinity, while requiring they satisfy a functional equation. I mean, that's kind of obvious. If \((x_k,y_k) \to \infty\) and \(h(x_k) = y_k\), then the function \(b(z) = h(z) + W(z)\) is another solution to the same interpolation.
This only doesn't happen in a local scenario (or if you add some extraneous constraint (like the bounds for the fractional calculus)).
So just keep in mind, when I was saying \(f(x_k) = y_k\) as \((x_k,y_k) \to 0\) completely determines a holomorphic function, I am also assuming that \(f\) is holomorphic at \(z=0\). And that's a crucial requirement. I think this is what you were forgetting for a moment. So unfortunately, this doesn't really help in the sense of "interpolating" tetration. lol.
Quote:We have injectivity of for well behaved, it decays in a good manner at zero. But is also surjective? In other words, do every come as a restriction at of some holomorphic function ?
I'll try to find a reference, because I don't remember exactly. But, I'm pretty sure the answer to this is yes, if you restrict "what kind of" accumulation points. As in, as long as the accumulation points accumulate fast enough then yes. I'll try to find the answer for this, I can't remember off the top of my head.