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
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 \[{\mathfrak e}_x(y)\]
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.
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.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)