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 Universal uniqueness criterion II bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/16/2008, 05:27 PM (This post was last modified: 11/18/2008, 11:32 AM by bo198214.) Lets summarize what we have so far: Proposition. Let $S$ be a vertical strip somewhat wider than $1$, i.e. $S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z) for some $x_1\in\mathbb{R}$ and $\epsilon>0$. Let $D=\mathbb{C}\setminus (-\infty,x_0]$ for some $x_0 and let $G\subseteq G'$ be two domains (open and connected) for values, and let $F$ be holomorphic on $G'$. Then there is at most one function $f$ that satisifies (1) $f$ is holomorphic on $D$ and $f(S)\subseteq G\subseteq f(D)=G'$ (2) $f$ is real and strictly increasing on $\mathbb{R}\cap S$ (3) $f(z+1)=F(f(z))$ for all $z\in D$ and $f(x_1)=y_1$ (4) There exists an inverse holomorphic function $f^{-1}$ on $G$, i.e. a holomorphic function such that $f(f^{-1}(z))=z=f^{-1}(f(z))$ for all $z\in G$. Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (4)) and satisfies $h(z)=g(\delta(z))$. By (3) and (4) $\delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1$ and $\delta(0)=0$. So $\delta$ can be continued from $S$ to an entire function and is real and strictly increasing on the real axis. $\delta(\mathbb{C})=\mathbb{C}$ by our previous considerations. By Big Picard every real value of $\delta$ is taken on infinitely often if $\delta$ is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still $\delta(\mathbb{C}\setminus\mathbb{R})=\mathbb{C}$. But this is in contradiction to $g^{-1}:G\to D=\mathbb{C}\setminus [x_0,\infty)$. So $\delta$ must be a polynomial that takes on every real value at most once. This is only possible for $\delta(x)=x+c$ with $c=0$ because $\delta(0)=0$.$\boxdot$. In the case of tetration one surely would chose $x_0=-2$ and $x_1=0$ or $x_1=-1$. However I am not sure about the domain $G$ which must contain $f(S)$ and hence give some bijection $f:G\leftrightarrow S'$, with some $S'\supseteq S$. Of course in the simplest case one just chooses $G=f(S)$ if one has some function $f$ in mind already. However then we can have a different function $f_2$ with $f(S)\neq f_2(S)$ but our intention was to have a criterion that singles out other solutions. So we need an area $G$ on which every slog should be defined at least and satisfy $\text{sexp}(\text{slog})=\text{id}=\text{slog}(\text{sexp})$ as well as $\text{sexp}(S)\subseteq G$. « Next Oldest | Next Newest »

 Messages In This Thread Universal uniqueness criterion II - by bo198214 - 11/16/2008, 05:27 PM Simplified Universal Uniqueness Criterion - by bo198214 - 11/16/2008, 06:13 PM RE: Simplified Universal Uniqueness Criterion - by Kouznetsov - 11/19/2008, 03:14 AM RE: Simplified Universal Uniqueness Criterion - by bo198214 - 11/19/2008, 01:16 PM RE: Universal uniqueness criterion II - by Kouznetsov - 11/17/2008, 01:27 AM RE: Universal uniqueness criterion II - by bo198214 - 11/17/2008, 02:45 PM RE: Universal uniqueness criterion II - by Kouznetsov - 11/18/2008, 01:25 AM RE: Universal uniqueness criterion II - by bo198214 - 11/18/2008, 11:31 AM RE: Universal uniqueness criterion II - by Kouznetsov - 11/19/2008, 02:59 AM

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