# Uniqueness of Tetration

This is work in progress, dont rely on it!

## Proposition

For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$: $$\sigma_b(0)=1, \sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b({\rm i}t)\right| <\infty$$.

#### Proof.

We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$. Then we know that $$\sup_{t\to\infty}\alpha_b(\sigma_b({\rm i}t))<\infty$$ and the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1: $$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$ and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$, $\phi_k$ in $\R$):

$$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$

$$\delta(z)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}kz-{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}kz+{\rm i}\varphi_{k}\right)\right)$$ $$\delta(x+{\rm i}y)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left({\rm i}(2\pi kx-\varphi_{k})-2\pi k y\right)+\exp\left({\rm i}(-2\pi kx+\varphi_{k})+2\pi k y\right)\right) =\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\cos(2\pi kx-\varphi_{k})\left(\exp\left(-2\pi k y\right)+\exp(2\pi ky)\right)+{\rm i}\sin(2\pi kx-\varphi_{k})\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\right)$$

$$\Im(\delta(x+{\rm i}y)+x+{\rm i}y)=y+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\sin(2\pi kx-\varphi_{k})$$

$$\alpha_b(\sigma_b({\rm i}t))=\delta({\rm i}t)+{\rm i}t={\rm i}t+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right)$$

But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.