I have a two part question about Louville theorem, and identitiy principle:
Loville theorem states that if function $f$ is holomorphic on $\mathbb{C}$ and $|f|$ is bounded $\Rightarrow$ $f$ is constant.
Identity principle states that if functions $f, g$ are holomorphic on area $D$ and if they match on set $A \subseteq D$ with limit in $D$ $\Rightarrow$ $f=g$ on $D$.
- Does there exist function $f$ such that it is holomorphic on $\mathbb{C}\backslash\left\{ 0 \right\}$ for every $n \in \mathbb{N}$ and that the following condition is met $f(\frac{1}{n})=\cos(\frac{n\pi}{2})$.
I can define a function $g(z) = \cos(\frac{\pi}{2z})$ which exists on $\mathbb{C} \backslash \left\{0\right\}$ and does met mentioned condition. But is it also unique? I think that we can not use the the identitiy principle, because the interval where our functions are holomorphic $\left\{ \frac{1}{n}; n\in\mathbb{N}\right\} \backslash \left\{ 0 \right\}$ does not conatin the limit $0$. Also I cna define another function $h(z) = f(z) + \lambda\sin(\frac{\pi}{z} )$ which also satisfies all required conditions ($\lambda \in \mathbb{R}$). Is my reasoning corect?
- Does there exist function $f$ such that it is holomorphic on $\mathbb{C}\backslash\left\{ 0 \right\}$ for every $n \in \mathbb{N}$ and that the following condition is met $n^{-5/2} < |f(\frac{1}{n})| < 3n^{-5/2}$.
I can define function $z^{5/2} < |g(z)| < 3z^{5/2}$ so $g(z)$ could be $2z^{5/2}$ and by identity principle $g(z) = f(\frac{1}{n})$. The function does therefore exist is unique and is holomorphic on $\mathbb{C}$. Is my reasonin here also correct?
I would like to get some feedback beacuse I find this chapter quite challenging and I do not know if I understand it properly yet. Thanks!