Conclusion about a holomorphic function which is not bounded

Question

This question is part of an assignment which I am trying.

Question: 1.Show that there is no bounded homomorphic function f on the right-half plane which is 0 at point 1,2,...and 1 at √2.

2.What if ' bounded ' is ommited?

I have done 1 by Lioville theorem but can there exists a such a function in 2?

I am not sure about that. Can anyone give example or tell which result to use to prove its existence?

Thanks!!

Answer 1

The function $\frac{\sin(\pi z)}{\sin(\pi \sqrt{2})}$ is equal to $0$ for $z = 1, 2, ...$ and equal to $1$ for $z=\sqrt{2}.$ Furthermore, it is holomorphic everywhere and unbounded. To see the latter, let $z=it$ and let $t$ be an arbitrary large real number.

Answer 2

I do another part since one part is already answered. I show there is no non-zero holomorphic function $f$ defined on open right half plane such that $f(n)=0$ for all $n\in\Bbb N$.

Let $\Bbb D:=\{z\in\Bbb C:|z|<1\}$ and $g:\Bbb D\to \Bbb D$ be non-constant holomorphic function and let $\{a_j\}_{j=1}^\infty$ be the zeros of $g$, counting multiplicities. Then, $\displaystyle \sum_{j=1}^\infty\big(1-|a_j|\big)<\infty$.

To prove this, without loss of generality assume each $a_j\not=0$. Write $g(z)=c_jz^j+c_{j+1}z^{j+1}+\cdots$ with $c_j\not=0$. For $n\geq 1$ define $$g_n(z):=\frac{g(z)}{\displaystyle z^j\prod_{k=1}^n\frac{a_k-z}{1-a_kz}}\text{ for }z\in\Bbb P\backslash\{a_j\}_{j=1}^\infty.\text{ Here}, \Bbb P=\Bbb D\backslash\{0\}.$$ By Riemann removable singularity theorem we can extend each $g_n$ on $\Bbb D$. By maximum modulus principal, $|g_n|\leq 1$. In particular, $\displaystyle |c_j|=\big|g_n(0)\big|\leq \prod_{k=1}^n|a_k|$ for each $n\geq 1$. Since $c_j\not=0$ we have $\displaystyle \prod_{j=1}^\infty|a_j|$, hence $\displaystyle \sum_{j=1}^\infty\big(1-|a_j|\big)<\infty$. converges.

Now, to conclude first part of your question consider the fact that $\psi:\{z\in\Bbb C|\text{Re}(z)>0\}\ni z\longmapsto \frac{1-z}{1+z}\in \Bbb D$ is bi-holomorphic and $\psi^{-1}=\psi$.

That is say consider $\psi\circ f\circ \psi^{-1}:\Bbb D\to\Bbb D$ after multiplying $f$ by suitable constant so that $f:\{z\in\Bbb C|\text{Re}(z)>0\}\to\Bbb D$.