# Prove that a bounded analytic function in the right half-plane which vanishes at each positive integer is identically zero.

I want to solve the example below, but I can not.

Prove that a bounded analytic function in the right half-plane which vanishes at each positive integer is identically zero.

Please, you will be very grateful if someone would help resolve the communities in this example. Previously thank you.

• Is it only bounded in the right half plane or is it bounded and only takes values in the right half plane May 3, 2013 at 17:12
• bounded in the right half-plane May 3, 2013 at 17:17
• We need to find an entire function. Do we know that $f(z)$ is entire and just bounded in the RHP? Then we could use $$e^{-f(z)} \implies |e^{-f(z)}| = e^{-Re(f(z)} \leq 1$$ which is also entire so by Louisville is constant. However, this problem I think requires us to find the analytic continuation of $f$ to the LHP. To do this we need an accumulation point... May 3, 2013 at 17:43

$\textbf{Claim:}$ $f$ analytic and bounded in the half plane: $Re(z)>0$ having zeros at $z=1,2,3,...$ implies that $f$ is identically zero.

Consider $g(z)=f(\frac{1}{1+z}-\frac{1}{2})$ for $|z|<1$ (note: $\frac{1}{1+z}-\frac{1}{2}$ maps $|z|<1$ onto the right half plane ... it is nothing but a Moebius transformation)

$g$ is analytic in the (open) unit disk and bounded there and $g(z)=0$ whenever $\frac{1}{1+z}-\frac{1}{2}=n$ for $n$ a positive integer, or:

$$z=-1+\frac{1}{n+\frac{1}{2}}$$

Let me call $$-1+\frac{1}{n+\frac{1}{2}}=z_n$$ and note that the product $\prod(|z_n|)$ diverges to zero!

We want to show that $g(z)$ is identically zero. As it is analytic at zero, we will show that it has a zero of "infinite mutliplicity" there ( if $g$ is analytic at zero and is not identically zero, it has a zero of some finite order ... that is what we will contradict).

Let $B_k(z)=\prod_{n=1}^{k}(1-z\cdot z_n')/(z-z_n)$ where $'=$conjugate (of course, $z_n$ is real, so I don't really need the prime).

$g(z)\cdot B_k(z)$ is analytic (as $g$ has zeros at $z_n$) and by the maximum modulus principal we have:

$$|g(z)\cdot B_k(z)|<=\textrm{max_value_of_this_for_(|z|=r)}$$

whenever $|z|<r$.

Take the $\sup$ of the right hand side as $r\rightarrow 1$. $|B_k(z)|$ has a limit of $1$ as $|z|\rightarrow1$ and $|g|$ is bounded by, let's say, M (the original bound we have for the original function on the half plane).

We have $|g(z)|\leq\frac{M}{|B_k(z)|}$ for all $|z|<1$ and all $k$.

Put in $z=0$:

$|g(z)|<=M\cdot\prod_{n\leq k}|z_n|$ and take the limit as $k\rightarrow\infty$.

The product diverges to zero, so $g(0)=0$.

Well, then, $\frac{g(z)}{z}$ is analytic so do this again, replacing $g$ by $\frac{g}{z}$

$|\frac{g(z)}{z}\cdot B_k(z)|\leq M$ (since we use maximum modulus and as $|z|\rightarrow 1$, both $|B_k(z)|$ AND $|z|$ tend to one). Again, since the product of $|z_n|$ diverges to zero, $\frac{g(z)}{z}$ has a "value" (value of $h(z)=\frac{g(z)}{z}$, the analytic function with the first zero at $z=0$ removed) of $0$... or $g$ has a double zero at $z=0$.

... a triple zero! Etc. etc.

(if $g$ is not identically zero, there exists $h(z)$ with $h(0)<>0$, $g(z)=z^a\cdot h(z)$ for some positive integer $a$ ... replace $g$ by $h$ in the argument above to show that $h(0)=0$, a contradiction).

• How did you get the idea to use the products $B_k(z)$? May 3, 2013 at 20:14
• @AntonioVargas en.wikipedia.org/wiki/Blaschke_product May 3, 2013 at 20:17
• Interesting, thanks for the link. May 3, 2013 at 20:35