existence of holomorphic function in the unit disc such that $f\Big(\frac{1}{2n}\Big)=f\Big(\frac{1}{2n+1}\Big)$

Question

Prove or Disprove

There exists a non constant function $f$ that is holomorphic in $D=\{z:|z|<1\}$ such that

$$\forall n\in\mathbb{N}:f\Big(\frac{1}{2n}\Big)=f\Big(\frac{1}{2n+1}\Big)$$

I tried to disprove the statement.

Let $G=D\cap\{z:\Re(z)>-\frac{1}{2}\}$.

If we look at $g(z)=f(z)-f\big(\frac{z}{z+1}\big)$, then

• $g$ is holomorphic in $G$
• $\forall z\in\big\{\frac{1}{2n}:n\in\mathbb{N}\big\}\cup\{0\}:g(z)=0$

So, by the identity theorem $g\equiv 0$ in $G$, i.e. $\forall z\in G:f(z)=f\big(\frac{z}{z+1}\big)$.

But, I don't know how to proceed to obtain that $f$ is constant.

Answer 1

By derivating, we have that for all $z\in G$: $f'(z)=\displaystyle{\frac{1}{(z+1)^2}f'(\frac{z}{z+1})}$ and one more time: $f''(z)=\displaystyle{\frac{1}{(z+1)^4}f''(\frac{z}{z+1})-\frac{2}{(z+1)^3}f'(\frac{z}{z+1})}$. Applying at $0$: $f''(0)=f''(0)-2f'(0)$, therefore $f'(0)=0$. By calculating higher order derivatives, you get that $f''(0)=0, f'''(0)=0,\dots$. But this implies that $f(z)=f(0)$ for all $z\in\mathbb{D}$, since non-constant holomorphic functions have no zeros of infinite order.

Answer 2

Let $\phi(z)= \frac{z}{z+1}$. Take $a\in (0,1)$. $f$ takes equal values at all points $\phi^n(a)$. The sequence $\phi^n(a)$ is decreasing, with limit $0$. What can you conclude?