Using the identity theorem for $h\left(\left( \frac{n}{n+1} \right)^{2}\right) = 1 - \frac{1}{n+1}$

Question

Question: Does there exist a holomorphic function $h$ on $\mathbb{C}$ such that $$h\left(\left( \frac{n}{n+1} \right)^{2}\right) = 1 - \frac{1}{n+1}$$ for all $n \geq 1$.

So far I have that if $S = \left\{ \left( \frac{n}{n+1} \right)^{2} \mid n \in \mathbb{N} \right\}$, then $1$ is a limit point of $S$ in $\mathbb{C}$.

Also $h(z) = \sqrt{z}$ in $S$ which is not a continuous (and thus not an analytic) function in $\mathbb{C}$. Is this sufficient to conclude that there can be no analytic $h$ on $\mathbb{C}$ satisfying this property? I am struggling to see why this might be the case, any help would be appreciated.

Answer 1

There is no such function. If there was, we would have$$(\forall n\in\mathbb N):h\left(\left(1-\frac1{n+1}\right)^2\right)=1-\frac1{n+1}$$and therefore, by the identity theorem, $h(z^2)=z$. So $h^2(z^2)=z^2$ and so $(\forall z\in\mathbb{C}):h^2(z)=z$. But there is no continuous square root function in $\mathbb C$.

Answer 2

With out "square root functions": as Jose Carlos Santos has shown: if a function $h$ with the above properties would exist, then $h(z^2)=z$. But then we would have (by differentiation):

$1=2zh'(z^2)$ for all $z$. With $z=0$ we get a contradiction.