Is there an analytic function (on the open unit disk) satisfying $\forall k \in \mathbb{Z}^+\left[f(\frac1{2k})=f(\frac1{2k+1})=\frac1{2k}\right]$?

Question

Although I could be wrong, I suspect the question in my title will be answered in the negative.

Let $f:D \to \mathbb{C}$, where $D$ is the open unit disk. Assume that $\forall k \in \mathbb{Z}^+\left[f(\frac1{2k})=f(\frac1{2k+1})=\frac1{2k}\right].$ Assume further that $f$ is analytic on $D$. I will try to derive a contradiction.

By continuity of $f$, I see that $f(0)=\lim\limits_{k \to \infty}f(1/2k)=\lim\limits_{k \to \infty}(1/2k)=0.$

I also see that $f'(0)=\lim\limits_{h \to 0} \frac {f(h)} h=\lim\limits_{k \to \infty} \frac {f(1/2k)} {1/2k}=1$.

Since the zeroes of a non-identically-zero function are isolated, $\exists r>0[0<|z|<r \to f(z)\neq 0] $

Is there a way to derive a contradiction from the above statements?

Answer 1

Suppose so. Then $f(\frac{1}{2k}) - \frac{1}{2k} = 0$ for all $k$ so by the uniqueness theorem $f(z) = z$ for all $z$ the unit disc. Hence, we cannot have that $f(\frac{1}{2k+1}) = \frac{1}{2k}$. In other words if the two conditions hold simultaneously there cannot exist such an analytic function.