Is there an analytic function $f$, such that $ f(\frac1n)=f(-\frac1n)=\frac1{2n+1}$, for all $n \in \mathbb N$? 
Let $ f:\{z|\; |z| \lt 1\} \rightarrow \mathbb C $ be a non constant analytic function. Which of the following conditions can possibly be satisfied by $f$ ?
  
  
*
  
*$f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{n^2}, \;\forall n \in \mathbb N$ 
  
*$f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{2n+1}, \;\forall n \in \mathbb N$ 
  
*$|f(\frac{1}{n})|\lt 2^{-n}, \;\forall n \in \mathbb N $ 
  
*$\frac{1}{\sqrt n}\lt |f(\frac{1}{n})|\lt \frac{2}{\sqrt n}, \;\forall n \in \mathbb N $

My attempt: 


*by Liouville's theorem $f(z)$ is constant, hence option 3 is false. 

*by maximum modulus principle $f(z)$ is constant, hence option 4 is false. 
1) it is clear that $f(z)=z^2$ is a polynomial and hence analytic, also f(z)=f(-z) is true, therefore option 1 is true.
2) $f(z)$ and $f(-z)$ are not equal, hence I can conclude option 2 is false. 
But I don't know how to use identity theorem to prove 1 is true and 2 is false. please explain me how to use identity theorem here.
 A: The fact that $f(\frac 1 n)=\frac1{2n+1}$ for all $n\ge 1$ implies that $f(z)-\frac{z}{z+2}$ has an accumulated zero; by the identity theorem, it follows $f(z)=\frac{z}{z+2}$. This invalidates $f(-\frac 1 n)=\frac1{2n+1}.$ So, there is no such analytic function. By the way, I can't see how Liouville's theorem and maximum modulus principle can show $f$ is constant. In 3, $f$ is not assumed to be an entire function and $|f(\frac 1 n)|<\frac1{2^n}$ does not mean $f$ is globally bounded. In 4, maximum modulus principle leads to nowhere.
A: You don't need anything as fancy as the identify theorem to reject possibility 2 -- not even anything specific to complex analysis.
First, the only way for $f$ to be even continuous is if $f(0)=0$.
And then, if you want $f$ to be differentiable at $0$, the limit
$$ \lim_{z\to 0} \frac{f(z)}{z} $$
must exist. How does this fraction behave at the points where you know $f(z)$?
A: *

*The analytic function $f(z)=z^2$ satisfies this.

*If 
$$
f\Big(\frac{1}{n}\Big)=\frac{1}{2n+1},
$$
then $f$ agrees with $g(z)=\dfrac{z}{2+z}$ at $z=\dfrac{1}{n}$, for all $n\in\mathbb N$, and since their limit point $0$ lies in the unit disc, then by Uniqueness Theorem, $f(z)=\dfrac{z}{2+z}$. But then
$$
f\Big(-\frac{1}{n}\Big)=\frac{-\frac1n}{2-\frac1n}=\frac{1}{1-2n}\ne\frac{1}{1+2n},
$$
and hence such $f$ DOES NOT exist.

*If $\,\Big|\,f\Big(\frac1n\Big)\Big|<2^{-n}$, then $f(0)=0$, since $f$ is continuous at $z=0$. As $f\not\equiv 0$, there exist $m\in\mathbb N$ and $g$ analytic in the unit disc, such that
$$
f(z)=z^mg(z), \quad g(0)\ne 0.
$$
But then
$$
\Big|f\Big(\frac1n\Big)\Big|=\bigg|\frac{g\big(\frac1n\big)}{n^m}\bigg|<2^{-n}
$$
and hence
$$
|g\Big(\frac1n\Big)|\le 2^{-n}n^m, \quad n\in\mathbb N.
$$
Now, as $n\to\infty$, the left hand side tends to $|g(0)|\ne 0$, while the right hand side tends to $0$.

*If $g=f^2$, then $g$ would satisfy
$$
\frac1n< \Big|g\Big(\frac1n\Big)\Big|<\frac4n
$$
Hence $g(0)=0$, by continuity, and $g(z)=z^mh(z)$, with $h$ analytic and $h(0)\ne 0$. Thus
$$
\frac1n< \frac{\big|h\big(\frac1n\big)\big|}{n^m}<\frac4n
$$
This implies that $m=1$. Hence, there exists an analytic function $h$, such that
$$
f^2(z)=zh(z), \quad h(0)\ne 0.
$$
This is impossible since, $f(0)=$, and hence $f(z)=zf_1(z)$, with $f_1$ analytic, and thus
$$
zf_1^2(z)=h(z)
$$
which implies that $h(0)=0$.
