# 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 $$f$$ ?

1. $$f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{n^2}, \;\forall n \in \mathbb N$$
2. $$f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{2n+1}, \;\forall n \in \mathbb N$$
3. $$|f(\frac{1}{n})|\lt 2^{-n}, \;\forall n \in \mathbb N$$
4. $$\frac{1}{\sqrt n}\lt |f(\frac{1}{n})|\lt \frac{2}{\sqrt n}, \;\forall n \in \mathbb N$$

My attempt:

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

2. 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.

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

2. 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.

3. 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 $$0$$.

4. 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 $$h(0)=0$$.

• I didn't understand Why you have written $f^2(z)=zh^2(z)$ instead of $f^2(z)=zh(z)$ as $f^2=g$ and $g=zh(z)$ – Priyanka Mar 20 '19 at 0:13
• Also $1/n<|g(1/n)|<4/n$ right? – Priyanka Mar 20 '19 at 1:15
• @Priyanka Correct. – Yiorgos S. Smyrlis Mar 20 '19 at 6:38

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.

• Option 3 itself implies that f is bounded so I thought liouville's theorem is applicable here. – Priyanka Mar 14 '19 at 15:08
• Even if $f$ is bounded, Liouville's theorem can be applied to non-entire function? – Song Mar 14 '19 at 15:11
• No function should be entire – Priyanka Mar 14 '19 at 15:13
• Then please tell me how can I approach for options 3 and 4 – Priyanka Mar 14 '19 at 15:14
• A general principle that can be applied is, if $f(0)$ is $0$, then its rate of decay at $0$ should be of polynomial order, that is, there is some $k\ge 1$ such that $c<\frac{|f(z)|}{|z|^k}<C$ for some constants $c,C>0$. This (unique) $k$ is called the order of zero. Both $3,4$ do not satisfy this condition. – Song Mar 14 '19 at 15:17

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 $$f(z)$$?