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 $h(0)=0$.

  • $\begingroup$ 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) $ $\endgroup$ – Priyanka Mar 20 '19 at 0:13
  • $\begingroup$ Also $ 1/n<|g(1/n)|<4/n $ right? $\endgroup$ – Priyanka Mar 20 '19 at 1:15
  • $\begingroup$ @Priyanka Correct. $\endgroup$ – 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.

  • $\begingroup$ Option 3 itself implies that f is bounded so I thought liouville's theorem is applicable here. $\endgroup$ – Priyanka Mar 14 '19 at 15:08
  • $\begingroup$ Even if $f$ is bounded, Liouville's theorem can be applied to non-entire function? $\endgroup$ – Song Mar 14 '19 at 15:11
  • $\begingroup$ No function should be entire $\endgroup$ – Priyanka Mar 14 '19 at 15:13
  • $\begingroup$ Then please tell me how can I approach for options 3 and 4 $\endgroup$ – Priyanka Mar 14 '19 at 15:14
  • $\begingroup$ 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. $\endgroup$ – 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 $f(z)$?


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