Can a non-constant analytic function satisfy this particular condition

Question

I am trying quiz questions in complex analysis and I could not think on how should I do this particular question.

Let $f:\{z:|z|<1\}\to\mathbb{C}$ be a non-constant analytic function. Then can it satisfy the condition $|f(1/n)|<2^{-n}$for all $n\in\mathbb{N}$.

The answer is that it can't satisfy. But I am unable to think which result in complex analysis to use.

Can you please help?

Answer 1

I will just put some sugar on top of the arguments from Halbaroth.

You can find a neighborhood around the origin $B_r(0)$ where $f$ factors into $$ f(z) = z^m g(z) \quad \forall |z| < r \,. $$ Note that $g$ is analytic and non-zero on $B_r(0)$. Thus, $$ \exists M >0: \quad 0 < M < |g(z)| \quad \forall |z| < r \,. $$

We can find a lower limit on $|f|$ by $$ \left| f\left(\frac{1}{n}\right) \right| = \left| \left(\frac{1}{n}\right)^m \; g\left(\frac{1}{n}\right) \right| > M n^{-m} $$ which holds for all $n>n_0, \frac{1}{n_0} \in B_r(0)$.

Now, the contradiction arises from knowing that exponential growth is stronger than polynomial. Thus, there exists $n_1$ s.t. $$ M n_1^{-m} > 2^{-n_1} \\ \implies \left| f\left(\frac{1}{n}\right) \right| > 2^{-n} \quad \forall n \geq n_1 \,. $$

Answer 2

Suppose for contradiction that there is such an $f$.

Writing the Taylor expansion of $f$ about $0$, suppose that the first non-zero coefficient is the $N$th (there must exist such an $N$ since $f$ is nonconstant).

I.e., for all $\,z\in D$ we may write

$$f(z) = \sum_{n=N}^∞ a_n z^n = z^N \underbrace{\sum_{n=0}^∞ a_{n+N} z^n}_{=:g(z)} = z^N g(z)$$

Then

• $g$ is also a holomorphic function on the disk, and
• $g(0) = a_N$; In particular, these together imply $$\lim_{z\to 0}|g(z)| = |a_N|\neq 0$$

(The first bullet point holds because $g$ has the same radius of convergence as $f$:

$$\limsup_{n\to ∞} |a_{n+N}|^{1/n} = \limsup_{n\to ∞} \left(|a_{n+N}|^{1/(n+N)} \underbrace{|a_{n+N}|^{-(N+n)/N}}_{\to 1}\right) = \limsup_{n\to ∞} |a_{n+N}|^{1/(n+N)} =\limsup_{n\to ∞} |a_{n+N}|^{1/n}.)$$

Now, $\lim_{z\to 0}|g(z)| = |a_N|$ implies that for large enough $n$, $|g(1/n)|\geq \frac{a_N}2,$ and so for $n$ sufficiently large,

$$2^n \left|f\left(\frac1n\right)\right| = \frac{2^n}{n^N} \left|g\left(\frac1n\right)\right| \geq \frac{2^{n-1}}{n^N}|a_N|.$$

The Ratio test tells you that the right hand side diverges to ∞, therefore so too must the left hand side, So there is no way $(2^n \left|f\left(1/n\right)\right|)_{n=1}^∞$ is bounded, and so your original inequality cannot hold for all $n \in \mathbb N$.