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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?

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    The function $f$ has a zero at the origin of finite order, say $m$. Then $f = z^m g$ where $g$ is a non-vanishing analytic function defined on a neighbourhood of the origin. Now, you should get a contradiction. – halbaroth Oct 15 '20 at 04:21
  • @Halbaroth Can you please tell what contradiction? I am not able to deduce any! –  Oct 17 '20 at 09:19
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    Let $u_n = (\frac{1}{n})^m g(\frac{1}{n})$. This sequence converges to zero but its rate of convergence is far slower than the rate of converge of $2^{-n}$. – halbaroth Oct 17 '20 at 10:10
  • This has been asked and answered before: https://math.stackexchange.com/q/1564554/42969, https://math.stackexchange.com/q/1287448/42969, https://math.stackexchange.com/q/3221059/42969, https://math.stackexchange.com/q/209198/42969, https://math.stackexchange.com/q/542950/42969. – Martin R Oct 22 '20 at 06:45

2 Answers2

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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 $n \in \mathbb N$.

Good Boy
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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. $$ M n_1^{-m} > 2^{-n_1} \\ \implies \left| f\left(\frac{1}{n}\right) \right| > 2^{-n} \quad \forall n \geq n_1 \,. $$

jack
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