Let $f(z)=\sum a_nz^n$ be a Taylor series that converges for $|z|<1$ and satisfies $$ |f(z)|\le \frac{1}{(1-|z|)^{k}} $$ for some fixed $k>0$.

Question: What can I deduce about the growth of the Taylor coefficients $a_n$?

Partial result: By judiciously selecting the location of the contour in the formula $a_n=\oint z^{-n}f(z)\tfrac{dz}{2\pi i z}$, namely, by performing the integration over the contour $|z|=\tfrac{n}{n+k}$ [which is the minimum of $|z|^{-n}(1-|z|)^{-k}$], I can get the "trivial bound" $|a_n|< c\cdot n^k$.

But I suspect that this is not sharp.
In particular, the growth of the Taylor coefficients of $(1-z)^{-k}$ is only $n^{k-1}$. Not $n^k$.

More precise formulation of the question:
What is the optimal $k'>0$ such that $$ |f(z)|\le \frac{1}{(1-|z|)^{k}}\quad\Rightarrow\quad |a_n|< c\cdot n^{k'} $$ for all $f(z)=\sum a_nz^n$.

From the above arguments, I know that $k-1\le k'\le k$.

  • $\begingroup$ @ChristianRemling. Yes, I'm also interested in various variants of this question, such as $|f(z)|\le 1$, and $|f(z)|\le -\log(1-|z|)$. $\endgroup$ Nov 24 at 16:24
  • $\begingroup$ For example $f(z)=\sum_{n \ge 2}\frac{e^{in \log n}}{\sqrt n \log^2n}z^n$ is continuous hence bounded in the closed unit disc but its coefficients are only $o(1/\sqrt n)$ and not $o(n^{-1/2-\epsilon})$ and similar examples should be manufactured for $k >0$ $\endgroup$
    – Conrad
    Nov 24 at 16:39
  • $\begingroup$ @Conrad: This is interesting. Can you please explain why $f$ is continuous? $\endgroup$ Nov 24 at 16:51
  • $\begingroup$ actually I may be wrong as there is a result of Pommerenke (?) proved on page 71 (Thm 3.3) in Hayman Multivalent Functions (2nd edition) that gives the required estimate $|a_n| \le C_pn^{-k-1}$ for $k >1/2$ and mean $p$-multivalent functions satisfying the estimate in the OP, while for $0 \le k<1/2$ one can get only $o(n^{-1/2})$ but not better; of course this doesn't prove the result in general for $k >1/2$ as one has to deal with functions that may be $\infty$ valent, but it provides counterxamples for $0 \le k <1/2$ $\endgroup$
    – Conrad
    Nov 24 at 16:56
  • 1
    $\begingroup$ @Christian By partial summation using that $s_N(t)=\sum_{n=2}^N e^{in \log n}e^{int}$ are $O(\sqrt N)$ uniformly in $t$, the estimate coming from the standard Van der Corput second derivative test applied to $f(x)=(x \log x+xt)/(2\pi)$ $\endgroup$
    – Conrad
    Nov 24 at 17:00

The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a_n$ will not satisfy $|a_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases fast enough.

For positive $k$, we can similarly consider something like $$ f(z)=\sum n^{-2} \left(n^n\right)^k z^{n^n} . $$ Using calculus to find the maximum, we see that $$ x^k(1-\delta)^x \le C\delta^{-k} . $$ Thus $f$ satisfies the desired bound, but the coefficients do not satisfy $|a_n|\lesssim n^{k-\epsilon}$ for any $\epsilon>0$.

  • 1
    $\begingroup$ that's why one needs some conditions on $f$ to get non-trivial results (eg $p$ valence for some finite $p$ where area considerations come into effect so give better bounds) $\endgroup$
    – Conrad
    Nov 24 at 17:52
  • $\begingroup$ Thank you Christian, that's great [though not the answer I wanted :-)]. I wonder what would happen if I were to add the further condition that the $a_n$ form a decreasing sequence... $\endgroup$ Nov 24 at 21:11
  • $\begingroup$ @ChristianRemling Yes, I meant $|a_n|$ decreasing. But don't worry about that. This was just a thought prompted by Conrad's statement that "one needs some condition on $f$ to get non trivial results". $\endgroup$ Nov 25 at 10:17

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