# Estimating the growth of the Taylor coefficients given the growth of the function at the boundary

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 $$k-1\le k'\le k$$.

• @ChristianRemling. Yes, I'm also interested in various variants of this question, such as $|f(z)|\le 1$, and $|f(z)|\le -\log(1-|z|)$. Nov 24 at 16:24
• 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 $k >0$ Nov 24 at 16:39
• @Conrad: This is interesting. Can you please explain why $f$ is continuous? Nov 24 at 16:51
• 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 $0 \le k <1/2$ Nov 24 at 16:56
• @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 $f(x)=(x \log x+xt)/(2\pi)$ 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 $$\epsilon>0$$.
• 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) Nov 24 at 17:52
• 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... Nov 24 at 21:11
• @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". Nov 25 at 10:17