# Proof of a series having infinitely many zeros in the unit disc

I have read an interesting question recently. This is about a series having infinitely many zeros inside the unit disc $$U$$. The series is $$\begin{equation*} f(z)=\sum_{k=1}^\infty 5^k z^{n_k}, \end{equation*}$$ where $$\{n_k\}$$ is a sequence of integers such that $$n_1 \ge 2$$ and $$n_{k+1}>2kn_k$$. I have shown that $$f \in H(U)$$ and it does not have radial limit at any point of the unit circle $$\partial U$$. However, I can't show that it has infinitely many zeros inside $$U$$. The tools that I can apply are graduate level complex analysis such as Rouche's theorem, the open mapping theorem, the maximum modulus theorem, Schwarz lemma, Runge's theorem, the Mittag-Leffler theorem and etc. Anyone can help?

This function has the property that on the circle $$|z|=1-\frac{1}{n_k}, k \ge 10$$, $$|h(z)|>C5^k$$, where we can take for example $$C=\frac{1}{100}$$

(note that $$(1-\frac{1}{n_k})^{n_k} \ge \frac{1}{3}$$ as that sequence converges increasingly to $$\frac{1}{e}$$ with the inequality already happening for $$k \ge 10$$ and $$n_k>k$$, while for $$m>k,(1-\frac{1}{n_k})^{n_m} \le (\frac{1}{e})^{2^{m-k}(m-1)...k} < 6^{-m}$$ so the corresponding terms form a geometric sum that converges to a small finite number, while the smaller terms sum in absolute value to obviously less than $$\frac{1}{4}5^{k}$$ by the trivial estimate and the corresponding geometric series)

Assume now that $$h$$ has finitely many zeroes only (could be none here of course). Let $$B$$ a finite Blaschke product with the same zeroes (including multiplicities etc, where we take $$B=1$$ if $$h$$ has no zeroes). Then $$g=\frac{h}{B}$$ has no zeroes in the unit disc and is analytic and still satisfying $$|g(z)| > C5^k, |z|=1-\frac{1}{n_k}$$, since $$|B(z)| <1, |z| <1$$.

But then $$\frac{1}{g}$$ is analytic in the unit disc and by maximum modulus, $$|\frac{1}{g}| < \frac{1}{C5^k}, |z| \le 1-\frac{1}{n_k}$$. If we let $$k \to \infty$$ we get $$\frac{1}{g}=0$$ in the unit disc and that is a contradiction.

Note that $$h(z)-w$$ has the same properties as $$h$$ (taking all $$k$$ large enough so $$5^k > 200|w|$$ say, so we can use $$C=\frac{1}{200}$$ say, so the same proof applies to show that $$h$$ takes every complex value infinitely many times in the unit disc!

• This is a very nice solution. – KK Wong Mar 6 at 4:47

Here is another solution to the problem following the hint in W. Rudin's Real and Complex analysis boo for problem 6, chaser 13.

The key, as also pointed put in the OP and by others is to show that there is $$c>0$$ such that along any circle of radius $$r_m:=1-\tfrac{1}{n_m}$$ \begin{align} |f(z)|\geq |5^mz^{n_m}|-|f(z) - 5^mz^{n_m}|\geq 5^m(r_m)^{n_m} -\sum_{\substack{k\geq1\\ k\neq m}}5^k(r_m)^{n_k}> 5^mc\tag{1}\label{one} \end{align}

Zeroes of $$f$$: Once this is proven, one can appeal to Rouché's theorem. Let $$p_m(z):=5^m z^{n_m}$$. The polynomial $$p_m$$ has only one zero $$z=0$$ of multiplicity $$n_m$$ in $$\mathcal{C}$$. From \eqref{one}, $$|f(z)-p_m(z)|<|p_m(z)|,\qquad |z|=r_m$$ By Rouché's theorem $$f$$ has the same number of zeroes, $$n_m$$, as $$g$$ counted according to their multiplicities inside the disk $$D(0;r_m)$$. This implies that the number of zeroes of $$f$$ in $$U$$ us countably infinite ($$r_m\xrightarrow{m\rightarrow\infty}1$$ and $$n_m>m$$).

Proof of \eqref{one}: Here is a rather pedestrian approach. Notice that from $$n_1>1$$ and $$n_{k+1}>2k n_k$$, it follows that \begin{align} n_{k+m}> 2^k\Big((k-1+m)\cdot\ldots\cdot m \Big) n_m >2kmn_m \tag{2}\label{two} \end{align} for all $$k,\, m\geq1$$. Along the circle $$|z|=r_m$$ \begin{align} \Big|\sum^{m-1}_{k=1}5^kz^{n_k}\Big|\leq \sum^{m-1}_{k=1}5^k(r_m)^{n_k}\leq \frac{5^m}{4}\tag{3}\label{three} \end{align} Since $$g(x)=\big(1-\tfrac{1}{x}\Big)^x$$ is strictly monotone increasing in $$[1,\infty)$$ and $$\lim_{x\rightarrow1-}g(x)=e^{-1}$$, \begin{align} \Big|\sum^\infty_{k=m+1}5^kz^{n_k}\Big|&\leq 5^m\sum^\infty_{k=1}5^k(r_m)^{n_{k+m}}< 5^m\sum^\infty_{k=1} 5^k(r_m)^{2mkn_m}\\ &< 5^m\sum^\infty_{k=1} 5^k e^{-2mk}= 5^m\frac{5e^{-2m}}{1-5e^{-2m}}<5^m\frac{5e^{-6}}{1-5e^{-6}}\tag{4}\label{four} \end{align} for $$m\geq3$$. Combining \eqref{three} and \eqref{four} we obtain $$5^m(r_m)^{n_m}-\sum_{\substack{k\geq1\\ k\neq m}}5^k(r_m)^{n_k}>5^m\Big((r_m)^{n_m}-\frac{1}{4}-\frac{5e^{-6}}{1-5e^{-6}}\Big)\geq 5^m\Big((r_3)^{n_3}-\frac{1}{4}-\frac{5e^{-6}}{1-5e^{-6}}\Big)$$ for all $$n\geq 3$$. $$n_3>8n_1>16$$. A numeric calculation gives $$(r_3)^{n_3}-\frac{1}{4}-\frac{5e^{-6}}{1-5e^{-6}}>\big(1-\frac{1}{17}\big)^{17} -\frac14-\frac{5e^{-6}}{1-5e^{-6}}>0.09$$ Inequality \eqref{one} follows from this.

• From \eqref{two} we habe that the the radius of convergcen of $$f$$ is $$\lim_{k\rightarrow\infty}\sqrt[n_k]{5^k}=1$$. That $$\limsup_{r\rightarrow1-}|f(re^{i\theta}|=\infty$$ for all $$\theta$$ shows that $$f$$ cannot be extended analytically to any region containing $$U$$.
• As pointed out by others, one can also show that for any $$w\in\mathbb{C}$$, there are infinitely many solutions to the equation $$f(z)=w$$, $$z\in U$$. Indeed, along $$|z|=r_m$$ $$|f(z)-w|\geq |p_m(z)|-|f(z)-w-p_m(z)|\geq |p_m(z)|-|f(z)-p_n(z)|-|w|> 5^mc-|w|>0$$ for all sufficiently large $$m$$. Another application of Rouche's theorem shows that $$f-w$$ has $$n_m$$ solutions (counted according to their multiplicities) inside the disk $$D(0;r_m)$$ (for each $$m$$ large enough).