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This question is about getting a concrete example for this question on bounded holomorphic functions posed by @user122916 (something that he really expected as explained in the comments).

Give an example of a sequence of complex numbers $(a_n)_{n\ge 0}$ so that \begin{eqnarray} |\sum_{n\ge 0} {a_n z^n} | &\le &1 \text{ for all }z \in \mathbb{C}, |z| < 1 \\ \sum_{n\ge 0} |a_n| &=& \infty \end{eqnarray}

Such sequences exist because there exist bounded holomorphic functions on the unit disk that do not have a continuous extension to the unit circle ( one finds a bounded Blaschke product with zero set that contains the unit circle in its closure). However, a concrete example escapes me. Note that all this is part of the theory of $H^{\infty}$ space, so the specialists might have one at hand.

Community
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orangeskid
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3 Answers3

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An example is $f(z) = \exp {(-\frac{1+z}{1-z})}.$ As $-\frac{1+z}{1-z}$ is a conformal map of the open unit disc $\mathbb {D}$ onto the left half plane, $f$ is bounded and holomorphic in $\mathbb {D}.$ We have $f$ continuous on $\overline {\mathbb {D}} \setminus \{1\}.$ Check that on $\partial \mathbb {D}\setminus \{1\},$ we have $|f| = 1.$ However $f(r) \to 0$ (to say the least) as $r\to 1^-.$ It follows that $f$ does not have a continuous extension to $\overline {\mathbb {D}}.$ Hence for this $f, \sum_{n=1}^{\infty}|a_n| = \infty.$

zhw.
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  • I find the function simple, yet very subtle. Perhaps one cannot get a simpler function. The power series expansion is rather complicated, due to the exponential. I notice that the denominators are part of the sequence http://oeis.org/A256402. I wonder if one can get a formula in finite terms. Thanks! – orangeskid May 21 '15 at 00:57
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    The series is $f(z) = e^{-1} \sum_{n=0}^\infty a_n z^n$ where $a_n =\displaystyle \sum_{k=0}^n \dfrac{(-2)^k}{k!} {n-1 \choose n-k}$. – Robert Israel May 21 '15 at 01:28
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    Maple tells me $a_n = 2,{\frac {{\it LaguerreL} \left( n,2 \right) -{\it LaguerreL} \left( n,1,2 \right) }{n}} $ for $n \ge 1$. – Robert Israel May 21 '15 at 01:30
  • @RobertIsrael: That is a lovely formula! Does is have to do with the "A=B" method? Tested it with Mathematica, oscillatory behavior and the asymptotics. Laguerre polynomials, why this is just so nice. – orangeskid May 21 '15 at 05:02
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    @Robert Israel: I looked at the Laguerre polynomials in wikipedia. From your first formula we get a simpler form for the coefficients ( omitting the $\frac{1}{e}$ factor) LaguerreL[n,-1,2] ( in Mathematica notation). Now this has a nice asymptotic formula for large $n$ ( from wikipedia again). Note that the remainder gets us an $l^1$ sequence. Hence the sequence $n^{-\frac{3}{4}} \cos ( \sqrt{8 n} + \frac{\pi}{4})$ does the job ($\alpha = -1 $ here). – orangeskid May 22 '15 at 06:30
  • @RobertIsrael: Still puzzled by this example. I had a feeling that this sequence, being the Fourier coefficients of an $L^{\infty}$ function on $S^1$ should be in all the $L^p$, $p>1$. I got the Hausdorff-Young theorem too strong, can only conclude that they are in $L^2$ at best. Now, if somebody could prove directly that this analytic functions is bounded by an explicit constant ( should come from an effective constant in Laguerre estimates but still interested). – orangeskid May 22 '15 at 09:09
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Consider a Blaschke product $$ f(z) = \prod_{n=1}^\infty \dfrac{z - r_n}{1 - r_n z}$$ where $0 \le r_n < 1$ and $r_n \to 1$ as $n \to \infty$; this converges in the open unit disk if $\sum_n (1 - r_n) < \infty$. If its Maclaurin series is $\sum_n a_n z^n$ and $\sum_n |a_n|$ converges, then $\sum_n a_n = S$ converges and Abel's theorem says $f(x) \to S$ as $x \to 1-$. Of course $f(r_n) = 0$, so all we have to do is ensure that $f(s_n)$ does not go to $0$ for some other sequence $s_n \to 1-$.

Take $t_n = 1 - b^{n}$, $r_n = t_{2n}$ and $s_n = t_{2n+1}$, where $0 < b < 1/2$. Note that if $m < n$, $(1-t_n)/(1-t_m) = b^{n-m}$ and $$ \dfrac{t_n - t_m}{1-t_n t_m} = 1 - \dfrac{(2-b^m) b^{n-m}}{1 + b^{n-m} - b^{n}} > 1 - 2 b^{n-m}$$ Thus we get $$|f(s_n)| > \prod_{j=1}^\infty (1 - 2 b^{1+2j})^2 > 0$$

Robert Israel
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  • Great ! Wow, a subtle use of the Blaschke product. I will have to chew on the estimates for a while. Thanks! – orangeskid May 21 '15 at 00:54
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From the answer of @zhw. and the comments of @Robert Israel I think we can provide an explicit example as follows:

$$z \mapsto e^{-\frac{1+z}{1-z}}$$ is holomorphic on the unit disk and bounded in absolute value by $1$ ( @zhw. answer)

From the comment of @Robert Israel we get the explicit series expansion

$$e^{-\frac{1+z}{1-z}} = e^{ -1 -\frac{2z}{1-z}} = e^{-1} \cdot \sum_{n\ge 0} L^{(-1)}_n(2)\, z^n$$

where $L^{(\alpha)}_n(x)$ are the Laguerre polynomials.

We have asymptotics for the Laguerre polynomials. Some numerical testings suggest the following explicit bound:

$$ \frac{L^{(-1)}_n(2)}{e} - \frac{2^{1/4} }{\sqrt{\pi}} \frac{\cos(\sqrt{8n} + \frac{\pi}{4}) }{n^{3/4}} = \theta_n\cdot n^{-5/4}$$ with $|\theta_n| \le \frac{1}{7.2}$ ($n \ge 1$).

We conclude that for all $z$ in $\mathbb{C}$, $|z| <1$ we have

$$\left|\sum_{n\ge 1} \frac{\cos(\sqrt{8n} + \frac{\pi}{4}) }{n^{3/4}}\, z^n\right | < \frac{\sqrt{\pi}}{2^{1/4}}\cdot (\, 1/e + 1 + \frac{5}{36} \zeta(5/4)\,) = 2.9899..<3$$

orangeskid
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