0

I have a complex-valued function defined as $$\psi(z) = \sum_{j\in\Bbb Z} \psi_jz^j$$

We of course know that $\sum_j\lvert\psi_j\rvert < \infty$ implies $\psi(z)$ is well-defined (finite) on the unit circle.

What I want to know is whether the converse is true. In other words, if $\psi(z)$ is finite for any $z$ on the unit circle, does it mean that $\sum_j\lvert\psi_j\rvert < \infty$? If yes, can someone post a proof of this?

As pointed out by @user1952009 this is not true. Now I strengthen the argument a bit. Suppose $\psi(z)$ is analytic on an annulus containing the unit circle. How do I then show the absolute summability above?

David C. Ullrich
  • 89,985
Calculon
  • 5,725
  • if the series converges at some $|z| > 1$ then clearly $\sum_j |\psi_j|$ converges, but that it converges only on $|z|=1$ it is not enough, see https://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Uniform_convergence "If $f$ is absolutely continuous, then the Fourier series converges uniformly, but not necessarily absolutely, to $f$." – reuns May 02 '16 at 13:03
  • @user1952009 thanks. I edited my question but now I realize with the new condition it becomes trivial. I will delete the question soon. – Calculon May 02 '16 at 13:09
  • in my opinion, an interesting question would be constructing an example of such a power series converging conditionaly on $|z|=1$. I'd say that (the Taylor series at $z=0$ of) $f(z) = (z+1)^{1/2}$ would be a candidate, or maybe $f(z) = e^{-1/(1+z)}$, or a function with a sequence of poles converging to $z=1$ but analytic on $|z| \le 1$ – reuns May 02 '16 at 13:11
  • @user1952009 So these candidates have power series representations that converge on the unit circle but their coefficients are not absolutely summable? – Calculon May 02 '16 at 13:15
  • if it reduces to checking that $g(t) = \lim_{r \to 1^-} f(r e^{it})$ is continuous, but not differentiable, then $(z+1)^{1/2}$ and $e^{-1/(1+z)}$ should be (that's the point which is interesting) – reuns May 02 '16 at 13:17
  • @user1952009 "If $f$ is absolutely continuous, then the Fourier series converges uniformly, but not necessarily absolutely, to $f$." That's very interesting, but here it's irrelevant. The "not necessarily absolutely" is for Fourier series of the form $\sum_{n=-\infty}^\infty$. Here (as is clear from the context) we're talking about $\sum_{n=0}^\infty$, and if a Fourier series of that form represents an absolutely continuous function the series does converge absolutely! This follows from Hardy's inequality. – David C. Ullrich May 02 '16 at 13:19
  • @DavidC.Ullrich Actually in this case $j$ is integer-valued, not necessarily nonnegative. – Calculon May 02 '16 at 13:21
  • @user1952009 Oops. When I google "Hardy's inequality" all I find is a different Hardy's inequality. Say $f=\sum_{n=0}^\infty$ is absolutely continuous. Write $g=f'$. Then $g$ lies in the classical Hardy space $H^1$, and "Hardy's inequality" shows that $\sum |\hat g(n)|/n<\infty$. Hence $\sum|\hat f(n)|<\infty$. – David C. Ullrich May 02 '16 at 13:22
  • @Calculon Oh. I think writing it in terms of powers of $z$ was a bad idea then - it certainly looked to me like you were talking about a power series. – David C. Ullrich May 02 '16 at 13:24
  • @DavidC.Ullrich I literally copied the notation from the lecture notes I am following. I didn't know the term power series was reserved for positive powers. The question will disappear soon anyway. – Calculon May 02 '16 at 13:26
  • @user1952009 Never mind - I was misled by the notation. I think I'll leave my comments up anyway, since I think they're interesting. But it turns out it's my comment that's irrelevant, not yours. – David C. Ullrich May 02 '16 at 13:31
  • @Calculon I don't see any reason to delete the question! It led to an interesting discussion. For example I didn't know that the Fourier series of an absolutely continuous function converges uniformly; I'm taking that as today's exercise. You should I think at least replace $\sum_j$ by $\sum_{j=-\infty}^\infty$ to clarify things. And definitely change the title! Your series is not a power series. – David C. Ullrich May 02 '16 at 13:33
  • @user1952009 In my head it was trivial that if $\psi(z)$ is well-defined for some $z$ outside the unit circle, then the absolute summability would follow. But I sat down to write it explicitly and now I don't see why the implication is true. Could you post a proof of this if it won't be too much trouble? – Calculon May 02 '16 at 13:40
  • 1
    @Calculon If it converges for some $z$ with $|z|=r>1$ then $r^j\psi_j\to0$ as $j\to+\infty$, hence $\sum_{j=1}^\infty|\psi_j|<\infty$. Similary convergence for some $z$ with $|z|<1$ shows that $\sum_{-\infty}^0|\psi_j|<\infty$. – David C. Ullrich May 02 '16 at 13:46
  • @DavidC.Ullrich : wait, I'm lost. I know the Hardy space norms, and it should be clearly related to this problem. why wouldn't "If $f$ is absolutely continuous, then the Fourier series converges uniformly, but not necessarily absolutely, to $f$." wouldn't apply to $f(t) = \lim_{r \to 1^-} G(re^{it})$ with $G(z)$ analytic on $|z| < 1$ ? (and hence to $G(z) = (z+1)^{1/2}$ with the correct branch cut such that it is analytic on the unit disk) (and from the beginning I was thinking only to power series, not Laurent series) – reuns May 02 '16 at 14:56
  • (and note that I never studied the Hölder modulus of continuity and how it determinates the convergence of the Fourier series, if you think it is useful here ) – reuns May 02 '16 at 15:01
  • @user1952009 Because it doesn't. If $f$ is AC on the circle and $f$ is the boundary value of a holomorphic function in the disk then the Fourier series for $f$ converges absolutely. I sketched a proof in two comments already - I could repeat myself, or you could say what part you don't get... – David C. Ullrich May 02 '16 at 18:24
  • @DavidC.Ullrich : I didn't notice that AC meant $f$ was almost everywhere differentiable (it is very rarely used). so with taking in account that, it is obvious that the Fourier series of $f$ converges absolutely. – reuns May 02 '16 at 19:38
  • now you didn't give an example yet of a function answering to the question. I can only propose $f(z) = \displaystyle\sum_n \frac{a_n}{n} z^n$ with $a_n = \pm 1$ i.i.d. randomly, because with such a random sequence $\sum_{n \le N} a_n = o(\sqrt{N}^{1+\epsilon})$, but I'm not able to give much more examples. – reuns May 02 '16 at 19:40
  • @user1952009 Yes, an AC function is ae differentiable. No, the Fourier series of an AC function need not converge absolutely! – David C. Ullrich May 02 '16 at 19:52

0 Answers0