16
$\begingroup$

Let $f(z)=\sum_{n=0}^{\infty} a_n z^n$ be a power series, $a_n, z\in \mathbb{C}$. Suppose the radius of convergence of $f$ is $1$, and $f$ is convergent at every point of the unit circle.

Question:If $f(z)=0$ for every $|z|=1$, then can we draw the conclusion that $a_n=0$ for all nonnegative integer $n$?

I think the answer is yes, but I failed to prove it. My approach is concerning about the function $F_\lambda(z):=f(\lambda z)$ for $0\leq\lambda\leq 1$, $|z|=1$. Abel's theorem shows that $F_\lambda$ converge to $F_1$ pointwisely as $\lambda\rightarrow 1$ on the unit circle. If I have the property that $f$ is bounded in the unit disk, then I can apply Lebesgue's dominated convergence theorem to prove $a_0=0$, and by induction I can prove $a_n=0$ for all $n$. However, I cannot prove $f(z) $ is bounded in the unit disk.

Any answers or comments are welcome. I'll really appreciate your help.

$\endgroup$
7
  • $\begingroup$ Maybe I'm making some very stupid mistake, but isn't $f$ continuous on the unit disk since it's the pointwise limit of uniformly continuous functions? $\endgroup$
    – k.stm
    Commented Oct 8, 2012 at 11:27
  • $\begingroup$ Can you use the Maximum modulus theorem? $\endgroup$
    – PAD
    Commented Oct 8, 2012 at 11:37
  • 6
    $\begingroup$ A pointwise limit of uniformly continuous functions need not be continuous. $\endgroup$
    – GEdgar
    Commented Oct 8, 2012 at 12:55
  • $\begingroup$ @jerrysciencemath : Are you satisfied with my answer below? $\endgroup$ Commented Oct 12, 2012 at 20:37
  • $\begingroup$ @MalikYounsi: I'm sorry that I'm busy doing other things these days, I haven't check the document in your reply, but it seems to be a satisfactory answer~ If I have time, I will check it through details, thanks for your answer! $\endgroup$
    – Yuchen Liu
    Commented Oct 13, 2012 at 8:28

1 Answer 1

10
$\begingroup$

It seems to me that this is a particular case of an old Theorem from Cantor (1870), called Cantor's uniqueness theorem. The theorem says that if, for every real $x$, $$\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n e^{inx}=0,$$ then all the complex numbers $c_n$'s are zero.

You can google "Uniqueness of Representation by Trigonometric Series" for more information. See e.g. this document for a proof and some history of the result.

$\endgroup$
4
  • 1
    $\begingroup$ Yes, I think this correct: +1. By the way, you seem to know a lot about holomorphic functions: are they a field you specialize in ? $\endgroup$ Commented Oct 8, 2012 at 21:33
  • $\begingroup$ @GeorgesElencwajg : Thank you, I also think it is correct. This question really is more about trigonometric series than anything else : the fact that $f(x)$ converges on the unit circle implies that the radius of convergence is $\geq 1$, and if the radius is $>1$ then the result is trivial... So the hypothesis about the radius of convergence does not mean much. About holomorphic functions, I am a PhD student in complex analysis, so I've read a lot about this subject. I'm interested in other fields too though. $\endgroup$ Commented Oct 9, 2012 at 12:42
  • 1
    $\begingroup$ Thanks for your explanations, Malik, and best wishes for your PhD. $\endgroup$ Commented Oct 9, 2012 at 14:49
  • $\begingroup$ I'm removing an answer I gave which was way off. Thanks to those who pointed that out... $\endgroup$
    – coffeemath
    Commented Oct 16, 2012 at 2:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .