Problem 4 in Chapter 4 of Stein's book "Real Analysis" says $\sum_{n\geqslant 0}z^{2^n}$ doesn't have radial limit as $z$ approaches the unit circle from inside almost everywhere. It's fairly easy to find a dense set s.t. the radial limit doesn't exist ($=\infty$, actually), but is there a simple way to prove the set of divergence has positive measure?
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2 Answers
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Put $f(z) = \sum_{n=0}^{\infty} z^{2^n}$. First note that the set of $\theta \in [0,1)$ such that the binary expansion of $\theta$ has arbitrarily large strings of $100$ consecutive zeros is a set of measure $1$. Now take such a $\theta$, and let $N$ be such that $\{2^{N} \theta\} \le 2^{-100}$ (here $\{x\}$ stands for the fractional part of $x$). Consider $f(e^{-1/2^N + 2\pi i\theta})$ and $f(e^{-1/2^{N+50}+2\pi i \theta})$. The difference of these two quantities is in size
$$
|f(e^{-1/2^N+2\pi i\theta}) - f(e^{-1/2^{N+50} +2\pi i\theta})|
$$
and using the triangle inequality appropriately this
is
$$
\ge \Big|\sum_{n=N+1}^{N+50} e^{-2^n/2^{N+50}} e^{2\pi i 2^n \theta} \Big| - \sum_{n=0}^{N} (e^{-2^{n}/2^{N+50}} - e^{-2^n/2^N}) - \sum_{n=N+51}^{\infty} e^{-2^{n}/2^{N+50}} -\sum_{n=N+1}^{\infty} e^{-2^n/2^N}. \tag{1}
$$
The third term and fourth terms in (1) are together in size at most $$ 2(e^{-2} + e^{-4} + \ldots )\le 1. $$ Since $|e^{-x}-e^{-y}| \le |x-y|$ for $x$ and $y$ in $[0,1]$, the second term in (1) is bounded in size by $$ \sum_{n=0}^{N} \Big( \frac{2^n}{2^N} - \frac{2^n}{2^{N+50}}\Big) \le 2. $$ Finally, since $\{2^n \theta\} \le 2^{-50}$ for $N+1\le n \le N+50$, we see easily that the first term in (1) is $$ \ge \sum_{n=N+1}^{N+50} e^{-2^n/2^{N+50}} - \frac{1}{2^{20}} \ge 40. $$ It follows that the difference in the two values of $f$ is at least $30$ in size, so that the radial limit cannot exist for this $\theta$.
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2$\begingroup$ @GHfromMO: Thanks for pointing that out -- I'll correct it. For the second part, my idea would be to compute moments on the circle with radius $r$. Since the powers of $2$ don't have too many linear relations, one would get an approximately (complex) Gaussian with variance tending to infinity (this is because of the assumption on coefficients). Then it follows that on each circle of radius sufficiently close to $1$, most points (measure close to $1$) will have a large value of the function. From that the result follows. $\endgroup$– LuciaCommented May 26, 2015 at 18:59
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2$\begingroup$ In my undergrad senior thesis I showed something along these lines. That is, the distribution function of the partial sums of a lacunary Fourier series have Gaussian moments. I assume this is what @GHfromMO might be referring to. $\endgroup$ Commented May 26, 2015 at 21:02
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1$\begingroup$ @MattYoung: Slightly more general version -- lacunary Fourier series with coefficients whose $\ell^2$ norm is infinite. Is that what you looked at, or with coefficients being $1$? $\endgroup$– LuciaCommented May 26, 2015 at 21:09
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3$\begingroup$ I think the final result worked for coefficients whose $l^2$ norm is infinite. I originally looked at the case where all the coefficients are $1$ but Hejhal showed me how to do the more general case. My only copy of the thesis is a printout. I typed it up in Microsoft Word before anyone told me about tex. $\endgroup$ Commented May 26, 2015 at 21:25
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1$\begingroup$ @GHfromMO I scanned the paper and put it on my website at link. My only electronic copy was on a 3.5" floppy that is long gone, so this is the best I can do. I wouldn't be surprised if the proof could be greatly simplified, but it's what I came up with at that time. $\endgroup$ Commented May 28, 2015 at 17:43
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I don't know any simply way, but I would be interested in one, too.
In fact $\sum_{n\geqslant 0}z^{2^n}$ has no radial limit anywhere on the unit circle. This follows from a 1928 Tauberian theorem of Ananda-Rau (see review here). The result is included as Theorem 104 in Hardy: Divergent series (Oxford Clarendon Press, 1948); the proof appears in the notes on Chapter VII.
For part (b) of the problem, see Theorem 6.4 in Chapter V (on Page 203) in Zygmund: Trigonometric Series I. Alternately, see Zygmund's original paper.
answered May 25, 2015 at 22:08
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1$\begingroup$ Thanks. The Tauberian theorem might be involved and actually Theorem 6.4 of Zygmund seems to be very similar to the high-indices theorem, which directly implies the stated result. However I feel like these theorems are an over-kill and there might be a way to prove it just using results in Stein's book. $\endgroup$– Erika LCommented May 25, 2015 at 22:47
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$\begingroup$ Sorry, posted a comment by mistake -- have no idea how to touch this. $\endgroup$ Commented May 26, 2015 at 0:54
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2$\begingroup$ Note that almost every real number $\theta$ will have the property that its binary expansion will have arbitrarily long strings of zeros. Now suppose $N$ is such that $\Vert 2^N \theta \Vert \le 2^{-h}$, and consider $f(e^{-1/2^N} e^{2\pi i\theta})$ and $f(e^{-1/2^{N+h}} e^{2\pi i \theta})$. They should differ by $\gg h$, showing that radial limits don't exist. $\endgroup$– LuciaCommented May 26, 2015 at 1:02
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1$\begingroup$ @Lucia: I had similar ideas initially, but then I abandoned them. For some reason I thought that $h$ must grow sufficiently fast with $N$. I am now convinced that your idea works, and I suggest that you give it as an answer (so that it can be voted for and accepted officially). $\endgroup$ Commented May 26, 2015 at 1:06