I made a throwaway comment on math stackexchange the other day that got me thinking about the following question. Let $$ f_n (\alpha) = \frac1n \sum_{k=0}^{n-1} e(2^k\alpha),$$ where $e(x) = \exp(i2\pi x)$.

Can you determine the set of possible limits $$ L = \{z : f_n(\alpha)\to z~\text{for some}~\alpha\}?$$

Of course by the ergodic theorem $f_n(\alpha)\to 0$ for almost every $\alpha$. Also $f_n(\alpha)\to 1$ for every dyadic rational, so for a comeagre set of $\alpha$ the sequence $f_n(\alpha)$ has a subsequence converging to $1$. So we're interested in a particular meagre null set of $\alpha$. :)

Here is everything I know about $L$:

  1. $L$ is disjoint from a neighbourhood of $S^1\setminus\{1\}$. Indeed suppose $f_n(\alpha)\approx z$, where $|z|=1$. Then $e(2^k\alpha) \approx z$ for almost all $k\leq n$, so also $e(2^{k+1}\alpha)\approx z$ for almost all $k\leq n$, so $z^2 \approx z$, which implies $z\approx 1$.

  2. $L$ is a closed, convex subset of $\{z: |z|\leq 1\}$. Hand-waving argument for convexity: Given $f_n(\alpha)\to z$ and $f_n(\beta)\to w$, with $\alpha,\beta\in[0,1]$, find $N$ such that $f_N(\alpha)\approx z$ and $f_N(\beta) \approx w$, write $\alpha_N$ and $\beta_N$ for $\alpha$ and $\beta$ rounded to the nearest multiple of $2^{-N}$, and then consider $$\gamma_N = \alpha_N + \beta_N/2^N + \alpha_N/2^{2N} + \beta_N/2^{3N} + \cdots.$$ Then the sequence $f_n(\gamma_N)$ hovers around the point $(z+w)/2$ for all sufficiently large $n$. Now patch the sequence $(\gamma_N)$ together in a similar way. Other convex combinations $pz + (1-p)w$ can be obtained similarly. Closedness is also similar.

  3. Obviously $f_n(\alpha)$ converges for every rational $\alpha$, and the limits tend to be interesting. For example, the points $1,-1/2,(1\pm i\sqrt{7})/6$ are in $L$, and so also by 2 their convex hull is contained in $L$. Thus at least $L$ contains a neighbourhood of $0$. By continuing in this way we begin to see the following picture.

Limits of $f_n(\alpha)$ for rational $\alpha$

Most, but not all, of the points on the boundary appear to be obtained from $\alpha$ of the form $1/(2^n-1)$.

  1. In fact $L = \{\int z\,d\mu : \mu~\text{a}~\times 2\text{-invariant measure on}~S^1\}.$ To see the inclusion $\subset$, take $\mu$ to be any weak* limit point of the sequence of measures $\frac1n \sum_{k=0}^{n-1} \delta_{2^k\alpha}$. To see the inclusion $\supset$, from the ergodic decomposition and the convex of $L$ observe that we may assume $\mu$ is ergodic for $\times 2$, and in this take $\alpha$ to be a $\mu$-random point and apply the ergodic theorem. This observation suggests the points $\int z \,d\mu_p$, where $\mu_p$ is the $\times 2$-invariant measure in which binary digits are independent and come up $1$ with probability $p$. However, this appears to constitute only a strict subset of the above picture, as shown below.

The integrals $\int z\,d\mu_p$

To me these observations strongly suggest the following question.

Is $L$ the convex hull of $\{\lim_{n\to\infty} f_n(\alpha) : \alpha\in\mathbf{Q}\}$?

  • 1
    $\begingroup$ I think the answer is known, but don't have the time to dig up the details. Here are some pointers. There are many more invariant measures than the "Bernoullian" ones you describe. You can construct some by taking any $k$-step Markov chain on $\{0,1\}^\mathbb{N}$ and taking the distribution of a realization starting with the invariant distribution. The "thermodynamical formalism" provides a way to construct ergodic "Gibbs" invariant measures. Then, the multifractal analysis of dynamical systems typically aims at finding the Hausdorff dimension of the level sets of fonction such as $f_n$. $\endgroup$ – Benoît Kloeckner Feb 19 '16 at 14:44
  • $\begingroup$ I added a toplevel tag, that is one with a two-letter prefix corresponding to arXiv categories. It is recommended to use at least on such tag (in addition to more specific tags). In case I did not pick the best one, you could change it easily via an edit. $\endgroup$ – user9072 Feb 19 '16 at 15:02
  • 1
    $\begingroup$ A trivial remark, improving upon 1: $L$ is contained in the convex hull of the image of $f_n$ for any $n$ (just cut up a long sum $\frac{1}{N} \sum_{k=0}^{N-1} e(2^k \alpha)$ as essentially an average of a bunch of short sums $\frac{1}{n} \sum_{k=0}^{n-1} e(2^k \beta)$). I suspect in fact that $L$ is precisely the intersection of these convex hulls, using the same sort of handwavy arguments you already have in your post. $\endgroup$ – Terry Tao Feb 19 '16 at 17:24
  • 1
    $\begingroup$ As Terence Tao notes, the set of all limits for rational $\alpha$ is a dense subset of $L$. From an abstract perspective, this is due to the fact that invariant measures supported on periodic orbits are weak-* dense in the simplex of invariant measures. The fact that the averages over Bernoulli measures lie in a smaller shape reflects the fact that the closure of this set of measures is much smaller. As is proved in Bousch's paper mentioned in Anthony's answer, the boundary points arise from numbers $\alpha$ whose binary expansion is a Sturmian sequence. $\endgroup$ – Ian Morris Feb 20 '16 at 18:03
  • 1
    $\begingroup$ On the other hand if you were instead to ask about $3^k$ in place of $2^k$, then while it would correspondingly still be true that $L$ is the closure of the barycentres associated to periodic orbits of $x \mapsto 3x \mod 1$, I believe that it is currently unknown which measures correspond to boundary points. $\endgroup$ – Ian Morris Feb 20 '16 at 18:04

See the paper "Le poisson n'a pas d'arêtes" by Thierry Bousch. This is a joke that was explained to me much later. The set is considered to resemble a fish. The French word arête means both bone and edge (of a polygon); so the English title of the paper would be "The fish has no bones/The fish has no edges". The paper proves that every point on the boundary is an extreme point. (so that the answer to your question is No). The set was also studied in works of Oliver Jenkinson.

  • 1
    $\begingroup$ On the other hand it seems likely to me that the set $L$ is the closed convex hull of the limit points of rational $\alpha$, basically by approximating any point in $L$ by a finite average $\frac{1}{n} \sum_{k=0}^{n-1} e^{2\pi i 2^k \alpha}$ and then rounding $\alpha$ to the nearest multiple of $1/(2^n-1)$. $\endgroup$ – Terry Tao Feb 19 '16 at 18:25
  • $\begingroup$ The second paper to which Anthony refers is "Frequency-locking on the boundary of the barycentre set" by Oliver Jenkinson (Experimental Mathematics 9 (2000) 309-317). $\endgroup$ – Ian Morris Feb 20 '16 at 13:31
  • 2
    $\begingroup$ Cool, thanks. @TerryTao True. Looks like you don't even need the convex hull part, just closure. $\endgroup$ – Sean Eberhard Feb 20 '16 at 14:34
  • $\begingroup$ The paper you mention proves some rather strange things about "le poisson" $L$. For almost all $u\in S^1$, indeed for all $u\in S^1$ apart from a set of Hausdorff dimension zero, the maximizer of $\langle x, u\rangle$ over $x\in L$ is a point of the form $\lim f_n(\alpha)$ with rational $\alpha$, but still of course almost all of the boundary is made up of points not of the form $\lim f_n(\alpha)$ with rational $\alpha$. $\endgroup$ – Sean Eberhard Feb 22 '16 at 16:28
  • $\begingroup$ Indeed there is a Devil's Staircase relationship between the direction $u$ and the maximising measure. This locking onto periodic orbits seems to be fairly common when maximising an integral (or integral-like object) over a rich set of invariant measures, and occurs in a recent preprint by Jenkinson and Pollicott. See also Contreras' recent Inventiones paper, Ground states are generically a periodic orbit. $\endgroup$ – Ian Morris Feb 25 '16 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.