Additive subgroups of the reals and irrational rotation

We present a useful characterization of the additive subgroups of $\mathbb{R}$.

Theorem. Let $G$ be a subgroup of $(\mathbb{R}, +)$. Then precisely one of the following holds:

(i) $G = d\mathbb{Z}$ for some $d \in \mathbb{R}$;

(ii) $G$ is dense in $\mathbb{R}$.

Proof: If $G = \{0\}$, then $G = 0\mathbb{Z}$. Assume $G$ is nontrivial. Let $$d = \inf\{g: 0 < g \in G\}.$$  Note that $d \geq 0$ since for every nonzero $y \in G$, $ky \in G$ and $ky > 0$ for some $k \in \mathbb{Z}$.
Case 1: $d > 0$. We claim that $d \in G$. If not, then for every $\epsilon > 0$ there exists $g \in G$ with $g \in (d, d + \epsilon)$. It follows that for every $\epsilon > 0$ there are $g, h \in G$ such that $0 < g - h < \epsilon$, thus there exists $x \in G$ with $0 < x < \epsilon$ (let $x = g - h$). This contradicts to $d > 0$. We also have $d = \min\{|g|: 0 \neq g \in G\}$. Now fix any $0< g \in G$ and consider $S = \{g - nd: n = 0, 1, 2, \ldots\}$. The set $S$ is discrete, we can find $n \geq 0$ such that $g - nd \geq 0$ and $g - (n + 1)d < 0$. Since $g - nd, g - (n +1)d \in G$, $g - (n + 1)d \leq -d$, so $g - nd \leq 0$ and $g = nd$. Therefore $G = d\mathbb{Z}$.
Case 2: $d = 0$. Arguing as in the beginning of case 1, we find that 0 is a cluster point of $G \cap \mathbb{R}_+$. Let $(a, b)$ be any nonempty bounded open interval of $\mathbb{R}$. Take $\epsilon = b - a$ and pick some $g \in G \cap (0, \epsilon)$. For sufficiently large $n$, $ng \geq a$, and we take the smallest such $n$. Then $ng - g < a$, i.e. $ng \in (a - g, a]$, thus $ng + g \in (a, a + g] \subseteq (a, b)$. Therefore $G \cap (a, b) \neq \emptyset$. It follows that $G$ is dense in $\mathbb{R}$.

The above result can be used to prove the density of the orbit of an irrational rotation in the unit circle.

Corollary. (i) For an irrational $t$, $\mathbb{Z} + t\mathbb{Z}$ is dense in $\mathbb{R}$.
(ii) For an irrational $t$, $\{e^{2\pi int}: n \in \mathbb{Z}\}$ is dense in the unit circle.

Proof: (i) Note that $\mathbb{Z} + t\mathbb{Z}$ is a subgroup of $(\mathbb{R}, +)$. If $\mathbb{Z} + t\mathbb{Z} = a\mathbb{Z}$ for some $a \in \mathbb{R}$, then $1 = an$ for some $n \in \mathbb{Z}$, so $a \in \mathbb{Q}$. But $t = am$ for some $m \in \mathbb{Z}$, so $t \in \mathbb{Q}$, contradiction. By Theorem, $\mathbb{Z} + t\mathbb{Z}$ must be dense in $\mathbb{R}$.
(ii)  Let $e^{2\pi i x}$ be a point on the unit circle. By (i), there exist sequences $m_k, n_k$ of integers such that $|m_k + n_k t - x| \rightarrow 0$. Then $|e^{2 \pi i n_k t} - e^{2 \pi i x}| = |e^{2\pi i x}||e^{2 \pi i (n_k t - x)} - 1| = |e^{2\pi i (m_k + n_k t - x)} - 1| = O(|m_k + n_k t - x|) \rightarrow 0$. Hence $\{e^{2\pi int}\}_{n \in \mathbb{Z}}$ is dense in the unit circle.

Hadamard gap theorem

Definition. A sequence $\{n_k\}$ of nonnegative integers is said to be lacunary if there is some $\delta > 0$ such that $n_{k + 1}/n_k \geq 1 + \delta$ for all $k \geq 1$.

A typical example of a lacunary sequence is $n_k = 2^k$.

Theorem (Hadamard Gap Theorem).  Suppose $\{n_k\}$ is a lacunary sequence and the power series $f(z) = \sum_{n = 0}^\infty a_n z^n$ with $a_n$ nonzero precisely when $n = n_k$ for some $k$ and has radius of convergence 1. Then $f(z)$ cannot be analytically extended to any larger region containing a point on the unit circle.

Proof: Suppose $f(z)$ extends to an analytic function on a region containing the open unit disk $B$ and a point $w_0$ on the unit circle. WLOG, we may assume $w_0 = 1$. Then $f(z)$ is analytic on a region $\Omega$ containing $B \cup \{1\}$. Define $g(w) = f(w^m(1 + w)/2)$ where $m$ is a positive integer. Since $1^m(1 + 1)/2 = 1$ and $|w^m(1+w)/2| = |w|^m|1+w|/2 < |w|^m \leq 1$ when $|w| \leq 1$ and $w \neq 1$ because $|1 + w| < 2$, so $g(w)$ is well-defined when $|w| \leq 1$, hence when $|w| < 1 + \epsilon$ for some $\epsilon > 0$. Write $g(w) = \sum_{m = 0}^\infty b_m w^m$ for $|w| < 1 + \epsilon$. Note that the powers of $w$ in the terms of $[w^m(1 + w)/2]^{n_k}$ range from $n_k m$ to $n_k (m + 1)$. Since $\{n_k\}$ is lacunary, we can choose $m$ to be so large that $n_{k + 1} m > n_k (m + 1)$ for all $k$, then the powers of $w$ in all of $[w^m(1 + w)/2]^{n_k}$ are distinct. Now
$$\sum_{n = 0}^{n_k} a_n z^n = \sum_{m = 0}^{n_k (m + 1)} b_m w^m.$$
When $|w| < 1 + \epsilon$, RHS converges as $k \rightarrow \infty$. But LHS is the partial sums of $\sum_{k = 1}^\infty a_{n_k} z^{n_k}$ as $k$ varies, so the power series converges for all $z = w^m(1 + w)/2$ where $|w| < 1 + \epsilon$, which includes the case $z = 1$. The image of $B(0, 1 + \epsilon)$ under the map $w^m(1 + w)/2$ is an region $W$ containing 1, so the power series of $f(z)$ at 0 converges on $W$, contradicting its radius of convergence being 1. QED.

Reference:
Gamelin - Complex Analysis
Rudin - Real and Complex Analysis (3rd Ed.)

Some results in measure theory

Proposition. Let $\mu$ be a positive Borel measure on $\mathbb{R}$ and $\epsilon > 0$. Then for almost every $x$, we have $$\int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} = +\infty.$$
Proof: For each positive integer $N$, define $E_N = \{x \in \mathbb{R}: \int_\mathbb{R} d\mu(t) / |x - t|^{1 + \epsilon} \leq N\}$. Let $I$ be any bounded closed interval with $E_N \cap I$ nonempty. Pick $x \in E_N \cap I$ and write $I = [x - \delta_1, x + \delta_2]$. We have $$N \geq \int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \int_I \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \frac{\mu(I)}{\max\{\delta_1, \delta_2\}^{1 + \epsilon}} \geq \frac{\mu(E_N \cap I)}{(\delta_1 + \delta_2)^{1 + \epsilon}},$$ i.e. $\mu(E_N \cap I) \leq Nm(I)^{1 + \epsilon}$ where $m(I)$ is the length of the interval $I$. Now let $[a, b] \subseteq \mathbb{R}$ and $n \in \mathbb{N}$. Decompose $[a, b]$ into a union of $n$ closed subintervals of length $(b - a)/n$, then $$\mu(E_N \cap [a, b]) \leq nN\left(\frac{b - a}{n}\right)^{1 + \epsilon} \rightarrow 0$$ as $n \rightarrow \infty$. Hence $\mu(E_N \cap [a, b]) = 0$. It follows that $\mu(E_N) = 0$ for all $N \in \mathbb{N}$ and our result follows.

(To be continued)