Exponential sum and the equidistribution theorem

Consider the sequence x \Mod 1, 2x \Mod 1, 3x \Mod 1,\ldots, where \Mod 1 gives the fractional part of a real number. The equidistribution theorem states that x is irrational iff for any sub-interval [a, b) of the unit interval, for sufficiently large N, roughly (b-a)N of the numbers x\Mod 1, \ldots, Nx\Mod 1 fall inside [a, b). More precisely,

(Equidistributed sequence) A sequence \{x_n\}_{n=1}^{\infty} with x_n \in \mathbb{R} is equidistributed mod 1 if for every 0 \leq a \leq b \leq 1,

    \[ \lim_{N \rightarrow \infty} \frac{\abs{\{x_n : a \leq x_n \Mod 1 < b \}}}{N}  = b - a. \]

(Equidistribution theorem) The sequence \{nx\}_{n=1}^{\infty} is equidistributed mod 1 iff x is irrational.

Let us try to understand why the sequence behaves differently depending on whether x is rational or not. If x = p/q for some integers p and q, then roughly N/q of the first N multiples of x are integers, i.e., 0 \Mod 1. In fact, it is not difficult to see that for i \in \{0, 1, \ldots, q-1\}, 1/q fraction of the sequence equals i/q \Mod 1. So equidistribution does not occur when x is rational.

Now for k\neq 0 \in \mathbb{Z}, consider the exponential function e_k(x):=e^{2\pi i k x}, which has period 1. As discussed above, roughly 1/q of the sequence \{e_k(nx)\}_{n=1}^{\infty} is 1, so the exponential sum \sum_{n=1}^{\infty} e_k(nx) contains an infinite number of ones. For example if q=2, the exponential sum evaluates to -1 + 1 - 1 + 1 - \ldots, which is Grandi’s series and diverges. For an arbitrary q > 1, the exponential sum is a repeating sum of all q-th roots of unity, and one can use the same technique showing that Grandi’s series diverges to establish that this exponential sum diverges too. However, the case when x is irrational differs: every nonzero integer multiple of x is an non-integer, and thus e_k(nx)\neq 1 for any n\neq 0. Writing r=e^{2\pi i x}, the exponential sum \sum_{n=1}^{\infty} e_k(nx) converges to the geometric series \sum_{n=1}^{\infty} r^n = \frac{r}{r-1}, since r\neq1.

In fact, the above intuition can be summarized into the following criterion due to Weyl:

(Weyl’s criterion) A sequence \{x_n\}_{n=1}^{\infty} is equidistributed mod 1 iff for all integers k\neq 0,

    \[ \lim_{N\rightarrow \infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i kx_n} = 0. \]

From the above discussion, Weyl’s criterion immediately implies the equidistribution theorem.

(of equidistribution theorem) Suppose x is rational, i.e., x=p/q for some integers p and q. Without loss of generality, we assume q > 1. Then for all integer n, by Euclid’s algorithm, nx \Mod 1 is in \{0, 1/q, \ldots, (q-i)/q\}. So the sequence \{nx\}_{n=1}^{\infty} is not equidistributed, e.g., no multiples of x fall inside [1/4q, 1/2q).

Now suppose x is irrational. Since r:= e^{2\pi i kx}\neq 1, we have

    \[ \frac{1}{N} \sum_{n=1}^N e^{2\pi i kx_n} = \frac{1}{N} \frac{r^{N+1} - 1}{r - 1}, \]

which tends to 0 as N \rightarrow \infty.

Before proving Weyl’s criterion, it is instructive to state the notion of equidistribution analytically. For 0\leq a \leq b \leq 1, let \chi_{[a,b)}(x)=1 iff x \Mod 1 \in [a, b). Note that \chi_{[a,b)} is a periodic function on all of \mathbb{R}.

(Weyl’s criterion restated) Let \{x_n\}_{n=1}^{\infty} be a sequence in \mathbb{R}. The following two statements are equivalent:
1) For 0\leq a < b \leq 1,

    \[ \lim_{N\rightarrow \infty} \frac{1}{N} \sum_{n=1}^N \chi_{[a, b)}(x_n) = \int_0^1 \chi_{[a, b)}. \]

2) For all integers k\neq 0,

    \[ \lim_{N\rightarrow \infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i kx_n} = \int_0^1 e^{2\pi i k}. \]

Written in this form, Weyl’s criterion is asserting that if the sum of every interval function at a sequence converges to its integral, then every exponential sum converges to its integral as well, and vice versa. The reason is that both the space of interval functions and the space of exponential functions can approximate one another. More precisely, an exponential function is clearly periodic and continuous (and thus uniformly continuous), and every uniformly continuous function can be approximated by a finite sum of interval functions. For the other direction, a periodic, interval function can be approximated by a periodic, continuous function, which by Fejér’s Theorem can also be approximated by a finite sum of exponential functions. Now we have all the machinery to write down a proof.

(of Weyl’s criterion) For the forward direction (1) => (2), fix k\neq 0 and \epsilon > 0. Since the exponential function e_k(x):=e^{2\pi i kx} continuous on the closed interval [0, 1], it is also uniformly continuous, i.e., there exists some \delta > 0 such that for every x, x', |e_k(x) - e_k(x')| < \epsilon. Let f_S be a step function where f_S(x) = e_k(m \delta) if x \in [m\delta, (m+1)\delta). By construction, f_S is a finite sum of interval functions with \norm{f_S - e_k}_{\infty} < \epsilon.

By linearity, we have

    \[ \lim_{N\rightarrow \infty}\frac{1}{N} \sum_{n=1}^N f_S(x_n) = \int_0^1 f_S. \]

By taking N sufficiently large and repeatedly applying triangle inequality, we have

    \begin{eqnarray*} \abs{\frac{1}{N} \sum_{n=1}^N e_k(x_n) - \int_0^1 e_k} & \leq & \abs{\frac{1}{N} \sum_{n=1}^N e_k(x_n) - \frac{1}{N} \sum_{n=1}^N f_S(x_n)} + \\ & & \abs{\frac{1}{N} \sum_{n=1}^N f_S(x_n) - \int_0^1 e_k} \\ & \leq & \epsilon + \abs{\frac{1}{N} \sum_{n=1}^N f_S(x_n) - \int_0^1 f_S} + \abs{\int_0^1 f_S - \int_0^1 e_k} \\ & \leq & 3\epsilon, \end{eqnarray*}

implying that \frac{1}{N} \sum_{n=1}^N e_k(x_n) converges to \int_0^1 e_k as N\rightarrow \infty.

Now we consider the reverse direction (2) => (1). Fix 0\leq a < b \leq 1 and let \epsilon > 0. Define

    \[ f^+(x) =  \begin{cases} 1 & \text{if } x \in [a, b) \\ 1/\epsilon (x - (a - \epsilon)) & \text{if } x \in [a - \epsilon, a) \\ -1/\epsilon (x - (b - \epsilon)) + 1 & \text{if } x \in [b, b + \epsilon) \\ 0 & \text{everywhere else.} \end{cases} \]

Note that f^+ is continuous, periodic, and dominates \chi_{[a, b)}. By Fejér’s Theorem, there exists P(x) = \sum_{k=1}^d p_k e^{2\pi i kx} such that \norm{f^+ - P}_{\infty} < \epsilon. By linearity, we have

    \[ \lim_{N\rightarrow \infty}\frac{1}{N} \sum_{n=1}^N P(x_n) = \int_0^1 P. \]

By taking N sufficiently large and repeatedly applying triangle inequality, we have

    \begin{eqnarray*} \abs{\sum_{n=1}^N f^+(x_n) - \int_0^1 f^+} & \leq & \abs{\sum_{n=1}^N f^+(x_n) - \sum_{n=1}^N P(x_n)} + \abs{\sum_{n=1}^N P(x_n) - \int_0^1 f^+} \\ & \leq & \epsilon + \abs{\sum_{n=1}^N P(x_n) - \int_0^1 P} + \abs{\int_0^1 P - \int_0^1 f^+} \\ & \leq & 3\epsilon, \end{eqnarray*}

implying that

    \[ \lim_{N\rightarrow \infty}\frac{1}{N} \sum_{n=1}^N f^+(x_n) = \int_0^1 f^+. \]

Since f^+ dominates \chi_{[a, b)}, we have

    \begin{eqnarray*} \limsup_{N \rightarrow \infty}  \frac{1}{N} \sum_{n=1}^N \chi_{[a,b)}(x_n) & \leq & \limsup_{N \rightarrow \infty}  \frac{1}{N} \sum_{n=1}^N f^+(x_n) \\ & = & \int_0^1 f^+ \\ & = & \int_0^1 \chi_{[a, b)} + \epsilon. \end{eqnarray*}

One can similarly define a periodic, continuous function f^- that approximates \chi_{[a, b)} from below, showing that \liminf_{N \rightarrow \infty}  \frac{1}{N} \sum_{n=1}^N \chi_{[a,b)}(x_n) is bounded below by \int_0^1 \chi_{[a, b)} - \epsilon, which together with the proceeding argument will show that the limit of \frac{1}{N} \sum_{n=1}^N \chi_{[a,b)}(x_n) must exist and converge to \int_0^1 \chi_{[a, b)}.

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