Fall 2010 - Problem 5

Solution.

1. For notation, define

   $$E_N\p{f} = \frac{1}{N} \sum_{n=0}^{N-1} f\p{n\alpha + \Z}.$$

   We follow the hint, and we first establish the conclusion for $f\p{x} = e^{2\pi ikx}$:

   $$\begin{aligned} \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi ikn\alpha} &= \frac{1}{N} \sum_{n=0}^{N-1} \p{e^{2\pi ik\alpha}}^n \\ &= \frac{1}{N} \frac{e^{2\pi ikN\alpha} - 1}{e^{2\pi ik\alpha} - 1}, \end{aligned}$$

   which tends to $0$ as $N \to \infty$ for any $k \in \Z$. Indeed, $\alpha$ is irrational, so the denominator is never $0$. On the other hand,

   $$\int_0^1 e^{2\pi ikx} \,\diff{x} = 0$$

   since $e^{2\pi ikx}$ is $1$-periodic. Thus, $\lim_{N\to\infty} E_N\p{e^{2\pi ikx}} = \int_0^1 e^{2\pi ikx} \,\diff{x}$. By linearity of integration, the equality is established for any trigonometric polynomial, i.e., any real linear combination of $\set{e^{2\pi ikx} \mid k \in \Z}$. Let $A$ denote the set of trigonometric polynomials.

   Notice that by definition, $A$ is a vector space over $\R$, $A$ is closed under multiplication since $e^{2\pi ikx} e^{2\pi i\ell x} = e^{2\pi i\p{k+\ell}x} \in A$, and $1 \in A$. Hence, $A$ is a subalgebra of $C\p{\R/\Z}$ which separates points, so by Stone-Weierstrass, $A$ is dense in $C\p{\R/\Z}$ with respect to the uniform norm.

   Let $f \in C\p{\R/\Z}$ and $\epsilon > 0$. By density, we may pick $g \in A$ such that $\norm{f - g}_{L^\infty} < \epsilon$. By our preliminary result, for $N$ large enough, we also have $\abs{E_N\p{g} - \int_0^1 g \,\diff{x}} < \epsilon$, so

   $$\begin{aligned} \abs{E_N\p{f} - \int_0^1 f\p{x + \Z} \,\diff{x}} &\leq \abs{E_N\p{f} - E_N\p{g}} + \abs{E_N\p{g} - \int_0^1 g\p{x + \Z} \,\diff{x}} + \abs{\int_0^1 g\p{x + \Z} \,\diff{x} - \int_0^1 f\p{x + \Z} \,\diff{x}} \\ &\leq \frac{1}{N} \sum_{n=0}^{N-1} \norm{f - g}_{L^\infty} + \epsilon + \int_0^1 \norm{f - g}_{L^\infty} \,\diff{x} \\ &= 2\norm{f - g}_{L^\infty} + \epsilon \\ &\leq 3\epsilon. \end{aligned}$$

   Sending $\epsilon \to 0$, we see

   $$\lim_{N\to\infty} \sum_{n=0}^{N-1} f\p{n\alpha + \Z} = \int_0^1 f\p{x + \Z} \,\diff{x},$$

   which completes the proof.

2. Let $f = \chi_{\br{a,b}}$, and pick sequences of continuous functions $g_k, h_k$ which converge to $f$ almost everywhere and such that $0 \leq g_k \leq f \leq h_k \leq 1$. For example, we can pick $g_k$ to be

   • $0$ outside of $\br{a, b}$ (possibly nowhere),
   • linear on $\br{a, a + \frac{1}{k}}$,
   • $1$ on $\br{a + \frac{1}{k}, b - \frac{1}{k}}$,
   • linear on $\br{b - \frac{1}{k}, b}$.

   Similarly, $h_k$ can be

   • $0$ outside of $\br{a - \frac{1}{k}, b + \frac{1}{k}}$ (possibly nowhere),
   • linear on $\br{a - \frac{1}{k}, a}$,
   • $1$ on $\br{a, b}$,
   • linear on $\br{b, b + \frac{1}{k}}$.

   By monotonicity and (1) applied to $g_k, h_k$, we have

   $$\begin{gathered} E_N\p{g_k} \leq E_N\p{f} \leq E_N\p{h_k} \\ \implies \int_0^1 g_k\p{x + \Z} \,\diff{x} \leq \liminf_{N\to\infty} E_N\p{f} \leq \limsup_{N\to\infty} E_N\p{f} \leq \int_0^1 h_k\p{x + \Z} \,\diff{x}. \end{gathered}$$

   Moreover, $g_k$ and $h_k$ are dominated by $1 \in L^1\p{\R/\Z}$ and converge to $f$ almost everywhere, so by dominated convergence and the above inequality,

   $$\int_0^1 f\p{x + \Z} \,\diff{x} \leq \liminf_{N\to\infty} E_N\p{f} \leq \limsup_{N\to\infty} E_N\p{f} \leq \int_0^1 f\p{x + \Z} \,\diff{x}.$$

   Thus, $\lim_{N\to\infty} E_N\p{f} = \int_0^1 f\p{x + \Z} \,\diff{x}$, as required.