In this post, we will prove the Weyl’s Equidistribution theorem. A sequence of real numbers $x_1, x_2, \\cdots$ is said to be equidistributed (mod 1) if for every sub-interval $(a,b)\\subset [0,1]$, we have\n$$\\lim_{N\\to \\infty}\\frac{|\\{1\\leq n\\leq N:\\; \\langle x_n \\rangle\\in (a,b)\\}|}{N} = b-a$$\nwhere $\\langle x \\rangle$ denotes the fractional part of $x$. Weyl’s equidistribution criteria states that the following statements are equivalent:\n
\nProof: (1) ⇒ (3)<\/b><\/p>\n
Let $I=[a,b)\\subseteq [0,1]$ and note that\n$$ \\frac{|\\{1 \\leq n \\leq N: \\langle x_n \\rangle\\in [a,b) \\}|}{N}=\\frac{1}{N}\\sum_{n=1}^N \\chi_{[a,b)}(\\langle x_n \\rangle) $$\nwhere $\\chi_{[a,b)}(x)$ equals 1 if $x\\in [a,b)$ and 0 otherwise. This shows that (3) holds for the case when $f$ is a characteristic function. Now, let $\\lambda_1, \\lambda_2\\in \\mathbb{R}$ and $f_1$, $f_2$ be functions for which (3) holds. Then,\n$$ \\begin{aligned}\\lim_{N\\to \\infty}\\sum_{n=1}^N (\\lambda_1 f_1 + \\lambda_2 f_2)(\\langle x_n \\rangle) &= \\lim_{N\\to \\infty} \\frac{\\lambda_1}{N}\\sum_{n=1}^N f_1(\\langle x_n\\rangle) + \\lim_{N\\to \\infty}\\frac{\\lambda_2}{N}\\sum_{n=1}^Nf_2(\\langle x_n\\rangle) \\\\ &= \\lambda_1\\int_0^1 f_1(x) dx + \\lambda_2 \\int_0^1 f_2(x) dx \\\\ &= \\int_0^1 (\\lambda_1 f_1 + \\lambda_2 f_2)(x) dx\\end{aligned} $$\nThus, (3) holds for all linear combinations of characteristic functions of subintervals of $[0,1]$. <\/p>\n
\nNow, let $f:[0,1]\\to \\mathbb{R}$ be an integrable function, and let $\\epsilon >0$. Choose step functions $f_1$ and $f_2$ such that:\n
(2) ⇒ (3)<\/b><\/p>\n
\nLet $f:[0,1]\\to \\mathbb{R}$ be continuous, and let $\\epsilon > 0$. The Stone-Weierstrass Theorem<\/a> allows us to choose a trigonometric polynomial $p$ such that:\n$$\n\\sup_{x\\in [0,1]} |f(x) – p(x)| < \\frac{\\epsilon}{3}\n$$\nAlso, (2) implies the existence of an $N_0$ such that for $N\\geq N_0$, we have\n$$\\left|\\frac{1}{N}\\sum_{n=1}^N p(\\langle x_n \\rangle)-\\int_0^1 p(x) dx \\right| < \\frac{\\epsilon}{3}$$\nNow,\n$$\n\\begin{aligned}\n&\\; \\left|\\frac{1}{N}\\sum_{n=1}^N f(\\langle x_n\\rangle) – \\int_0^1 f(x) dx\\right| \\\\ &= \\left|\\frac{1}{N}\\sum_{n=1}^N (f(\\langle x_n \\rangle) – p(\\langle x_n \\rangle)) + \\int_0^1 (p(x)-f(x))dx + \\frac{1}{N}\\sum_{n=1}^N p(\\langle x_n \\rangle) – \\int_0^1 p(x) dx\\right| \\\\\n&< \\frac{1}{N}\\sum_{n=1}^N\\left|f(\\langle x_n \\rangle) – p(\\langle x_n \\rangle) \\right| + \\int_0^1 \\left|p(x)-f(x) \\right| dx + \\left|\\frac{1}{N}\\sum_{n=1}^N p(\\langle x_n \\rangle) – \\int_0^1 p(x) dx \\right| \\\\\n&< \\frac{\\epsilon}{3}+\\frac{\\epsilon}{3}+\\frac{\\epsilon}{3} \\\\\n&= \\epsilon\n\\end{aligned}\n$$\nfor all $N\\geq N_0$. Thus, (3) holds for continuous functions on $[0,1]$. By the proof of (1) ⇒ (3), it is sufficient to show that (3) holds for all step functions on $[0,1]$. If $g$ is a step function on $[0,1]$, we can find continuous functions $g_1, g_2$ such that $g_1\\leq g\\leq g_2$ and $\\int_0^1 (g_1(x)-g_2(x))dx < \\epsilon$. We again conclude that (3) holds for $g$.\n<\/p>\n The implications (3) ⇒ (1)<\/b> and (3) ⇒ (2)<\/b> are obvious.<\/p>\n \n In this post, we will prove the Weyl’s Equidistribution theorem. A sequence of real numbers $x_1, x_2, \\cdots$ is said to be equidistributed (mod 1) if for every sub-interval $(a,b)\\subset [0,1]$, we have $$\\lim_{N\\to \\infty}\\frac{|\\{1\\leq n\\leq N:\\; \\langle x_n \\rangle\\in … Continue reading References<\/h2>
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