“Good afternoon, dear friends!” Senn took his seat. “We are back after the two month hiatus. Things weren’t the easiest for us but your support has been phenomenal - we dearly appreciate it. We took this time to set up our own studio - so that in the future we can stream from home whenever necessary. Pura is doing better now; all your good wishes shall definitely be conveyed to Estella as she sees the light of day. Thank you again, and let’s begin!”
Since we were engaged in so much, my good friend Tensou 天蒼 has made me the problems this time. It is pretty advanced this time around, but at least you’ll get a new flavour. We shall look at number sequences and the various properties they exhibit.
For a real number $x$, let $[x]$ denote the integral part of $x$, and let ${x}=x-[x]$ denote the fractional part of $x$ (or the residue of $x$ modulo 1). We note that ${x}\in I=[0,1)\forall x\in \mathbb R$.
Let $\omega=(x_n)_{n=1}^\infty$ be a given sequence of real numbers. For a positive integer $N$ and a subset $E$ of $I$, let the counting function $A(E; N; \omega)$ be defined as the number of terms $x_n$, $1 \le n \le N$, for which $x_n\in E$. If there is no risk of confusion, we shall often write $A(E; N)$ instead of $A(E; N; \omega)$.
Definition 1: The sequence $\omega = (x_n)_{n=1}^\infty$ of real numbers is said to be uniformly distributed modulo 1 (abbreviated ud mod 1) if for every pair of real numbers $a,b$, $0\le a<b\le 1$ we have
\[\lim_{N\to \infty} \frac{A([a,b);N;\omega)}{N}=b-a. \tag{1}\]So, in simple terms, the sequence $\omega$ is ud mod 1 if every half-open subinterval of $I$ eventually gets its “proper share” of fractional parts.
Problem 1: Show that $(1)$ can be reframed as
\[\lim_{N\to \infty} \frac 1N\sum_{n=1}^N c_{[a,b)}(\{x_n\})=\int_0^1 c_{[a,b)}(x) \:\text{d}x, \tag{2}\]where $c_{[a,b)}$ is the characteristic function of the interval $[a,b)\subset I$.
Problem 2.1: Show that $\omega=(x_n)$ is ud mod 1 if and only if for every real valued continuous function $f$ defined on the closed unit interval $\overline I=[0,1]$, we have
\[\lim_{N\to \infty}\frac 1N\sum_{n=1}^N f(\{x_n\})=\int_0^1 f(x)\text{d}x \tag{3}\]Hint: Recall how step functions (and hence, characteristic functions) are used to approximate Riemann Integrals.
Problem 2.2: Does $(3)$ hold if $f$ is a complex-valued continuous function on $\mathbb R$?
Problem 2.3: By tweaking few parts of the proof of Problem 2.1 (or otherwise if you dare) show that a sequence $(x_n)$ is ud mod 1 if and only if we have
\[\lim_{N\to \infty} \frac 1N \sum_{n=1}^N f(x_n)=\int_0^1 f(x_n)\:\text{d}x \tag{4}\]for every complex valued continuous function $f$ on $\mathbb R$ with period 1.
Problem 3: Using $(1)$ or $(3)$, (just to cheer you up) construct any sequence that is not ud mod 1.
Now, prove that the sequence
\[\frac 01, \frac 02, \frac 12, \frac 03, \frac 13, \frac 23,\dots ,\frac 0k,\frac 1k,\dots ,\frac{k-1}{k},\dots\]is ud mod 1.
Now, let $(x_n)$ be a sequence in $I$. For a subinterval $[a, b)$ of $I$ and $N \ge 1$, let $S([a, b); N)$ be the sum of the elements from $x_1, x_2,\dots ,x_N$ that are in $[a, b)$. Show that $(x_n)$ is ud mod 1 if and only if
\[\lim_{N\to \infty}\frac{S([a, b); N)}{N}=\frac 12(b^2-a^2)\]for all subintervals $[a,b)$ of $I$.
Problem 4: Deduce from $(1)$ that if the sequence $(x_n)_{n=1}^\infty$ is ud mod 1, then so is $(x_n+\alpha)_{n=1}^\infty$ for a real constant $\alpha$. Now, show that if a sequence $(x_n)$ is ud mod 1, and if $(y_n)$ is a sequence such that
\[\lim_{n\to \infty}(x_n-y_n)=\alpha,\]where $\alpha$ is a real constant, then $(y_n)$ is ud mod 1.
Now, we want to generalize our ideas of distribution modulo 1.
Let $\Delta$ be a subdivision of the interval $[0,\infty)$ so that
\[\begin{align*} &\Delta : 0=z_0<z_1<z_2<\dots\\ &\lim_{k\to \infty} z_k=\infty \end{align*}\]For $z_{k-1}\le x<z_k$, let us define
\[\begin{align*} &[x]_{\Delta}=z_{k-1}\\ &\{x\}_{\Delta}=\frac{x-z_{k-1}}{z_k-z_{k-1}} \end{align*}\]so that $0\le {x}_{\Delta}<1$.
Definition 2: The sequence $(x_n)_{n=1}^\infty$ of nonnegative real numbers is said to be uniformly distributed modulo $\Delta$ (abbreviated ud mod $\Delta$) if the sequence $({x_n}_{\Delta})_{n=1}^\infty$ is ud mod 1.
Note that if $\Delta$ is the subdivision for which $z_k=k$, this concept reduces to that of ud mod 1.
Problem 5: Let $(x_n)$ be an increasing sequence of nonnegative real numbers with
\[\lim_{n\to \infty} x_n=\infty.\]Let $A(x,\alpha)$ be the number of $x_n<x$ with ${x_n}_{\Delta}<\alpha$ and let $A(x)=A(x,1)$. Check that the sequence $(x_n)$ is ud mod $\Delta$ if and only if we have
\[\lim_{x\to \infty}\frac{A(x,\alpha)}{A(x)}=\alpha\]for all $\alpha\in(0,1)$.
Now, prove that
\[\lim_{k\to \infty}\frac{A(z_{k+1})}{A(z_k)}=1\]is a necessary condition for $(x_n)$ to be ud mod $\Delta$.
“Wow, Problem 5 is pretty hard at first glance! Instead of the general approximating a continuous function by step functions, we here need to approximate it with piece-wise linear components. Can you try it out?”
Bonus question: Use $(4)$ to show that the sequence $(x_n)$ is ud mod 1 if and only if
\[\lim_{N\to \infty}\frac 1N\sum_{n=1}^N e^{2\pi ihx_n}=0\]for all integers $h\neq 0$.
Hint: You may want to use the Weierstrass Approximation Theorem. Using this or otherwise, classify the values of $\theta$ for which the sequence $(n\theta)$ is ud mod 1. If you can’t provide a proof, you may also want to make an intuitive guess and write an intuitive explanation down.
Senn quietly closed the scribble book. “Pura doesn’t approve of this at all - she thinks the problems are cut out for senior undergrads - but I’m sure some of you are math majors and were pining to get hands on something non-trivial. Here it is! And for the rest, we’ll go back to our own flavour from next month, when we get more time to interact!”
Submitting
Send in your solutions at maths.club@iiserkol.ac.in.