$L^1$ norms of products of sines

1 Introduction

A trigonometric polynomial of degree $n$ is an expression of the form

$$\sum _{k=-n}^n{c}_k{e}^{ikt},c_k\in ℂ.$$

Using the identity $e^{it}=\cos t+i\sin t$, we can write a trigonometric polynomial of degree $n$ in the form

$$a_0+\sum _{k=1}^n{a}_k\cos kt+\sum _{k=1}^n{b}_k\sin kt,a_k,b_k\in ℂ.$$

The trigonometric functions $\cos kt$ and $\sin kt$, $k\in \mathbb{Z}$, are the building blocks for $2\pi$-periodic functions (cf. [6]). To formalize the idea of the size of a $2\pi$-periodic function and to formalize the idea of approximating $2\pi$-periodic functions using trigonometric polynomials, we introduce $L^p$ norms.

For and for a $2\pi$-periodic function $f$, we define the $L^p$ norm of $f$ by

$${\parallel f\parallel }_p={\left(\frac{1}{2\pi }{\int }_0^{2\pi }{|f(t)|}^p𝑑t\right)}^{1/p}.$$

For a continuous $2\pi$-periodic function $f$, we define the $L^{\mathrm{\infty }}$ norm of $f$ by

$${\parallel f\parallel }_{\mathrm{\infty }}=\underset{0\le t\le 2\pi }{\max }|f(t)|.$$

If $f$ is a continuous $2\pi$-periodic function, then there is a sequence of trigonometric polynomials $f_n$ such that ${\parallel f-f_n\parallel }_{\mathrm{\infty }}\to 0$ as $n\to \mathrm{\infty }$ [11, p. 54, Corollary 5.4].

The Dirichlet kernel $D_n$ is defined by

$$D_n(t)=\sum _{k=-n}^n{e}^{ikt}=1+2\sum _{k=1}^n\cos kt.$$

One can show [5, p. 71, Exercise 1.1] that

$${\parallel D_n\parallel }_1=\frac{4}{{\pi }^2}\cdot \log n+O(1).$$

(On the other hand, it can quickly be seen that ${\parallel D_n\parallel }_{\mathrm{\infty }}=2n+1$, and it follows immediately from Parseval’s identity that ${\parallel D_n\parallel }_2=\sqrt{2n+1}$.)

Pólya and Szegő [8, Part VI] present various problems about trigonometric polynomials together with solutions to them. A result on $L^{\mathrm{\infty }}$ norms of trigonometric polynomials that Pólya and Szegő present is for the sum $A_n(t)={\sum }_{k=1}^n\frac{\sin kt}{k}$. The local maxima and local minima of $A_n$ can be explicitly determined [8, p. 74, no. 23], and it can be shown that [8, p. 74, no. 25]

$${\parallel A_n\parallel }_{\mathrm{\infty }}\sim {\int }_0^{\pi }\frac{\sin t}{t}𝑑t.$$

In [2, p. 532, Theorem 2], the author proves the following.

Theorem 1.

Let $F_n\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}{\mathrm{\prod }}_{k\mathrm{=}\mathrm{1}}^n\sin \mathit{}\mathrm{(}k\mathit{}t\mathrm{)}$, let $M$ be the maximum value of

$$\frac{1}{w}{\int }_0^w\log \sin tdt$$

for $w\mathrm{\in }\mathrm{(}\mathrm{0}\mathrm{,}\pi \mathrm{)}$, let $A\mathrm{=}{e}^M$, and let $B\mathrm{=}\mathrm{4}\mathit{}{e}^M\mathit{}{\mathrm{\left(}\mathrm{1}\mathrm{-}{e}^{\mathrm{2}\mathit{}M}\mathrm{\right)}}^{\mathrm{-}\frac{\mathrm{1}}{\mathrm{4}}}$. We have

$${\parallel F_n\parallel }_1\sim \frac{B}{n}\cdot A^n.$$

We compute that $M=-0.49452\mathrm{\dots }$ and $A=0.60985\mathrm{\dots }$.

When I was working on this problem, I first found simpler weaker estimates that apply to a larger class of products.

For $k\ge 1$, let $a_k$ be a positive integer, and let

$$F_n^a(t)=\prod _{k=1}^n\sin (a_{k}t),a=(a_k).$$

In this paper we show that we can use simpler methods to obtain nontrivial upper and lower bounds on ${\parallel F_n^a\parallel }_1$. The results are substantially weaker than Theorem 1, but hold for any sequence $a$. As well, their proofs can be more readily understood. We present an asymptotic result showing that the $L^1$ norm of $\sin (t)\sin (q^{m}t)\mathrm{\cdots }\sin (q^{m(n-1)}t)$ approaches ${\left(\frac{2}{\pi }\right)}^n$ as $m\to \mathrm{\infty }$, for $q\ge 2$ an integer. We present inequalities for the norms of trigonometric polynomials.

In Figure 1 we plot ${\prod }_{k=1}^8|\sin (kt)|$ for $0\le t\le 2\pi$. In Figure 2 we plot ${\prod }_{k=1}^4|\sin (p_{k}t)|$ for $0\le t\le 2\pi$, where $p_k$ is the $k$th prime. In Figure 3 we plot ${\prod }_{k=1}^{10}|\sin (\lceil \sqrt{k}\rceil t)|$ for $0\le t\le 2\pi$, where $\lceil x\rceil$ is the least integer $\ge x$. We want upper and lower bounds on the areas under these graphs.

Figure 1: ${\prod }_{k=1}^8|\sin (kt)|$ for $0\le t\le 2\pi$

Figure 2: ${\prod }_{k=1}^4|\sin (p_{k}t)|$ for $0\le t\le 2\pi$, where $p_k$ is the $k$th prime

Figure 3: ${\prod }_{k=1}^{10}|\sin (\lceil \sqrt{k}\rceil t)|$ for $0\le t\le 2\pi$

2 Upper and lower bounds

Hölder’s inequality is the first tool for which we reach when we want to bound the norm of a product.

Theorem 2.

For any sequence $a$, we have

$${\parallel F_n^a\parallel }_1=O\left(\frac{1}{\sqrt{n}}\right).$$

Proof.

Hölder’s inequality [7, p. 45, Theorem 2.3] (cf. [10, p. 151, Exercise 9.9]) states that if ${\sum }_{k=1}^n\frac{1}{p_k}=1$ then

$${\parallel f\parallel }_1\le \prod _{k=1}^n{\parallel f\parallel }_{p_k}.$$

As $1={\sum }_{k=1}^n\frac{1}{n}$, this implies that

$${\parallel F_n^a\parallel }_1\le \prod _{k=1}^n{\parallel \sin (a_{k}t)\parallel }_n.$$

For each $k$,

${\displaystyle {\int }_0^{2\pi }}{|\sin (a_{k}t)|}^n𝑑t$ $=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle {\int }_0^{2\pi a_k}}{|\sin t|}^n𝑑t$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_{2\pi (j-1)}^{2\pi j}}{|\sin t|}^n𝑑t$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_0^{2\pi }}{|\sin (t+2\pi (j-1))|}^n𝑑t$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_0^{2\pi }}{|\sin t|}^n𝑑t$

$=$ ${\displaystyle {\int }_0^{2\pi }}{|\sin t|}^n𝑑t$

$=$ $2{\displaystyle {\int }_0^{\pi }}{\sin }^{n}tdt.$

Let $G_n={\int }_0^{\pi }{\sin }^{n}tdt$. Doing integration by parts, for $n\ge 2$ we have

$G_n$ $=$ ${\displaystyle {\int }_0^{\pi }}{\sin }^{n-1}t\sin tdt$

$=$ $-{{\sin }^{n-1}t\cos t|}_0^{\pi }+(n-1){\displaystyle {\int }_0^{\pi }}{\sin }^{n-2}t{\cos }^{2}tdt$

$=$ $(n-1){\displaystyle {\int }_0^{\pi }}{\sin }^{n-2}t(1-{\sin }^{2}t)𝑑t$

$=$ $(n-1)G_{n-2}-(n-1)G_n.$

Thus,

$$G_n=\frac{n-1}{n}{G}_{n-2}.$$

Say $n=2m+1$, $m\ge 1$. For $m=1$, we have $G_{2m+1}=G_3=\frac{2}{3}{G}_1=\frac{4}{3}$. Assume that for some $m\ge 1$ we have

$$G_{2m+1}=\frac{2^{2m+1}m!m!}{(2m+1)!}.$$ (1)

Then

$G_{2m+3}$ $=$ ${\displaystyle \frac{2m+2}{2m+3}}{G}_{2m+1}$

$=$ ${\displaystyle \frac{2m+2}{2m+3}}{\displaystyle \frac{2^{2m+1}m!m!}{(2m+1)!}}$

$=$ ${\displaystyle \frac{2^{2m+3}(m+1)!(m+1)!}{(2m+3)!}}.$

Therefore, by induction (1) holds for all $m\ge 1$. By applying Stirling’s approximation to (1) we get

$G_{2m+1}$ $\sim$ ${\displaystyle \frac{2^{2m+1}\sqrt{2\pi m}{\left(\frac{m}{e}\right)}^m\sqrt{2\pi m}{\left(\frac{m}{e}\right)}^m}{\sqrt{2\pi (2m+1)}{\left(\frac{2m+1}{e}\right)}^{2m+1}}}$

$=$ ${\displaystyle \frac{2^{2m+1}\cdot 2\pi }{\sqrt{2\pi (2m+1)}}}\cdot {\left({\displaystyle \frac{m}{m+\frac{1}{2}}}\right)}^{2m+1}\cdot {\displaystyle \frac{1}{2^{2m+1}}}\cdot {\displaystyle \frac{e^{2m+1}}{e^{2m}}}$

$=$ ${\displaystyle \frac{e\sqrt{2\pi }}{\sqrt{2m+1}}}\cdot {\left({\displaystyle \frac{m}{m+\frac{1}{2}}}\right)}^{2m+1}$

${\displaystyle \frac{e\sqrt{2\pi }}{\sqrt{2m+1}}}.$

Thus,

$$G_{2m+1}=O\left(\frac{1}{\sqrt{2m+1}}\right).$$

Say $n=2m$, $m\ge 1$. For $m=1$, we have $G_{2m}=G_2=\frac{1}{2}{G}_0=\frac{\pi }{2}$. Assume that for some $m\ge 1$ we have

$$G_{2m}=\pi \frac{(2m)!}{2^{2m}m!m!}.$$ (2)

Then

$G_{2m+2}$ $=$ ${\displaystyle \frac{2m+1}{2m+2}}{G}_{2m}$

$=$ ${\displaystyle \frac{2m+1}{2m+2}}\cdot \pi {\displaystyle \frac{(2m)!}{2^{2m}m!m!}}$

$=$ $\pi {\displaystyle \frac{(2m+2)!}{2^{2m+2}(m+1)!(m+1)!}}.$

Therefore, by induction (2) holds for all $m\ge 1$. Like for $n=2m+1$, by applying Stirling’s approximation to (2) we get $G_{2m}\sim \frac{\sqrt{2\pi }}{\sqrt{2m}}$, and so

$$G_{2m}=O\left(\frac{1}{\sqrt{2m}}\right).$$

Hence $G_n=O\left(\frac{1}{\sqrt{n}}\right)$. It follows that

${\parallel F_n^a\parallel }_1$ $\le$ ${\displaystyle \prod _{k=1}^n}{\left({\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|\sin (a_{k}t)|}^n𝑑t\right)}^{1/n}$

$=$ ${\displaystyle \prod _{k=1}^n}{\left({\displaystyle \frac{G_n}{\pi }}\right)}^{1/n}$

$=$ ${\displaystyle \frac{G_n}{\pi }}$

$=$ $O\left({\displaystyle \frac{1}{\sqrt{n}}}\right).$

∎

In the proof of Theorem 2, we saw that for any $a_k\ge 1$, we have

$${\int }_0^{2\pi }{|\sin (a_{k}t)|}^n𝑑t=2G_n,G_n={\int }_0^{\pi }{\sin }^{n}tdt,$$

We showed that

$$G_{2m+1}\sim \frac{e\sqrt{2\pi }}{\sqrt{2m+1}}\cdot {\left(\frac{m}{m+\frac{1}{2}}\right)}^{2m+1}.$$

We can check, by taking logarithms and using L’Hospital’s rule, that

$$\underset{m\to \mathrm{\infty }}{lim}{\left(\frac{m}{m+\frac{1}{2}}\right)}^{2m+1}=e^{-1}.$$

Therefore, there is some $C_1>0$ such that for all $m\ge 1$ we have

$$G_{2m+1}\ge \frac{C_1}{\sqrt{2m+1}}.$$

We also showed that

$$G_{2m}\sim \frac{\sqrt{2\pi }}{\sqrt{2m}},$$

and hence there is some $C_2>0$ such that for all $m\ge 1$ we have

$$G_{2m}\ge \frac{C_2}{\sqrt{2m}}.$$

If $C=\min \{C_1,C_2\}$, then for all $n\ge 2$ we have

$$G_n\ge \frac{C}{\sqrt{n}}.$$

It follows that for any positive integer $a$, if $a_k=a$ for all $k\ge 1$ then

$${\parallel F_n^a\parallel }_1=\frac{G_n}{\pi }\ge \frac{C}{\sqrt{\pi }}\cdot \frac{1}{\sqrt{n}}.$$

In other words, if all the terms in the sequence $a$ are the same then the inequality given by Theorem 2 is sharp.

In the above theorem we gave an upper bound on ${\parallel F_n^a\parallel }_1$, and in the following theorem we give a lower bound on ${\parallel F_n^a\parallel }_1$.

Theorem 3.

For any sequence $a$, we have

$${\parallel F_n^a\parallel }_1>\frac{1}{2^n}.$$

Proof.

Since $-\log$ is a convex function on $(0,\mathrm{\infty })$, by Jensen’s inequality [7, p. 44, Theorem 2.2] we have for any nonnegative function $f$ with that

$$-\log \left(\frac{1}{2\pi }{\int }_0^{2\pi }f(t)𝑑t\right)\le \frac{1}{2\pi }{\int }_0^{2\pi }-\log (f(t))dt,$$

and the two sides are equal if and only if $f$ is constant almost everywhere (for continuous $f$ this is equivalent to $f$ being constant). Hence, as there is no sequence $a$ of positive integers such that $F_n^a$ is constant,

$$\log \left(\frac{1}{2\pi }{\int }_0^{2\pi }|F_n^a(t)|𝑑t\right)>\frac{1}{2\pi }{\int }_0^{2\pi }\log (|F_n^a(t)|)𝑑t.$$ (3)

The left-hand side of (3) is $\log {\parallel F_n^a\parallel }_1$, and the the right-hand side is equal to

$$\frac{1}{2\pi }{\int }_0^{2\pi }\sum _{k=1}^n\log |\sin (a_{k}t)|dt=\frac{1}{2\pi }\sum _{k=1}^n{\int }_0^{2\pi }\log |\sin (a_{k}t)|dt.$$

But for each $k$,

${\displaystyle {\int }_0^{2\pi }}\log |\sin (a_{k}t)|dt$ $=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle {\int }_0^{2\pi a_k}}\log |\sin t|dt$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_{2\pi (j-1)}^{2\pi j}}\log |\sin t|dt$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_0^{2\pi }}\log |\sin (t+2\pi (j-1))|dt$

$=$ ${\displaystyle \frac{1}{a_k}}{\displaystyle \sum _{j=1}^{a_k}}{\displaystyle {\int }_0^{2\pi }}\log |\sin t|dt$

$=$ ${\displaystyle {\int }_0^{2\pi }}\log |\sin t|dt.$

Hence (3) is

$$\log {\parallel F_n^a\parallel }_1>\frac{n}{2\pi }{\int }_0^{2\pi }\log |\sin t|dt.$$

We calculate ${\int }_0^{2\pi }\log |\sin t|dt$ in the following way. (The earliest evaluation of this integral of which the author is aware is by Euler [4], who gives two derivations, the first using the Euler-Maclaurin summation formula, the power series expansion for $\log \left(\frac{1+x}{1-x}\right)$, and the power series expansion of $x\cot (x)$, and the second using the Fourier series of $\log |\sin t|$.) First,

$${\int }_0^{2\pi }\log |\sin t|dt=4{\int }_0^{\frac{\pi }{2}}\log \sin tdt.$$

We have

${\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \sin tdt$ $=$ ${\displaystyle {\int }_{-\frac{\pi }{2}}^0}\log \left(\sin \left(t+{\displaystyle \frac{\pi }{2}}\right)\right)𝑑t$

$=$ ${\displaystyle {\int }_{-\frac{\pi }{2}}^0}\log \left(\sin t\cos {\displaystyle \frac{\pi }{2}}+\sin {\displaystyle \frac{\pi }{2}}\cos t\right)𝑑t$

$=$ ${\displaystyle {\int }_{-\frac{\pi }{2}}^0}\log \cos tdt$

$=$ ${\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \cos tdt.$

Therefore,

$2{\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \sin tdt$ $=$ ${\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \sin tdt+{\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \cos tdt$

$=$ ${\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \left(2\sin t\cos t\right)-\log 2dt$

$=$ ${\displaystyle {\int }_0^{\frac{\pi }{2}}}\log \sin (2t)dt-{\displaystyle \frac{\pi }{2}}\log 2$

$=$ ${\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi }}\log \sin tdt-{\displaystyle \frac{\pi }{2}}\log 2.$

Because ${\int }_0^{\pi }\log \sin tdt=2{\int }_0^{\frac{\pi }{2}}\log \sin tdt$, we have

$$2{\int }_0^{\frac{\pi }{2}}\log \sin tdt={\int }_0^{\frac{\pi }{2}}\log \sin tdt-\frac{\pi }{2}\log 2,$$

and so

$${\int }_0^{\frac{\pi }{2}}\log \sin tdt=-\frac{\pi }{2}\log 2.$$

Thus

$${\int }_0^{2\pi }\log |\sin t|dt=-2\pi \log 2.$$

Therefore we have

$$\log \parallel F_n^a{\parallel }_1>\frac{n}{2\pi }\cdot -2\pi \log 2=-n\log 2=\log (2^{-n}),$$

and thus

$${\parallel F_n^a\parallel }_1>2^{-n}.$$

∎

In the above theorem we gave a lower bound for ${\parallel F_n^a\parallel }_1$. In the following theorem we give another lower bound for ${\parallel F_n^a\parallel }_1$, and we then construct examples where one lower bound is better than the other.

Theorem 4.

For any sequence $a$, if

$$A_n=\underset{1\le k\le n}{\max }{a}_k,$$

then

$${\parallel F_n^a\parallel }_1>\frac{1}{4}\cdot \frac{1}{n+1}\cdot \frac{1}{A_n^{n+1}}\prod _{k=1}^n{a}_k.$$

Proof.

If $0\le t\le \frac{\pi }{2}$ then $\sin t\ge \frac{2}{\pi }t$ [13]. In words, if , then $(t,\sin t)$ lies above the line joining $(0,0)$ and $(\frac{\pi }{2},1)$.

${\displaystyle {\int }_0^{2\pi }}|F_n^a(t)|𝑑t$ $=$ ${\displaystyle {\int }_0^{2\pi }}{\displaystyle \prod _{k=1}^n}|\sin (a_{k}t)|dt$

$>$ ${\displaystyle {\int }_0^{\frac{\pi }{2A_n}}}{\displaystyle \prod _{k=1}^n}\sin (a_{k}t)dt$

$\ge$ ${\displaystyle {\int }_0^{\frac{\pi }{2A_n}}}{\displaystyle \prod _{k=1}^n}\left({\displaystyle \frac{2}{\pi }}{a}_{k}t\right)dt$

$=$ ${\left({\displaystyle \frac{2}{\pi }}\right)}^n\left({\displaystyle \prod _{k=1}^n}{a}_k\right){\displaystyle {\int }_0^{\frac{\pi }{2A_n}}}{t}^n𝑑t$

$=$ ${\left({\displaystyle \frac{2}{\pi }}\right)}^n\left({\displaystyle \prod _{k=1}^n}{a}_k\right){\displaystyle \frac{1}{n+1}}{\left({\displaystyle \frac{\pi }{2A_n}}\right)}^{n+1}$

$=$ ${\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{1}{n+1}}\cdot {\displaystyle \frac{1}{A_n^{n+1}}}{\displaystyle \prod _{k=1}^n}{a}_k.$

∎

If, for instance, $a_k=2^k$, the above inequality is

$${\parallel F_n^a\parallel }_1>\frac{1}{4}\cdot \frac{1}{n+1}\cdot \frac{1}{2^{n(n+1)}}\cdot 2^{\frac{n(n+1)}{2}}=\frac{1}{4}\cdot \frac{1}{n+1}\cdot 2^{-\frac{n(n+1)}{2}},$$

which is worse than (i.e. less than) the inequality given by Theorem 3.

If $a_k=k$, the inequality is

$${\parallel F_n^a\parallel }_1>\frac{1}{4}\cdot \frac{1}{n+1}\cdot \frac{1}{n^{n+1}}\cdot n!,$$

and applying Stirling’s approximation we get

$${\parallel F_n^a\parallel }_1\gg \frac{1}{n+1}\cdot \frac{1}{\sqrt{n}}\cdot \frac{1}{e^n},$$

which is also worse than the inequality given by Theorem 3.

But if $a_k=\lceil \sqrt{k}\rceil$, the inequality is

$${\parallel F_n^a\parallel }_1>\frac{1}{4}\cdot \frac{1}{n+1}\cdot \frac{1}{{(n+1)}^{\frac{n+1}{2}}}\prod _{k=1}^n{k}^{\frac{1}{2}}=\frac{1}{4}\cdot \frac{1}{n+1}\cdot \frac{1}{n^{\frac{n+1}{2}}}\cdot \frac{1}{{\left(1+\frac{1}{n}\right)}^{\frac{n+1}{2}}}\prod _{k=1}^n{k}^{\frac{1}{2}}.$$

Taking logarithms and using L’Hospital’s rule, we get

$$\underset{n\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{n}\right)}^{\frac{n+1}{2}}=e^{\frac{1}{2}}.$$

Then using Stirling’s approximation we obtain

$${\parallel F_n^a\parallel }_1\gg n^{-\frac{5}{4}}{e}^{-\frac{n}{2}},$$

and since , this lower bound is better than (i.e. greater than) the lower bound $2^{-n}$ in Theorem 3.

3 Mixing

This section talks about measure spaces and mixing. These topics take repeated exposure to become comfortable with, but Theorem 5 is a pretty result whose statement can be understood without understanding its proof. The notion of mixing is related to independent random variables, for which the expectation of their product is equal to the product of their expectations.

Let $X$ be a measure space with probability measure $\mu$. Following [9, p. 21, Definition 3.6], we say that a measure preserving map $T:X\to X$ is $r$-fold mixing if for all $g,f_1,\mathrm{\dots },f_r\in L^{r+1}(X)$ we have

(4)

If for each $r$ the map $T$ is $r$-fold mixing, we say that $T$ is mixing of all orders.

Let $\lambda$ be Lebesgue measure on $[0,1]$. Let $q\ge 2$ be an integer, and define $T_q:[0,1]\to [0,1]$ by $T_q(t)=R(qt)$; $R(x)=x-[x]$, where $[x]$ is the greatest integer $\le x$. $T_q$ is mixing of all orders. This can be proved by first showing that the dynamical system $([0,1],\lambda ,T_q)$ is isomorphic to a Bernoulli shift (cf. [3, p. 17, Example 2.8]). This implies that if the Bernoulli shift is $r$-fold mixing then $T_q$ is $r$-fold mixing. One then shows that a Bernoulli shift is mixing of all orders [3, p. 53, Exercise 2.7.9]. Using that $T_q$ is mixing of all orders gets us the following result.

Theorem 5.

Let $q\mathrm{\ge }\mathrm{2}$ be an integer. For each $n\mathrm{\ge }\mathrm{1}$ we have

$$\underset{m\to \mathrm{\infty }}{lim}{\int }_0^1|\sin (2\pi t)|\cdot \prod _{k=1}^n\left|\sin \left(2\pi q^{km}t\right)\right|dt={\left(\frac{2}{\pi }\right)}^{n+1}.$$

Proof.

Define $g(t)=f_1(t)=\mathrm{\cdots }=f_n(t)=|\sin (2\pi t)|$. For any nonzero integer $N$ we have

$${\int }_0^1|\sin (2\pi Nt)|𝑑t=\frac{2}{\pi },$$

and it follows from (4), using $m=\max \{m_1,\mathrm{\dots },m_n\}$, that

$$\underset{m\to \mathrm{\infty }}{lim}{\int }_0^1|\sin (2\pi t)|\cdot \prod _{k=1}^n\left|\sin \left(2\pi q^{km}t\right)\right|dt={\left(\frac{2}{\pi }\right)}^{n+1}.$$

∎

In other words, if for $k\ge 1$ we set $a_k(m)=q^{(k-1)m}$, then for each $n$ we have

$$\underset{m\to \mathrm{\infty }}{lim}{\parallel F_n^{a(m)}\parallel }_1={\left(\frac{2}{\pi }\right)}^n,a(m)=(a_k(m)).$$

It would be overwhelming for a reader without experience in ergodic theory to work out the details of the reasoning that we indicated above Theorem 5 for why $T_q$ is mixing of all orders. In the following we explain a more understandable derivation of the case $n=1$ of Theorem 5. For a measure space $X$ with probability measure $\mu$, we say that a measure preserving transformation $T:X\to X$ is mixing (in other words, $1$-fold mixing), if for all $f,g\in L^2(X)$ we have

$$\underset{m\to \mathrm{\infty }}{lim}{\int }_{X}f\left(T^m(t)\right)g(t)𝑑\mu (t)=\left({\int }_{X}f(t)𝑑\mu (t)\right)\left({\int }_{X}g(t)𝑑\mu (t)\right).$$

Stein and Shakarchi [12, p. 305] prove that $T_2:[0,1]\to [0,1]$ (for $T_q(t)=R(qt)$) is mixing, and their argument works to show that $T_q:[0,1]\to [0,1]$ is mixing for $q\ge 2$ an integer. Hence, taking $f(t)=g(t)=|\sin (2\pi t)|$, we get

$$\underset{m\to \mathrm{\infty }}{lim}{\int }_0^1|\sin (2\pi q^{m}t)|\cdot |\sin (2\pi t)|𝑑t={\left(\frac{2}{\pi }\right)}^2.$$

4 Sequences of powers

If $S_n={\sum }_{k=1}^n{a}_k$, then

$F_n^a(t)$ $=$ ${\displaystyle \prod _{k=1}^n}\sin (a_{k}t)$

$=$ ${\displaystyle \prod _{k=1}^n}{\displaystyle \frac{e^{i{a}_{k}t}-e^{-i{a}_{k}t}}{2i}}$

$=$ ${\displaystyle \prod _{k=1}^n}\left({\displaystyle \frac{-e^{-i{a}_{k}t}}{2i}}\right)\left(1-e^{2i{a}_{k}t}\right)$

$=$ ${\displaystyle \frac{1}{2^n}}\exp \left({\displaystyle \frac{in\pi }{2}}-i{S}_{n}t\right){\displaystyle \prod _{k=1}^n}\left(1-e^{2i{a}_{k}t}\right).$

Thus $|F_n^a(t)|=2^{-n}{\prod }_{k=1}^n\left|1-e^{2i{a}_{k}t}\right|$. Define $P_n^a(t)={\prod }_{k=1}^n\left(1-e^{i{a}_{k}t}\right)$. Bell, Borwein and Richmond [1] estimate ${\parallel P_n^a\parallel }_{\mathrm{\infty }}$ when $a_k$ is a power of $k$ or is quadratic in $k$. They prove that if $a_k=k^m$, with $m\ge 2$ an integer, then there exists a constant such that ${\parallel P_n^a\parallel }_{\mathrm{\infty }}>c^n$ for all sufficiently large $n$. The product $P_n^a(t)={\prod }_{k=1}^n(1-e^{i{a}_{k}t})$ can be written as a sum,

$$\prod _{k=1}^n(1-e^{i{a}_{k}t})=\sum _{k=0}^{S_n}{a}_k{e}^{ikt},a_k=\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{-ikt}{P}_n^a(t)𝑑t,$$

for $S_n={\sum }_{k=1}^n{a}_k$. We have

$$|P_n^a(t)|=\left|\sum _{k=0}^{S_n}{a}_k{e}^{ikt}\right|\le \sum _{k=0}^{S_n}|e^{ikt}{a}_k|=\sum _{k=0}^{S_n}|a_k|.$$

But

$$|a_k|\le \frac{1}{2\pi }{\int }_0^{2\pi }\left|e^{-ikt}{P}_n^a(t)\right|𝑑t=\frac{1}{2\pi }{\int }_0^{2\pi }|P_n^a(t)|𝑑t={\parallel P_n^a\parallel }_1.$$

Hence

$${\parallel P_n^a\parallel }_{\mathrm{\infty }}\le (S_n+1){\parallel P_n^a\parallel }_1.$$

It follows that

$${\parallel F_n^a\parallel }_1=\frac{1}{2^n}\cdot {\parallel P_n^a\parallel }_1\ge \frac{1}{2^n}\cdot \frac{{\parallel P_n^a\parallel }_{\mathrm{\infty }}}{S_n+1}>\frac{1}{S_n+1}\cdot {\left(\frac{c}{2}\right)}^n.$$

As $S_n={\sum }_{k=1}^n{k}^m=O(n^{m+1})$, the above lower bound on ${\parallel F_n^a\parallel }_1$ is better than the one given by Theorem 3.