Average Mahler’s measure and $L_p$ norms of Littlewood polynomials

Abstract

Littlewood polynomials are polynomials with each of their coefficients in the set $\{-1,1\}$. We compute asymptotic formulas for the arithmetic mean values of the Mahler’s measure and the $L_p$ norms of Littlewood polynomials of degree $n-1$. We show that the arithmetic means of the Mahler’s measure and the $L_p$ norms of Littlewood polynomials of degree $n-1$ are asymptotically $e^{-\gamma /2}\sqrt {n}$ and $\Gamma (1+p/2)^{1/p}\sqrt {n}$, respectively, as $n$ grows large. Here $\gamma$ is Euler’s constant. We also compute asymptotic formulas for the power means $M_{\alpha }$ of the $L_p$ norms of Littlewood polynomials of degree $n-1$ for any $p > 0$ and $\alpha > 0$. We are able to compute asymptotic formulas for the geometric means of the Mahler’s measure of the “truncated” Littlewood polynomials $\hat{f}$ defined by $\hat{f}(z) := \min \{|f(z)|,1/n\}$ associated with Littlewood polynomials $f$ of degree $n-1$. These “truncated” Littlewood polynomials have the same limiting distribution functions as the Littlewood polynomials. Analogous results for the unimodular polynomials, i.e., with complex coefficients of modulus $1$, were proved before. Our results for Littlewood polynomials were expected for a long time but looked beyond reach, as a result of Fielding known for means of unimodular polynomials was not available for means of Littlewood polynomials.

1. Introduction and main results

The Mahler’s measure $M(f)$ of a polynomial $f$ with complex coefficients is defined by

$$\begin{equation*} \log M(f) = \frac{1}{2\pi }\int _0^{2\pi } \log \left\lvert f(e^{it})\right\rvert dt. \end{equation*}$$

It is well known that if the polynomial $f$ is of the form

$$\begin{equation*} f(z)=a_{m}\prod _{k=1}^{m}{(z-\alpha _k)}, \end{equation*}$$

then

$$\begin{equation*} M(f) = |a_m|\prod _{k=1}^{m-1} \max \{1,|\alpha _k|\} \end{equation*}$$

and

$$\begin{equation*} M(f) = \lim _{p\to 0^+} \left\lVert f\right\rVert _p, \end{equation*}$$

where $\left\lVert f\right\rVert _p$ is defined by

$$\begin{equation*} \left\lVert f\right\rVert _p = \left(\frac{1}{2\pi }\int _0^{2\pi } \left\lvert f(e^{it})\right\rvert ^p dt\right)^{1/p}, \qquad p > 0. \end{equation*}$$

A polynomial $f$ is called Littlewood polynomial if each of its coefficients is in the set $\{-1,1\}$. A polynomial $f$ is called unimodular if each of its coefficients is a complex number of modulus $1$. Let $\mathfrak{L}_n$ denote the set of all Littlewood polynomials of degree $n-1$. Let $\mathfrak{U}_n$ denote the set of all unimodular polynomials of degree $n-1$. Note that Parseval’s formula implies $\|f\|_2 = \sqrt {n}$ for all $f \in {\mathfrak{U}}_n$. We also introduce the set ${\mathfrak{L}}:=\bigcup _{n=1}^\infty {\mathfrak{L}}_n$ of all Littlewood polynomials and the set ${\mathfrak{U}}:= \bigcup _{n=1}^\infty {\mathfrak{U}}_n$ of all unimodular polynomials.

Littlewood posed a number of problems regarding the behavior of Littlewood and unimodular polynomials on the unit circle; see, for instance, 11, Problem 19. He asked if there exist absolute constants $c_1 > 0$ and $c_2 > 0$ and a sequence $(f_n)$ of polynomials $f_n\in \mathfrak{U}_n$ (or perhaps $f_n\in \mathfrak{L}_n$) such that

$$c_1\sqrt {n} \leq |f_n(z)| \leq c_2\sqrt {n}, \qquad z \in {\mathbb{C}}, \quad |z|=1.$$

Kahane 9 proved that there is such a sequence $(f_n)$ of polynomials $f_n\in \mathfrak{U}_n$, showing in fact that for every $\varepsilon > 0$ there is a sequence $(f_n)$ of polynomials $f_n\in \mathfrak{U}_n$ such that

$$(1-\varepsilon )\sqrt {n} \leq |f_n(z)| \leq (1+\varepsilon ) \sqrt {n}, \qquad z \in {\mathbb{C}}, \quad |z|=1,$$

for all sufficiently large $n$. Whether or not there is such a sequence $(f_n)$ of polynomials $f_n\in \mathfrak{L}_n$ is still open. The Rudin-Shapiro polynomials of degree $n = 2^k-1$ satisfy the upper bound with $c_2 = \sqrt 2$ and no sequence $(f_n)$ of polynomials $f_n\in \mathfrak{L}_n$ is known that satisfies just a lower bound with an absolute constant $c_1 > 0$. Erdős conjectured that there is an absolute constant $\varepsilon > 0$ such that the maximum modulus of any Littlewood polynomial $f_n\in \mathfrak{L}_n$ on the unit circle is at least $(1 + \varepsilon )\sqrt {n}$.

Problems regarding the existence of unimodular or Littlewood polynomials with certain flatness properties on the unit circle also arise in the context of the Mahler’s measure. In 1963, Mahler 12 proved that the maximum value of the Mahler’s measure of a polynomial of degree at most $m$ having all its coefficients in the closed unit disk is attained by a unimodular polynomial. Mahler posed the problem of determining the mean value of the Mahler’s measure of these polynomials to Fielding, who proved in 1970 that

$$\begin{equation} \lim _{n\to \infty } \sup _{f\in \mathfrak{U}_n} \frac{M(f)}{\sqrt {n}} = 1 {\tag{1.1}} \end{equation}$$

and

$$\begin{equation} \underset{f\in \mathfrak{U}_n}{\operatorname {mean}} \, \frac{M(f)}{\sqrt {n}} \geq e^{-\gamma /2}\left(1+O\left(n^{-1/4+\varepsilon }\right)\right), {\tag{1.2}} \end{equation}$$

where $\gamma$ denotes Euler’s constant and $\varepsilon$ is an arbitrarily small positive constant. See 8. Here

$$\underset{f \in \mathfrak{U}_n}{\operatorname {mean}} \left( F(f) \right) := \frac{1}{(2\pi )^n}\int _0^{2\pi }\int _0^{2\pi } \cdots \int _{0}^{2\pi } F(f) \,d\theta _0 \, d\theta _1 \,\cdots \, d\theta _{n-1},$$

where

$$f(z)=\sum _{j=0}^{n-1}{e^{i\theta _j}z^j}\,.$$

In 4, the authors strengthen (1.2) by proving that the limiting value of both the geometric and the arithmetic means of the normalized Mahler’s measure of unimodular polynomials $f \in \mathfrak{U}_n$ is exactly $e^{-\gamma /2}= 0.749306\ldots .$

Theorem 1.1.

Let $\mathfrak{U}_n$ denote the set of unimodular polynomials of degree $n-1$. Then

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{U}_n}{\operatorname {mean}} \, \log \left(\frac{M(f)}{\sqrt {n}}\right) = -\gamma /2 \end{equation*}$$

and

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{U}_n}{\operatorname {mean}} \, \log \left(\frac{\left\lVert f\right\rVert _p}{\sqrt {n}}\right) = \frac{1}{p}\log \Gamma (1+p/2), \qquad p > 0. \end{equation*}$$

Also

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{U}_n}{\operatorname {mean}} \, \frac{M(f)}{\sqrt {n}} = e^{-\gamma /2} \end{equation*}$$

and

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{U}_n}{\operatorname {mean}} \, \frac{\left\lVert f\right\rVert _p^p}{\sqrt {n}} = \Gamma (1+p/2), \qquad p > 0. \end{equation*}$$

In this paper first we determine the limiting values of the arithmetic means of the normalized Mahler’s measure and $L_p$ norms of Littlewood polynomials $f\in \mathfrak{L}_n$. The values we obtain are the same as those found in 4 in the unimodular case. To do so we will employ a result of Konyagin and Schlag (Lemma 2.4 below) in the place of Fielding’s result about the unimodular polynomials. We prove the following theorem in the next section. For any $f\in {\mathfrak{L}}_n$, we define ${\hat{f}}(z):=\max \{ |f(z)|, n^{-1} \}$.

Theorem 1.2.

We have

$$\begin{equation} \lim _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left(\frac{M({\hat{f}})}{\sqrt {n}}\right)\right) = e^{-\gamma /2} {\tag{1.3}} \end{equation}$$

and

$$\begin{equation} \lim _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \frac{M(f)}{\sqrt {n}} = e^{-\gamma /2}, {\tag{1.4}} \end{equation}$$

where $\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\left( \cdot \right):=\frac{1}{2^n}\sum _{f\in {\mathfrak{L}}_n}\left( \cdot \right)$. We also have

$$\begin{equation} \lim _{n\to \infty } \exp \left( \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \log \left( \frac{\| {\hat{f}}\|_p}{\sqrt {n}}\right) \right)=\Gamma (1+p/2)^{1/p}, \qquad p > 0, {\tag{1.5}} \end{equation}$$

and

$$\begin{equation} \lim _{n\to \infty } \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \frac{\left\lVert f\right\rVert _p}{\sqrt {n}} = \Gamma (1+p/2)^{1/p}, \qquad p > 0. {\tag{1.6}} \end{equation}$$

Equality 1.6 is proved in Theorem 1 of 2, while 1.3, 1.4 and 1.5 are new results. Hopefully, the results 1.3 and 1.5 will shed some light on how to compute the limiting value of the geometric means of the normalized Mahler’s measure of Littlewood polynomials $f \in \mathfrak{L}_n$ as well in the future.

Although we cannot offer an asymptotic formula for the geometric means of the Mahler’s measures of Littlewood polynomials $f \in \mathfrak{L}_n$, we can prove the following result.

Corollary 1.3.

We have

$$\begin{equation*} e^{-\gamma /2} \ge \limsup _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M(f)}{\sqrt {n}}\right) \right). \end{equation*}$$

For every fixed $\varepsilon > 0$, we also have

$$\begin{equation*} \liminf _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M(f)}{n^{1/2-\varepsilon }}\right) \right) \ge e^{-\gamma /2}. \end{equation*}$$

The problem of determining whether there exists a positive number $\varepsilon$ such that $M(f)/\left\lVert f\right\rVert _2<1-\varepsilon$ for every Littlewood polynomial of positive degree is commonly known as Mahler’s problem (see for instance 1). The largest known value of $M(f)/\left\lVert f\right\rVert _2$ for a Littlewood polynomial of positive degree is $0.98636\ldots ,$ achieved by $f(x)=x^{12}+x^{11}+x^{10}+x^9+x^8-x^7-x^6+x^5+x^4-x^3+x^2-x+1$. The best known asymptotic result in Mahler’s problem is due to Erdélyi and Lubinsky 6, who recently proved that the Fekete polynomials, defined by $f_p(x)=\sum _{k=1}^{p-1} \left(\frac{k}{p}\right) x^{k-1}$, where $p$ is a prime number and $\left(\frac{\cdot }{p}\right)$ denotes the usual Legendre symbol, satisfy

$$\begin{equation} \frac{M(f_p)}{\left\lVert f_p\right\rVert _2} > \frac{1}{2}-\varepsilon {\tag{1.7}} \end{equation}$$

for arbitrarily small $\varepsilon >0$ when $p$ is sufficiently large. Recently, in 3, we constructed Littlewood polynomials $f_n$ of degree $n$ satisfying 1.7 for all positive integers $n$, not only for prime numbers. The Rudin-Shapiro polynomials $P_n$ are also known to satisfy $M(P_n) > c\|P_n\|_2$ for every $n$ with an absolute constant $c>0$; see 5.

An analogue of Fielding’s result (1.2) for Littlewood polynomials as well as an improved lower bound in Mahler’s problem follows immediately.

Corollary 1.4.

For every $\varepsilon >0$ there exist infinitely many Littlewood polynomials $f$ satisfying

$$\begin{equation*} \frac{M(f)}{\left\lVert f\right\rVert _2} > e^{-\gamma /2} - \varepsilon , \end{equation*}$$

and infinitely many Littlewood polynomials $f$ satisfying

$$\begin{equation*} \frac{M(f)}{\left\lVert f\right\rVert _2} < e^{-\gamma /2} + \varepsilon . \end{equation*}$$

Since the arithmetic means and geometric means of the normalized Mahler’s measure of ${\hat{f}}$ associated with Littlewood polynomials $f \in \mathfrak{L}_n$ converge to the same limiting value as $n \to \infty$, we can deduce that the values $M({\hat{f}})$ are close to each other for almost all $f\in \mathfrak{L}_n$, and hence almost all of the values $M({\hat{f}})$ are close to the mean. More precisely we can prove the following result.

Theorem 1.5.

The normalized Mahler’s measure $M({\hat{f}})/\sqrt {n}$ converges to $e^{-\gamma /2}$ in probability. That is,

$$\begin{equation} \lim _{n\to \infty } \frac{1}{2^n}\left| \left\{ f \in \mathfrak{L}_n : \left| \frac{M({\hat{f}})}{\sqrt {n}} - e^{-\gamma /2} \right| > \varepsilon \right\} \right| =0 {\tag{1.8}} \end{equation}$$

for every $\varepsilon >0$, where $|A|$ denotes the number of the elements in the set $A$. Also, the normalized $L_p$ norms $\|f\|_p/\sqrt {n}$ converge to $\Gamma (1+p/2)^{1/p}$ in probability. That is,

$$\begin{equation} \lim _{n\to \infty } \frac{1}{2^n}\left|\left\{ f \in \mathfrak{L}_n : \left| \frac{\| f\|_p}{\sqrt {n}} - \Gamma (1+p/2)^{1/p} \right| > \varepsilon \right\} \right| =0 {\tag{1.9}} \end{equation}$$

for every $\varepsilon > 0$ and $p > 0$.

Result 1.9 is proved in Theorem 1 of 2, while 1.8 is a new result.

We may obtain more information on the distribution of the Mahler’s measure and the $L_p$ norms by studying their moments. We have the following result.

Theorem 1.6.

We have

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \left(\frac{\left\lVert f\right\rVert _p}{\sqrt {n}}\right)^\alpha =\left( \lim _{n\to \infty } \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \frac{\left\lVert f\right\rVert _p}{\sqrt {n}}\right)^\alpha \end{equation*}$$

and

$$\begin{equation*} \lim _{n\to \infty } \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \left(\frac{M({\hat{f}})}{\sqrt {n}}\right)^\alpha =\left( \lim _{n\to \infty } \underset{f\in \mathfrak{L}_n}{\operatorname {mean}} \, \frac{M({\hat{f}})}{\sqrt {n}}\right)^\alpha \end{equation*}$$

for every $\alpha >0$ and $p>0$.

We think that similar techniques permit the extension of our main results to the derivatives of Littlewood polynomials.

2. Proofs of the main results

For $f \in \mathfrak{L}_n$ of the form

$$f(z):=\sum _{j=0}^{n-1}{a_jz^j}, \qquad z:=e^{i\theta }, \quad \theta \in [0,2\pi ),$$

we define

$$\begin{equation*} U_n:=U_n(f, \theta ) := \frac{|f(z)|^2}{n} = \frac{1}{n} \left| \sum _{j=0}^{n-1}a_je^{ij\theta } \right|^2. \end{equation*}$$

For a fixed $\theta \in [0,2\pi )$, P. Borwein and Lockhart 2 studied the distribution of $U_n$ as a random variable. Let

$$\begin{equation*} \sigma _{n,c}^2 (\theta ) :=\sum _{j=0}^{n-1} \cos ^2(j\theta ) \;\; \text{ and } \;\; \sigma _{n,s}^2 (\theta ) :=\sum _{j=0}^{n-1} \sin ^2(j\theta ) \end{equation*}$$

and write

$$\begin{equation*} a_{k,n}(\theta ):=\frac{\cos (k\theta )}{\sigma _{n,c}(\theta )} \end{equation*}$$

and, for $\theta$ not an integer multiple of $\pi$,

$$\begin{equation*} b_{k,n}(\theta ):=\frac{\sin (k\theta )}{\sigma _{n,s}(\theta )}. \end{equation*}$$

We also define the random variables

$$\begin{equation*} C_n(\theta ):=\frac{\sum _{j=0}^{n-1}a_j\cos (j\theta )}{\sigma _{n,c}(\theta )} \;\; \text{ and } \;\; S_n(\theta ):=\frac{\sum _{j=0}^{n-1}a_j\sin (j\theta )}{\sigma _{n,s}(\theta )}. \end{equation*}$$

Note that

$$\begin{equation*} U_n(f,\theta ) = \frac{|f(z)|^2}{n} = \frac{\sigma _{n,c}^2(\theta )C_n^2(\theta )+\sigma _{n,s}^2(\theta )S_n^2(\theta )}{n}. \end{equation*}$$

Let $Z, Z_1, Z_2$ be the standard normal distributions, and let ${\mathbb{E}}(X)$ be the expected value of the random variable $X$. We need the following lemma, which records some facts observed in the proof of Theorem 1 of 2.

Lemma 2.1.

For any fixed $\theta$ not an integer multiple of $\pi$, we have

(i)

$(C_n(\theta ) , S_n(\theta ))$ converges in distribution to $(Z_1, Z_2)$ as $n\to \infty$;

(ii)

$|f(e^{i\theta })|^2/n$ converges in distribution to $(Z_1^2+Z_2^2)/2$ as $n\to \infty$;

(iii)

${\mathbb{E}}(|C_n(\theta )|^p)$ and  ${\mathbb{E}}(|S_n(\theta )|^p)$ converge uniformly on $(0,\pi )\cup (\pi ,2\pi )$ to${\mathbb{E}}(|Z|^p)$ as $n\to \infty$.

Proof of Lemma 2.1.

See pages 1466 and 1469 of 2.

■

For a fixed $n \in {\mathbb{N}}$ and and a fixed $z=e^{i\theta } \in {\mathbb{C}}$, let

$$\begin{equation*} F_n^z(x):= \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : \frac{|f(z)|^2}{n} \le x \right\} \right| \end{equation*}$$

be the distribution function of $U_n$, and let $F(x)=1-e^{-x}$ be the distribution function of the standard exponential function on $[0,\infty )$. The following is a Berry-Esseen type central limit theorem for $F_n^z(x)$. Let $\| y \|:=\min _{m\in {\mathbb{Z}}} |2\pi m -y|$.

Proposition 2.2.

Let $0< A< 1$ and $z:=e^{i\theta }$ with $\| 2\theta \| \ge 1/n^A$. We have

$$\begin{equation} F_n^z(x) = F(x)+O(n^{-1/2}+n^{-(1-A)}) {\tag{2.10}} \end{equation}$$

for any $0 \le x < \infty$, where the implicit constant in the $O$ symbol is absolute, independent of $\theta$ and $x$.

Proof of Proposition 2.2.

For a fixed $\theta \in [0,2\pi )$, let $g_{n,c}(t)$ and $g_{n,s}(t)$ be the probability density functions of $C_n(\theta )$ and $S_n(\theta )$ and let $G_{n,c}(t)$ and $G_{n,s}(t)$ be the corresponding distribution functions, respectively. In view of (iii) of Lemma 2.1, we have ${\mathbb{E}}(|C_n(\theta )|^p)$ and ${\mathbb{E}}(|S_n(\theta )|^p)$ converge uniformly on $(0,\pi )\cup (\pi ,2\pi )$ to

$${\mathbb{E}}(|Z|^p) =\int _{-\infty }^{\infty } |x|^p d\Phi (x)=\frac{2^{p/2}\Gamma \left( (p+1)/2 \right)}{\sqrt {\pi }},$$

where

$$\Phi (x) =(1/\sqrt {2\pi })\int _{-\infty }^x {e^{-t^2/2} \,dt}.$$

By (i) of Lemma 2.1, both $G_{n,c}(t)$ and $G_{n,s}(t)$ converge to the standard normal distributions. So using the Berry-Esseen’s central limit theorem (see Theorem 1 on p. 542 of 7) with either $X_n = C_n(\theta )$ or $X_n = S_n(\theta )$, where $\theta \not \in \{0,\pi \}$, we have

$$\begin{equation*} \left| G_{n,c}(x)-\Phi (x)\right| \le c_3 \frac{ \rho _n}{\sigma _n^3 \sqrt {n}} \quad \text{and} \quad \left| G_{n,s}(x)-\Phi (x)\right| \le c_3 \frac{ \rho _n}{\sigma _n^3 \sqrt {n}}, \end{equation*}$$

where $c_3 > 0$ is an absolute constant,

$$\rho _n := {\mathbb{E}}(|X_n|^3), \qquad \lim _{n \rightarrow \infty } {\rho _n} = \frac{2^{3/2} \, \Gamma (2)}{\sqrt {\pi }} = 2\sqrt {\frac{2}{\pi }},$$

and

$$\sigma _n^2 := {\mathbb{E}}(|X_n|^2), \qquad \lim _{n \rightarrow \infty } {\sigma _n^2} = \frac{2 \, \Gamma (3/2)}{\sqrt {\pi }} = 1.$$

It follows that

$$\begin{equation*} G_{n,c}(x) = \Phi (x) + O(n^{-1/2})\;\; \text{and}\;\; G_{n,s}(x) = \Phi (x) + O(n^{-1/2}) \end{equation*}$$

for all $x$ in $(-\infty , \infty )$. Note that

$$\begin{eqnarray*} \frac{\sigma _{n,c}^2}{n} & = & \frac{1}{n} \sum _{j=0}^{n-1}\cos ^2 (j\theta ) = \frac{1}{2} + \frac{1}{2n} \sum _{j=0}^{n-1}\cos (2j\theta ) = \frac{1}{2} + \frac{1}{2n}\Re \left( \sum _{j=0}^{n-1}e^{2ij\theta }\right) \\ & = & \frac{1}{2} + O\left( \frac{1}{n\| 2 \theta \|} \right) = \frac{1}{2} + O\left( n^{-(1-A)} \right) \end{eqnarray*}$$

for any $\theta \in [0,2\pi )$ with $\| 2\theta \| \ge 1/n^A$, where the implicit constant in the $O$ symbol is absolute. Similarly, we have

$$\begin{equation*} \frac{\sigma _{n,c}^2}{n} = \frac{1}{2} + O\left( n^{-(1-A)} \right) \end{equation*}$$

for any $\theta \in [0,2\pi )$ with $\| 2\theta \| \ge 1/n^A$, where the implicit constant in the $O$ symbol is absolute. Therefore, for any $\theta \in [0,2\pi )$ with $\| 2\theta \| \ge 1/n^A$ and for any $x \in (-\infty , \infty )$, we have

$$\begin{eqnarray*} & & \left| \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : \frac{(\sigma _{n,c}C_n)(\theta )}{\sqrt {n}} \leq x \right\}\right| - \Phi (x) \right| \\ & \le & \left| \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : \frac{(\sigma _{n,c}C_n)(\theta )}{\sqrt {n}} \leq x \right\}\right| - \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : C_n(\theta ) \leq x \right\}\right| \right| \\ & & \qquad + \left|\frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : C_n(\theta ) \leq x\right\}\right| - \Phi (x) \right| \\ & \le & \left| \Phi \left( x\left(\frac{1}{2}+\frac{C}{n^{1-A}}\right)^{-1/2}\right) - \Phi \left( x\left(\frac{1}{2}-\frac{C}{n^{1-A}}\right)^{-1/2}\right) \right| + \left| G_{n,c}(x)-\Phi (x) \right| \\ & \ll & \frac{x}{n^{1-A}}e^{-c_4x^2} + n^{-1/2} \ll n^{-1/2}+n^{-(1-A)} \end{eqnarray*}$$

with some absolute constants $C>0$ and $c_4>0$, where the implicit constant in the $\ll$ symbol is also absolute. Similarly, we have

$$\begin{equation*} \left| \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : \frac{(\sigma _{n,s}S_n)(\theta )}{\sqrt {n}} \leq x \right\}\right| - \Phi (x) \right| \ll n^{-1/2}+n^{-(1-A)}. \end{equation*}$$

We recall that

$$\begin{equation*} \frac{|f(z)|^2}{n}=\frac{\sigma _{n,c}^2(\theta )C_n(\theta )^2}{n} + \frac{\sigma _{n,s}^2(\theta )S_n(\theta )^2 }{n},\qquad z=e^{i\theta }. \end{equation*}$$

Therefore, we have

$$\begin{equation*} F_n^z(x)=F(x) + O(n^{-1/2}+n^{-(1-A)}). \end{equation*}$$ ■

For any $z\in {\mathbb{C}}$ and $|z|=1$ we define the corresponding distribution function

$$\begin{equation*} G_n^z(t):=\frac{1}{2^n}\left| \left\{ f \in {\mathfrak{L}}_n : \frac{|{\hat{f}}(z)|^2}{n} \le t\right\}\right|. \end{equation*}$$

Observe that $|{\hat{f}}(z)|^2/n \geq n^{-3}$; hence $G_n^z(t)=0$ for $t < n^{-3}$. Observe also that $|f(z)|<n^{-1}$ and $n^{-3} \le t$ imply $|f(z)|^2/n < n^{-3} \le t$, and hence $\chi _{[0,t)}\left( \frac{|f(z)|^2}{n}\right) =1$. Hence, if $t \ge n^{-3}$, then we have

$$\begin{eqnarray*} G_n^z(t) & = & F_n^z(t). \end{eqnarray*}$$

Therefore

$$\begin{equation} G_n^z(t)= \begin{cases} F_n^z(t) & \text{ if $t \ge n^{-3},$} \\ 0 & \text{ if $t < n^{-3}.$} \end{cases} {\tag{2.11}} \end{equation}$$

Theorem 2.3.

Suppose $z=e^{i\theta }$ with $\theta \in [0,2\pi )$ and $\| 2\theta \| > 1/n^A$ for some $0< A < 1$. There is an absolute constant $c_5 > 0$ such that

$$\begin{equation} \left| \int _{0^+}^\infty \log t \; d \left( G_n^z(t)-F(t)\right) \right| \le c_5 (n^{-1/2}+n^{-(1-A)}) \log n {\tag{2.12}} \end{equation}$$

for every $n \in {\mathbb{N}}$. Hence

$$\begin{equation} \left| \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right) + \gamma \right| \le c_5 (n^{-1/2}+n^{-(1-A)}) \log n {\tag{2.13}} \end{equation}$$

for every $n \in {\mathbb{N}}$. In particular,

$$\begin{equation} \lim _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right)= - \gamma , {\tag{2.14}} \end{equation}$$

where the convergence is uniform for $\theta \in [0,2\pi )$ such that $\| 2\theta \| > 1/n^A$.

Proof of Theorem 2.3.

We divide the interval of integration into three subintervals $[0,n^{-3}], [n^{-3},n], [n, \infty )$ and estimate on them separately.

First we observe that by 2.11

$$\begin{eqnarray*} & & \left| \int _n^\infty \log t \; d \left( G_n^z(t)-F(t)\right) \right| \\ & = & \left| \int _n^\infty \log t \; d \left( F_n^z(t)-F(t)\right) \right| \\ & = & \left| \left[ \left( F_n^z(t)-F(t)\right) \log t \right]_n^\infty - \int _n^\infty \frac{F_n^z(t)-F(t)}{t}\; dt \right| \\ & \le & \left| \left[ \left( e^{-t} \right) \log t \right]_n^\infty \right| + \int _n^\infty {\frac{e^{-t}}{t}\; dt}, \end{eqnarray*}$$

since $F_n^z(t)=1$ for $t\ge n$ and hence $F_n^z(t)-F(t) =1 - (1-e^{-t})=e^{-t}$ for $t\ge n$. Therefore,

$$\begin{eqnarray} & & \left| \int _n^\infty \log t \; d \left( G_n^z(t)-F(t)\right) \right| \le \frac{\log n}{e^n}+ \frac{1}{n} \int _n^\infty e^{-t} \; dt \ll \frac{\log n}{e^n}, {\tag{2.15}} \end{eqnarray}$$

where the implicit constant in the $\ll$ symbol is absolute.

Now we estimate on the interval $[n^{-3},n]$. Using 2.11 and 2.10, we obtain

$$\begin{align} & \left| \int _{1/n^3}^n \log t \; d \left( G_n^z(t)-F(t)\right) \right| = \left| \int _{1/n^3}^n \log t \; d \left( F_n^z(t)-F(t)\right) \right|{\tag{2.16}}\\ & = \left| \left[ \left( F_n^z(t)-F(t)\right) \log t \right]_{1/n^3}^n - \int _{1/n^3}^n \frac{F_n^z(t)-F(t)}{t}\; dt \right| \\ & \ll \left| F_n^z(n)-F(n)\right| \log n + \left| F_n^z(n^{-3})-F(n^{-3})\right| \log n +\int _{1/n^3}^n \frac{|F_n^z(t)-F(t)|}{t}\; dt \\ & \ll (n^{-1/2}+n^{-(1-A)})\log n + (n^{-1/2}+n^{-(1-A)}) \int _{1/n^3}^n\frac{1}{t} \; dt \\ & \ll (n^{-1/2}+n^{-(1-A)})\log n. \end{align}$$

Finally we estimate on $[0,n^{-3}]$. We have

$$\begin{equation*} \left| \int _{0^+}^{1/n^3} \log t \; d \left( G_n^z(t)-F(t)\right) \right| = \left| \left[ -F(t) \log t \right]_{0^+}^{1/n^3} + \int _{0^+}^{1/n^3} \frac{F(t)}{t}\; dt \right| \end{equation*}$$

as $G_n^z(t)\equiv 0$ when $t< 1/n^3$. Observe that $\lim _{t \to 0^+}F(t)\log t =0,$ so using 2.10 again we obtain

$$\begin{eqnarray*} & & \left| \int _{0^+}^{1/n^3} \log t \; d \left(-F(t)\right) \right| \ll (n^{-1/2}+n^{-(1-A)})\log n + \int _{0^+}^{1/n^3} {\frac{F(t)}{t}\; dt}. \end{eqnarray*}$$

Estimating the second term on the right hand side gives

$$\begin{equation*} \int _{0^+}^{1/n^3} \frac{F(t)}{t}\; dt = \int _{0^+}^{1/n^3} \frac{1-e^{-t}}{t}\; dt \le \int _{0^+}^{1/n^3} 1\; dt = \frac{1}{n^3} , \end{equation*}$$

as $1-e^{-t} \le t$ for $t\ge 0$. Hence

$$\begin{equation} \left| \int _{0^+}^{1/n^3} \log t \; d \left( G_n^z(t)-F(t)\right) \right| \ll (n^{-1/2}+n^{-(1-A)})\log n. {\tag{2.17}} \end{equation}$$

Now, 2.12 follows from 2.15, 2.16 and 2.17.

■

In the case of unimodular polynomials in 4 a result of Fielding in 8 helped the authors to succeed, but an analogue of Fielding’s result is not available in the case of Littlewood polynomials. However, to prove 1.4 of our Theorem 1.2 we can use a recent result of Konyagin and Schlag in 10. From Theorem 1.2 of 10 (with $\phi \equiv 1$), for any $\varepsilon >0$, we have

$$\begin{equation*} \limsup _{n\to \infty } \frac{1}{2^n} \left| \left\{ f \in {\mathfrak{L}}_n : \min \left\{ \left| f(z) \right|: z\in {\mathbb{C}}\text{ and } ||z|-1|<\varepsilon n^{-2}\right\} < \varepsilon n^{-1/2}\right\}\right| \le c_6 \varepsilon \end{equation*}$$

with some absolute constant $c_6>0$. Hence we have

Lemma 2.4.

For $n \in {\mathbb{N}}$ and $\varepsilon > 0$ we define

$$\begin{equation*} A_n(\varepsilon ) = \bigg \{ f\in {\mathfrak{L}}_n : \min _{|z|=1}|f(z)| < \varepsilon n^{-1/2} \bigg \}. \end{equation*}$$

There is an absolute constant $c_6 > 0$ such that

$$\begin{equation} \limsup _{n\to \infty } \frac{1}{2^n}|A_n(\varepsilon )| \le c_6 \varepsilon {\tag{2.18}} \end{equation}$$

for every $\varepsilon > 0$.

Proof of Theorem 1.2.

Integrating 2.13 over the set

$$H_n := \{ \theta \in [0,2\pi ): \, \| 2\theta \| \ge n^{-1/2} \}$$

and using Theorem 2.3 with $A=1/2$, we obtain

$$\begin{equation*} \lim _{n \to \infty } {\frac{1}{2\pi } \int _{H_n} {\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right) \; d\theta }} = -\gamma . \end{equation*}$$

Observe that

$$-3 \log n = \log (n^{-3}) \leq \log (|{\hat{f}}(z)|^2/n) \leq \log n, \quad z = e^{i\theta }, \quad \theta \in [0,2\pi ).$$

Hence

$$\begin{equation*} -3\log n \leq \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right) \leq \log n, \quad \theta \in [0,2\pi ), \end{equation*}$$

and as the measure of the set $[0,2\pi ) \setminus H_n$ is $2n^{-1/2}$, we have

$$\begin{equation*} \frac{-6\log n}{\sqrt {n}} \leq \int _{[0,2\pi ) \setminus H_n}{\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} {\, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right) \, d\theta }} \leq \frac{2\log n}{\sqrt {n}}. \end{equation*}$$

This yields

$$\begin{equation*} \lim _{n \to \infty } {\frac{1}{2\pi } \int _{[0,2\pi ) \setminus H_n} {\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{|{\hat{f}}(z)|^2}{n} \right) \; d\theta }} = 0, \end{equation*}$$

which finishes the proof of 1.3.

To prove 1.4, let $\varepsilon >0$ be fixed and suppose $n \ge \varepsilon ^{-2}$. We first observe that if $f \in {\mathfrak{L}}_n \setminus A_n(\varepsilon )$, then $\min _{|z|=1}|f(z)| \ge \varepsilon n^{-1/2}$. Hence $|f(z)| \ge \varepsilon n^{-1/2} \ge n^{-1}$ whenever $|z|=1$. It then follows that ${\hat{f}}(z)=\max \{ |f(z)|, n^{-1} \} = |f(z)|$ for all $z \in {\mathbb{C}}$ with $|z|=1$. Therefore, if $f \in {\mathfrak{L}}_n \setminus A_n(\varepsilon )$, then ${\hat{f}}(z)\equiv |f(z)|$ for $z \in {\mathbb{C}}$ with $|z|=1$. Thus

$$\begin{eqnarray} \underset{f \in {\mathfrak{L}}_n}{\operatorname {mean}}\frac{M(f)}{\sqrt {n}} & = & \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\frac{M({\hat{f}})}{\sqrt {n}}+ \frac{1}{2^n} \sum _{f\in A_n(\varepsilon )} \frac{M(f)- M({\hat{f}})}{\sqrt {n}}. {\tag{2.19}} \end{eqnarray}$$

Combining 2.18 with the inequalities

$$\begin{equation} 1 \leq M(f) \leq M({\hat{f}}) \leq \|{\hat{f}}\|_2 \leq \|f\|_2 + \|1/n\|_2 = n^{1/2} + n^{-1/2} \leq 2n^{1/2} {\tag{2.20}} \end{equation}$$

valid for all Littlewood polynomial $f \in \mathfrak{L}_n$, we obtain

$$\begin{equation} \limsup _{n\to \infty } \left| \frac{1}{2^n} \sum _{f\in A_n(\varepsilon )} \frac{M(f)- M({\hat{f}})}{\sqrt {n}} \right| \le 2\limsup _{n\to \infty }\frac{1}{2^n}|A_n(\varepsilon )|\le 2c_6\varepsilon . {\tag{2.21}} \end{equation}$$

Using the inequality between the geometric and arithmetic means (or Jensen’s inequality) and 1.3 we obtain

$$\begin{equation} \liminf _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \frac{M({\hat{f}})}{\sqrt {n}} \ge \liminf _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \log \left( \frac{M({\hat{f}})}{\sqrt {n}} \right) \right) = e^{-\gamma /2}. {\tag{2.22}} \end{equation}$$

Observe that for $p\in (0,1)$, we have

$$\begin{equation*} 1 \leq (M(f))^p \leq \|f\|_p^p \leq \|{\hat{f}}\|_p^p \le \||f| + 1/n\|_p^p \leq \|f\|_p^p + n^{-p}, \end{equation*}$$

and this gives $1 \leq \|{\hat{f}}\|_p/\| f\|_p \leq (1 + n^{-p} )^{1/p} \le (1 + n^{-p}+n^{-1} )^{1+1/p}$. For $p > 1$,

$$\begin{equation*} 1 \leq M(f) \leq \|f\|_p \leq \|{\hat{f}}\|_p \le \||f| + 1/n\|_p \leq \|f\|_p + n^{-1}, \end{equation*}$$

and this gives $1 \leq \|{\hat{f}}\|_p/\| f\|_p \leq 1 + n^{-1} \le (1 + n^{-p}+n^{-1} )^{1+1/p}$. Hence, we have

$$\begin{equation} 1 \leq \|{\hat{f}}\|_p/\| f\|_p \le (1 + n^{-p}+n^{-1} )^{1+1/p} {\tag{2.23}} \end{equation}$$

hold for all Littlewood polynomials $f \in {\mathfrak{L}}_n$. In 2, it was shown that

$$\begin{equation*} \lim _{n\to \infty }{\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}{\frac{\| f\|_p}{\sqrt {n}}}}= \Gamma (1+p/2)^{1/p}. \end{equation*}$$

Combining this with 2.17, we also have

$$\begin{equation*} \lim _{n\to \infty }{\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}{\frac{\|{\hat{f}}\|_p}{\sqrt {n}}}} = \Gamma (1+p/2)^{1/p}. \end{equation*}$$

Therefore,

$$\begin{equation*} \limsup _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \frac{M({\hat{f}})}{\sqrt {n}} \le \limsup _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \frac{\|{\hat{f}}\|_p}{\sqrt {n}} = \Gamma (1+p/2)^{1/p}. \end{equation*}$$

As this holds for all $p>0$, and

$$\begin{equation*} \lim _{p\to 0^+} \Gamma (1+p/2)^{1/p} = \exp \left( \frac{1}{2} \frac{\Gamma '(1)}{\Gamma (1)}\right) = e^{-\gamma /2}, \end{equation*}$$

we have

$$\begin{equation} \limsup _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \frac{M({\hat{f}})}{\sqrt {n}} \leq e^{-\gamma /2}. {\tag{2.24}} \end{equation}$$

Combining 2.22 and 2.24 gives

$$\begin{equation} \lim _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \frac{M({\hat{f}})}{\sqrt {n}} = e^{-\gamma /2}. {\tag{2.25}} \end{equation}$$

Combining 2.19, 2.21, and 2.25 gives 1.4.

Formula 1.6 has already been proved as Theorem 1 in 2.

To prove 1.5 we first recall that in 2 it has been proved that $\|f\|_p/\sqrt {n}$ converges to $\Gamma (1+p/2)^{1/p}$ in probability. Hence for any $\varepsilon >0$, we have

$$\begin{equation*} \lim _{n\to \infty } \frac{1}{2^n}\left| \left\{ f \in {\mathfrak{L}}_n : \left| \frac{\| f \|_p}{\sqrt {n}} -\Gamma (1+p/2)^{1/p} \right| >\varepsilon \right\}\right| =0. \end{equation*}$$

Let $h_p := \Gamma (1+p/2)^{1/p},\, p > 0$, and $h_0 := e^{-\gamma /2}$. Let $0 < \varepsilon < h_0/20$. We define

$$B_{n,\varepsilon } = \left\{f \in {\mathfrak{L}}_n: \left| \left\| \frac{{\hat{f}}}{\sqrt {n}} \right\|_p - h_p \right| < \varepsilon , \hskip .5em\left| \frac{M({\hat{f}})}{\sqrt {n}} - h_0 \right| < \varepsilon \right\}.{}$$

Using the convergence in probability stated in Theorem 1.5 (that we prove later in this paper) and 2.23, we get

$$\begin{equation} 1-\varepsilon < m_{n,\varepsilon } : = \frac{|B_{n,\varepsilon }|}{2^n} \leq 1 {\tag{2.26}} \end{equation}$$

for all sufficiently large $n$. Therefore

$$\begin{eqnarray} & & \bigg ( \prod _{f \in {\mathfrak{L}}_n}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \\ & = & \bigg ( \prod _{f \in B_{n,\varepsilon }}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \bigg ( \prod _{f \in {\mathfrak{L}}_n \setminus B_{n,\varepsilon }} {\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \\ & \geq & (h_p - \varepsilon )^{m_{n,\varepsilon }} \bigg ( \prod _{f \in {\mathfrak{L}}_n \setminus B_{n,\varepsilon }} {\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \\ & \geq & \min \{ h_p - \varepsilon , (h_p - \varepsilon )^{1-\varepsilon } \} \bigg ( \prod _{f \in {\mathfrak{L}}_n \setminus B_{n,\varepsilon }} {\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \\ & \geq & \min \{ h_p - \varepsilon , (h_p - \varepsilon )^{1-\varepsilon } \} \bigg ( \prod _{f \in {\mathfrak{L}}_n \setminus B_{n,\varepsilon }} {(M({\hat{f}})/\sqrt {n})} \bigg )^{1/2^n}. {\tag{2.27}} \end{eqnarray}$$

Now we can use 2.26 and the convergence of the geometric means of the Mahler’s measure of ${\hat{f}}$ of Littlewood polynomials to $h_0$ to estimate the second factor in this lower bound as

$$\begin{eqnarray} \bigg ( \prod _{f \in {\mathfrak{L}}_n \setminus B_{n,\varepsilon }} {(M({\hat{f}})/\sqrt {n})} \bigg )^{1/2^n} & = & \frac{\bigg ( \prod _{f \in {\mathfrak{L}}_n} {(M({\hat{f}})/\sqrt {n})} \bigg )^{1/2^n}}{\bigg ( \prod _{f \in B_{n,\varepsilon }} {(M({\hat{f}})/\sqrt {n})} \bigg )^{1/2^n}} \\ & \geq & \frac{h_0-\varepsilon }{(h_0+\varepsilon )^{m_{n,\varepsilon }}} \geq \frac{h_0-\varepsilon }{(h_0+\varepsilon )^{1-\varepsilon }} {\tag{2.28}} \end{eqnarray}$$

for all sufficiently large $n$. Combining 2.27 and 2.28 we obtain

$$\begin{equation} \bigg ( \prod _{f \in {\mathfrak{L}}_n}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} \geq \min \{ h_p - \varepsilon , (h_p - \varepsilon )^{1-\varepsilon } \}\frac{h_0-\varepsilon }{(h_0+\varepsilon )^{1-\varepsilon }} {\tag{2.29}} \end{equation}$$

for all sufficiently large $n$. On the other hand, using the inequality between the geometric and arithmetic means and 2.23, and then recalling 1.6, we obtain

$$\begin{eqnarray} \lim _{n\to \infty } \bigg ( \prod _{f \in {\mathfrak{L}}_n}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n} & \leq & \lim _{n\to \infty } \frac{1}{2^n} \bigg ( \sum _{f \in {\mathfrak{L}}_n}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg ) \\ & = & \lim _{n\to \infty } \frac{1}{2^n} \bigg ( \sum _{f \in {\mathfrak{L}}_n}{\|f/\sqrt {n}\|_p} \bigg ) \\ & \leq & h_p + \varepsilon . {\tag{2.30}} \end{eqnarray}$$

Combining 2.29 and 2.30 and letting $n \to \infty$ and then $\varepsilon \to 0^+$, we get

$$\lim _{n \rightarrow \infty }{\bigg ( \prod _{f \in {\mathfrak{L}}_n}{\|{\hat{f}}/\sqrt {n}\|_p} \bigg )^{1/2^n}} = \lim _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left(\frac{\|{\hat{f}}\|_p}{\sqrt {n}}\right) \right) =h_p.$$

This proves 1.5 and completes the proof of Theorem 1.2.

■

Proof of Corollary 1.3.

Let

$$\begin{equation*} A_n(\varepsilon ) = \left\{ f\in {\mathfrak{L}}_n : \min _{|z|=1}|f(z)| < \varepsilon n^{-1/2} \right\} \end{equation*}$$

be the same as in Lemma 2.4. Observe that if $f \in {\mathfrak{L}}_n \setminus A_n(\varepsilon )$ and $n \geq \varepsilon ^{-2}$, then $|f(z)| = {\hat{f}}(z)$ for all $z \in {\mathbb{C}}$ with $|z| = 1$. Hence,

$$\begin{eqnarray*} \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M(f)}{\sqrt {n}}\right) & = & \frac{1}{2^n}\sum _{f \in {\mathfrak{L}}_n \setminus A_n(\varepsilon )}\log \left( \frac{M(f)}{\sqrt {n}}\right) + \frac{1}{2^n}\sum _{f\in A_n(\varepsilon )}\log \left( \frac{M(f)}{\sqrt {n}}\right) \\ & = & \frac{1}{2^n}\sum _{f \in {\mathfrak{L}}_n \setminus A_n(\varepsilon )}\log \left( \frac{M({\hat{f}})}{\sqrt {n}}\right) + \frac{1}{2^n}\sum _{f\in A_n(\varepsilon )}\log \left( \frac{M(f)}{\sqrt {n}}\right) \\ & = & \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M({\hat{f}})}{\sqrt {n}}\right) + \frac{1}{2^n}\sum _{f\in A_n(\varepsilon )}\log \left( \frac{M(f)}{M({\hat{f}})}\right). \\ \end{eqnarray*}$$

Recalling 2.20, we obtain

$$\begin{equation*} 1 \ge \exp \left( \frac{1}{2^n}\sum _{f\in A_n(\varepsilon )}\log \left( \frac{M(f)}{M({\hat{f}})}\right) \right) \ge \exp \left( -\frac{1}{2^n}|A_n(\varepsilon )|\log (2\sqrt {n}) \right). \end{equation*}$$

Combining this with 1.4 and Lemma 2.4, we obtain

$$\begin{equation*} e^{-\gamma /2} \ge \limsup _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M(f)}{\sqrt {n}}\right) \right) \end{equation*}$$

and

$$\begin{equation*} \liminf _{n\to \infty } \exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\log \left( \frac{M(f)}{n^{1/2-c_6\varepsilon }}\right) \right) \ge e^{-\gamma /2}. \end{equation*}$$ ■

Proof of Theorem 1.5.

Result 1.9 has been proved in Theorem 1 of 2.

To prove 1.8, we need a quantitative form of the inequality between the geometric and arithmetic means. Let $s_k$ be the arithmetic mean and $g_k$ be the geometric mean of the positive numbers $a_1, a_2, \ldots , a_k$. Let

$$m = m(\varepsilon ) = \left| \left\{j \in \{1,2,\ldots ,k\}: |a_j/s_k - 1| > \varepsilon \right\} \right|.$$

We claim that for every $0 < \varepsilon < 1$, we have

$$\left(\frac{e^{\varepsilon }}{1+\varepsilon } \right)^{m/k} \le s_k/g_k.$$

Indeed, we first assume $s_k=1$. Following Pólya’s proof of the inequality between the geometric and arithmetic means, we have $e^{a_j-1}/a_j \ge 1$ for all $j=1,2,\ldots ,k$. But when $|a_j-1|>\varepsilon$, we can gain a non-trivial factor $>1$. Consider the function $f(x):=e^{x-1}/x$ for $x>0$. The function $f(x)$ is decreasing on $(0,1]$ and increasing on $[1,\infty )$, and takes its absolute minimum value $1$ on $(0,\infty )$ at $1$. So if $|x-1|>\varepsilon$, then $x> 1+\varepsilon$ or $x< 1-\varepsilon$, and hence

$$\frac{e^{x-1}}{x} \ge \min \left\{ \frac{e^{\varepsilon }}{1+\varepsilon }, \frac{e^{-\varepsilon }}{1-\varepsilon }\right\} = \frac{e^{\varepsilon }}{1+\varepsilon } > 1.$$

Therefore $\sum _{j=1}^ka_j=s_kn=n$ implies

$$\begin{equation*} \frac{1}{a_1a_2\cdots a_k}=\frac{e^{\sum _{j=1}^ka_j -n}}{a_1a_2\cdots a_k}=\prod _{ j=1}^k \frac{e^{a_j-1}}{a_j} \ge \left(\frac{e^{\varepsilon }}{1+\varepsilon } \right)^m. \end{equation*}$$

By taking the $k$th root of both sides and recalling that $s_k=1$, we have

$$\begin{equation*} s_k/g_k \ge \left(\frac{e^{\varepsilon }}{1+\varepsilon } \right)^{m/k}. \end{equation*}$$

For the case that $s_k\neq 1$, we replace $a_j$ by $a_j' =a_j/s_k$ so that $s_k'=\frac{1}{k}\sum _{j=1}^k a_j/s_k =1$. Then from above, we have

$$\begin{equation*} \frac{s_k^k}{a_1a_2\cdots a_k}=\frac{1}{a_1'a_2'\cdots a_k'}=\prod _{ j=1}^k \frac{e^{a_j'-1}}{a_j'} \ge \left(\frac{e^{\varepsilon }}{1+\varepsilon } \right)^{m}. \end{equation*}$$

This also gives

$$\begin{equation*} s_k/g_k\ge \left(\frac{e^{\varepsilon }}{1+\varepsilon } \right)^{m/k}. \end{equation*}$$

This proves the claim.

Applying the claim to our case, we have $a_j=M({\hat{f}})/\sqrt {n}$ with $f \in \mathfrak{L}_n$, $k=2^n$, $s_k = \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \, \left( M({\hat{f}})/\sqrt {n}\right)$ and $g_k=\exp \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \,\log \left( M({\hat{f}})/\sqrt {n} \right)\right).$ Note that $\lim _{n\to \infty } \log (s_k/g_k) = 0$ from Theorem 1.2 and 2.25. Therefore, for any $\varepsilon >0$,

$$\begin{equation*} \lim _{n\to \infty } \frac{1}{2^n}\left| \left\{ f \in \mathfrak{L}_n : \left| \frac{M({\hat{f}})}{\sqrt {n}} - e^{-\gamma /2} \right| > \varepsilon \right\} \right| =0. \end{equation*}$$

This completes the proof.

■

Proof of Theorem 1.6.

Let $p>0$ and $\alpha >0$. Let $h_p := \Gamma (1+p/2)^{1/p}$ as before. In view of 1.9 in Theorem 1.5, $\| f \|_p/\sqrt {n}$ converges to $h_p$ in probability and hence $\left(\| f \|_p/\sqrt {n}\right)^\alpha$ also converges to $h_p^\alpha$ in probability. We let

$$A_{n,\varepsilon }(\alpha ) := \{ f \in {\mathfrak{L}}_n: \left| \left(\|f \|_p/\sqrt {n}\right)^{\alpha } - h_p^\alpha \right| < \varepsilon \}$$

so that for any $\varepsilon >0$, we have

$$\begin{equation*} \lim _{n\to \infty } \frac{|A_{n,\varepsilon }(\alpha )|}{2^n} =1. \end{equation*}$$

We now consider

$$\begin{align} &\left| \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha \right) - h_p^\alpha \right| \le \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left| \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha - h_p^\alpha \right| \right) \\ &\qquad = \frac{1}{2^n} \sum _{f \in A_{n,\varepsilon }(\alpha )} \left| \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha - h_p^\alpha \right| + \frac{1}{2^n} \sum _{f \in {\mathfrak{L}}_n \setminus A_{n,\varepsilon }(\alpha )} \left| \left( \frac{\|f\|_p}{\sqrt {n}}\right)^\alpha - h_p^\alpha \right|. {\tag{2.31}} \end{align}$$

The first term in 2.31 is

$$\frac{1}{2^n} \sum _{f\in A_{n,\varepsilon }(\alpha )} \left| \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha - h_p^\alpha \right| \le \varepsilon \frac{|A_{n,\varepsilon }(\alpha )|}{2^n} \to \varepsilon$$

as $n \to \infty$. In view of 2, we know that

$$\begin{equation} \lim _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\left( \frac{\| f \|_p}{\sqrt {n}}\right)^{2p} = \Gamma (1+p/2)^2, {\tag{2.32}} \end{equation}$$

and using a similar proof, one can show that

$$\lim _{n\to \infty } \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}}\left( \frac{\| f \|_p}{\sqrt {n}}\right)^{mp} = \Gamma (1+p/2)^m,$$

for any integer $m\ge 1$ by considering

$$\begin{equation*} g_n^{(*)}(\theta _1,\theta _2, \cdots , \theta _m) = \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left|\frac{f(e^{i\theta _1})}{\sqrt {n}} \, \frac{f(e^{i\theta _2})}{\sqrt {n}} \, \cdots \, \frac{f(e^{i\theta _m})}{\sqrt {n}}\right|^p\right) \end{equation*}$$

and showing that $g_n^{(*)}(\theta _1,\theta _2, \cdots , \theta _m)$ converges almost everywhere to $\Gamma (1+p/2)^m$ on $[0,2\pi )^m$ by using Lindeberg’s central limit theorem as in the proof of 2.32 in 2. Then for $p \ge 2$, by Hölder’s inequality, the second term in 2.31 is at most

$$\begin{eqnarray*} & & \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right)^{1-\frac{1}{p}} \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left(\frac{\| f\|_p}{\sqrt {n}} \right)^{p\alpha } \right)\right)^{\frac{1}{p}} + \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right) h_p^\alpha \\ & \le & \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right)^{1-\frac{1}{p}} \left( \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left( \frac{\| f\|_p}{\sqrt {n}} \right)^{p([\alpha ]+1)} \right)\right)^{\frac{1}{p}} + \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right) h_p^\alpha , \end{eqnarray*}$$

where the right hand side tends to $0$ as $n\to \infty$. For $0< p< 2$, then $\| f \|_p \le \| f \|_2 =\sqrt {n}$ for all $f\in {\mathfrak{L}}_n$. Hence the second term in 2.31 is also

$$\begin{equation*} \le \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right) + \left( 1 - \frac{|A_{n,\varepsilon }(\alpha )|}{2^n}\right) h_p^\alpha \end{equation*}$$

where the right hand side tends to $0$ as $n\to \infty$. Therefore, we have

$$\begin{equation*} \limsup _{n\to \infty } \left| \underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha \right) - h_p^\alpha \right| \le \varepsilon , \end{equation*}$$

and hence

$$\begin{equation*} \lim _{n\to \infty }\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \left( \frac{\| f\|_p}{\sqrt {n}}\right)^\alpha \right) = h_p^\alpha = \left( \lim _{n\to \infty }\underset{f\in {\mathfrak{L}}_n}{\operatorname {mean}} \left( \frac{\| f\|_p}{\sqrt {n}}\right) \right)^\alpha . \end{equation*}$$

The case of the Mahler’s measure can be proved in the same way. Now we can use 1.8 rather than 1.9 of Theorem 1.5 and observe that $M({\hat{f}})/\sqrt {n} \le \| {\hat{f}}\|_2/\sqrt {n} \le 1 + 1/n^{3/2}$ for all $f\in {\mathfrak{L}}_n$.

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