The Subspace of Almost Convergent Sequences

Abstract

A bounded sequence of reals is called almost convergent if all Banach limits take the same value at it. We study the space of almost convergent sequences and operators related to this space.

⠀

Denote by $\ell_{\infty}$ the space of bounded sequences $x=(x_{1},x_{2},\dots)$ with the norm $\|x\|_{\ell_{\infty}}=\sup_{n\in 𝕅}|x_{n}|$, where $𝕅$ is the set of naturals with the usual semiorder. The linear functional $B\in\ell^{*}_{\infty}$ is called the Banach limit if

(1) $B\geq 0$, i.e., $Bx\geq 0$ for all $x\in\ell_{\infty}$, $x\geq 0$,

(2) $B{𝟙}=1$, where ${𝟙}=(1,1,\dots)$,

(3) $Bx=BTx$ for all $x\in\ell_{\infty}$, where $T$ is the shift operator; i.e., $T(x_{1},x_{2},\dots)=(x_{2},x_{3},\dots)$.

The existence of Banach limits was announced by Mazur in [1] and the proof was presented by Banach in [2]. The definition directly yields

$$ \liminf\limits_{n\to\infty}x_{n}\leq Bx\leq\limsup\limits_{n\to\infty}x_{n} $$

for all $x\in\ell_{\infty}$ and $B\in{\mathfrak{B}}$, where ${\mathfrak{B}}$ is the set of Banach limits. This means that every $B\in{\mathfrak{B}}$ is a norm-preserving extension of the functional $x\to\lim\nolimits_{n\to\infty}x_{n}$ from the subspace of convergent sequences to the whole $\ell_{\infty}$. According to Lorentz [3], a sequence $x\in\ell_{\infty}$ almost converges to $\lambda\in 𝕉^{1}$ if $Bx=\lambda$ for all $B\in{\mathfrak{B}}$. The set of such sequences is denoted by $ac_{\lambda}$; while the set of all almost convergent sequences, by $ac$. In this case we write $\operatorname{Lim}x_{n}=\lambda$. For example, if $x_{i+kj}=x_{i}$ for all $1\leq i<j$ and $k\in 𝕅$, then

$$ \operatorname{Lim}x_{n}=\frac{1}{j}\sum\limits_{i=1}^{j}x_{i}. $$

(1)

In particular, $\operatorname{Lim}(-1)^{n}=0$. It was proven in [3] that $x\in ac_{\lambda}$ if and only if

$$ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=m+1}^{m+n}x_{k}=\lambda $$

uniformly in $m\in 𝕅$. Sucheston in [4] refined the Lorentz Theorem by showing that

$$ q(x)\leq Bx\leq p(x) $$

for all $x\in\ell_{\infty}$ and $B\in{\mathfrak{B}}$, where

$$ q(x)=\lim\limits_{n\to\infty}\inf_{m\in 𝕅}\frac{1}{n}\sum\limits_{k=m+1}^{m+n}x_{k},\quad p(x)=\lim\limits_{n\to\infty}\sup_{m\in 𝕅}\frac{1}{n}\sum\limits_{k=m+1}^{m+n}x_{k}. $$

It is clear that ${\mathfrak{B}}$ is a closed convex subset of the unit sphere of $\ell^{*}_{\infty}$; therefore, by the Krein–Milman Theorem, ${\mathfrak{B}}=\overline{\operatorname{conv}}\operatorname{ext}{\mathfrak{B}}$, where $\operatorname{ext}{\mathfrak{B}}$ is the set of extreme points of ${\mathfrak{B}}$ and the closure of the convex hull is taken in the weak${}^{*}$ topology.

The set $ac$ is a nonseparable and uncomplemented closed subspace of $\ell_{\infty}$ [5] which was studied in [6,7,8] and other articles.

Given $x\in\ell_{\infty}$, we have

$$ \inf_{y\in ac}\|x-y\|_{\ell_{\infty}}=\frac{1}{2}(p(x)-q(x)). $$

This article is devoted to the further study of $ac$.

Theorem 1

1. If $m$ is odd, then

$$ \operatorname{Lim}\sin^{m}nt=0\quad\text{for }t\in[-\pi,\pi]. $$

2. If $m$ is even, then

$$ \operatorname{Lim}\sin^{m}nt=\begin{cases}0,\quad t=0,\pm\pi,&\\ {\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-1}}\sum\limits_{j\in Q_{m}}(-1)^{(m-2j)/2}\cdot C^{j}_{m}},&\end{cases} $$

where

$$ Q_{m}=\bigg{\{}j:0\leq j<\frac{m}{2},\ \frac{(m-2j)|t|}{2\pi}\in 𝕅\bigg{\}}. $$

Proof

If $t=0,\pm\pi$, then the equality $\operatorname{Lim}\sin^{m}nt=0$ for all $m\in 𝕅$ is obvious.

Let $r\in 𝕅$, $t\neq 0,\pm\pi$, and let $m$ be odd. Then

$$ \begin{gathered}\displaystyle\operatorname{Lim}\sin^{m}nt=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=r+1}^{r+n}\sin^{m}kt\\ \displaystyle=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=r+1}^{r+n}\biggl{[}\frac{1}{2^{m-1}}\sum\limits_{j=0}^{(m-1)/2}(-1)^{(m-2j-1)/2}\cdot C^{j}_{m}\cdot\sin(m-2j)kt\biggr{]}\\ \displaystyle=\frac{1}{2^{m-1}}\cdot\sum\limits_{j=0}^{(m-1)/2}(-1)^{(m-2j-1)/2}\cdot C^{j}_{m}\cdot\biggl{[}\lim\limits_{n\to\infty}\frac{1}{n}\cdot\sum\limits_{k=r+1}^{r+n}\sin(m-2j)kt\biggr{]}\\ \displaystyle=\frac{1}{2^{m-1}}\sum\limits_{j=0}^{(m-1)/2}(-1)^{(m-2j-1)/2}\cdot C^{j}_{m}\cdot\operatorname{Lim}\sin(m-2j)nt.\end{gathered} $$

Since $\operatorname{Lim}\sin(m-2j)nt=0$ for every $t\in 𝕉^{1}$; therefore, $\operatorname{Lim}\sin^{m}nt=0$.

Let $r\in 𝕅$, $t\neq 0,\pm\pi$, and let $m$ be even. Then

$$ \begin{gathered}\displaystyle\operatorname{Lim}\sin^{m}nt=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=r+1}^{r+n}\sin^{m}kt\\ \displaystyle=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=r+1}^{r+n}\biggl{[}\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-1}}\sum\limits_{j=0}^{(m-2)/2}(-1)^{(m-2j)/2}\cdot C^{j}_{m}\cdot\cos(m-2j)kt\biggr{]}\\ \displaystyle=\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-2}}\sum\limits_{j=0}^{(m-2)/2}(-1)^{(m-2j)/2}\cdot C^{j}_{m}\cdot\biggl{[}\lim\limits_{n\to\infty}\frac{1}{2n}\cdot\sum\limits_{k=r+1}^{r+n}\cos(m-2j)kt\biggr{]}\\ \displaystyle=\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-2}}\cdot\sum\limits_{j=0}^{(m-2)/2}(-1)^{(m-2j)/2}\cdot C^{j}_{m}\cdot\biggl{[}\frac{1}{2}-\lim\limits_{n\to\infty}\frac{1}{n}\cdot\sum\limits_{k=r+1}^{r+n}\sin^{2}\frac{(m-2j)kt}{2}\biggr{]}\\ \displaystyle=\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-2}}\sum\limits_{j=0}^{(m-2)/2}(-1)^{(m-2j)/2}\cdot C^{j}_{m}\cdot\left[\frac{1}{2}-\operatorname{Lim}\sin^{2}\frac{(m-2j)nt}{2}\right],\end{gathered} $$

where

$$ \operatorname{Lim}\sin^{2}\frac{(m-2j)nt}{2}=\begin{cases}0&\text{if }\frac{(m-2j)|t|}{2\pi}\in 𝕅,\\ \frac{1}{2}&\text{for the remaining }t\in[-\pi,\pi].\end{cases} $$

Introduce the set

$$ Q_{m}=\bigg{\{}j:0\leq j<\frac{m}{2},\ \frac{(m-2j)|t|}{2\pi}\in 𝕅\bigg{\}}. $$

Then

$$ \operatorname{Lim}\sin^{m}nt=\frac{C^{m/2}_{m}}{2^{m}}+\frac{1}{2^{m-1}}\sum\limits_{j\in Q_{m}}(-1)^{(m-2j)/2}\cdot C^{j}_{m}.\quad\text{☐} $$

Theorem 2

If $f$ is a continuous function on $[-1,1]$, then $f(\sin nt)\in ac$ for all $t\in 𝕉^{1}$.

Proof

By the Weierstrass Theorem, for every continuous $f$ on $[-1,1]$ and for every $\varepsilon>0$ there exists a polynomial $\sum\nolimits_{m=0}^{k}a_{m}t^{m}$ such that

$$ \biggl{|}f(t)-\sum\limits_{m=0}^{k}a_{m}t^{m}\biggr{|}\leq\varepsilon $$

for all $t\in[-1,1]$. Hence,

$$ \bigg{|}f(\sin nt)-\sum\limits_{m=0}^{k}a_{m}\sin^{m}nt\bigg{|}\leq\varepsilon $$

(2)

for all $t\in[-\pi,\pi]$. By Theorem 1,

$$ \sum\limits_{m=0}^{k}a_{m}\sin^{m}nt\in ac $$

(3)

for every $t\in[-\pi,\pi]$. From (2), (3), and the closedness of $ac$ in $\ell_{\infty}$ it follows that $f(\sin nt)\in ac$ for every $t\in[-\pi,\pi]$.  ☐

By Theorem 2, the formula $Af(t)=\operatorname{Lim}f(\sin nt)$ defines some linear operator from $C[-1,1]$ into the space of periodic functions with period $2\pi$.

Theorem 3

The operator $A$ acts from $C[-1,1]$ to $L_{\infty}[-\pi,\pi]$ and the norm of $A$ equals $1$.

Proof

Let $\varepsilon>0$, let $P_{\varepsilon}$ be a polynomial on $[-1,1]$ such that $\|f-P_{\varepsilon}\|_{C}<\varepsilon$, and $B\in{\mathfrak{B}}$. Then

$$ |Af(t)-AP_{\varepsilon}(t)|=|B(f-P_{\varepsilon})(t)|\leq\|f-P_{\varepsilon}\|_{C}<\varepsilon $$

for every $t\in[-\pi,\pi]$. By Theorem 1, $AP_{\varepsilon}(t)$ is a measurable function; therefore, $Af(t)$ is a measurable function as the uniform limit of a sequence of measurable functions.

The operator $A$ is positive and the norm of $A$ is therefore attained at $f(t)=1$. Hence, $\|A\|=1$.  ☐

Theorem 3 is exact in the sense that $A$ does not act from $C[-1,1]$ to $C[-\pi,\pi]$. Indeed, if $f_{0}(t)=|t|$, then

$$ Af_{0}(t)=\begin{cases}\frac{1}{j}\sum\limits_{k=1}^{j}\sin\frac{k}{j}\pi&\text{if }t=\frac{i}{j}\pi,\ i,j\in 𝕅,\ i<j,\\ \frac{2}{\pi}&\text{if }\frac{t}{\pi}\text{ is irrational},\\ 0&\text{if }t=k\pi,\ k\in 𝕅\end{cases} $$

(see [9, 10]). The function $Af_{0}(t)$ is discontinuous at each point $t\in[-\pi,\pi]$ where $t/\pi$ is rational. For the power functions $f_{m}(t)=t^{m}$ we obtain some more exact result.

Assertion 1 of Theorem 1 can be strengthened as follows:

Theorem 4

If $f$ is an odd continuous function on $[-1,1]$, then $\operatorname{Lim}f(\sin nt)=0$ for all $t\in[-\pi,\pi]$.

It is well known that every odd continuous function on $[-1,1]$ can be approximated however closely by polynomials of odd degrees. This and Theorem 1 imply Theorem 4.

Theorem 4 states that the kernel of $A$ contains all odd continuous functions. It turns out that the kernel of $A$ also contains some even functions. Indeed, if

$$ f_{0}(t)=\begin{cases}0,&t=0,\\ t^{2}-\frac{1}{2},&t\neq 0,\end{cases} $$

then

$$ f_{0}(\sin nt)=\sin^{2}nt-\frac{1}{2}=\frac{\sin^{2}nt-\cos^{2}nt}{2}=-\frac{\cos 2nt}{2}\in ac_{0} $$

for all $t\neq 0,\pm\pi$ and $f_{0}(\sin nt)\in ac_{0}$ for $t=0,\pm\pi$.

The asymptotic behavior of the sequence $\operatorname{Lim}\sin^{m}nt$ is described by

Theorem 5

If $-\pi\leq t\leq\pi$, then

$$ \lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}nt=\begin{cases}\frac{1}{j}&\text{if }t=\frac{i}{j}\pi,\ |i|,j\in 𝕅,\ |i|<j\text{ and }j\text{ is even,}\\ 0&\text{for the remaining }t\in[-\pi,\pi].\end{cases} $$

Proof

Let $t=\frac{i}{j}\pi$. Then $\sin nt$ is a periodic sequence with period $2j$. If $j$ is odd, then $\bigl{|}\sin n\frac{i}{j}\pi\bigr{|}<1$ for all $1\leq n\leq 2j$. Hence,

$$ \lim\limits_{m\to\infty}\sin^{m}n\frac{i}{j}\pi=0. $$

Using (1), we obtain

$$ \lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}n\frac{i}{j}\pi=0. $$

If $j$ is even, then

$$ \sin n\frac{i}{j}\pi=1\quad\text{for }n=\frac{j}{2},\frac{3j}{2}, $$

$$ \bigg{|}\sin n\frac{i}{j}\pi\bigg{|}<1\quad\text{for }n\neq\frac{j}{2},\frac{3j}{2}. $$

Applying (1) again, we get

$$ \lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}n\frac{i}{j}\pi=\frac{1}{j}. $$

Consider the case when $t/\pi$ is irrational. In this case $\{nt\ (\operatorname{mod}2\pi)$, $n\in 𝕅\}$ is a dense uniformly distributed subset of $[0,2\pi)$. Let $0<\varepsilon<\pi/2$. Consider

$$ Q=Q_{\varepsilon,t}=\bigg{\{}n:n\in 𝕅,\ \bigg{|}nt-\bigg{(}\frac{\pi}{2}+k\pi\bigg{)}\bigg{|}>\varepsilon\ \text{for all }(k\in 𝕅)\bigg{\}} $$

and the complement $\complement Q$ of $Q$. Denote the density of $\complement Q$ by $q$. By construction,

$$ q=\frac{4\varepsilon}{2\pi}=\frac{2\varepsilon}{\pi}. $$

(4)

If $n\in Q$, then

$$ 1-\sin nt\geq 1-\sin\bigg{(}\frac{\pi}{2}-\varepsilon\bigg{)}=1-\cos\varepsilon=2\sin^{2}\frac{\varepsilon}{2}\geq\frac{\varepsilon^{2}}{5},\quad\sin nt\leq 1-\frac{\varepsilon^{2}}{5}. $$

Hence,

$$ \sin^{m}nt\leq\bigg{(}1-\frac{\varepsilon^{2}}{5}\bigg{)}^{m}\leq\bigg{(}1-\frac{\varepsilon^{2}}{5}\bigg{)}^{1/\varepsilon^{3}} $$

(5)

for $m>1/\varepsilon^{3}$.

Estimate the sequence $\sin^{m}nt$ on $\complement Q$. To this end, consider the Cesàro operator

$$ (Cx)_{j}=\frac{1}{j}\sum\limits_{i=1}^{j}x_{i}. $$

It is well known that $C$ acts in $\ell_{\infty}$ and the norm of $Q$ equals 1. There exists $B\in{\mathfrak{B}}$ such that $B$ is invariant under $C$, i.e., $Bx=BCx$ for all $x\in\ell_{\infty}$ [8]. The set of such Banach limits is denoted by ${\mathfrak{B}}(C)$. Put

$$ (\chi_{\complement Q})_{k}=\begin{cases}1,&k\in\complement Q,\\ 0,&k\in Q,\end{cases}\quad k\in 𝕅. $$

Then $\lim\nolimits_{j\to\infty}(C\chi_{\complement Q})_{j}$ exists and is equal to the density of $\complement Q$. From this and (4) we obtain

$$ \begin{gathered}\displaystyle\operatorname{Lim}\sin^{m}nt\chi_{\complement Q}=B\sin^{m}nt\chi_{\complement Q}=BC\sin^{m}nt\chi_{\complement Q}\\ \displaystyle\leq\limsup\limits_{j\to\infty}(C\sin^{m}\chi_{\complement Q})_{j}\leq\limsup\limits_{j\to\infty}(\chi_{\complement Q})_{j}=q=\frac{2\varepsilon}{\pi},\end{gathered} $$

where $B\in{\mathfrak{B}}(C)$. From the above estimate and (5) it follows that

$$ \begin{gathered}\displaystyle\lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}nt\leq\lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}nt\chi_{Q}+\lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}nt\chi_{\complement Q}\\ \displaystyle\leq\bigg{(}1-\frac{\varepsilon^{2}}{5}\bigg{)}^{1/\varepsilon^{3}}+\frac{2\varepsilon}{\pi}.\end{gathered} $$

By arbitrariness of $\varepsilon>0$,

$$ \lim\limits_{m\to\infty}\operatorname{Lim}\sin^{m}nt=0.\quad\text{☐} $$

The authors are grateful to S. S. Kutateladze, A. S. Usachev, and N. N. Avdeev for their valuable remarks.