Invariant vector means and complementability of Banach spaces in their second duals

Abstract

Let X be a Banach space. Fix a torsion-free commutative and cancellative semigroup S whose torsion-free rank is the same as the density of $X^{**}$. We then show that X is complemented in $X^{**}$ if and only if there exists an invariant mean $M:\ell _\infty (S,X)\rightarrow X$. This improves upon previous results due to Bustos Domecq (J Math Anal Appl 275(2):512–520, 2002), Kania (J Math Anal Appl 445:797–802, 2017), Goucher and Kania (Studia Math 260:91–101, 2021).

1 Introduction

Invariant means on amenable groups are an important tool in many parts of Mathematics, especially in Harmonic analysis (see [8, 9]). For the basic properties of invariant means, we refer the reader to [8]. Invariant means and their generalizations for vector-valued functions also play an important role in the stability of functional equations and selections of set-valued functions (see [1, 5, 6, 16]).

The space of all bounded functions from a set S into a Banach space X is denoted by $\ell _\infty (S,X)$. Let us recall the definition of an amenable semigroup (see [3]).

Definition 1.1

A semigroup $(S,+)$ is called left [resp. right] amenable if and only if there exists a linear map $L:\ell _\infty (S,{\mathbb {R}}) \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned}&\inf f(S)\le L(f)\le \sup f(S),\ f\in \ell _\infty (S,{\mathbb {R}})\\&L(_a f) = L(f),\ a\in S,\ f\in \ell _\infty (S,{\mathbb {R}}),\\&[L(f_a) = L(f),\ a\in S,\ f\in \ell _\infty (S,{\mathbb {R}})], \end{aligned}$$

where

$$\begin{aligned}&_a f(x) = f(a+x),\ a,x\in S,\ f\in \ell _\infty (S,{\mathbb {R}}),\\&[f_a(x) = f(x+a),\ a,x\in S,\ f\in \ell _\infty (S,{\mathbb {R}})]. \end{aligned}$$

If both left and right invariant means exist, then S is called amenable.

Remark 1.2

In the above definition the first condition

$$\begin{aligned} \inf f(S)\le L(f)\le \sup f(S) \end{aligned}$$

is equivalent to conditions $L(\mathbbm {1}_S)=1$ and $|L(f)|\le \Vert f \Vert :=\sup |f(S)|$.

It is known that every commutative semigroup is amenable (an easy consequence of the Markov–Kakutani fixed point theorem, see [15, Theorem 5.23]).

Certain generalizations of invariant means were investigated for vector-valued functions in [5] and the existence thereof appears to be related to properties such as reflexivity.

Some generalized definition of an invariant mean has been used by many mathematicians as a folklore (e.g. by Pełczyński [14]). The explicit form of this definition can be found e.g. in the work of Ger [6].

Definition 1.3

Let $(S,+)$ be a left [right] amenable semigroup and X be a Banach space. A linear map ${M:\ell _{\infty } (S,X) \rightarrow X}$ is called a left [right] X-valued invariant mean if

$$\begin{aligned}&\Vert M \Vert \le 1,\\&M(c\mathbbm {1}_S)=c,\ c\in X,\\&M(_a f)=M(f),\ a\in S,f\in \ell _\infty (S,X),\\&[M(f_a)=M(f),\ a\in S,f\in \ell _\infty (S,X),] \end{aligned}$$

where

$$\begin{aligned}&{} _a f(x) = f(a+x),\ a,x\in S,\ f\in \ell _\infty (S,X),\\&\quad [f_a(x) = f(x+a),\ a,x\in S,\ f\in \ell _\infty (S,X).] \end{aligned}$$

If M is a left and right invariant mean, then M is called an X-valued invariant mean.

If in the above definition the norm of map M is equal to at most $\lambda \ge 1$, then M is called an X-valued invariant $\lambda$-mean.

The existence of such invariant means for a fixed Banach space and for all amenable semigroups has been studied by Domecq [4, Theorem 1 and 2] and by the author in [12]. However, as observed by Lipecki in his Mathematical Review (MR1943762) of Bustos Domecq’s paper, the proof of Theorem 2 contains a gap (a flawed choice of the semigroup, so we cannot use the Principle of Local Reflexivity). This gap was corrected by Kania [10].

Goucher and Kania [7] consider the following question (communicated privately to T. Kania by J.M.F. Castillo).

Suppose that a Banach space X admits an invariant mean with respect to every/some commutative group. Must X be complemented in $X^{**}$?

They proved the following (see [7, Theorem A], [10, Theorem 1.2])

Theorem 1.4

Let X be a Banach space and $\lambda \ge 1$. Then the following assertions are equivalent.

• 1. X is complemented in $X^{**}$ by a projection of norm at most $\lambda$;

• 2. for every amenable semigroup S there exists an X-valued invariant $\lambda$-mean on S;

• 3. for every commutative semigroup S there exists an X-valued invariant $\lambda$-mean on S;

• 4. for every free commutative group G of rank $\vert X^{**}\vert$ there exists an X-valued invariant $\lambda$-mean on G;

• 5. there exists an X-valued invariant $\lambda$-mean on the additive group of $X^{**}$.

It is also demonstrated ( [7, Remark 1.1]) that there exists a commutative noncancellative semigroup S (that could be chosen as large as one wishes) such that there exists an X-valued invariant mean on S.

In this paper we will prove that if X is a Banach space and there exists an invariant X-valued mean on any arbitrary commutative cancellative semigroup S of torsion-free rank ${\text {dens}}{X^{**}}$, then X is $\lambda$-complemented in $X^{**}$.

2 Preliminaries

First we recall the definition of torsion-free rank (see [2]).

Definition 2.1

Let S be a commutative cancellative semigroup. A set $A\subset S$ is independent if $\sum _{i=1}^n k_i a_i=\sum _{i=1}^n m_i a_i$ for any $n\in {\mathbb {N}}$ and $a_i\in A$, $k_i,m_i \in {\mathbb {N}}_0$, $i\in \{ 1,\ldots ,n \}$ implies $k_i=m_i$ for $i\in \{ 1,\ldots ,n \}$.

Let further ${\mathcal {A}}_0$ be the family of all independent sets L in S consisting only of elements whose order is infinite and such that L is maximal with respect to these properties. The cardinal number of any set in ${\mathcal {A}}_0$ is called a torsion-free rank of S and is denoted by $r_0 (S)$ (all the sets in ${\mathcal {A}}_0$ have the same cardinal number).

The density character of a Banach space X, denoted ${\text {dens}}X$, is the smallest cardinal $\kappa$ for which X has a dense subset of cardinality $\kappa$.

Lemma 2.2

Let X be an infinite-dimensional Banach space. Then

• 1. if ${\mathcal {B}}$ is a linearly independent subset of X and ${\mathbb {F}}$ is a countable dense subfield of a scalar field of X, then $\vert {\text {span}}_{{\mathbb {F}}}{\mathcal {B}}\vert =\vert {\mathcal {B}}\vert$;

• 2. for every closed subspace Y of X there exists a linearly independent subset ${\mathcal {B}}$ of X such that $\vert {\mathcal {B}}\vert ={\text {dens}}X$, $\overline{{\text {span}}}\,{\mathcal {B}}=X$, $\overline{{\text {span}}}\,({\mathcal {B}}\cap Y)=Y$. Moreover, we can assume that the norm of each $x\in {\mathcal {B}}$ is equal to 1.

Proof

• 1. We observe that

  $$\begin{aligned} \vert {\mathcal {B}}\vert&\le \vert {\text {span}}_{\mathbb {F}} {\mathcal {B}}\vert =\vert \bigcup _{n\in {\mathbb {N}}} ({\mathbb {F}}\cdot {\mathcal {B}})^n\vert \le \vert {\mathbb {N}}\vert \cdot \sup _{n\in {\mathbb {N}}} \vert ({\mathbb {F}}\cdot {\mathcal {B}})^n\vert \\&=\vert {\mathbb {N}}\vert \cdot \vert {\mathbb {F}}\cdot {\mathcal {B}}\vert =\vert {\mathbb {N}}\vert \cdot \vert {\mathbb {F}}\vert \cdot \vert {\mathcal {B}}\vert =\vert {\mathcal {B}}\vert . \end{aligned}$$

• 2. Let Y be a closed subspace of X, D be a dense subset of X such that $\vert D\vert ={\text {dens}}X$ and K be a dense subset of Y such that $\vert K\vert \le {\text {dens}}X$. Let further

  $$\begin{aligned}&D_1 := \Big \{ \frac{x}{\Vert x \Vert }:\, x\in K\setminus \{0\} \Big \},\\&\quad D_2 := \Big \{ \frac{x}{\Vert x \Vert }:\, x\in D\setminus \{0\} \Big \}. \end{aligned}$$

  Let further ${\mathcal {B}}_1$ be a maximal linearly independent subset of $D_1$ and ${\mathcal {B}}$ be a maximal linearly independent subset of $D_1 \cup D_2$ such that ${\mathcal {B}}_1 \subset {\mathcal {B}}$. We have $\overline{{\text {span}}}\,{\mathcal {B}}_1 =Y$, $\overline{{\text {span}}}\,{\mathcal {B}} =X$ and $\vert {\mathcal {B}}\vert \le \vert D_1\vert +\vert D_2\vert ={\text {dens}}X$. Let ${\mathbb {F}}={\mathbb {Q}}$ when X is a real space or ${\mathbb {F}}={\mathbb {Q}}(i)$ when X is a complex space. Since ${\text {span}}_{\mathbb {F}} {\mathcal {B}}$ is dense in X, $\vert {\mathcal {B}}\vert =\vert {\text {span}}_{\mathbb {F}} {\mathcal {B}}\vert \ge {\text {dens}}X$. We also note that the norm of each $x\in {\mathcal {B}}$ is equal 1.

$\square$

We will also require the version of the principle of local reflexivity due to Lindenstrauss and Rosenthal [11]. We denote by $\kappa :X\rightarrow X^{**}$ the canonical embedding from a Banach space X into the second dual.

Theorem 2.3

Let X be a Banach space. Then for every finite-dimensional subspace $F\subset X^{**}$ and each $\varepsilon \in (0, 1]$ there exists a linear map $P_F^\varepsilon :F\rightarrow \kappa (X)$ such that

• 1. $(1-\varepsilon )\Vert x \Vert \le \Vert P_F^\varepsilon (x) \Vert \le (1+\varepsilon )\Vert x \Vert$, $x\in F$;

• 2. $P_F^\varepsilon (x) =x$ for $x\in F\cap \kappa (X)$.

It is a standard fact that subgroups and quotients of amenable groups are amenable. Using exactly the same ideas one can prove that if a Banach space admits an invariant mean with respect to a group, then it also does so with respect to subgroups and quotients of the group (see [12, Theorem 3.12] and [7, Lemma 2.3]). We would like to get a similar result for quotients of semigroups (subsemigroups of an amenable group need not be amenable) but first we must say something about normal semigroups and quotients of semigroups (see also [17]). Let $(S,+)$ be a semigroup, G be a subsemigroup of S. Then G is called a normal subsemigroup if $x+G=G+x$ for every $x\in S$. Of course in a commutative semigroup each subsemigroup is normal.

Let further S be a semigroup and G be a normal subsemigroup of S. We define the quotient semigroup $S/G := S/\overset{G}{\sim }$, where $x\overset{G}{\sim } y$ iff $(x+G)\cap (y+G)\ne \emptyset$. It is easy to notice that for any $g\in G$ the set $[g]_{\overset{G}{\sim }}$ is a neutral element of S/G. Moreover, if G is a group, then G is a neutral element of S/G.

Lemma 2.4

Let S be an amenable semigroup and G be a normal subsemigroup of S. If there exists an X-valued invariant $\lambda$-mean $M:\ell _{\infty } (S,X)\rightarrow X$, then there exists an X-valued invariant $\lambda$-mean $M:\ell _{\infty } (S/G,X)\rightarrow X$.

Proof

We define a map $M_1 :\ell _{\infty } (S/G,X)\rightarrow X$ by the formula

$$\begin{aligned} M_1 (f) := M(\psi (f)),\ f\in \ell _{\infty } (S/G,X), \end{aligned}$$

where $\psi (f)(s)=f({[s]_{\overset{G}{\sim }}})$ for $s\in S$ and $f\in \ell _{\infty } (S/G,X)$. Since $\psi$ is linear, $\Vert \psi (f) \Vert =\Vert f \Vert$ and

$$\begin{aligned}&\psi (_{[t]_{\overset{G}{\sim }}} f)(s)=_{[t]_{\overset{G}{\sim }}} f({[s]_{\overset{G}{\sim }}}) = f({[t+s]_{\overset{G}{\sim }}})=\psi (f)(t+s)= (_t \psi (f))(s),\\&\quad \psi ( f_{[t]_{\overset{G}{\sim }}} )(s)= f_{[t]_{\overset{G}{\sim }}} ({[s]_{\overset{G}{\sim }}}) = f({[s+t]_{\overset{G}{\sim }}})=\psi (f)(s+t)= (\psi (f)_t )(s), \end{aligned}$$

for all $s,t\in S$, $f\in \ell _{\infty } (S/G,X)$, then $M_1$ is an X-valued invariant $\lambda$-mean on S/G. $\square$

3 Main results

Throughout this section we fix an infinite-dimensional Banach space X, $\lambda \ge 1$. Let $\gamma$ be a cardinal number. We denote by $S_\gamma$ the commutative semigroup comprising all finite subsets of $\gamma$ endowed with the operation of taking the union of sets. It is easy to observe that $\vert S_\gamma \vert =\gamma$.

Theorem 3.1

Let $\gamma$ be an infinite cardinal number. If there exists an X-valued invariant $\lambda$-mean $M:\ell _{\infty } (S_\gamma ,X)\rightarrow X$, then for every subspace E of $X^{**}$ such that ${\text {dens}}E=\gamma$ there exists a linear map $P:E\rightarrow X$ such that $\Vert P \Vert \le \lambda$ and $P(x)=x$ for $x\in \kappa (X)\cap E$.

Proof

Let ${\mathbb {K}}$ be a scalar field of X. In view of Lemma 2.2 there exists a linearly independent subset ${\mathcal {B}}$ of E such that $\overline{{\text {span}}}\,{\mathcal {B}}=E$, $\overline{{\text {span}}}\,({\mathcal {B}}\cap \kappa (X)) =\kappa (X)\cap E$, $\vert {\mathcal {B}}\vert ={\text {dens}}E=\gamma$. Let $T:\gamma \rightarrow {\mathcal {B}}$ be a bijection and $M:\ell _{\infty } (S_\gamma ,X)\rightarrow X$ be an X-valued invariant $\lambda$-mean .

For $A\in S_\gamma$ we define $\varepsilon _A :=\frac{1}{\vert A\vert +1}$ and $P_{{\text {span}}T(A) }^{\varepsilon _A}$ is a fixed linear operator satisfying the conditions of Theorem 2.3.

We define the map $P:E\rightarrow X$ in the following way (on the dense subspace ${\text {span}}{\mathcal {B}}$, the map is simply continuously extended to the closure): for $x\in {\text {span}}{\mathcal {B}}$ we put $P(x):=M(\phi _x)$, where

$$\begin{aligned} \phi _x (A):=\left\{ \begin{array}{ll} P_{{\text {span}}T(A) }^{\varepsilon _A} (x),&{}x\in {\text {span}}T(A)\\ 0,&{}x\notin {\text {span}}T(A) \end{array},\, A\in S_\gamma \right. \end{aligned}$$

when $x\in {\mathcal {B}}$ and

$$\begin{aligned} \phi _x (A):=\sum _{i=1}^{n} \lambda _i \phi _{x_i} (A),\, A\in S_\gamma , \end{aligned}$$

when $x=\sum \limits _{i=1}^n \lambda _i x_i$, $\lambda _1,\ldots ,\lambda _n \in {\mathbb {K}}$, $x_1,\ldots ,x_n\in {\mathcal {B}}$.

For $x,y\in {\text {span}}{\mathcal {B}}$ and $\alpha \in {\mathbb {K}}$ we notice that $\phi _{\alpha x+y}=\alpha \phi _x +\phi _y$. Thus

$$\begin{aligned}&P(\alpha x+y)=M(\phi _{\alpha x+y})=\alpha M(\phi _x) +M(\phi _y)=\alpha P(x)+P(y), \end{aligned}$$

so P is linear on ${\text {span}}{\mathcal {B}}$.

Let $x=\sum \limits _{i=1}^n \lambda _i x_i$ for some $\lambda _1,\ldots ,\lambda _n \in {\mathbb {K}}$, $x_1,\ldots ,x_n\in {\mathcal {B}}$. Let further $A_0\in S_\gamma$ be such that $x_1,\ldots x_n\in T(A_0)$.

We observe that

$$\begin{aligned} \Vert P(x) \Vert&=\Vert M(\phi _x) \Vert =\Vert M(\phi _x (\cdot \cup A_0 )) \Vert \le \lambda \sup _{A\in S_\gamma } \Vert \phi _x (A\cup A_0 ) \Vert \\&= \lambda \sup _{A\in S_\gamma } \Vert \sum _{i=1}^n \lambda _i \phi _{x_i} (A\cup A_0 ) \Vert =\lambda \sup _{A\in S_\gamma } \Vert \sum _{i=1}^n \lambda _i P_{{\text {span}}(A\cup A_0 )}^{\varepsilon _{A\cup A_0 }} (x_i) \Vert \\&= \lambda \sup _{A\in S_\gamma } \Vert P_{{\text {span}}T(A\cup A_0) }^{\varepsilon _{A\cup A_0}} \Big ( \sum _{i=1}^n \lambda _i x_i \Big ) \Vert = \lambda \sup _{A\in S_\gamma } \Vert P_{{\text {span}}T(A\cup A_0) }^{\varepsilon _{A\cup A_0}} (x) \Vert \\&\le \lambda \sup _{A\in S_\gamma } (1+\varepsilon _{A\cup A_0 }) \Vert x \Vert \le \lambda \Big ( 1+\frac{1}{1+\vert A_0\vert }\Big ) \Vert x \Vert . \end{aligned}$$

Since $A_0$ is arbitrary, we get $\Vert P(x) \Vert \le \lambda \Vert x \Vert$.

Moreover, if $x\in \kappa (X)$, then from the properties of ${\mathcal {B}}$ we get $x_1,\ldots ,x_n\in \kappa (X)$ and

$$\begin{aligned} \phi _x (A\cup A_0 )&=\sum _{i=1}^{n} \lambda _i \phi _{x_i} (A\cup A_0 )=\sum _{i=1}^{n} \lambda _i P_{{\text {span}}T(A\cup A_0) }^{\varepsilon _{A\cup A_0}} (x_i )\\&= \sum _{i=1}^{n} \lambda _i x_i =x,\ A\in S_\gamma . \end{aligned}$$

Hence

$$\begin{aligned}&P(x)=M(\phi _x )=M(\phi _x (\cdot \cup A_0 ))=x. \end{aligned}$$

$\square$

Theorem 3.2

Let S be a commutative cancellative semigroup of torsion-free rank $\delta$, $\gamma = \max (\delta ,\omega )$. If there exists an X-valued invariant $\lambda$-mean $M_S:\ell _{\infty } (S,X)\rightarrow X$, then there exists an X-valued invariant $\lambda$-mean $M:\ell _{\infty } (S_{\gamma },X)\rightarrow X$.

Proof

First we observe that we can assume that S contains only elements of infinite order. Indeed the set G of all elements of finite order is a group and a torsion-free rank of S/G is equal to $\gamma$. In view of Lemma 2.4 there exists an X-valued invariant $\lambda$-mean on S/G.

Let $A\subset S$ be a maximal linearly independent set. Hence $\vert A\vert =\delta$.

• First assume that $\vert A\vert =\gamma$ and let $A=\{ x_\alpha :\; \alpha <\gamma \}$. For each $x\in S$ we define a set

  $$\begin{aligned} D_x := \{ x_1,\ldots ,x_n \in A:\, \exists _{k, k_1,\ldots ,k_n \in {\mathbb {N}}} \;\exists _{I\subset \{1,\ldots ,n\} }\; kx + \sum _{i\in I} k_i x_i = \sum _{i\notin I} k_i x_i \}. \end{aligned}$$

  First, we show that the above set is well-defined. If there exist $k,m\in {\mathbb {N}}$, $k_1,\ldots k_n,m_1,\ldots m_n\in {\mathbb {N}}\cup \{0\}$, $x_1,\ldots ,x_n\in A$, and $I,J\subset \{ 1,\ldots ,n \}$ such that $k_i \ne 0$ for $i\in I$, $m_i\ne 0$ for $i\in J$ and

  $$\begin{aligned}&kx+\sum _{i\in I} k_i x_i = \sum _{i\notin I} k_i x_i,\\&\quad mx+\sum _{i\in J} m_i x_i = \sum _{i\notin J} m_i x_i, \end{aligned}$$

  then

  $$\begin{aligned}&mkx + \sum _{i\in I} m k_i x_i + \sum _{i\notin J} k m_i x_i = kmx + \sum _{i\in J} k m_i x_i + \sum _{i\notin I} m k_i x_i, \end{aligned}$$

  whence

  $$\begin{aligned}&\sum _{i\in I} m k_i x_i + \sum _{i\notin J} k m_i x_i = \sum _{i\in J} k m_i x_i + \sum _{i\notin I} m k_i x_i, \end{aligned}$$

  so

  $$\begin{aligned}&\sum _{i\in I\cap J} m k_i x_i +\sum _{i\in I\setminus J} (m k_i + k m_i) x_i + \sum _{i\notin I\cup J} k m_i x_i \\&= \sum _{i\in I\cap J} k m_i x_i +\sum _{i\in J\setminus I} (k m_i +m k_i) x_i + \sum _{i\notin I\cup J} m k_i x_i. \end{aligned}$$

  As A is linearly independent, we have $I\setminus J=J\setminus I=\emptyset$, which means that $I=J$. Thus we get that $km_i =mk_i$ for $i\in \{ 1,\ldots ,n \}$, so $D_x$ is well-defined.

  We define a map $\varphi :\ell _{\infty } (S_\gamma ,X)\rightarrow \ell _{\infty } (S ,X)$ by the formula

  $$\begin{aligned} \varphi (f)(x) := f(\{ \alpha <\gamma :\; x_\alpha \in D_x \}),\ x\in S,\ f\in \ell _{\infty } (S_\gamma ,X). \end{aligned}$$

  It is easy to observe that $\varphi$ is linear, $\Vert \varphi (f) \Vert =\Vert f \Vert$ for $f\in \ell _{\infty } (S_\gamma ,X)$ and $\varphi (c \mathbbm {1}_{S_\gamma } )= c \mathbbm {1}_{S}$ for $c\in X$.

  Let $M_S:\ell _{\infty } (S,X)\rightarrow X$ be an X-valued invariant $\lambda$-mean. We define $M:\ell _{\infty } (S_\gamma ,X) \rightarrow X$ by the formula

  $$\begin{aligned} M(f):=M_S (\varphi (f)),\ f\in \ell _{\infty } (S_\gamma ,X). \end{aligned}$$

  From the properties of $\varphi$ we obtain that M is linear, $M (c\mathbbm {1}_{S_\gamma } )=c$ for $c\in X$, and $\Vert M \Vert \le \Vert M_S \Vert \le \lambda$.

  Now we show that M is invariant. Let $f\in \ell _{\infty } (S_\gamma ,X)$ and $A\in S_\gamma$. Since $A=\{ \alpha _1,\ldots ,\alpha _n \}$, from the invariance on each singleton $\{ \alpha _i \}$ we obtain

  $$\begin{aligned} M(_A f)&= M( _{\{\alpha _1 \}} ( _{\{ \alpha _2,\ldots ,\alpha _n \}} f))=M( _{\{ \alpha _2,\ldots ,\alpha _n \}} f )=\ldots \\&= M( _{\{ \alpha _n \}} f )=M(f),\ f\in \ell _{\infty } (S_\gamma ,X). \end{aligned}$$

  Hence we need to prove the invariance on each singleton, so we can assume that $A=\{\beta \}$ for some $\beta <\gamma$. Let $Z:=\{ x\in S:\; x_\beta \notin D_{x+x_\beta } \}$. We show that

  $$\begin{aligned} Z\cap (mx_\beta +Z)=\emptyset ,\ m\in {\mathbb {N}}. \end{aligned}$$

  (3.1)

  Suppose that $x\in Z\cap (mx_\beta +Z)$ for some $m\in {\mathbb {N}}$. Then there exists $y\in Z$ such that $x=mx_\beta +y$. Hence $x_\beta \notin D_{y+x_\beta } \cup D_{y+(m+1)x_\beta }$ but on the other hand, if $x_\beta \notin D_{y+x_\beta }$, then $x_\beta \in D_{y+(m+1)x_\beta }$, so we have a contradiction.

  Since S is cancellative, from (3.1) we obtain that

  $$\begin{aligned} (nx_\beta +Z)\cap (mx_\beta +Z)=\emptyset ,\ m,n\in {\mathbb {N}}_0 ,\ m>n. \end{aligned}$$

  (3.2)

  Let $g\in \ell _{\infty } (S ,X)$ be such that $g(x)=0$ for $x\in S\setminus Z$. From (3.2) we get

  $$\begin{aligned} n \Vert M_S (g) \Vert&=\Vert \sum _{i=1}^n M_S (_{ix_\beta } g) \Vert =\Vert M_S ({\sum _{i=1}^n} _{ix_\beta } g) \Vert \\&\le \lambda \Vert {\sum _{i=1}^n} _{ix_\beta } g \Vert \le \lambda \Vert g \Vert ,\ n\in {\mathbb {N}}, \end{aligned}$$

  so $M_S (g)=0$.

  For each $y\in S$ we have

  • if $x_\beta \notin D_y$, then $D_{y+x_\beta }=D_y \cup \{ x_\beta \}$, so

    $$\begin{aligned} \varphi ( _{ \{ \beta \} } f)(y)&=f(\{ \alpha<\gamma :\; x_\alpha \in D_{y} \} \cup \{ \beta \}) \\&=f(\{ \alpha <\gamma :\; x_\alpha \in D_{y+x_\beta } \} ) = \big ( _{x_\beta } \varphi (f)\big ) (y); \end{aligned}$$

  • if $x_\beta \in D_y$ and $x_{\beta }\in D_{y+x_\beta }$, then $D_{y+x_\beta }=D_y$, so

    $$\begin{aligned} \varphi ( _{ \{ \beta \} } f)(y)&=f(\{ \alpha<\gamma :\; x_\alpha \in D_{y} \} \cup \{ \beta \}) = f(\{ \alpha<\gamma :\; x_\alpha \in D_{y} \} )\\&=f(\{ \alpha <\gamma :\; x_\alpha \in D_{y+x_\beta } \} )= \big ( _{x_\beta } \varphi (f)\big ) (y); \end{aligned}$$

  • if $x_\beta \notin D_{y+x_\beta }$, then $y\in Z$.

  Hence

  $$\begin{aligned}&\big ( \varphi ( _{ \{ \beta \} } f) - _{x_\beta } \varphi (f)\big ) (y)=0,\ y\in S\setminus Z, \end{aligned}$$

  so

  $$\begin{aligned}&M(_{ \{ \beta \} } f)= M_S (\varphi ( _{ \{ \beta \} } f )=M_S (_{x_\beta } \varphi (f) )=M_S (\varphi (f) )= M(f). \end{aligned}$$

• Now assume that $\vert A\vert <\gamma$. Hence $\gamma = \omega$. Let $N=\vert A\vert$, $A=\{ x_1,\ldots ,x_N\}$. Since S can be embedded in a group, for each $x\in S$ there exist $k(x)\in {\mathbb {N}}$, $k_1(x),\ldots ,k_N(x)\in {\mathbb {Z}}$ such that $k(x)x=\sum \limits _{i=1}^{N} k_i(x) x_i$. We define a map $\varphi :\ell _{\infty } (S_\omega ,X)\rightarrow \ell _{\infty } (S ,X)$ by the formula

  $$\begin{aligned}&\varphi (f)(x) := f\Big ( \big \{ \alpha \in \omega :\; \alpha k(x)\le |k_1(x)| \big \} \Big ),\ x\in S,\ f\in \ell _{\infty } (S_\omega ,X). \end{aligned}$$

  It is easy to observe that $\varphi$ is linear, $\Vert \varphi (f) \Vert \le \Vert f \Vert$ for $f\in \ell _{\infty } (S_\omega ,X)$ and $\varphi (c \mathbbm {1}_{S_\omega } )= c \mathbbm {1}_{S}$ for $c\in X$.

  Let $M_S:\ell _{\infty } (S,X)\rightarrow X$ be an X-valued invariant $\lambda$-mean. We define $M:\ell _{\infty } (S_\omega ,X) \rightarrow X$ by the formula

  $$\begin{aligned} M(f):=M_S (\varphi (f)),\ f\in \ell _{\infty } (S_\omega ,X). \end{aligned}$$

  From the properties of $\varphi$ we obtain that M is linear, $M (c\mathbbm {1}_{S_\gamma } )=c$ for $c\in X$, and $\Vert M \Vert \le \lambda$.

  Now we show that M is invariant. Let $f\in \ell _{\infty } (S_\omega ,X)$ and $A\in S_\omega$. Similarly as in the previous case we need only to prove the invariance on each singleton, so we can assume that $A=\{\beta \}$ for some $\beta \in \omega$. Let

  $$\begin{aligned} Z:=\bigg \{ x\in S:\; |k_1(x)|<\beta k(x) \bigg \} . \end{aligned}$$

  We show that

  $$\begin{aligned} Z\cap (2m\beta x_1 +Z)=\emptyset ,\ m\in {\mathbb {N}}. \end{aligned}$$

  (3.3)

  Suppose that $x\in Z\cap (mx_\beta +Z)$ for some $m\in {\mathbb {N}}$. Then there exists $y\in Z$ such that $x=2m\beta x_1 +y$. Hence

  $$\begin{aligned}&k(y) [y+2m\beta x_1] = [k_1(y) +2mk(y)\beta ] x_1 + \sum \limits _{i=2}^N k_i (y) x_i, \end{aligned}$$

  which gives us

  $$\begin{aligned}&\beta k(y)> k_1(y)+2m\beta k(y) > -\beta k(y) +2\beta k(y) =\beta k(y), \end{aligned}$$

  so we have a contradiction. Since S is cancellative, from (3.3) we obtain that

  $$\begin{aligned} (2n\beta x_1 +Z)\cap (2m\beta x_1 +Z)=\emptyset ,\ m,n\in {\mathbb {N}}_0 ,\ m>n. \end{aligned}$$

  (3.4)

  Now observe that for $x\in S\setminus Z$ we have

  $$\begin{aligned} \varphi (f_{\{ \beta \}}(x))&=f_{\{ \beta \}} \Big ( \big \{ \alpha \in \omega :\; \alpha k(x)\le |k_1(x)| \big \} \Big )\\&=f\Big ( \big \{ \alpha \in \omega :\; \alpha k(x)\le |k_1(x)| \big \} \cup \{ \beta \}\Big )\\&=f\Big ( \big \{ \alpha \in \omega :\; \alpha k(x)\le |k_1(x)| \big \} \Big ) =\varphi (f(x)), \end{aligned}$$

  so from (3.4) we obtain that

  $$\begin{aligned}&n\Vert M(f-f_{\{ \beta \}} ) \Vert = \Vert n M_S \big ( \varphi (f)-\varphi (f_{\{ \beta \}} )\big ) \Vert \\&=\Vert \sum _{i=1}^{n} M_S\big ( (\varphi (f)-\varphi (f_{\{ \beta \}} ))_{2i\beta x_1} \big ) \Vert \\&=\Vert M_S\Big ( \sum _{i=1}^{n} \big ( \varphi (f)-\varphi (f_{\{ \beta \}} )\big ) _{2i\beta x_1} \Big ) \Vert \le \lambda \Vert \varphi (f)-\varphi (f_{\{ \beta \}} ) \Vert \end{aligned}$$

  for every $n\in {\mathbb {N}}$, which means that $M(f_{\{ \beta \}}) =M(f)$.

$\square$

Using Theorems 1.4, 3.1 and 3.2 we obtain the following

Corollary 3.3

The following assertions are equivalent:

• 1. X is complemented in $X^{**}$ by a projection of norm at most $\lambda$;

• 2. for every amenable semigroup S there exists an X-valued invariant $\lambda$-mean on S;

• 3. for any cancellative semigroup S of torsion-free rank $\delta$, ${\text {dens}}{X^{**}}=\max (\delta , \omega )$, there exists an X-valued invariant $\lambda$-mean on S.

The following example shows that in general in the third assertion of the previous corollary the torsion-free rank of semigroup S cannot be less than the density of X.

Example 3.4

Let $\Gamma$ be an uncountable set such that $\vert \Gamma \vert$ is a regular cardinal number. We define the set

$$\begin{aligned} X:=\{ f\in \ell _\infty (\Gamma ):\, \vert \{ \alpha \in \Gamma :\, f(\alpha )\ne 0 \} \vert < \vert \Gamma \vert \}. \end{aligned}$$

It is easy too see that X is a Banach space. Since $\mathbbm {1}_{ \{ \alpha \} }\in X$ for $\alpha \in \Gamma$, ${\text {dens}}X = \vert \Gamma \vert$.

Let S be an amenable semigroup, $\vert S\vert <{\text {dens}}X$ and $L:\ell _\infty (S,{\mathbb {R}}) \rightarrow {\mathbb {R}}$ be an invariant mean. We define $M:\ell _{\infty } (S,X)\rightarrow X$ by the formula

$$\begin{aligned} M(g)(\alpha ):=L(g(\cdot )(\alpha )),\ g\in \ell _{\infty } (S,X),\ \alpha \in \Gamma . \end{aligned}$$

First, we observe that

$$\begin{aligned} \{ \alpha \in \Gamma :\, M(g)(\alpha )\ne 0 \}&=\{ \alpha \in \Gamma :\, L(g(\cdot )(\alpha ))\ne 0 \} \\&\subset \bigcup _{s\in S} \{ \alpha \in \Gamma :\, g(s)(\alpha )\ne 0 \} \end{aligned}$$

and since $\vert \Gamma \vert$ is regular, we have

$$\begin{aligned}&\vert \{ \alpha \in \Gamma :\, M(g)(\alpha )\ne 0 \}\vert \le \vert S\vert \cdot \sup _{s\in S} \vert \{ \alpha \in \Gamma :\, g(s)(\alpha )\ne 0 \}\vert <\vert \Gamma \vert , \end{aligned}$$

so M is well-defined.

It is easy to see that M is linear. We have also

$$\begin{aligned} \Vert M(g) \Vert&=\sup _{\alpha \in \Gamma } |M(g)(\alpha )| =\sup _{\alpha \in \Gamma } |L(g(\cdot )(\alpha )) |\\&\le \sup _{\alpha \in \Gamma } \sup _{s\in S} |g(s)(\alpha )| =\sup _{s\in S} \Vert g(s) \Vert =\Vert g \Vert ,\ g\in \ell _{\infty } (S,X), \end{aligned}$$

and

$$\begin{aligned} M(c\mathbbm {1}_S)(\alpha )=L(c(\alpha )\mathbbm {1}_S )=c(\alpha ),\ c\in X,\ \alpha \in \Gamma . \end{aligned}$$

Finally, we observe that

$$\begin{aligned} M(_a g )(\alpha )&= L(g(a+\cdot )(\alpha ))= L(g(\cdot )(\alpha ))\\&=M(g)(\alpha ),\ g\in \ell _{\infty } (S,X),\ a\in S,\ \alpha \in \Gamma , \end{aligned}$$

so M is an X-valued invariant mean.

In the paper of Pełczyński and Sudakov [13, Theorem 1] it is shown that X isn’t complemented in its bidual.