Internal characterization of Brezis–Lieb spaces

Abstract

In order to find an extension of Brezis–Lieb’s lemma to the case of nets, we replace the almost everywhere convergence by the unbounded order convergence and introduce the pre-Brezis–Lieb property in normed lattices. Then we identify a wide class of Banach lattices in which the Brezis–Lieb lemma holds true. Among other things, it gives an extension of the Brezis–Lieb lemma for nets in $L^p$ for $p\in [1,\infty )$.

1 Introduction

Let $(\Omega ,\Sigma ,\mu )$ be a measure space in which, for every set $A\in \Sigma$, $\mu (A)>0$, there exists $\Sigma \ni A_0\subseteq A$, such that $0<\mu (A_0)<\infty$. Given $p\in (0,\infty )$, denote by ${\mathcal {L}}^p=\{f: \int \limits _{\Omega }|f|^p\mu <\infty \}$ the vector space of p-integrable functions from $\Omega$ into ${\mathbb {C}}$. The Brezis–Lieb lemma [3, Thm.1] is known as the following useful refinement of the Fatou lemma.

Theorem 1

(Brezis–Lieb’s lemma for ${\mathcal {L}}^p\,(0<p<\infty )$) Suppose $f_n{{\,\mathrm{\xrightarrow []{a.e.}}\,}}f$ and $\int \limits _{\Omega }|f_n|^pd\mu \leqslant C<\infty$ for all n and some $p\in (0,\infty )$. Then

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\int \limits _{\Omega }(|f_n|^p-|f_n-f|^p)d\mu =\int \limits _{\Omega }|f|^pd\mu . \end{aligned}$$

(1.1)

As the following example shows, Theorem 1 does not have a reasonable direct generalization for nets.

Example 1

Consider $[0,1]\subset {\mathbb {R}}$ with the Lebesgue measure $\mu$. Let $\Delta$ be the family of all finite subsets of [0, 1] ordered by inclusion, and ${\mathbb {I}}_F$ be the indicator function of $F\in \Delta$. Then ${\mathbb {I}}_F{{\,\mathrm{\xrightarrow []{a.e.}}\,}}{\mathbb {I}}_{[0,1]}$ and $\int \limits _0^1|{\mathbb {I}}_F|d\mu =0$, however

$$\begin{aligned} \lim \limits _{F\rightarrow \infty }\int \limits _0^1(|{\mathbb {I}}_F|-|{\mathbb {I}}_F-{\mathbb {I}}_{[0,1]}|)d\mu = \lim \limits _{F\rightarrow \infty }\int \limits _0^1(-|{\mathbb {I}}_{[0,1]}|)d\mu =-1\ne 1=\int \limits _0^1|{\mathbb {I}}_{[0,1]}|d\mu . \end{aligned}$$

In order to avoid the collision, we restate Theorem 1 in the case of $1\leqslant p<\infty$ in terms of the Banach space $L^p$ of equivalence classes of functions from ${\mathcal {L}}^p(\mu )$ w.r. to $\mu$ (cf. [12, Thm.2]).

Theorem 2

(Brezis–Lieb’s lemma for $L^p$$(1\leqslant p<\infty )$) Let ${\mathbf {f}}_n{{\,\mathrm{\xrightarrow []{a.e.}}\,}}{\mathbf {f}}$ in $L^p(\mu )$ and $\Vert {\mathbf {f}}_n\Vert _p\rightarrow \Vert {\mathbf {f}}\Vert _p$, where $\Vert {\mathbf {f}}_n\Vert _p:=\bigg [{\int \limits _{\Omega }|f_n|^pd\mu }\bigg ]^{1/p}$ with $f_n\in {\mathcal {L}}^p(\mu )$ and $f_n\in {\mathbf {f}}_n$. Then $\Vert {\mathbf {f}}_n-{\mathbf {f}}\Vert _p\rightarrow 0$.

Anton Schep kindly provided us with the reference [13, p.59] showing that Theorem 2 for $p>1$ is due to Frigyes Riesz. Although in Theorem 2, we still have a.e.-convergent sequences in $L^p$, it is possible now (e.g. due to [7, Prop.3.1]) to replace the a.e.-convergence by the uo-convergence and restate Theorem 1 once more (cf. also [5, Prop.2.2] and [10, Prop.1.5]) as follows.

Theorem 3

(Brezis–Lieb’s lemma for uo-convergent sequences in $L^p$) Let $x_n{{\,\mathrm{\xrightarrow []{uo}}\,}}x$ in $L^p$, where $p\in [1,\infty )$. If $\Vert x_n\Vert _p\rightarrow \Vert x\Vert _p$ then $\Vert x_n-x\Vert _p\rightarrow 0$.

Notice that Theorem 3 is a result of the Banach lattice theory which does not involve the measure theory directly. This observation motivates us to investigate those Banach lattices in which the statement of Theorem 3 holds true. We call them by the $\sigma$-Brezis–Lieb spaces. After introducing a geometrical property of normed lattices in Definition 2, we prove Theorem 4 which is the main result of the present paper. Theorem 4 gives an internal geometric characterization of $\sigma$-Brezis–Lieb’s spaces and implies immediately the following result.

Proposition 1

Let ${\mathbf {f}}_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}{\mathbf {f}}$ in $L^p(\mu )$$(1\leqslant p<\infty )$, and $\Vert {\mathbf {f}}_\alpha \Vert _p\rightarrow \Vert {\mathbf {f}}\Vert _p$. Then $\Vert {\mathbf {f}}_\alpha -{\mathbf {f}}\Vert _p\rightarrow 0$.

It is worth mentioning that Proposition 1 may serve as a net-extension of the Brezis–Lieb lemma (in its form of Theorem 3).

2 Brezis–Lieb spaces

In this section all normed lattices are considered over the real field ${\mathbb {R}}$. Recall that a net $v_\alpha$ in a vector lattice Euo-converges to $v\in E$ whenever, for every $u\in E_+$, the net $|v_\alpha -v|\wedge u$ converges in order to 0. For the further theory of vector lattices, we refer to [1, 2] and, for the unbounded order convergence, to [7, 8].

Definition 1

A normed lattice $(E,\Vert \cdot \Vert )$ is said to be a Brezis–Lieb space (shortly, BL-space) (resp. $\sigma$-Brezis–Lieb space$\sigma$-BL-space)) if, for any net $x_\alpha$ (resp. for any sequence $x_n)$ in X such that $\Vert x_\alpha \Vert \rightarrow \Vert x_0\Vert$ (resp. $\Vert x_n\Vert \rightarrow \Vert x_0\Vert )$ and $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x_0$ (resp. $x_n{{\,\mathrm{\xrightarrow []{uo}}\,}}x_0 )$, we have $\Vert x_\alpha -x_0\Vert \rightarrow 0$ (resp. $\Vert x_n-x_0\Vert \rightarrow 0)$.

Trivially, any BL-space is a $\sigma$-BL-space, and any finite-dimensional normed lattice is a BL-space. Furthermore, by [7, Thm.3.2], any regular sublattice F of any BL-space ($\sigma$-BL-space) E is itself a BL-space ($\sigma$-BL-space). Taking into account the fact that the a.e.-convergence for sequences in $L^p$ coincides with the uo-convergence [7, Prop.3.1], Theorem 3 says exactly that $L^p$ is a $\sigma$-BL-space for $1\leqslant p<\infty$. The following result is due to Vladimir Troitsky who also kindly provided us with its proof.

Proposition 2

A Banach lattice E with countable sup property and a weak unit w is a BL-space iff E is a $\sigma$-BL-space.

Proof

By [7, Cor.3.5], $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x$ iff $|x_\alpha -x|\wedge w{{\,\mathrm{\xrightarrow []{o}}\,}}0$. Suppose that E is a $\sigma$-Brezis–Lieb space. Suppose that $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x$ and $\Vert x_\alpha \Vert \rightarrow \Vert x\Vert$, yet $\Vert x_\alpha -x\Vert \not \rightarrow 0$. Then there exists $\varepsilon >0$ such that for every $\alpha$ one can find $\beta \ge \alpha$ with $\Vert x_\beta -x\Vert \ge \varepsilon$.

It follows from $|x_\alpha -x|\wedge w{{\,\mathrm{\xrightarrow []{o}}\,}}0$ that there is a net $(u_\gamma )_{\gamma \in \Gamma }$ such that $u_\gamma \downarrow 0$ and for every $\gamma$ there exists $\alpha _0$ such that $|x_\alpha -x|\wedge w\le u_\gamma$ whenever $\alpha \ge \alpha _0$. Since E has countable sup property, we can find an increasing sequence $\gamma _n$ in $\Gamma$ such that $u_{\gamma _n}\downarrow 0$. For each n, find $\alpha _n$ such that $|x_\alpha -x|\wedge w\le u_{\gamma _n}$ for all $\alpha \ge \alpha _n$.

Since $\Vert x_\alpha \Vert \rightarrow \Vert x\Vert$, we have $\Bigl |\Vert x_\alpha \Vert -\Vert x\Vert \Bigr |<1$ for all sufficiently large $\alpha$. So we can choose $\beta _1\ge \alpha _1$ such that $\Bigl |\Vert x_{\beta _1}\Vert -\Vert x\Vert \Bigr |<1$ and $\Vert x_{\beta _1}-x\Vert \ge \varepsilon$. Similarly, choose $\beta _2$ such that $\beta _2>\beta _1$ and $\beta _2\ge \alpha _2$ with $\Bigl |\Vert x_{\beta _2}\Vert -\Vert x\Vert \Bigr |<\frac{1}{2}$ and $\Vert x_{\beta _2}-x\Vert \ge \varepsilon$. Proceeding inductively, we get a strictly increasing sequence $\beta _n$ such that $\beta _n\ge \alpha _n$, $\Vert x_{\beta _n}-x\Vert \ge \varepsilon$, and $\Bigl |\Vert x_{\beta _n}\Vert -\Vert x\Vert \Bigr |<\frac{1}{n}$ for every n. It follows that $\Vert x_{\beta _n}\Vert \rightarrow \Vert x\Vert$. Also, it follows from $\beta _n\ge \alpha _n$ that $|x_{\beta _n}-x|\wedge w\le u_{\gamma _n}$, so that $x_{\beta _n}{{\,\mathrm{\xrightarrow []{uo}}\,}}x$. Therefore, $\Vert x_{\beta _n}-x\Vert \rightarrow 0$, which is a contradiction. $\square$

Now, we consider examples of Banach lattices which are not $\sigma$-Brezis–Lieb spaces.

Example 2

The Banach lattice $(c_0,\Vert \cdot \Vert _{\infty })$ is not a $\sigma$-BL-space. To see this, take $x_n=e_{2n}+\sum _{k=1}^{n}\frac{1}{k}e_k$ and $x=\sum _{k=1}^{\infty }\frac{1}{k}e_k$ in $c_0$. Clearly, $\Vert x\Vert =\Vert x_n\Vert =1$ for all n and $x_n{{\,\mathrm{\xrightarrow []{uo}}\,}}x$, however $1=\Vert x-x_n\Vert$ does not converge to 0.

We do not know whether or not for an arbitrary lattice norm $\Vert \cdot \Vert$ in $c_0$, which is equivalent to $\Vert \cdot \Vert _{\infty }$, the Banach lattice $(c_0,\Vert \cdot \Vert )$ is not a $\sigma$-BL-space.

Example 3

Since $c_0$ is an order ideal in c and in $\ell ^{\infty }$, $c_0$ is regular there, and hence, both Banach lattices $(c,\Vert \cdot \Vert _{\infty })$ and $(\ell ^{\infty },\Vert \cdot \Vert _{\infty })$ are not $\sigma$-BL-spaces. Accordingly to the fact, that $c_0$ is a regular sublattice of c and to the last sentence of Example 2, it is also unknown whether or not the Banach lattice $(c,\Vert \cdot \Vert )$ is not a $\sigma$-BL-space for an arbitrary lattice norm $\Vert \cdot \Vert$ that is equivalent to $\Vert \cdot \Vert _{\infty }$.

In opposite to c, the Banach lattice $\ell ^{\infty }$ is Dedekind complete. Let $\Vert \cdot \Vert$ be any lattice norm in $\ell ^{\infty }$ that is equivalent to $\Vert \cdot \Vert _{\infty }$. Clearly, the norm $\Vert \cdot \Vert$ is not order continuous. Therefore, by Theorem 4, $(\ell ^{\infty },\Vert \cdot \Vert )$ is not a $\sigma$-BL-space.

A slight change of an infinite-dimensional BL-space can turn it into a normed lattice which is not even a $\sigma$-BL-space.

Example 4

Let E be an infinite-dimensional normed lattice. Let $F={\mathbb {R}}\oplus _{\infty }E$. Take a disjoint sequence $y_n$ in E such that $\Vert y_n\Vert _E=1$ for all n. Then $y_n{{\,\mathrm{\xrightarrow []{uo}}\,}}0$ in E [7, Cor.3.6]. Let $x_n=(1,y_n)\in F$. Then $\Vert x_n\Vert _F=\sup (1,\Vert y_n\Vert _E)=1$ and $x_n=(1,y_n){{\,\mathrm{\xrightarrow []{uo}}\,}}(1,0)=:x$ in F, however

$$\begin{aligned} \Vert x_n-x\Vert _{F}=\Vert (0,y_n)\Vert _F=\Vert y_n\Vert _E=1, \end{aligned}$$

and so $x_n$ does not converge to x in $(F,\Vert \cdot \Vert _F)$. Therefore $F={\mathbb {R}}\oplus _{\infty }E$ is not a $\sigma$-BL-space.

In order to characterize BL-spaces, we introduce the following definition.

Definition 2

A normed lattice $(E,\Vert \cdot \Vert )$ is said to have the pre-Brezis–Lieb property (shortly, pre-BL-property), whenever $\limsup _{n\rightarrow \infty }\Vert u_0+u_n\Vert >\Vert u_0\Vert$ for any disjoint normalized sequence $(u_n)_{n=1}^{\infty }$ in $E_+$ and for any $u_0\in E_+$, $u_0>0$.

Every finite dimensional normed lattice E has the pre-BL-property. It is easy to see that the Banach lattices $c_0$, c, and $\ell ^{\infty }$ w.r. to the supremum norm $\Vert \cdot \Vert _{\infty }$ do not have the pre-BL-property. The modification of the norm in an infinite-dimensional Banach lattice E with the pre-BL-property, as in Example 4, turns it into the Banach lattice $F={\mathbb {R}}\oplus _{\infty }E$ without the pre-BL-property. Indeed, take a disjoint normalized sequence $(y_n)_{n=1}^{\infty }$ in $E_+$. Let $u_0=(1,0)$ and $u_n=(0,y_n)$ for $n\geqslant 1$. Then $(u_n)_{n=0}^{\infty }$ is a disjoint normalized sequence in $F_+$ with $\limsup _{n\rightarrow \infty }\Vert u_0+u_n\Vert =1=\Vert u_0\Vert$.

Remarkably, it is not a coincidence. The following theorem identifies BL-spaces among $\sigma$-Dedekind complete Banach lattices.

Theorem 4

For a $\sigma$-Dedekind complete Banach lattice E, the following conditions are equivalent:

• (1) E is a Brezis–Lieb space;

• (2) E is a $\sigma$-Brezis–Lieb space;

• (3) E has the pre-Brezis–Lieb property, and the norm in E is order continuous.

Proof

The implication $(1)\Rightarrow (2)$ is trivial.

$(2)\Rightarrow (3)$: We show first that E has the pre-BL-property. If not, then there exist a disjoint normalized sequence $(u_n)_{n=1}^{\infty }$ in $E_+$ and $0<u_0\in E_+$ with $\limsup _{n\rightarrow \infty }\Vert u_0+u_n\Vert =\Vert u_0\Vert$. Since $\Vert u_0+u_n\Vert \geqslant \Vert u_0\Vert$, then $\lim _{n\rightarrow \infty }\Vert u_0+u_n\Vert =\Vert u_0\Vert$. Denote $v_n:=u_0+u_n$. By [7, Cor.3.6], $u_n{{\,\mathrm{\xrightarrow []{uo}}\,}}0$ and hence $v_n{{\,\mathrm{\xrightarrow []{uo}}\,}}u_0$. Since E is a $\sigma$-BL-space and $\lim _{n\rightarrow \infty }\Vert v_n\Vert =\Vert u_0\Vert$, then $\Vert v_n-u_0\Vert \rightarrow 0$, a contradiction to $\Vert v_n-u_0\Vert =\Vert u_0+u_n-u_0\Vert =\Vert u_n\Vert =1$. Notice that, in this part of the proof of $(2)\Rightarrow (3)$, the $\sigma$-Dedekind completeness of E was not used.

Assume that the norm in E is not order continuous. Then, by the Fremlin–Meyer–Nieberg theorem (see e.g. [2, Thm.4.14]) there exists $y\in E_+$ and a disjoint sequence $e_k\in [0,y]$ such that $\Vert e_k\Vert \not \rightarrow 0$. Without lost of generality, we may assume $\Vert e_k\Vert =1$ for all $k\in {\mathbb {N}}$. By the $\sigma$-Dedekind completeness of E, for any sequence $\alpha _n\in {\mathbb {R}}_+$, there exist the following vectors

$$\begin{aligned} x_0=\bigvee \limits _{k=1}^{\infty }e_k, \quad x_n=\alpha _{2n}e_{2n}+\bigvee \limits _{k=1,k\ne n,k\ne 2n}^{\infty }e_k \quad (\forall n\in {\mathbb {N}}). \end{aligned}$$

(2.1)

Now, we choose $\alpha _{2n}\geqslant 1$ in (2.1) such that $\Vert x_n\Vert =\Vert x_0\Vert$ for all $n\in {\mathbb {N}}$. Clearly, $x_n{{\,\mathrm{\xrightarrow []{uo}}\,}}x_0$. Since E is a $\sigma$-BL-space, then $\Vert x_n-x_0\Vert \rightarrow 0$, violating

$$\begin{aligned} \Vert x_n-x_0\Vert =\Vert (\alpha _{2n}-1)e_{2n}-e_n\Vert =\Vert (\alpha _{2n}-1)e_{2n}+e_n\Vert \geqslant \Vert e_n\Vert =1. \end{aligned}$$

The obtained contradiction shows that the norm in E is order continuous.

$(3)\Rightarrow (1)$: If E is not a Brezis–Lieb space, then there exists a net $(x_\alpha )_{\alpha \in A}$ in E such that $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x$ and $\Vert x_\alpha \Vert \rightarrow \Vert x\Vert$, but $\Vert x_\alpha -x\Vert \not \rightarrow 0$. Then $|x_\alpha |{{\,\mathrm{\xrightarrow []{uo}}\,}}|x|$ and $\Vert |x_\alpha |\Vert \rightarrow \Vert |x|\Vert$.

Notice that $\Vert |x_\alpha |-|x|\Vert \not \rightarrow 0$. Indeed, if $\Vert |x_\alpha |-|x|\Vert \rightarrow 0$, then, for any $\varepsilon >0$, $(|x_\alpha |)_{\alpha \in A}$ is eventually in $[-|x|,|x|]+\varepsilon B_E$. Thus $(|x_\alpha |)_{\alpha \in A}$ is almost order bounded. Since E is order continuous and $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x$, then by [8, Pop.3.7.], $\Vert x_\alpha -x\Vert \rightarrow 0$, that is impossible. Therefore, without lost of generality, we may assume $x_\alpha \in E_+$ and, by normalizing, also $\Vert x_\alpha \Vert =\Vert x\Vert =1$ for all $\alpha$.

Passing to a subnet, denoted by $x_\alpha$ again, we may assume

$$\begin{aligned} \Vert x_\alpha -x\Vert>C>0 \quad (\forall \alpha \in A). \end{aligned}$$

(2.2)

Notice that $x\geqslant (x-x_\alpha )^+=(x_\alpha -x)^-{{\,\mathrm{\xrightarrow []{uo}}\,}}0$, and hence $(x_\alpha -x)^-{{\,\mathrm{\xrightarrow []{o}}\,}}0$. The order continuity of the norm ensures

$$\begin{aligned} \Vert (x_\alpha -x)^-\Vert \rightarrow 0. \end{aligned}$$

(2.3)

Denoting $w_\alpha =(x_\alpha -x)^+$ and using (2.2) and (2.3), we may also assume

$$\begin{aligned} \Vert w_\alpha \Vert =\Vert (x_\alpha -x)^+\Vert >C \quad (\forall \alpha \in A). \end{aligned}$$

(2.4)

In view of (2.4), we obtain

$$\begin{aligned} 2=\Vert x_\alpha \Vert +\Vert x\Vert \geqslant \Vert (x_\alpha -x)^+\Vert =\Vert w_\alpha \Vert >C \quad (\forall \alpha \in A). \end{aligned}$$

(2.5)

From $x_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}x$ we have $w_\alpha {{\,\mathrm{\xrightarrow []{uo}}\,}}(x-x)^+=0$. It follows from [4, Thm.3.2] that there exists an increasing sequence of indices $\alpha _n$ and a disjoint sequence $z_n$ such that

$$\begin{aligned} \Vert w_{\alpha _n}-z_n\Vert \rightarrow 0 \end{aligned}$$

(2.6)

Without loss of generality, replacing $z_n$ with $|z_n|$, we may assume $z_n\ge 0$. Passing to further increasing sequence of indices, we may assume that

$$\begin{aligned} \Vert w_{\alpha _n}\Vert \rightarrow M\in [C,2] \quad (n\rightarrow \infty ). \end{aligned}$$

Now

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\bigg \Vert M^{-1}x+\Vert z_{n}\Vert ^{-1}z_{n}\bigg \Vert = M^{-1}\lim \limits _{n\rightarrow \infty }\Vert x+z_{n}\Vert =[\text {by} \ (2.6)]=\\&M^{-1}\lim \limits _{n\rightarrow \infty }\Vert x+w_{\alpha _n}\Vert =[\text {by} \ (2.3)]= M^{-1}\lim \limits _{n\rightarrow \infty }\Vert x+(x_{\alpha _n}-x)\Vert =\\&M^{-1}\lim \limits _{n\rightarrow \infty }\Vert x_{\alpha _n}\Vert =M^{-1}=\Vert M^{-1}x\Vert , \end{aligned}$$

violating the pre-Brezis–Lieb property for $u_0=M^{-1}x$ and $u_n=\Vert z_{n}\Vert ^{-1}z _{n}$, $n\geqslant 1$. The obtained contradiction completes the proof. $\square$

Anton Schep kindly informed us that a special case of Theorem 4 is due to Nakano [11, Thm.33.6].

Since every order continuous Banach lattice is Dedekind complete, the following result is a direct consequence of Theorem 4.

Corollary 1

For an order continuous Banach lattice E, the following conditions are equivalent:

• (1) E is a BL-space;

• (2) E is a $\sigma$-BL-space;

• (3) E has the pre-BL-property.

Corollary 1 applied to the order continuous Banach lattices $L^p$$(1\leqslant p<\infty )$ gives Proposition 1.

It follows from Theorem 4 that if E is a $\sigma$-Dedekind complete Banach lattice, then E is a $\sigma$-BL-space iff E has the pre-BL-property, and the norm in E is order continuous. This result is related to the following fact mentioned on page 28 of [9], where the same condition with weak convergence replaced with uo-convergence which is used in Definition 1: A Banach lattice $(E,\Vert \cdot \Vert )$ is order continuous iff there is an equivalent norm $\Vert \cdot \Vert _1$ on E so that

$$\begin{aligned} E \ni x_n \xrightarrow {w}x \ \ \text {and} \ \ \Vert x_n\Vert _1\rightarrow \Vert x\Vert _1\ \Rightarrow \ \Vert x_n-x\Vert _1\rightarrow 0. \end{aligned}$$

Relationship between weak and uo-convergence have been studied in [14].

We do not know whether or not implication $(2)\Rightarrow (3)$ of Theorem 4 holds true without the assumption that the Banach lattice E is $\sigma$-Dedekind complete. More precisely:

Question 1

Does every $\sigma$-Brezis–Lieb Banach lattice have an order continuous norm?

In the proof of $(2)\Rightarrow (3)$ of Theorem 4, the $\sigma$-Dedekind completeness of E has been used only for showing that E has an order continuous norm. So, any $\sigma$-Brezis–Lieb Banach lattice has the pre-BL-property. Therefore, for answering in positive the question of possibility to drop $\sigma$-Dedekind completeness assumption in Theorem 4, it suffices to answer in positive the following question which is formally weaker than Question 1.

Question 2

Does the pre-BL-property imply order continuity of the norm in the underlying Banach lattice?

In the end of the paper, we mention one more question closely related to the question in the last sentence of Example 2.

Question 3

Does the pre-BL-property of a Banach lattice E ensure that E is a KB-space?