Extreme points of the set of elements majorised by an integrable function: Resolution of a problem by Luxemburg and of its noncommutative counterpart

TL;DR

This paper characterizes the extreme points of the set of functions majorized by an integrable function, solving a long-standing problem by Luxemburg and extending the result to a noncommutative setting.

Contribution

It provides a complete characterization of extreme points in the majorization set and extends the classical result to noncommutative spaces.

Findings

Complete description of extreme points of majorized functions

Resolution of Luxemburg's 1967 problem

Extension to noncommutative integration theory

Abstract

Let $f$ be an arbitrary integrable function on a finite measure space $(X,\Sigma, \nu)$. We characterise the extreme points of the set $\Omega (f)$ of all measurable functions on $(X,\Sigma, \nu)$ majorised by $f$, providing a complete answer to a problem raised by W.A.J. Luxemburg in 1967. Moreover, we obtain a noncommutative version of this result.

Equations

$${\int }_{\{s;\lambda (s;x)=\lambda (t;x)\}}\lambda (s;y)ds=\lambda (t;x)\tau (E^x(\{\lambda (t;x)\})).$$

$${\int }_{\{s;\lambda (s;x)=\lambda (t;x)\}}\lambda (s;y)ds=\lambda (t;x)\nu (x^{-1}(\lambda (t;x))).$$

$$E_D(\{U{M}_{\beta }{U}^{\ast }:U\in M_n(\mathbb{C})\text{mbox}isunitary\})=\{M_{\alpha }\in D:\alpha \prec \beta \},\beta \in M_n(\mathbb{C}{)}_h,$$

$$\mathrm{extr}({\overline{E_{\mathcal{A}}({\mathcal{U}}_{\mathcal{M}}(y))}}^{\sigma \text{-sot}})=\mathrm{extr}(\{x\in {\mathcal{A}}_h:x\prec y\})=\{x\in {\mathcal{A}}_h:\lambda (x)=\lambda (y)\},$$

$$d(s;x)=\tau (E^x(s,\infty )),s\in \mathbb{R},$$

$$\mu (t;x)=\inf \{s⩾0:d(s;|x|)⩽t\},t\in [0,1).$$

$$\lambda (t;x)=\inf \{s\in \mathbb{R}:d(s;x)⩽t\},t\in [0,1),$$

$$\check{\lambda }(t;x)=\sup \{s\in \mathbb{R}:\tau (E^x(-\infty ,s])⩽t\},t\in [0,1).$$

$$d(x)=d(\lambda (x)).$$

$$\check{\lambda }(t;x)=\lambda ((1-t)-;x)=-\lambda (t;-x),\forall t\in [0,1),$$

$$\lambda (t;f)={\delta }_f(t)=\inf \{s\in \mathbb{R}:m(\{x\in X:f(x)>s\})⩽t\},t\in [0,1).$$

$$L_1(\mathcal{M},\tau )=\{x\in S(\mathcal{M},\tau ):\tau (|x|)<\infty \}.$$

$$\tau (x)={\int }_0^1\lambda (t;x)dt={\int }_0^1\check{\lambda }(t;x)dt.$$

$${\int }_0^s\lambda (t;g)dt⩽{\int }_0^s\lambda (t;f)dt$$

$${\int }_0^1\lambda (t;g)dt={\int }_0^1\lambda (t;f)dt.$$

$${\int }_0^s\mu (t;g)dt⩽{\int }_0^s\mu (t;f)dt,\forall s\in [0,1).$$

$$\Omega (y):=\{x\in L_1(\mathcal{M},\tau {)}_h:x\prec y\}.$$

$$\lambda (x+y)\prec \lambda (x)+\lambda (y),\forall x,y\in L_1(\mathcal{M},\tau ).$$

$$E^x(s,\infty )\le e\le E^x[s,\infty ),$$

$$\lambda (t;xe)=\lambda (t;x)\text{mbox}forallt\in [0,\tau (e)),$$

$$\lambda (t;x{e}^{\perp })=\lambda (t+\tau (e);x)\text{mbox}forallt\in [0,\tau (e^{\perp })).$$

$${\int }_0^1\lambda (s;x)\check{\lambda }(s;y)ds\le \tau (xy)\le {\int }_0^1\lambda (s;x)\lambda (s;y)ds.$$

$${\int }_0^t\lambda (s;x)ds$$

$$=\max \{\tau (xe);e\in \mathcal{P}(\mathcal{M}),\tau (e)=t,ex=xe\}.$$

$$u:=p_1-\frac{T_1}{T_2}{p}_2.$$

$$\delta <\min \{\frac{a_1-a_2}{2},\frac{(a_3-a_4)T_2}{2T_1},\frac{(2a_1-a_2-a_3)T_2}{2T_1},\frac{a_2+a_3-2a_4}{2}\}.$$

$${\int }_0^1\lambda (t;x_{\pm })dt=\tau (x_{\pm })=\tau (x\pm u)=\tau (x)={\int }_0^1\lambda (s;x)ds={\int }_0^1\lambda (t;y)dt.$$

$${\int }_{s_1}^{s_4}\lambda (t;x_{\pm })dt-{\int }_{s_1}^{s_4}\lambda (t;x)dt={\int }_0^1\lambda (t;x_{\pm })dt-{\int }_0^1\lambda (t;x)dt=0.$$

$${\int }_0^s\lambda (t;x_{\pm })dt={\int }_0^s\lambda (t;x)dt⩽{\int }_0^s\lambda (t;y)dt,s\notin [s_1,s_4],$$

$${\int }_0^s\lambda (t;x_{\pm })dt⩽{\int }_0^{s_1}\lambda (t;x)dt+\lambda (s_1;x)(s-s_1)\stackrel{(ii)}{⩽}{\int }_0^s\lambda (t;y)dt,s\in [s_1,s_4].$$

$$N=\max \left(-\lambda (s_1;y),0\right)+\max \left(-\lambda (s_1;x),0\right)+1.$$

Full text

Dedicated to the memory of Professor Wilhelmus Anthonius Josephus Luxemburg (1929–2018)

1 Introduction

In 1967, W.A.J. Luxemburg raised the following question (see [30, Problem 1]):

Determine all the extreme points of $\Omega(f)$, $f\in L_{1}(X,\Sigma,\nu)$, for an arbitrary finite measure space $(X,\Sigma,\nu)$, where $\Omega(f)$ is the set of all integrable functions on $(X,\Sigma,\nu)$ majorised by $f$ in the sense of Hardy–Littlewood–Pólya.

The case when $X=\{1,2,\cdots,n\}$ with counting measure is well-known. Let $\prec$ be the partial order of Hardy, Littlewood and Pólya for real $n$-vectors. It is known that $y\prec x$ if and only if $y$ belongs to the convex hull $\Omega(x)$ of the set of permutations of $x$ [19], i.e., the convex hull of $\{Px:P\mbox{ is a permutation matrix}\}$. Moreover, the extreme points of $\Omega(x)$ are precisely the permutations of $x$ (see e.g. [19]). If $(X,\Sigma,\nu)$ is an arbitrary finite measure space and $f,g\in L_{1}(X)$, then $g$ is called majorised by $f$ in the sense of Hardy–Littlewood–Pólya (denoted by $g\prec f$) provided that $\int_{0}^{s}\lambda(t;g)dt\leqslant\int_{0}^{s}\lambda(t;f)dt$ for all $s\in[0,1)$ and $\int_{0}^{1}\lambda(t;g)dt=\int_{0}^{1}\lambda(t;f)dt,$ where $\lambda(f)$ is the right-continuous equimeasurable nonincreasing rearrangement of $f$ (see [8, 33] for the definition, see also Section 2 below). In the particular case when $(X,\Sigma,\nu)$ is atomless, the set $\Omega(f):=\{g\in L_{1}(X):g\prec f\}$ is said to be the orbit of $f$ (with respect to doubly stochastic operators) [36, 34, 33]. It was first proved by Ryff [34] that if $f,g\in L_{1}(0,1)$ and $g\sim f$ (i.e. $g\prec f$ and $f\prec g$), then $g$ is an extreme point of $\Omega(f)$. In 1967, Luxemburg [30] extended the result by Ryff [34] to the setting of an arbitrary atomless finite measure space $(X,\Sigma,\nu)$. However, the converse implication was left unresolved in [34] and [30], and was later treated by Ryff [35] (see also [36]), who proved the following characterisation.

Let $(X,\Sigma,\nu)$ be a atomless finite measure space. Let $f=L_{1}(X,\Sigma,\nu)$ and $g\in\Omega(f)$. Then, $g$ is an extreme point of $\Omega(f)$ if and only $g$ is equimeasurable with $f$ (i.e. $g\sim f$).

However, the case of arbitrary finite measure spaces (e.g. $L_{1}(0,\frac{1}{2})\oplus\mathbb{C}e$) seems to have been left open and this case cannot be obtained as a corollary from the atomless case (see Remark 1.3). The main object of the present paper is to provide a complete answer to Luxemburg’s question. Moreover, we consider this question in a much more general setting, giving a characterisation for the extreme points of the set $\Omega(y)$ of all self-adjoint operators majorised by a self-adjoint operator $y$ in the noncommutative $L_{1}$-space affiliated to a finite von Neumann algebra (see e.g. [20, 21, 26] for related partial results in the noncommutative setting).

Let $\mathcal{M}$ be a von Neumann algebra equipped with a faithful normal finite trace $\tau$ and $L_{1}(\mathcal{M},\tau)_{h}$ (resp. $L_{1}(\mathcal{M},\tau)_{+}$) be the set of all self-adjoint (resp. positive) operators in the noncommutative $L_{1}$-space $L_{1}({\mathcal{M}},\tau)$. Petz [32] introduced the spectral scale $\lambda(x)$ of a $\tau$-measurable self-adjoint operator $x$. In the special case when ${\mathcal{M}}$ is commutative (and hence, the pair $({\mathcal{M}},\tau)$ can be identified with $L_{\infty}(X,\Sigma,\nu)$), the the notion of spectral scales coincides with the non-increasing rearrangements. For detailed discussions of this notion, we refer to [21] (see also [20, 16, 15]). Theorem 1.1 below is the main result of the present paper, which unifies Ryff’s theorem [35, 36] and the classic result for vectors [19] with significant extension. The following theorem yields the complete resolution of Luxemburg’s problem in the general setting.

Theorem 1.1

Assume that ${\mathcal{M}}$ is a von Neumann algebra equipped with a faithful normal tracial state $\tau$. Let $y\in L_{1}({\mathcal{M}},\tau)_{h}$ and let $\Omega(y)$ be defined as the set of all self-adjoint operators $x\in L_{1}({\mathcal{M}},\tau)$ satisfying $\lambda(x)\prec\lambda(y).$ Then, $x$ is an extreme point if and only if for each $t\in(0,1)$, one of the following options holds:

1. (1).

$\lambda(t;x)=\lambda(t;y)$; 2. (2).

$\lambda(t;x)\neq\lambda(t;y)$* with the spectral projection $E^{x}\{\lambda(t;x)\}$ being an atom in ${\mathcal{M}}$ and*

[TABLE]

Since any finite measure space is a von Neumann algebra equipped with a faithful normal finite trace, the complete resolution of Luxemburg’s problem is an immediate consequence of Theorem 1.1. Without loss of generality, we state the result for normalized measure spaces.

Corollary 1.2

Assume that $(X,\Sigma,\nu)$ is an arbitrary normalized measure space. Let $y\in L_{1}(X)$ and let $\Omega(y)$ be the set of all integrable functions on $(X,\Sigma,\nu)$ majorised by $y$ in the sense of Hardy–Littlewood–Pólya. Then, $x$ is an extreme point if and only if for each $t\in(0,1)$, one of the following options holds:

1. (1).

$\lambda(t;x)=\lambda(t;y)$; 2. (2).

$\lambda(t;x)\neq\lambda(t;y)$* with the preimage $x^{-1}(\lambda(t;x))$ being an atom in $(X,\Sigma,\nu)$ and*

[TABLE]

Remark 1.3

Let $f(t):=\frac{1}{\sqrt{t}}$, $t\in(0,\frac{1}{2})$. Hence, $f\oplus 0\in L_{1}(0,\frac{1}{2})\oplus\mathbb{C}e$, where $e$ is an atom with measure $\frac{1}{2}$. Even though $L_{1}(0,\frac{1}{2})\oplus\mathbb{C}e$ can be embedded into an atomless finite measure space, Ryff’s theorem only provides a sufficient condition for $g\in L_{1}(0,\frac{1}{2})\oplus\mathbb{C}e$ to be an extreme point of $\Omega(f)$. That is, every $g$ with $\lambda(g)=\lambda(f\oplus 0e)$ is an extreme point of $\Omega(f\oplus 0e)$. However, there exist extreme points of $\Omega(f\oplus 0e)$ which could not be described in this way. By Corollary 1.2, for every $a\in[0,\frac{1}{2})$, $g:=f\chi_{(0,a)}\oplus 2e\int_{a}^{\frac{1}{2}}f(t)dt$ is an extreme point of $\Omega(f\oplus 0e)$. In particular, $0\oplus 2\sqrt{2}e$ is an extreme point of $\Omega(f\oplus 0e)$.

Let $M_{n}(\mathbb{C})$ be the $n\times n$ matrices and $E_{D}$ be the compression onto $D$, the diagonal masa (see e.g. [38]) in $M_{n}(\mathbb{C})$. The celebrated Schur-Horn theorem [23, 37] sates that

[TABLE]

where $M_{\alpha}$ is the diagonal matrix with the entries of $\alpha$ in the main diagonal. Inspired by the Arveson–Kadison conjecture [2], several authors (e.g. [1, 25, 3, 4, 31, 27]) have worked on the analogues to the Schur–Horn theorem. In particular, Argerami and Massey [1] established a result for type II1 factors. As an application of Theorem 1.1, by applying results in [23, 37] and [1], we obtain the following.

Corollary 1.4

Given a type II1 factor ${\mathcal{M}}$ (resp. ${\mathcal{M}}=M_{n}(\mathbb{C})$) and a diffuse abelian von Neumann subalgebra ${\mathcal{A}}\subset{\mathcal{M}}$ (resp. ${\mathcal{A}}$ is the diagonal masa in $M_{n}(\mathbb{C})$), for every $y\in{\mathcal{M}}_{h}$, we have

[TABLE]

where $E_{\mathcal{A}}$ is the conditional expectation onto ${\mathcal{A}}$ and ${\mathcal{U}}_{\mathcal{M}}(y)$ is the unitary orbit of $y$ in ${\mathcal{M}}$.

In particular, by the Krein–Milman theorem, $\overline{E_{\mathcal{A}}({\mathcal{U}}_{\mathcal{M}}(y))}^{\sigma\text{-sot}}$ is the closed convex hull of $\{x\in{\mathcal{A}}_{h}:\lambda(x)=\lambda(y)\}$ in the $\sigma$-strong operator topology.

Finally, we comment briefly at the interconnection of spectral scales and singular value functions. For a given positive $\tau$-measurable operator $x$ affiliated to ${\mathcal{M}}$, the spectral scale $\lambda(x)$ of the operator is equal to the singular values function $\mu(x),$ see e.g. [18]. However, for a general self-adjoint $x\in L_{1}(\mathcal{M},\tau)_{h}$ the spectral scale $\lambda(x)$ and singular value function $\mu(x)$ are different, and so one can consider other majorisations given by singular value functions. Namely, we say that $y$ is submajorised by $x$ (denoted by $y\prec\prec x$) if $\int_{0}^{s}\mu(t;y)dt\leqslant\int_{0}^{s}\mu(t;x)dt$ for all $s\in[0,\tau(\mathbf{1})).$ In [39] and [40] (see also [5]), it is proved that for an arbitrary atomless finite von Neumann algebra with a faithful normal finite trace $\tau$ and for any $x\in L_{1}(\mathcal{M},\tau)_{h}$, the extreme points of $\{y\in L_{1}(\mathcal{M},\tau)_{h}\ :\ y\prec\prec x\}$ are exactly the operators $y$ satisfying $\mu(y)=\mu(x)$ (see [10, 11, 12, 26, 20] for more related results). The result stated in Theorem 1.1 does not follow from that of [5] even in the setting of atomless von Neumann algebras. Our proof of Theorem 1.1 is completely different from that in [5] and is based on a careful study of Hardy–Littlewood–Pólya majorisation in the commutative and the noncommutative setting.

2 Preliminaries

Throughout this paper, we denote by $\mathcal{M}$ a von Neumann algebra equipped with a faithful normal finite trace $\tau$. We denote by $\mathbf{1}$ the identity in $\mathcal{M}$ and by $\mathcal{P}(\mathcal{M})$ the collection of all orthogonal projections in $\mathcal{M}$. Without loss of generality, we assume that $\tau({\bf 1})=1$. A densely defined closed linear operator $x$ affiliated with $\mathcal{M}$ is called $\tau$-measurable if for each $\varepsilon>0$ there exists $e\in\mathcal{P}(\mathcal{M})$ with $\tau(e^{\perp})\leq\varepsilon$ such that $e(\mathcal{H})\subset\mathfrak{D}(x)$. Let us denote by $S(\mathcal{M},\tau)$ the set of all $\tau$-measurable operators. We note that, since we assume that the trace $\tau$ is finite, the space $S({\mathcal{M}},\tau)$ is the set of all densely defined closed linear operators $x$ affiliated with $\mathcal{M}$. However, if the trace $\tau$ is infinite, then there are densely defined closed linear operators which are not $\tau$-measurable. The set of all self-adjoint elements in $S(\mathcal{M},\tau)$ is denoted by $S(\mathcal{M},\tau)_{h}$, which is a real linear subspace of $S(\mathcal{M},\tau)$. The set of all positive elements in $S(\mathcal{M},\tau)_{h}$ is denoted by $S(\mathcal{M},\tau)_{+}$.

If $x\in S(\mathcal{M},\tau)_{h}$, then the spectral distribution function $d(x)$ of $x$ is defined by setting

[TABLE]

where $E^{x}(s,\infty)$ is spectral projection of $x$ on the interval $(s,\infty).$

It is clear that the function $d(x):\mathbb{R}\rightarrow[0,\tau(\mathbf{1})]$ is decreasing and the normality of the trace implies that $d(x)$ is right-continuous. If $x\in S(\mathcal{M},\tau)$, then the singular value function $\mu(x):[0,1)\rightarrow[0,\infty]$ [18, 17, 16] is defined to be the decreasing right-continuous inverse of the spectral distribution function $d(|x|)$, that is,

[TABLE]

We introduce the notion of spectral scales (see [32], see also [21, 22, 15, 16, 1]). If $x\in S(\mathcal{M},\tau)_{h}$, then the spectral scales (also called eigenvalue functions) $\lambda(x):[0,1)\rightarrow(-\infty,\infty]$ and $\check{\lambda}(x):[0,1)\rightarrow(-\infty,\infty]$ are defined by

[TABLE]

and

[TABLE]

The spectral scales $\lambda(x)$ (resp. $\check{\lambda}(x)$) are decreasing (resp. increasing) right-continuous functions. It is an immediate corollary of [32, Proposition 1] (see also [16, Chapter III, Remark 5.4]) that

[TABLE]

If $x\in S(\mathcal{M},\tau)_{+}$, then it is evident that $\lambda(t;x)=\mu(t;x)$ for all $t\in[0,1)$. Note [16, 15] that

[TABLE]

where $\lambda((\tau(\mathbf{1})-t)-;x)=\lim_{\varepsilon\rightarrow 0^{+}}\lambda((\tau(\mathbf{1})-t)-\varepsilon;x)$.

Assume that $\mathcal{M}=L_{\infty}(0,1)$ and $\tau(f)=\int_{0}^{1}fdm$, where $m$ the Lebesgue measure on $(0,1)$. In this case, $S(\mathcal{M},\tau)_{h}$ consists of all real measurable functions $f$ on $(0,1)$. For every $f$, $\lambda(f)$ coincides with the right-continuous equimeasurable nonincreasing rearrangement $\delta_{f}$ of $f$ (see e.g. [21]):

[TABLE]

Using the extended trace $\tau:S(\mathcal{M},\tau)_{+}\rightarrow[0;\infty]$ to a linear functional on $S(\mathcal{M},\tau),$ denoted again by $\tau$, the noncommutative $L_{1}$-space (written by $L_{1}(\mathcal{M},\tau)$) is defined by

[TABLE]

(see e.g. [13, p.84]). Let us denote $L_{1}(\mathcal{M},\tau)_{h}=\{x\in L_{1}(\mathcal{M},\tau):x=x^{\ast}\}$ and $L_{1}(\mathcal{M},\tau)_{+}=L_{1}(\mathcal{M},\tau)\cap S(\mathcal{M},\tau)_{+}.$ We note that for every $x\in L_{1}({\mathcal{M}},\tau)_{h}$, we have (see e.g. [15], [32, Proposition 1] and [16, Chapter III, Proposition 5.5])

[TABLE]

For every $f,g\in L_{1}({\mathcal{M}},\tau),$ $g$ is said to be majorised by $f$ in the sense of Hardy–Littlewood–Pólya (written by $g\prec f$) if

[TABLE]

for all $s\in[0,1)$ and

[TABLE]

$g$ is said to be submajorised by $f$ in the sense of Hardy–Littlewood–Pólya (written by $g\prec\prec f$) if and only if

[TABLE]

For a self-adjoint element $y\in L_{1}(\mathcal{M},\tau)_{h}$, we denote

[TABLE]

We note that

[TABLE]

For every $x\in S({\mathcal{M}},\tau)_{h}$ and $s\in\mathbb{R}$, if $e\in{\mathcal{P}}({\mathcal{M}})$ is such that

[TABLE]

then (see [15] or [16, Chapter III, Lemma 8.2]):

[TABLE]

and

[TABLE]

Suppose that $x,y\in S({\mathcal{M}},\tau)_{h}$. If $\lambda(|x|)\lambda(|y|)\in L_{1}(0,1)$, then (see [15, Proposition 2.3], see also [16, Chapter III, Theorem 6.6]) $xy\in L_{1}({\mathcal{M}},\tau)$ and

[TABLE]

If, in addition, ${\mathcal{M}}$ is atomless, then for any $x\in L_{1}({\mathcal{M}},\tau)$ and $t\in[0,1)$, we have (see e.g. [21, Proposition 1.1] and [16, Chapter III, Remark 8.6])

[TABLE]

3 Some technical results

Let ${\mathcal{M}}$ be a von Neumann algebra equipped with a finite faithful normal trace. Some of the results in this section are well-known for positive operators (see e.g. [16, 24, 6, 7]). However, the results in this section do not follow from those for positive operators.

We note that for any $y\in L_{1}(0,1)$, the notation $\Omega(y):=\{x\in L_{1}({\mathcal{M}},\tau):\lambda(x)\prec y\}$ makes sense. All results in the present section hold true for $y\in L_{1}(0,1)$ and $x\in L_{1}({\mathcal{M}},\tau)$. However, to avoid ambiguity, we always assume that $y\in L_{1}({\mathcal{M}},\tau)$.

The following is a noncommutative analogue of Ryff’s Proposition stated in [35].

Lemma 3.1

Let $y\in L_{1}({\mathcal{M}},\tau)_{h}.$ If $x\in\Omega(y)=\{x\in L_{1}(\mathcal{M},\tau)_{h}:\ x\prec y\}$ satisfies that

1. (i)

$\lambda(s_{i+1};x)<\lambda(s_{i};x)$* for some $0<s_{i}<s_{i+1}<1,$ $i=1,2,3$,* 2. (ii)

$\int_{0}^{s_{1}}\lambda(t;x)dt+\lambda(s_{1};x)(s-s_{1})\leq\int_{0}^{s}\lambda(t;y)dt,$* for all $s\in[s_{1},s_{4}],$*

then $x\not\in\emph{extr}(\Omega(y)).$

Proof 3.2**.**

For the sake of convenience, we denote $a_{i}=\lambda(s_{i};x)$, $i=1,2,3,4$. Let $p_{1}:=E^{x}[\frac{a_{2}+a_{3}}{2},\frac{a_{1}+a_{2}}{2})$ and $p_{2}:=E^{x}[\frac{a_{3}+a_{4}}{2},\frac{a_{2}+a_{3}}{2})$. We denote $T_{1}=\tau(p_{1})$ and $T_{2}=\tau(p_{2})$. Observe that $T_{1},T_{2}\in(0,1)$ and $p_{1}p_{2}=0$. Set

[TABLE]

It is clear that $\tau(u)=0$. Assume that $\delta>0$ such that

[TABLE]

Let $x_{\pm}=x\pm\delta u$. By the spectral theorem, $\lambda(s;x_{\pm})=\lambda(s;x)$ for $s\not\in[s_{1},s_{4}]$ and $\lambda(s;x_{\pm})\leqslant\lambda(s_{1};x)$ for $s\in[s_{1},s_{4}].$ We assert that $x_{\pm}\prec y$.

*Note that *

[TABLE]

Since $\lambda(s;x_{\pm})=\lambda(s;x)$ for $s\not\in[s_{1},s_{4}]$, it may be concluded that

[TABLE]

Hence,

[TABLE]

where the last inequality follows from the assumption that $x\in\Omega(y).$ On the other hand, since $\lambda(x)$ is decreasing, it follows that

[TABLE]

Hence, $x_{\pm}\in\Omega(y)$ and $x=\frac{1}{2}(x_{+}+x_{-}).$ That is, $x\not\in\emph{extr}(\Omega(y))$.

Lemma 1**.**

Let $y\in L_{1}({\mathcal{M}},\tau).$ If $x\in\emph{extr}(\Omega(y))$ and $\int_{0}^{s}\lambda(t;y)dt>\int_{0}^{s}\lambda(t;x)dt$ for all $s\in(t_{1},t_{2}),$ then $\lambda(x)$ is a step function on $(t_{1},t_{2})$.

Proof 3.3**.**

Assume by contradiction that there exists $s_{1}\in(t_{1},t_{2})$ such that $(s_{1}-\varepsilon,s_{1})$ and $(s_{1},s_{1}+\varepsilon)$ are not constancy intervals of $\lambda(x)$ for some $\varepsilon>0.$ Define

[TABLE]

Let $Y=y+N{\bf 1}$ and $X=x+N{\bf 1}$. Since $x\in{\rm extr}(\Omega(y))$, it follows that $X\in{\rm extr}(\Omega(Y))$. Moreover, by spectral theorem (see also [20, Remark 6.1 (5)] or [16, Chapter III, Proposition 5.7]), we have $\lambda(X)=\lambda(x+N{\bf 1})=\lambda(x)+N$ and $\lambda(Y)=\lambda(y+N{\bf 1})=\lambda(y)+N$. Therefore,

[TABLE]

Note that $\lambda(s_{1};Y),\lambda(s_{1};X)>0$. Since $\lambda(Y)$ and $\lambda(X)$ are decreasing functions, it follows that $\lambda(Y)$ and $\lambda(X)$ are strictly positive on $(0,s_{1}]$.

Since $\lambda(Y)$ and $\lambda(X)$ are right-continuous, one can choose $s_{2}>s_{1}$ such that $\lambda(s_{2};Y)>0$ and $\int_{0}^{s}\lambda(t;Y)-\lambda(t;X)dt>0$ for every $s\in(t_{1},s_{2})$ and

[TABLE]

Hence, for any $s\in[s_{1},s_{2}]$, we have

[TABLE]

Since $Y$ and $X$ satisfy the assumptions in Lemma 3.1, it follows that $X\not\in\text{extr}(\Omega(Y)).$ Consequently, $x\not\in\text{extr}(\Omega(y)).$

Let $x\in L_{1}({\mathcal{M}},\tau)_{h}$. Denote by ${\mathcal{N}}$ the abelian von Neumann algebra generated by all spectral projections of $x$. In particular, $x\in L_{1}({\mathcal{N}},\tau)_{h}$. Recall that (see [21, Proposition 1.1])

[TABLE]

Assume that $\tau(xe)=\int_{0}^{s}\lambda(t;x)dt$ for a projection $e\in{\mathcal{M}}$ with $\tau(e)=s$. Let $E_{\mathcal{N}}$ be a conditional expectation from $L_{1}({\mathcal{M}},\tau)$ onto $L_{1}({\mathcal{N}},\tau)$ (see e.g. [42], see also [14, Proposition 2.1]). In particular, $E_{\mathcal{N}}(e)\leq{\bf 1}$ and $\tau(E_{\mathcal{N}}(e))=\tau(e)=s$. Moreover,

[TABLE]

We note that $E_{\mathcal{N}}(f)$, $f\in{\mathcal{P}}({\mathcal{M}})$, is not necessarily a projection [40].

Proposition 2**.**

Under the above assumptions on $e$, we have

[TABLE]

Proof 3.4**.**

We present the proof for the first inequality and a similar argument yields that $E_{\mathcal{N}}(e)\leq E^{x}[\lambda(s;x),\infty)$.

Without loss of generality, we may assume that $E^{x}(\lambda(s;x),\infty)\neq 0$. Since ${\mathcal{N}}$ is a commutative algebra and $0\leq E_{\mathcal{N}}(e)\leq 1$, it follows that

[TABLE]

If $E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty)=E^{x}(\lambda(s;x),\infty)$, then

[TABLE]

which proves the first inequality of (9).

Now, we assume that

[TABLE]

This implies that

[TABLE]

Since $\tau(E^{x}(\lambda(s;x),\infty))\leq s$, it follows that $\lambda(t;E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty))=0$ when $t>s$. Hence,

[TABLE]

In particular,

[TABLE]

Assume that $\tau(E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty))=a_{1}\geq 0$ and $\tau(E_{\mathcal{N}}(e)E^{x}(-\infty,\lambda(s;x)])=a_{2}\geq 0$. Observe that $a_{1}+a_{2}=s=\tau(E_{\mathcal{N}}(e))$. Moreover, by (1), for every $0<t<\tau(E^{x}(\lambda(s;x),\infty))$, we have $\lambda(t;x)>\lambda(s;x)$. Hence,

[TABLE]

Recall that $\int_{0}^{s}\lambda(t;E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty))dt=\tau(E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty))$. The above inequality implies that

[TABLE]

Since $\lambda(s;x)a_{2}=\tau(\lambda(s;x)E_{\mathcal{N}}(e)E^{x}(-\infty,\lambda(s;x)])\geq\tau(E_{\mathcal{N}}(e)E^{x}(-\infty,\lambda(s;x)]x)$, it follows that

[TABLE]

which is a contradiction with (8). Hence, the equality $E_{\mathcal{N}}(e)E^{x}(\lambda(s;x),\infty)=E^{x}(\lambda(s;x),\infty)$ holds, and therefore, by (10), we have $E_{\mathcal{N}}(e)\geq E^{x}(\lambda(s;x),\infty)$.

Lemma 3**.**

Let $x\in L_{1}(\mathcal{M},\tau)_{h}$. Let $0<s<\tau(\mathbf{1})=1$ and let $a$ be in the unit ball of $\mathcal{M}_{+}$ such that $\tau(a)=s$ and $\tau(xa)=\int_{0}^{s}\lambda(t;x)dt.$ If $\lambda(x)$ is not a constant in any left neighborhood of $s,$ then $a=E^{x}(\lambda(s;x),\infty).$

Proof 3.5**.**

Let $\mathcal{N}$ be the commutative von Neumann subalgebra of $\mathcal{M}$ generated by the spectral projections of $x$. Clearly, the restriction of $\tau$ to $\mathcal{N}$ is finite. There exists a conditional expectation $E_{\mathcal{N}}$ from $L_{1}(\mathcal{M},\tau)$ to $L_{1}(\mathcal{N},\tau)$ [14, Proposition 2.1]. In particular, for any $z\in L_{1}({\mathcal{M}},\tau)$, we have $E_{\mathcal{N}}(z)\prec\prec z$ (see e.g. [14, Proposition 2.1 (g)]) and,

[TABLE]

Moreover, for every $z\in{\mathcal{M}}$ and $y\in L_{1}({\mathcal{N}},\tau)$, we have

[TABLE]

Since $a$ is positive, it follows that $\lambda(a)=\mu(a)$ (see Section 2 or [20]) and, therefore,

[TABLE]

for all $r\in(0,1)$ and,

[TABLE]

Moreover, since $E_{\mathcal{N}}$ is a contraction on ${\mathcal{M}}$ and $\left\|a\right\|_{\infty}\leq 1$, it follows that $\lambda(E_{\mathcal{N}}(a))\leqslant 1$ [14, Proposition 2.1 (g)]. We set $y:=(x-\lambda(s;x))_{+}.$ Note that

[TABLE]

Therefore,

[TABLE]

Since $\lambda(x)$ is a decreasing function, we have

[TABLE]

It follows from $\tau(a)=s$ and $\tau(xa)=\int_{0}^{s}\lambda(t;x)dt$ that

[TABLE]

*Thus, *

[TABLE]

Since $y$ is positive and $\lambda(E_{\mathcal{N}}(a))\leqslant 1,$ we conclude that $\lambda(t;y)(1-\lambda(t;E_{\mathcal{N}}(a)))\geqslant 0.$ Hence, $\lambda(t;y)(1-\lambda(t;E_{\mathcal{N}}(a)))=0$ for all $t\in(0,s).$ Recall that $\lambda(x)$ is not a constant in any left neighborhood of $s$. We obtain that $\lambda(y)>0$ on $(0,s)$. Recall that $E_{\mathcal{N}}(a)\geq 0$ with $\tau(E_{\mathcal{N}}(a))=\tau(a)=s$. We obtain that $\lambda(E_{\mathcal{N}}(a))=1$ on $(0,s]$ and $\lambda(E_{\mathcal{N}}(a))=0$ on $[s,1)$. This implies that $E_{\mathcal{N}}(a)$ is a projection in ${\mathcal{N}}$. Hence, $E_{\mathcal{N}}(a)=E_{\mathcal{N}}(a)E_{\mathcal{N}}(a)=E_{\mathcal{N}}(a\cdot E_{\mathcal{N}}(a))$ and $E_{\mathcal{N}}(a({\bf 1}-E_{\mathcal{N}}(a)))=0$. It follows that

[TABLE]

Therefore, $a^{1/2}({\bf 1}-E_{\mathcal{N}}(a))a^{1/2}=0$ and $a^{1/2}=E_{\mathcal{N}}(a)a^{1/2}$. By the assumption that $a\leq 1$, we have

[TABLE]

Recall that $\tau(E_{\mathcal{N}}(a))=\tau(a)$. Hence, $\tau(E_{\mathcal{N}}(a)-a)=0$. Due to the faithfulness of the trace $\tau$, we obtain that $a=E_{\mathcal{N}}(a)\in{\mathcal{P}}({\mathcal{N}})$. Since $\mathcal{N}$ is a commutative von Neumann subalgebra of $\mathcal{M}$ generated by all spectral projections of $x$, it follows that $a=E^{x}\{B\}$ for some Borel set $B\subset[0,1]$. By [32, Proposition 1] and the assumption that $\tau(ax)=\int_{0}^{s}\lambda(t;x)dt$, we have

[TABLE]

Moreover, since $\lambda$ is decreasing and is non-constant in any left neighborhood of $s$, it follows that $B=(\lambda(s;x),\infty)$.

Remark 3.6

Let $x\in L_{1}({\mathcal{M}},\tau)_{+}$. By (6), for any $a$ in the unit ball of ${\mathcal{M}}_{+}$ with $\tau(a)<s$, we have

[TABLE]

Since $\lambda(x)$ is a non-negative decreasing function, $0\leq\lambda(a)\leq 1$ and $\tau(a)=\int_{0}^{1}\lambda(t;a)dt<s$, it follows that

[TABLE]

Then, by (7), we obtain that

[TABLE]

Lemma 3 together with (18) implies that if $a$ is in the unit ball of $\mathcal{M}_{+}$ such that $\tau(a)\leq s$ and $\tau(xa)=\int_{0}^{s}\lambda(t;x)dt,$ and $\lambda(x)$ is not a constant in any left neighborhood of $s,$ then $a=E^{x}(\lambda(s;x),\infty).$

Corollary 4**.**

Let $x\in L_{1}(\mathcal{M},\tau)_{h}$. Let $0<s<\tau(\mathbf{1})=1$ and let $e\in{\mathcal{P}}(\mathcal{M})$ be such that $\tau(e)=s$ and $\tau(xe)=\int_{0}^{s}\lambda(t;x)dt.$ Then,

[TABLE]

Proof 3.7**.**

By Lemma 3, it suffices to prove the case when $\lambda(x)$ is a constant on a left neighbourhood of $s$. Denote by ${\mathcal{N}}$ the von Neumann algebra generated by all spectral projections of $x$. Let $E_{\mathcal{N}}$ be a conditional expectation from $L_{1}({\mathcal{M}},\tau)$ onto $L_{1}({\mathcal{N}},\tau)$. Let $\lambda:=\lambda(s;x)$ and $x_{1}:=(x-\lambda)E^{x}(\lambda,\infty)$. Recall that $E^{x}[\lambda,\infty)\geq E_{\mathcal{N}}(e)\geq E^{x}(\lambda,\infty)$ (see (9)). In particular, $E_{\mathcal{N}}(e)E^{x}(\lambda,\infty)=E^{x}(\lambda,\infty)$. Observing that $E^{x}(\lambda,\infty)$ is the support of $x_{1}$, we have

[TABLE]

Note that $0\leq E^{x}(\lambda,\infty)eE^{x}(\lambda,\infty)\leq\textbf{1}$ and $\tau(E^{x}(\lambda,\infty)eE^{x}(\lambda,\infty))\leq\tau(E^{x}(\lambda,\infty))$. Since $x_{1}\geq 0$, it follows from Remark 3.6 that

[TABLE]

That is, $e\geq E^{x}(\lambda,\infty)$. Let $e_{1}:=e-E^{x}(\lambda,\infty)\in{\mathcal{P}}({\mathcal{M}})$. We have

[TABLE]

Hence, by the assumption that $\lambda=\lambda(s;x)$, we obtain that

[TABLE]

We have

[TABLE]

Therefore, $e_{1}(\lambda-xE^{x}(-\infty,\lambda])e_{1}=0$. Since $\lambda-xE^{x}(-\infty,\lambda]\geq 0$, it follows that $(\lambda-xE^{x}(-\infty,\lambda])^{1/2}e_{1}=0$. Hence,

[TABLE]

and

[TABLE]

This implies that $e_{1}\leq E^{x}\{\lambda(s;x)\}$, which completes the proof.

The following proposition is similar to a well-known property of rearrangements of functions, see [28, property $9^{0}$, p. 65] and [20, Theorem 3.5].

Proposition 5**.**

Let $x,\ x_{1},\ x_{2}\in L_{1}(\mathcal{M},\tau)_{h}$ be such that $x=(x_{1}+x_{2})/2$ and $\lambda(x_{1})=\lambda(x_{2})=\lambda(x).$ Then, $x=x_{1}=x_{2}$.

Proof 3.8**.**

Fix $\theta\in(\lambda(1^{-};x),\lambda(0,x))$. Define $s$ by setting

[TABLE]

If $s>0$, then

[TABLE]

This means that $\lambda(x)$ is not constant in any left neighborhood of $s$ whenever $s>0$.

Fix a projection $e=E^{x_{1}+x_{2}}(\lambda(s;x_{1}+x_{2}),\infty)=E^{2x}(\lambda(s;2x),\infty)=E^{x}(\lambda(s;x),\infty)\in\mathcal{M}.$ Clearly, $\tau(e)=s$. By (7), we have

[TABLE]

Hence, $\int_{0}^{s}\lambda(t;x_{i})dt=\tau(ex_{i}),\;i=1,2$. From Lemma 3, we obtain that $e=E^{x}(\lambda(s;x),\infty)=E^{x_{1}}(\lambda(s;x_{1}),\infty)=E^{x_{2}}(\lambda(s;x_{2}),\infty).$ Hence, we have

[TABLE]

Note that (for convenience, we denote $E^{x}[\infty,\infty)=E^{x_{1}}[\infty,\infty)=E^{x_{2}}[\infty,\infty)=0$)

[TABLE]

which together with (3.8) implies that

[TABLE]

We note by $l(z)$ the left support of $z\in S({\mathcal{M}},\tau)$. Since $\lambda(1^{-};x)=\lambda(1^{-};x_{1})=\lambda(1^{-};x_{2})$ and $E^{x}\{\lambda(1^{-};x)\}={\bf 1}-l((x-\lambda(1^{-};x))_{+})={\bf 1}-l((x_{1}-\lambda(1^{-};x_{1}))_{+})=E^{x_{1}}\{\lambda(1^{-};x_{1})\}={\bf 1}-l((x_{2}-\lambda(1^{-};x_{2}))_{+})=E^{x_{2}}\{\lambda(1^{-};x_{2})\}$, it follows that

[TABLE]

The following lemma provides the proof of the implication “$\Leftarrow$” in Theorem 1.1.

Lemma 6**.**

Let $y\in L_{1}({\mathcal{M}},\tau)$. Let $x,x_{1},x_{2}\in\Omega(y)$ with $x=\frac{x_{1}+x_{2}}{2}$. If $x$ satisfies that for every $t\in(0,1)$, one of the followings holds:

1. (1).

$\lambda(t;y)=\lambda(t;x)$; 2. (2).

$\lambda(t;y)\neq\lambda(t;x)$* with $E^{x}\{\lambda(t;x)\}$ is an atom and*

[TABLE]

then $x_{1}=x_{2}=x$.

Proof 3.9**.**

By the definition of $x$ and assumption (2) above, for every $t$ such that $\lambda(t;x)=\lambda(t;y)$, we have $\int_{0}^{t}\lambda(s;x)ds=\int_{0}^{t}\lambda(s;y)ds$. For any $t$ such that $\lambda(t;y)\neq\lambda(t;x)$, we denote $[t_{1},t_{2})=\{s:\lambda(s;x)=\lambda(t;x)\}$, $t_{1}<t_{2}$. In particular, we have

[TABLE]

and

[TABLE]

Since $x_{1},x_{2}\in\Omega(y)$ and

[TABLE]

it follows that

[TABLE]

The same argument with $t_{1}$ replaced with $t_{2}$ yields that

[TABLE]

Let $e_{1}:=E^{x}(\lambda(t_{1};x),\infty)$ and $e_{2}:=E^{x}[\lambda(t_{1};x),\infty)$. In particular, $e_{2}-e_{1}=E^{x}\{\lambda(t_{1};x)\}$. Observe that $\tau(e_{1})=t_{1}$ and $\tau(e_{2})=t_{2}$ (due to the assumption that $[t_{1},t_{2})=\{s:\lambda(s;x)=\lambda(t;x)\}$). Note that, by (7) and the definition of spectral scales $\lambda(x)$, we have

[TABLE]

We obtain that $\tau(x_{1}e_{1})=\int_{0}^{t_{1}}\lambda(s;x_{1})ds=\int_{0}^{t_{1}}\lambda(s;x_{2})ds=\tau(x_{2}e_{1})$. By Corollary 4, we have

[TABLE]

and

[TABLE]

Similar argument with $t_{1}$ replaced with $t_{2}$ yields that

[TABLE]

and

[TABLE]

In particular, $e_{1}=E^{x_{1}}(\lambda(t_{1};x_{1}),\infty)+q$ for some subprojection $q$ of $E^{x_{1}}\{\lambda(t_{1};x_{1})\}$. Since $q$ commutes with $E^{x_{1}}\{\lambda(t_{1};x_{1})\}$, it follows that $q$ commutes with any spectral projection of $x_{1}$. Hence, $e_{1}$ commutes with $x_{1}$. The same argument implies that both $e_{1}$ and $e_{2}$ commute with $x_{1}$ and with $x_{2}$. Moreover, the atom $e:=e_{2}-e_{1}\in{\mathcal{P}}({\mathcal{M}})$ satisfies that

[TABLE]

By the spectral theorem, $\lambda(x_{1}e_{1})=\lambda(x_{1})$ and $\lambda(x_{2}e_{1})=\lambda(x_{2})$ on $(0,t_{1})$ (see (4)). On the other hand,

[TABLE]

and

[TABLE]

for all $t\in[0,t_{2}-t_{1})$. Since $e$ is an atom, it follows that $\lambda_{1}:=\lambda(t;x_{1}e)=\lambda(t+t_{1};x_{1})$ and $\lambda_{2}:=\lambda(t;x_{2}e)=\lambda(t+t_{1};x_{2})$ for every $t\in[0,t_{2}-t_{1})$. Indeed, combining (20) and (21), we have

[TABLE]

*Hence, $\lambda_{1}=\lambda_{2}=\lambda(t_{1};x)$. That is, $\lambda(x_{1})=\lambda(x_{2})=\lambda(x)$ on $[t_{1},t_{2})$. Since $t$ is arbitrary, it follows that $\lambda(x_{1})=\lambda(x_{2})=\lambda(x)$. By Proposition 5, we obtain that $x_{1}=x_{2}=x$. *

4 Proof of the main result

Now, we prove the main result of the present paper.

{proof*}

[of Theorem 1.1] “$\Rightarrow$”. Assume that $x\in\text{extr}(\Omega(y))$. We assert that for every $t\in(0,1)$, one of the followings holds:

1. (1).

$\lambda(t;y)=\lambda(t;x)$; 2. (2).

$\lambda(t;y)\neq\lambda(t;x)$ with $E^{x}\{\lambda(t;x)\}$ an atom and

[TABLE]

We set

[TABLE]

Since $\lambda(y),\lambda(x)\in L_{1}(0,1)$, it follows that the mapping $f:s\mapsto\int_{0}^{s}\lambda(t;y)-\lambda(t;x)dt$ is continuous. Moreover, noting that $f(0)=f(1)=0$, we infer that $A$ is an open set, i.e., $A=\cup_{i}(a_{i},b_{i})$, where $a_{i},b_{i}\not\in A$. By Lemma 1, $\lambda(x)$ is a step function on $(a_{i},b_{i})$.

We assert that

[TABLE]

for any $\varepsilon>0$. Note that $\lambda(a_{i}^{-};y)\leq\lambda(a_{i}^{-};x)$ (indeed, if $\lambda(a_{i}^{-};y)>\lambda(a_{i}^{-};x)$, then, by $y\succ x$, we obtain that $\int_{0}^{a_{i}}\lambda(t;y)-\lambda(t;x)dt>0$, which is a contradiction with $a_{i}\notin A$). Assume by contradiction that (22) does not hold, that is, $\lambda(a_{i}^{-};x)=\lambda(a_{i}+\varepsilon;x)$ for some $\varepsilon\in(0,b_{i}-a_{i})$. Then, $\lambda(a_{i}^{-};y)\leq\lambda(a_{i}^{-};x)=\lambda(a_{i}+\varepsilon;x)$. Since $\lambda(y)$ is decreasing, it follows that $\int_{a_{i}}^{a_{i}+\varepsilon}\lambda(t;y)-\lambda(t;x)dt\leq 0$. However, since $a_{i}\notin A$ and $a_{i}+\varepsilon\in A$, it follows that $\int_{0}^{a_{i}}\lambda(t;y)-\lambda(t;x)dt=0$ and $\int_{a_{i}}^{a_{i}+\varepsilon}\lambda(t;y)-\lambda(t;x)dt>0$, which is a contradiction.

For a given $\varepsilon>0$, we claim that $\lambda(b_{i}-\varepsilon_{1};y)\leq\lambda(b_{i}-\varepsilon_{1};x)$ for some $\varepsilon_{1}\in(0,\varepsilon)$. Assume by contradiction that $\lambda(b_{i}-\varepsilon_{1};y)>\lambda(b_{i}-\varepsilon_{1};x)$ for all $\varepsilon_{1}\in(0,\varepsilon)$. Then, by $x\prec y$, we obtain that $\int_{0}^{b_{i}}\lambda(t;y)-\lambda(t;x)dt>0$. That is, $b_{i}\in A$, which is a contradiction with the assumption. We assert that

[TABLE]

for any $\varepsilon>0$. Assume by contradiction that $\lambda(b_{i}-\varepsilon;x)=\lambda(b_{i};x)$ for some $\varepsilon\in(0,b_{i}-a_{i})$. Then, $\lambda(b_{i}-\varepsilon_{1};y)\leq\lambda(b_{i}-\varepsilon;x)=\lambda(b_{i}-\varepsilon_{1};x)=\lambda(b_{i};x)$ for some $\varepsilon_{1}\in(0,\varepsilon)$. However, since $x\in\Omega(y)$ and $b_{i}\notin A$, it follows that $\int_{b_{i}}^{b_{i}+\delta}\lambda(t;y)-\lambda(t;x)dt\geq 0$ for any $\delta>0$. By the right-continuity of spectral scales, we obtain that

[TABLE]

That is, $\lambda(b_{i}-\varepsilon_{1};y)\geq\lambda(b_{i};y)\geq\lambda(b_{i};x)=\lambda(b_{i}-\varepsilon_{1};x)\geq\lambda(b_{i}-\varepsilon_{1};y)$, which is a contradiction with the assumption that $b_{i}-\varepsilon_{1}\in A$.

Fix an index $i$, we will consider the step function $\lambda(x)$ on $(a_{i},b_{i})$ and construct operators $x_{\pm}$ such that $2x=x_{+}+x_{-}$.

Case 1. Suppose that $\lambda(x)$ takes three or more values on $(a_{i},b_{i})$. Hence, there exist $a_{i}<s_{1}<s_{2}<s_{3}\leq b_{i}$ such that $\lambda(x)|_{[s_{1},s_{2})}=C_{1}$ and $\lambda(x)|_{[s_{2},s_{3})}=C_{2}$ with $\lambda(s_{3};x)<C_{2}$, $C_{1}>C_{2}$. Let

[TABLE]

and

[TABLE]

It is clear that $2x=x_{+}+x_{-}$ and

[TABLE]

We assert that there exists sufficiently small $\delta$ such that

[TABLE]

Assume that $\delta>0$ satisfies that

[TABLE]

[TABLE]

and

[TABLE]

Note that $x$ can be represented in the form of

[TABLE]

Recalling that $u=E^{x}\{C_{1}\}-\frac{s_{2}-s_{1}}{s_{3}-s_{2}}\cdot E^{x}\{C_{2}\}$ and $x_{\pm}=x\pm\delta u$, we have

[TABLE]

By the definition of $\delta$, we obtain that $\lambda(s_{1}^{-};x)\stackrel{{\scriptstyle\eqref{delta1}}}{{>}}C_{1}\pm\delta\stackrel{{\scriptstyle\eqref{delta3}}}{{>}}C_{2}\mp\frac{s_{2}-s_{1}}{s_{3}-s_{2}}\delta\stackrel{{\scriptstyle\eqref{delta2}}}{{>}}\lambda(s_{3};x)$. Hence,

[TABLE]

In particular, for every $s\notin[s_{1},s_{3}]$, we have

[TABLE]

Recall that $f:s\mapsto\int_{0}^{s}\lambda(t;y)-\lambda(t;x)dt$ is continuous and so there exists $s^{\prime}\in[s_{1},s_{2}]$ such that

[TABLE]

Hence, for all $s\in[s_{1},s_{2}]$, there exists a sufficiently small $\delta$ such that

[TABLE]

On the other hand, since $\int_{0}^{s_{3}}\lambda(x)\geq\int_{0}^{s_{3}}\lambda(x)=\int_{0}^{s_{3}}\lambda(x_{\pm})$, $t\mapsto\int_{0}^{t}\lambda(y)$ is a concave function and $\lambda(x_{\pm})$ is a constant on $(s_{2},s_{3})$, it follows that $x_{\pm}\prec y$. Hence, $x$ is not an extreme point of $\Omega(y)$, which is a contradiction.

Case 2. Suppose that $\lambda(x)$ takes only two values on $(a_{i},b_{i})$. Hence, there exist $a_{i}<s_{0}<b_{i}$ such that $\lambda(x)|_{[a_{i},s_{0})}=C_{1}$ and $\lambda(x)|_{[s_{0},b_{i})}=C_{2}$, $C_{1}>C_{2}$. Let $C^{\prime}_{1}:=\frac{1}{s_{0}-a_{i}}\int_{a_{i}}^{s_{0}}\lambda(t;y)dt$. By the assumption that $\int_{a_{i}}^{s}\lambda(t;x)dt<\int_{a_{i}}^{s}\lambda(t;y)dt$, $s\in({a_{i}},b_{i})$, we have $C^{\prime}_{1}>C_{1}$. Let $C_{2}^{\prime}:=\frac{1}{b_{i}-s_{0}}\int_{s_{0}}^{b_{i}}\lambda(t;y)dt$. Recall that $\int_{a_{i}}^{b_{i}}\lambda(t;y)-\lambda(t;x)dt=0$ and $\int_{a_{i}}^{s_{0}}\lambda(t;y)-\lambda(t;x)dt>0$. We obtain that $\int_{s_{0}}^{b_{i}}\lambda(t;y)-\lambda(t;x)dt=\int_{a_{i}}^{b_{i}}\lambda(t;y)-\lambda(t;x)dt-\int_{a_{i}}^{s_{0}}\lambda(t;y)-\lambda(t;x)dt<0$. Hence,

[TABLE]

Let $\delta>0$ be such that

[TABLE]

Set

[TABLE]

and $x_{\pm}:=x\pm\delta u$. Note that $\lambda(x)$ can be represented in the form of

[TABLE]

Therefore,

[TABLE]

By the definition of $\delta$, we obtain that $\lambda(a_{i}^{-};x)\geq C_{1}\pm\delta>C_{2}\mp\frac{s_{0}-a_{i}}{b_{i}-s_{0}}\delta\geq\lambda(b_{i};x)$. Hence,

[TABLE]

and $\int_{0}^{1}\lambda(t;x_{\pm})dt=\int_{0}^{1}\lambda(t;x)dt=\int_{0}^{1}\lambda(t;y)dt$. Recall that every partial averaging operator $T$111Let ${\mathcal{A}}=\{A_{k}\subset(0,1)\}$ be a finite or infinite sequence of disjoint sets of positive measure. Denote by $A_{\infty}$ the complement of $\cup_{k}A_{k}$. The partial averaging operator is defined by (see [41, p. 198] or [43, Chapter 1.2.12])

$P(x\mid{\mathcal{A}})=\sum_{k}\frac{1}{m(A_{k})}\left(\int_{A_{k}}x(s)ds\right)\chi_{A_{k}}+x\chi_{A_{\infty}},~{}x\in L_{1}(0,1).$

is a doubly stochastic operator (see e.g. [43, Chapter 1.2.12] or [9, Section 2]), i.e., $Tf\prec f$, $f\in L_{1}(0,1)$ (see [34] for definition of doubly stochastic operators). Hence, for every $s\in[a_{i},s_{0}]$, we have

[TABLE]

Similarly, for every $s\in[s_{0},b_{i}]$, we have

[TABLE]

and therefore, for every $s\in[s_{0},b_{i}]$, we have

[TABLE]

Hence, $x_{\pm}\prec y$ and thus $x$ is not an extreme point of $\Omega(y),$ which is a contradiction.

*Case 3. * Now, we consider the case when $\lambda(x)|_{(a_{i},b_{i})}=C.$ Assume that $E^{x}\{C\}$ is not a minimal projection. Let $E^{x}\{C\}=p_{1}+p_{2}$, $p_{1}\perp p_{2}\in{\mathcal{P}}({\mathcal{M}})$. Let $\delta>0$. Set

[TABLE]

and

[TABLE]

It is clear that $\int_{0}^{1}\lambda(t;x_{\pm})dt=\int_{0}^{1}\lambda(t;x)dt=\int_{0}^{1}\lambda(t;y)dt$. The same argument in Case 2 yields that there exists sufficiently small $\delta$ such that $x_{\pm}\prec y$. That is, $x$ is not an extreme point of $\Omega(y)$. Hence, $E^{x}\{C\}$ is a minimal projection in ${\mathcal{M}}$. Moreover, by the assumption that $a_{i},b_{i}\notin A$, we have $\int_{0}^{a_{i}}\lambda(t;y)-\lambda(t;x)dt=0=\int_{0}^{b_{i}}\lambda(t;y)-\lambda(t;x)dt$. Hence, $\int_{a_{i}}^{b_{i}}\lambda(t;x)dt=\int_{a_{i}}^{b_{i}}\lambda(t;y)dt$. which completes the proof.

“$\Leftarrow$”. The converse implication follows from Lemma 6.

{proof*}

[of Corollary 1.4] Let ${\mathcal{M}}$ be a type II1 factor (resp. ${\mathcal{M}}=M_{n}(\mathbb{C})$) and ${\mathcal{A}}$ be a diffuse abelian von Neumann subalgebra of ${\mathcal{M}}$ (resp. ${\mathcal{A}}$ is the diagonal masa in $M_{n}(\mathbb{C})$). The first equality in Corollary 1.4 has been established in [1] (and the Schur-Horn theorem [37, 23], see also [1, (1.1)]). Since ${\mathcal{A}}$ is diffuse (resp. the diagonal masa in $M_{n}(\mathbb{C})$), it follows that for any $y\in{\mathcal{M}}_{h}$, there exists $z\in{\mathcal{A}}_{h}$ such that $\lambda(z)=\lambda(y)$. Applying Theorem 1.1, we obtain that

[TABLE]

Therefore, we have

[TABLE]

which establishes the second equality.

Acknowledgements The first author supported by the grant (No. AP08052004 and No. AP08051978) of the Science Committee of the Ministry of Education and Science of the Republic of Kazakhstan. The second author acknowledges the support of University International Postgraduate Award (UIPA). The third author was supported by the Australian Research Council (FL170100052).

The authors would like to thank Thomas Schekter and Dmitriy Zanin for helpful discussions and thank the reviewers for numerous useful comments and suggestions.