The case of equality in Young's inequality for the s-numbers in semi-finite von Neumann algebras

TL;DR

This paper investigates the conditions for equality in Young's inequality for s-numbers within semi-finite von Neumann algebras, establishing that equality implies a specific power relation between the operators.

Contribution

It provides a characterization of equality cases in Young's inequality for s-numbers in semi-finite von Neumann algebras, extending results to unbounded operators.

Findings

Equality in Young's inequality occurs only when |a|^p=|b|^q.

The results extend to unbounded operators affiliated with the algebra.

The work relates to other symmetric norm Young inequalities.

Abstract

For a semi-finite von Neumann algebra $\mathcal A$, we study the case of equality in Young's inequality of s-numbers for a pair of $\tau$-measurable operators $a,b$, and we prove that equality is only possible if $|a|^p=|b|^q$. We also extend the result to unbounded operators affiliated with $\mathcal A$, and relate this problem with other symmetric norm Young inequalities.

Equations

$$\alpha \beta \le \text{nicefrac}1p{\alpha }^p+\text{nicefrac}1q{\beta }^q$$

$${\mu }_s(a{b}^{\ast })\le {\mu }_s\left(\text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q\right),s>0$$

$$\Vert a{b}^{\ast }{\Vert }_E=\Vert \text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q{\Vert }_E$$

$$V(\epsilon ,\delta )=\{x\in \mathcal{A}:\exists p\in \mathcal{P}(\mathcal{A})\text{mbox}s.t.\tau (1-p)<\delta \text{mbox}and\Vert xp\Vert <\epsilon \}.$$

$${\mu }_s(x)=\inf \{\Vert xp\Vert :p\in \mathcal{P}(\mathcal{A})\text{mbox}with\tau (1-p)\le s\}.$$

$$\mu_{s}(x)=\inf_{\begin{subarray}{c}p\in\mathcal{P}(\mathcal{A})\\ \tau(1-p)\leq s\end{subarray}}\big{[}\sup_{\begin{subarray}{c}\xi\in{\rm{Ran\,}}(p)\\ \|\xi\|=1\end{subarray}}\langle x\xi,\xi\rangle\big{]}=\sup_{\begin{subarray}{c}p\in\mathcal{P}(\mathcal{A})\\ \tau(p)\geq s\end{subarray}}\big{[}\inf_{\begin{subarray}{c}\xi\in{\rm{Ran\,}}(p)\\ \|\xi\|=1\end{subarray}}\|x\xi\|\big{]}.$$

$${\mu }_t(a)=\min \{s\ge 0:\tau (p^x(s,+\infty ))\le t\}.$$

$${\mu }_s(x)=\sum _{k\in {\mathbb{N}}_0}{\lambda }_k{\chi }_{[k,k+1)}(s)$$

$$|a{b}^{\ast }|=\nu ||a||b||{\nu }^{\ast }\text{mbox}and||a||b||=\nu |a{b}^{\ast }|\nu .$$

$$\tau (p^a(t,+\infty ))=\tau (p^{a+b}(t,+\infty ))=\tau (p^a(t,+\infty ))+\tau (p^b(t,+\infty ))$$

$$\sup \{\tau (xp):p\in \mathcal{P}(\mathcal{A}),\tau (p)\le t\}={\int }_0^t{\mu }_s(x)ds.$$

$$x={\int }_0^{\infty }{\mu }_s(x)de(s).$$

$$e_t-e_s=e(s,t)=e[s,t)=e(s,t]=e[s,t].$$

$${\int }_0^t{\mu }_s(x)ds=\tau (x{e}_t).$$

$${\mu }_s(a{b}^{\ast })\le {\mu }_s\left(\text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q\right)$$

$${\mu }_s(a{b}^{\ast })={\mu }_s\left(\text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q\right)$$

$${\mu }_s(a{b}^{\ast })\le {\mu }_s\left(\text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q\right).$$

$${\mu }_s(x_n{y}_n)\le {\mu }_s\left(\text{nicefrac}1p{x}_n^p+\text{nicefrac}1q{y}_n^q\right)\le {\mu }_s\left(\text{nicefrac}1p|a{|}^p+\text{nicefrac}1q|b{|}^q\right)$$

$$\sigma (x)=\text{mbox}clos\{{\mu }_s(x):s>0\}$$

$$|ab|={\int }_0^{\infty }{\lambda }_{s}de(s)\text{ and }\frac{1}{p}{a}^p+\frac{1}{q}{b}^q={\int }_0^{\infty }{\lambda }_{s}dq(s),$$

$${\int }_0^{\delta }{\mu }_s(x{)}^{p}ds<\infty$$

$${\int }_0^{\delta }{\mu }_s(x{)}^{p}ds\ge {\mu }_{\delta }(x{)}^{p-1}{\int }_0^{\delta }{\mu }_s(x)ds$$

$$a=a{p}^a(r,+\infty )+a{p}^a[0,r].$$

$$a^p=a^p{p}^a(r,+\infty )+a^p{p}^a[0,r].$$

$$a^p\in {\mathcal{L}}_{loc}^1(\mathcal{A})\iff a^p\in {\mathcal{L}}^1(\mathcal{A})+\mathcal{A}\iff \text{mbox}thereexistsr>0\text{mbox}suchthat{a}^p{p}^a(r,+\infty )\in {\mathcal{L}}^1(\mathcal{A}).$$

$${\int }_0^t{\mu }_s(a{)}^{p}ds={\int }_0^t{\mu }_s(l+m{)}^{p}ds\le {\int }_0^t({\mu }_s(l)+{\mu }_s(m){)}^{p}ds$$

$${\left({\int }_0^t{\mu }_s(a{)}^{p}ds\right)}^{1/p}$$

$$\mu_{s}(ap^{a}(r,+\infty))=\left\{\begin{array}[]{ll}\mu_{s}(a)&0<s<t\\ 0&s\geq t\end{array}\right.,$$

$${\mu }_s(ab)={\mu }_s\left(\frac{1}{p}{a}^p+\frac{1}{q}{b}^q\right)\text{ for all }s>0$$

$${\int }_0^t{\lambda }_s^{2}de(s)+{\epsilon }^2{e}_{t}a(1-p_b)a{e}_t=e_t|ba{|}^2{e}_t+{\epsilon }^2{e}_{t}a(1-p_b)a{e}_t=e_t|b_{\epsilon }a{|}^2{e}_t.$$

Full text

1 Introduction

The well-known inequality, valid for $p>1$ and $1/p+1/q=1$, named after W. H. Young, is usually stated as

[TABLE]

for any $\alpha,\beta\in\mathbb{R}^{+}$, with equality if and only if $\alpha^{p}=\beta^{q}$.

In this paper, we establish an analogue for the case of equality in the setting of operators affiliated to semi-finite von Neumann algebras. For more references and further discussion on the subject of Young’s inequality for matrices and operators, we refer the reader to [12] where the proof is given for the particular case of compact operators in $\mathcal{B}(H)$ -the discrete (or atomic measure) case- of this fact. In particular, we remark that it was the fundamental paper by T. Ando [1] which initiated the study of Young’s inequality for the singular values of $n\times n$ matrices.

The emphasis in this paper is in the measure theoretic approach to operators affiliated with a semi-finite von Neumann algebra, since the approach by induction used in [12] is not at hand. The inequality for $s$-numbers of operators $a,b$ affiliated with a semi-finite von Neuman algebra $\mathcal{A}$, is stated as

[TABLE]

and extended here to unbounded operators; we are interested in the case of equality.

We remark that this result includes all semi-finite von Neumann algebras $\mathcal{A}$, since by a standard tensor product technique [8, p.286], we can always embed $\mathcal{A}$ into the diffuse algebra $\mathcal{A}\otimes\mathscr{L}^{\infty}([0,1],dt)$ without altering the $s$-numbers.

This paper is organized as follows: Section 2 presents the general facts about $s$-numbers recalling the well-known and establishing some simple lemmas used later. Section 3 deals with some simplifications and reductions of the problem to deal with it in full generality. Section 4, after certain technical propositions, contains the main result of this paper, Theorem 4.7, that states that equality holds for all $s$-numbers in (1) if and only if $|a|^{p}=|b|^{q}$, or equivalently, if equality of norms

[TABLE]

holds for some strictly increasing symmetric norm $\|\cdot\|_{E}$ (definition given in Section 4.1, just before the main theorem).

2 Singular numbers in von Neumann algebras

In this paper $\mathcal{A}$ stands for a finite or semi-finite von Neumann algebra with faithful normal trace $\tau$, which when convenient we will assume represented in a complex Hilbert space $\mathcal{H}$. The set of (self-adjoint) projections in $\mathcal{A}$ will be denoted by $\mathcal{P}(\mathcal{A})$.

We consider the topology of convergence in measure in $\mathcal{A}$: a neighbourhood of [math] is given by

[TABLE]

We will denote with $\widetilde{\mathcal{A}}$ the closure in measure of $\mathcal{A}$, therefore $\widetilde{\mathcal{A}}$ is the ring of $\tau$-measurable operators affiliated with $\mathcal{A}$. In the atomic case, convergence in measure reduces to the norm topology, therefore $\widetilde{\mathcal{A}}=\mathcal{A}$ in that case.

For $0\leq x\in\widetilde{\mathcal{A}}$ and $s>0$, we denote the $s$-th singular number of $x$ by $\mu_{s}(x)$:

[TABLE]

With $\mu_{s}(a)$ we denote the $s$-numbers of $|a|$, that is $\mu_{s}(a):=\mu_{s}(|a|)$. We remark that $\lim\limits_{s\to 0^{+}}\mu_{s}(x)=\|x\|$ including the posibility of $+\infty$ when $x$ is unbounded. The standard reference on the subject is the paper by Fack and Kosaki [8].

We comment here on some useful characterizations (Proposition 3.1 in [10], Proposition 2.2, Lemma 2.5, Proposition 3.1 in [8]).

• •

The variational (min-max) characterization:

[TABLE]

• •

The distribution characterization: if $B\subset\mathbb{R}_{\geq 0}$ is a Borelian set and we denote $p^{x}(B)=\chi_{B}(|x|)$ (the range projections of $|x|$), then

[TABLE]

From the very definition of $\widetilde{\mathcal{A}}$, the number $\tau(p^{x}(s,+\infty))$ is eventually finite, and moreover $\tau(p^{x}(s,+\infty))\to 0$ when $s\to\infty$.

• •

For $x\in\widetilde{\mathcal{A}}$, the following are equivalent:

$\tau(p^{x}(t,+\infty))<+\infty$ for all $t>0$. 2. 2.

$\lim\limits_{t\to\infty}\mu_{t}(x)\to 0$. 3. 3.

There exists a sequence of bounded operators $x_{n}\in{\mathscr{L}}^{1}({\mathcal{A}})$ such that $x_{n}\to x$ in the measure topology.

Remark 2.1**.**

With any of these three characterizations, we say that $x$ is $\tau$-compact; these operators form a complete bilateral ideal in $\widetilde{\mathcal{A}}$ that we will denote by ${\mathscr{K}}({\widetilde{\mathcal{A}}})$; note that a $\tau$-compact operator is not necessarily bounded. We will denote with ${\mathscr{K}}({\widetilde{\mathcal{A}}})^{+}$ the positive $(x\geq 0)$ $\tau$-compact operators.

In the atomic case (when $\mathcal{A}={\cal B}({\cal H})$), then we recover the ordinary compact operators ${\mathscr{K}}({\cal H})$. If $\{\lambda_{k}(x)\}_{k\in\mathbb{N}_{0}}$ denotes the usual singular values of $x$ (i.e. the eigenvalues of $|x|$), and we arrange them in a right-continuos decreasing function which is constant on $[k,k+1)$, then we obtain the distribution function $\mu_{s}(x)$ as follows:

[TABLE]

In this lemma we collect some other known facts on $s$-numbers that we will use later.

Lemma 2.2**.**

Let $x,y\in\mathcal{A}$, $a,b\in\widetilde{\mathcal{A}}$. Then for each $s>0$,

$\mu_{s}(xay)\leq\|x\|\|y\|\mu_{s}(a)$, and if $a\leq b$ then $\mu_{s}(a)\leq\mu_{s}(b)$. 2. 2.

$\mu_{s}(|ab^{*}|)=\mu_{s}(||a||b||)$. 3. 3.

$\mu_{s+t}(a+b)\leq\mu_{s}(a)+\mu_{t}(b)$, $s,t\geq 0$. 4. 4.

If $p\in\mathcal{P}(\mathcal{A})$ then $\mu_{s}(ap)=0$ for each $s\geq\tau(p)$. 5. 5.

$\tau(|a|)=\int_{0}^{\infty}\mu_{s}(a)ds$. 6. 6.

If $a,b\geq 0$, $a\in{\mathscr{K}}({\widetilde{\mathcal{A}}})$, $ab=0$ and $\mu_{s}(a+b)=\mu_{s}(a)$ for all $s>0$, then $b=0$.

Proof.

The first assertion is a consequence of the min-max characterization of the $s$-numbers. To prove the second, note that if $b=\nu|b|$ is the polar decomposition of $b$, a straitghtforward computation using the functional calculus shows that

[TABLE]

Then by the first item we obtain $\mu_{s}(|ab^{*}|)=\mu_{s}(||a||b||)$. The proof of the third, fourth and fifth assertion is due to Fack and Kosaki and can be found in their original paper [8, Lemmas 2.5, 2.6 and Proposition 2.7]. The final assertion seems evident, but requires some proof though. For $t>0$, let $p^{a}[t,+\infty)=\chi_{[t,+\infty)}(a)$ be the spectral projections of $a$, and likewise for $b,a+b$. Then $\mu_{s}(a+b)=\mu_{s}(a)$ for all $s>0$ implies (since $a$ is $\tau$-compact and $ab=0$) that

[TABLE]

for all $t>0$ (cf. [8, Corollary 2.9]). Therefore $\tau(p^{b}(t,+\infty))=0$ for all $t>0$, implying $b=0$. ∎

3 Diffuse algebras

Recall that an algebra is diffuse if it has no minimal projections. Following Fack and Kosaki [8, p.286], we can always embed $\mathcal{A}$ into the diffuse algebra $\mathcal{A}\otimes\mathscr{L}^{\infty}([0,1],dt)$ without altering the $s$-numbers. Then, the following [8, Lemma 2.1] will be useful later:

Remark 3.1**.**

If $x\geq 0$ is $\tau$-measurable, then for each $t\geq 0$

[TABLE]

3.1 Complete flags

If $0\leq x\in{\mathscr{K}}({\widetilde{\mathcal{A}}})$ and $\mathcal{A}$ is a diffuse von Neumann algebra, there exists an increasing assignment $\mathbb{R}_{\geq 0}\ni t\mapsto e_{t}\in\mathcal{P}(\mathcal{A})$ ($e_{s}\leq e_{t}$ for $s\leq t$) such that $\tau(e_{t})=t$ for all $t\geq 0$ and

[TABLE]

Note the analogy with the atomic case, where $x=\sum_{k\in\mathbb{N}}\lambda_{k}(x)p_{k}$ with $p_{k}$ the projection to the $\lambda_{k}$ eigenspace of $x$, and we assume the eigenvalues are arranged in decrasing order.

Since $e_{0}=0$, we denote $e(s,t)=e_{t}-e_{s}$ for $s\leq t\in\mathbb{R}_{\geq 0}$ and since $\mathcal{A}$ is diffuse,

[TABLE]

The spectral resolution $\{e_{t}\}_{t\geq 0}$ is called a complete flag for $x$; for more details on this useful constructions in diffuse semi-finite algebras, we refer the reader to the papers [2, 3] by Argerami and Massey. In particular, for each $t>0$,

[TABLE]

3.2 Equality of singular numbers, $\tau$-compact operators

Let $(\mathcal{A},\tau)$ be a semi-finite von Neumann algebra with semi-finite trace ($\tau(1)=+\infty$ here).

In [9, Theorem 1] Farenick and Manjegani proved the remarkable Young’s inequality for the $s$-numbers: if $p>1$, $1/p+1/q=1$, and $a,b\in\mathcal{A}$, then

[TABLE]

for all $s\geq 0$. The purpose of this paper is to attack the following conjecture:

Let $p,q>1$ with $1/p+1/q=1$. Does

[TABLE]

for all $s>0$ imply $|a|^{p}=|b|^{q}$?

Remark 3.2**.**

If the algebra $\mathcal{A}$ is atomic, we have already answered in the affirmative the conjecture in [12, Theorem 2.12]. There, we used the existence of eigenvectors for each non-trivial eigenvalue. In this paper we will be dealing with the continuous case (that contains the previous one, see Section 3), using continuous techniques.

3.3 Extension to unbounded operators

We extend the inequality and the conjecture to unbounded operators.

Theorem 3.3**.**

Let $a,b\in\widetilde{\mathcal{A}}$, then for each $s>0$

[TABLE]

Proof.

Let $a=u|a|$, $b=\nu|b|$ be the polar decompositions of $a,b$. Approximating $|a|,|b|$ in measure from below with bounded operators $x_{n},y_{n}\geq 0$, we have for each $s>0$

[TABLE]

by (3) applied to the pair $x_{n},y_{n}$ and Lemma 2.2.1. Since $x_{n}\leq|a|$, $y_{n}\leq|b|$, it is easy to check that $|x_{n}y_{n}|\leq||a||b||$; since $|ab^{*}|=\nu||a||b||\nu^{*}$, then $\mu_{s}(x_{n}y_{n})\leq\mu_{s}(ab^{*})$. Since $x_{n}y_{n}$ converges in measure to $|a||b|$, then by [8, Lemma 3.4], $\lim_{n}\mu_{s}(x_{n}yn)=\mu_{s}(ab^{*})$ for each $s>0$, proving the claim. ∎

3.4 Some restrictions and simplifications

To make sense out of the conjecture (4), we should ask for a complete description of an operator in terms of its $s$-numbers. We therefore think that it is natural to the confine the conjecture to the ideal ${\mathscr{K}}({\widetilde{\mathcal{A}}})$ of $\tau$-compact operators (Remark 2.1).

In fact, it is known that for $x\in{\mathscr{K}}({\widetilde{\mathcal{A}}})^{+}$,

[TABLE]

(see [16, Theorem 4.10]). On the other hand, if $e,f$ are disjoint and infinite projections ($\tau(e)=\tau(f)=\infty$), taking $x=e+\frac{1}{2}f$ shows that $\sigma(x)=\{1/2,1\}$ while $\mu_{s}(x)=1$ for all $s>0$, therefore it is hopeless to recover $x$ from the data in $\mu_{s}(x)$.

Exchanging $a$ with $b$, we can always assume that $1<p\leq 2$. Since $\lambda_{s}=\mu_{s}(|ab^{*}|)=\mu_{s}(|a||b|)$, we can safely assume that $a,b\geq 0$. Moreover, we can assume (see Section 3) that $\mathcal{A}$ is diffuse and there exist complete flags $e_{t},q_{t}\in\mathcal{P}(\mathcal{A})$ ($t\geq 0$, $\tau(e_{t})=\tau(q_{t})=t$) such that

[TABLE]

since $a,b\in{\mathscr{K}}({\widetilde{\mathcal{A}}})^{+}$ and $\tau$-compact operators form a (closed in measure) ideal of $\widetilde{\mathcal{A}}$.

Our arguments will be based on continuous majorization. We are therefore interested in those operators that are locally integrable. More precisely, let $1\leq p<\infty$, let $x\in\mathcal{A}$ and assume that there exists $\delta>0$ such that

[TABLE]

(hence the integral is finite for all finite $\delta>0$). We will denote the set containing all these operators by ${\mathscr{L}}^{p}_{loc}({\mathcal{A}})\subset\widetilde{\mathcal{A}}$. Note that in particular, all bounded operators $a\in\mathcal{A}$ are of this class. Moreover,

[TABLE]

shows that ${\mathscr{L}}^{p}_{loc}({\mathcal{A}})\subset{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ for each $p\geq 1$.

Lemma 3.4**.**

Let $a\in\widetilde{\mathcal{A}}$ and $p\geq 1$. Then $a\in{\mathscr{L}}^{p}_{loc}({\mathcal{A}})$ if and only if $a\in{\mathscr{L}}^{p}({\mathcal{A}})+\mathcal{A}$, and in that case the decomposition can be taken as follows for some $r>0$.

[TABLE]

Proof.

By polar decomposition, it suffices to consider $a\geq 0$. Note that $ap^{a}[0,r]\leq r\in\mathcal{A}$ and since $a\in\mathcal{A}$, eventually $\tau(p^{a}(r,+\infty))<\infty$ for some $r>0$. Likewise, for $p>1$,

[TABLE]

These expressions imply the following (see [11, Proposition 1.2]):

[TABLE]

Note that then $ap^{a}(r,+\infty)\in{\mathscr{L}}^{p}({\mathcal{A}})$ for the same $r$, therefore $a\in{\mathscr{L}}^{p}({\mathcal{A}})+\mathcal{A}$ by (7). On the other hand, if $a=l+m\in{\mathscr{L}}^{p}({\mathcal{A}})+\mathcal{A}$, then taking $f(x)=x^{p}$ which is continuous, convex and increasing in $[0,+\infty)$,

[TABLE]

by [8, Lemma 4.4.iii]. Therefore for any $t>0$

[TABLE]

by the classical Mikowkski inequality, therefore $a\in{\mathscr{L}}^{p}_{loc}({\mathcal{A}})$. Take $r=\mu_{t}(a)$, and note that for all $s>0$,

[TABLE]

therefore (7) gives the stated decomposition. ∎

4 Main results

We start by examining the ranges of $a,b$. Throughout, $p,q$ are positive with $1/p+1/q=1$.

Proposition 4.1**.**

Let $0\leq a,b\in{\mathscr{K}}({\widetilde{\mathcal{A}}})$ with $ab\in{\mathscr{L}}^{2}_{loc}({\mathcal{A}})$. If $p\neq 2$ and

[TABLE]

then $\overline{{\rm{Ran\,}}(a)}=\overline{{\rm{Ran\,}}(b)}$.

Proof.

Exchanging $a,b$ it will suffice to consider $1<p<2$. Let $p_{b}$ be the projection onto the closure of the range of $b$. Let $b_{\varepsilon}=b+\varepsilon(1-p_{b})$, then $b_{\varepsilon}^{q}=b^{q}+\varepsilon^{q}(1-p_{b})$ and $b_{\varepsilon}^{2}=b^{2}+\varepsilon^{2}(1-p_{b})$. Fix $t>0$, let $\{e_{s}\}$ be a complete flag for $|ba|$, then denoting $\lambda_{s}=\mu_{s}(ab)$ we have

[TABLE]

Taking the trace, it follows that

[TABLE]

by Remark 3.1. On the other hand, by (5) applied to $a,b_{\varepsilon}$,

[TABLE]

Note that in particular, all the integrals computed up to now are finite by the hypothesis on $ab$, and

[TABLE]

Cancelling $\int_{0}^{t}\lambda_{s}^{2}ds$ and dividing by $\varepsilon^{2}$, noting that $q>2$ and letting $\varepsilon\to 0$ gives us that $\tau(e_{t}a(1-p_{b})ae_{t})=0$. Since the trace is faithful, we conclude that $(1-p_{b})ae_{t}=0$ or equivalently, $ae_{t}=p_{b}ae_{t}$ for all $t>0$. Then

[TABLE]

that is $a({\rm{Ran\,}}|ba|)\subset\overline{{\rm{Ran\,}}(b)}$.

Now if $\xi\in\mathcal{H}$, then $a|ba|\xi\in\overline{{\rm{Ran\,}}(b)}$, therefore $a^{2}|ba|\xi=a(a|ba|\xi)\in a\overline{{\rm{Ran\,}}(b)}\subset\overline{{\rm{Ran\,}}(ab)}=\overline{{\rm{Ran\,}}|ba|}$, and $a^{3}|ba|\xi=a(a^{2}|ba|\xi)\in a\overline{{\rm{Ran\,}}|ba|}\subset\overline{{\rm{Ran\,}}(b)}$. Iterating this argument, we arrive to the conclusion that $a^{2n+1}({\rm{Ran\,}}|ba|)\subset\overline{{\rm{Ran\,}}(b)}$ for all $n\in\mathbb{N}_{0}$. Using an approximation of $f=\chi_{\sigma(a)}$ by odd functions, we conclude that $p_{a}({\rm{Ran\,}}|ba|)=f(a)({\rm{Ran\,}}|ba|)\subset\overline{{\rm{Ran\,}}(b)}$ where $p_{a}$ is the projection onto the closure of the range of $a$. Therefore $|ba|^{2}\xi=ab^{2}a\xi=p_{a}ab^{2}a\xi=p_{a}|ba|^{2}\xi\subset\overline{{\rm{Ran\,}}(b)}$, which gives $\overline{{\rm{Ran\,}}|ba|}=\overline{{\rm{Ran\,}}(|ba|^{2})}\subset\overline{{\rm{Ran\,}}(b)}$. But then

[TABLE]

which proves that the range of $b$ is invariant for $a$; since $a\geq 0$ the same is true for the kernel of $b$. Therefore we can write $a=a_{b}+a_{\perp}$, with $a_{b}=p_{b}ap_{b}\geq 0$ and $a_{\perp}=(1-p_{b})a(1-p_{b})\geq 0$. Note that $ba^{2}b=ba_{b}^{2}b$ and $a^{p}=a_{b}^{p}+a_{\perp}^{p}$, thus for all $s>0$,

[TABLE]

by the hypothesis and (5) applied to $a_{b},b$. This proves that for all $s>0$

[TABLE]

which (by Lemma 2.2.5) is only possible if $a_{\perp}=0$, proving the assertion of the proposition. ∎

The following will be used twice throughout the proof of the main theorem, therefore we preferred to state it as a separate lemma:

Lemma 4.2**.**

Let $0\leq x\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ and $p\in\mathcal{A}$ a projection with finite trace. Then

[TABLE]

implies $xp=px$.

Proof.

Since $p$ is a projection and $x\geq 0$,

[TABLE]

Since the square root is operator monotone, $pxp\leq|xp|$. Take the trace and invoke items 4 and 5 of Lemma 2.2, then

[TABLE]

thus by the hypothesis $pxp$ and $|xp|$ have equal (and finite) trace. Since the trace is faithful this is only possible if $pxp=|xp|$, or equivalently if $pxpxp=px^{2}p$. This implies that

[TABLE]

which by the case of equality in Cauchy-Schwarz inequality implies $xp=px$. ∎

We will also need the following classical result on operator ranges for [6, Theorem 2].

Remark 4.3**.**

**(Douglas’ Lemma). Let $x,y\in\widetilde{\mathcal{A}}$. If $xx^{*}\leq\lambda yy^{*}$ for some $\lambda\geq 0$, there exists a contraction $c$ such that $x=yc$, therefore ${\rm{Ran\,}}(x)\subset{\rm{Ran\,}}(y)$. **

With this tools at hand, we are now able to prove the main theorem.

Theorem 4.4**.**

Let $0\leq a,b\in{\mathscr{K}}({\widetilde{\mathcal{A}}})$ with $ab\in{\mathscr{L}}^{2}_{loc}({\mathcal{A}})$. If

[TABLE]

then $a^{p}=b^{q}$. When $p=q=2$ it suffices to assume $ab\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$.

Proof.

Exchanging $a,b$ it will suffice to consider $1<p\leq 2$. Denoting $\lambda_{s}=\mu_{s}(ab)$ we write $ba^{2}b=|ab|^{2}=\int_{0}^{\infty}\lambda_{s}^{2}de(s)$ with a complete flag $\{e_{s}\}_{s\geq 0}\subset\mathcal{P}(\mathcal{A})$. For $I=[s,t]\subset[0,+\infty)$ denote $e_{I}=e_{s}-e_{t}=e(s,t)$, then $e_{t}=e_{[0,t]}$. Since $\lambda_{s}$ is non-increasing,

[TABLE]

and the previous lemma ensures that ${\rm{Ran\,}}(e_{I})\subset{\rm{Ran\,}}(|ab|)={\rm{Ran\,}}(ba)\subset{\rm{Ran\,}}(b)$ for each interval $I=[s,t]$. Moreover, if $e=\bigvee_{s\geq 0}e_{s}$ is the join of the increasing projections, clearly $e=p_{|ab|}$, the projection onto the closure of the range of $|ab|$.

We now treat three cases separately.

Case $4/3\leq p<2$. By Proposition 4.1 we can consider $\mathcal{H}=\overline{{\rm{Ran\,}}(a)}=\overline{{\rm{Ran\,}}(b)}$, and the semi-finite von Neumann subalgebra ${\mathcal{M}}\subset\mathcal{A}$ generated by the (finitely supported) spectral projections of $a,b,ab$. We give $\mathcal{M}$ the inherited trace $\tau$ and identity $1=1_{\mathcal{M}}=P_{\mathcal{H}}=p_{b}$. All the operators involved $a,b,ab,a^{p},b^{q}$ are in $\widetilde{\mathcal{M}}$, and $\mathcal{M}$ can be faithfully represented in this ${\cal B}({\cal H})$. Then we can safely assume that $b$ is injective, $e,e_{I}\in\mathcal{M}$ for each interval $I$, and $\mathcal{M}\subset\mathcal{H}$ is a common core for all $x\in\widetilde{\mathcal{M}}$.

We remark that in what follows, we will only use that $b$ is injective, or equivalently, that the range of $b$ is dense.

Again for each interval $I$, let $\mathcal{H}_{I}$ be the closure of $b^{-1}{\rm{Ran\,}}(e_{I})\subset\mathcal{H}$. Let $f_{I}=P_{\mathcal{H}_{I}}$ and $f=\bigvee_{s\geq 0}f_{s}$ the closed join of all projections, where $f_{s}=f_{[0,s]}$. We divide the proof in several smaller claims.

Claim: $f=p_{a}$, the projection onto the closure of the range of $a$. Let $\eta\in{\rm{Ran\,}}(f_{s})$; then $\eta=\lim_{n}\eta_{n}$ with $b\eta_{n}=e_{s}\xi_{n}$; since ${\rm{Ran\,}}(e_{s})\subset{\rm{Ran\,}}(|ab|)={\rm{Ran\,}}(ba)$, it must be $b\eta_{n}=ba\psi_{n}$ for some $\psi_{n}\in\mathcal{H}$, and by the injectivity of $b$, we obtain $\eta_{n}\in{\rm{Ran\,}}(a)$, therefore $\eta\in\overline{{\rm{Ran\,}}(a)}$. This proves that $f\leq p_{a}$. On the other hand, if $\eta\in{\rm{Ran\,}}(a)$, then $b\eta\in{\rm{Ran\,}}(ba)={\rm{Ran\,}}|ab|\subset{\rm{Ran\,}}(e)$, therefore $b\eta=\lim\xi_{n}$ with $\xi_{n}\in{\rm{Ran\,}}(e_{s_{n}})$ for some $s_{n}>0$. Therefore $\xi_{n}=b\eta_{n}$ with $\eta_{n}\in{\rm{Ran\,}}(f_{s_{n}})\subset{\rm{Ran\,}}(f)$ for each $n$. Now

[TABLE]

and since the range of $b$ is dense, $\{\eta_{n}\}_{n}$ is a weak Cauchy sequence in ${\rm{Ran\,}}(f)$ which, being closed and linear, it is weakly closed. Therefore $\eta_{n}$ converges weakly to some $\eta_{0}\in{\rm{Ran\,}}(f)$. But for each $\xi\in\mathcal{H}$,

[TABLE]

which implies that $b\eta=b\eta_{0}$, and by the injectivity of $b$, we obtain $\eta=\eta_{0}\in{\rm{Ran\,}}(f)$, thus $p_{a}\leq f$. This proves that $f=p_{a}$.

Claim: there exists a closed operator $c_{I}="b^{-1}e_{I}b^{-1}"$ defined on ${\rm{Ran\,}}(b)$ such that $0\leq c_{I}\leq\frac{1}{\lambda_{t}^{2}}a^{2}$. Since $b$ is injective and ${\rm{Ran\,}}(e_{I})\subset{\rm{Ran\,}}(b)$, for each $\xi\in\mathcal{H}$ there exists a unique $\eta\in\mathcal{H}_{I}$ such that $b\eta=e_{I}\xi$. Define $c_{I}$ in ${\rm{Ran\,}}(b)$ as follows: $c_{I}b\xi=\eta$. We now compute

[TABLE]

which shows that $c_{I}$ is a symmetric operator on ${\rm{Ran\,}}(b)$. Moreover, since $e_{I}\leq\frac{1}{\lambda_{t}^{2}}ba^{2}b$ (recall $I=[s,t]$ and equation (8)), it follows that

[TABLE]

therefore $c_{I}$ has a self-adjoint extension (c.f. [15, Theorem 5.1.13], that we still denote $c_{I}$), and $0\leq c_{I}\leq\frac{1}{\lambda_{t}^{2}}a^{2}$.

Claim: $c_{I}\in\widetilde{\mathcal{M}}$. Let $u\in\mathcal{M}^{\prime}$, let $\xi\in\mathcal{H}$. Then there exists unique $\eta,\psi\in\mathcal{H}$ such that $e_{I}\xi=b\psi$ and $e_{I}(u\xi)=b\eta$. Now $e_{I}u\xi=ue_{I}\xi$ since $u\in\mathcal{M}^{\prime}$, therefore

[TABLE]

and since $b$ is injective, $u\psi=\eta$. We now compute

[TABLE]

which shows that $c_{I}u=uc_{I}$, proving that $c_{I}\in\widetilde{\mathcal{M}}$.

Claim: $bc_{I}b=e_{I}$, $e_{I}bf_{I}=bf_{I}$ and $f_{I}b^{2}f_{I}c_{I}=f_{I}$ for each $I$. From the very definition, $bc_{I}b=e_{I}$. On the other hand note that if $\eta\in f_{I}$ then $\eta=\lim_{n}\eta_{n}$ with $b\eta_{n}\in{\rm{Ran\,}}(e_{I})$ and for any $\xi\in\mathcal{H}$

[TABLE]

therefore $b\eta_{n}\in{\rm{Ran\,}}(e_{I})$ converges weakly to $b\eta$, therefore $b\eta\in{\rm{Ran\,}}(e_{I})$ and we obtain $e_{I}bf_{I}=bf_{I}$. Taking adjoints, $f_{I}b=f_{I}be_{I}$, hence for each $\xi\in\mathcal{H}$,

[TABLE]

which proves that $f_{I}b^{2}f_{I}c_{I}=f_{I}$ for any $I$.

Claim: $f_{I}c_{J}=c_{J}f_{I}$ for any $I,J$. Since when $e_{I}\xi=b\eta$, then $\eta\in{\rm{Ran\,}}(f_{I})$, clearly $f_{I}c_{I}=c_{I}=c_{I}f_{I}$. Moreover it is not hard to see that $f_{I}c_{J}=c_{I\cap J}$ for any pair of intervals $I,J$ by the injectivity of $b$, therefore $f_{I}c_{J}=c_{J}f_{I}$.

Claim: $f_{I}\sim e_{I}$. Inspection of the ranges shows that (again by the injectivity of $b$)

[TABLE]

and since $c_{I}b=(bc_{I})^{*}$, it follows that $e_{I}$ is von Neumann equivalent to $f_{I}$, which implies that $f,f_{I}\in\mathcal{M}$ and moreover $\tau(f_{I})=\tau(e_{I})$ for each $I$.

Summing up our findings: for any interval $I=[s,t]\subset[0,+\infty)$, we have $e,e_{I},f,f_{I}\in\mathcal{P}(\mathcal{M})$ with $f_{I}\simeq e_{I}$, $\tau(f_{I})=\tau(e_{I})=t-s$, $e_{I}bf_{I}=bf_{I}$, $f_{I}b=f_{I}be_{I}$. Moreover $c_{I}\in\widetilde{\mathcal{M}}$, $bc_{I}b=e_{I}$,

[TABLE]

Claim: $a^{p}f_{s}=f_{s}a^{p}$ for all $s>0$. Let $\pi=\{I_{i}\}_{i=1\cdots n}$ with $I_{i}=[s_{i},s_{i+1}]$ be a partition of $[0,+\infty)$, and denote $e_{i}=e_{I_{i}}$ and likewise with $f_{i},c_{i}$. We have

[TABLE]

which implies $a^{2}\geq\sum_{i}\lambda_{s_{i+1}}^{2}c_{i}$ since $b$ is injective with dense range. Now refining the partition

[TABLE]

for any $\xi\in\mathcal{H}$. Since the range of $b$ is dense and the operators involved are positive, we conclude that $\lim\limits_{|\pi|\to 0}\sum_{i}\lambda_{s_{i+1}}^{2}c_{i}=a^{2}$ in the strong operator topology. Since $f_{s}c_{i}=c_{i}f_{s}(=c_{[0,s]\cap I_{i}})$, we conclude that $f_{s}a^{2}=a^{2}f_{s}$ for all $s\geq 0$, which implies that $a^{p}f_{i}=f_{i}a^{p}$ for all $i$.

We now take the $p/2$-th root in (9), which is a monotone operator function since $1<p<2$. Thus

[TABLE]

Since $4/3\leq p<2$, this implies that $2<q\leq 4$. Then $t\mapsto t^{q/2}$ is operator convex [15, Theorem 2.4] and $(f_{i}b^{2}f_{i})^{q/2}\leq f_{i}b^{q}f_{i}$. By Young’s inequality in the commutative algebra generated by $c_{i},f_{i}$ [7, Lemma 2.2]

[TABLE]

where $D=\frac{1}{p}a^{p}+\frac{1}{q}b^{q}$ for short.

Claim: $f_{t}D=Df_{t}$ for all $t>0$. Assume that $\pi$ is a partition of $[0,t]$. Summing over $i$, we obtain $\sum_{i}\lambda_{s_{i+1}}f_{i}\leq\sum_{i}f_{i}Df_{i}$, and taking traces

[TABLE]

by (6), Remark 3.1 and the assumption on $a,b$ (recall ${\mathscr{L}}^{2}_{loc}({\mathcal{A}})\subset{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$). Refining the partition $\pi$, it follows that $\int_{0}^{t}\lambda_{s}ds=\tau(Df_{t})$. Since $\lambda_{s}=\mu_{s}(D)$ and $\tau(f_{t})=t$, Lemma 4.2 implies that $f_{t}D=Df_{t}$.

Claim: $D=\int_{0}^{\infty}\lambda_{s}df(s)$. Since $t$ was arbitrary, $f_{i}D=Df_{i}$ also holds. Returning to the previous inequality (4) we now sum over $i$ to obtain

[TABLE]

Let $\overline{D}=\int_{0}^{\infty}\lambda_{s}df(s)$, then $(1-f)\overline{D}=0$ and $\overline{D}f_{t}=\int_{0}^{t}\lambda_{s}df(s)$. Refining the partition $\pi$ of $[0,t]$ we obtain $\overline{D}f_{t}\leq f_{t}D$, and since $\tau(\overline{D}f_{t})=\tau(f_{t}D)=\int_{0}^{t}\lambda_{s}ds$, it must be $\overline{D}f_{t}=Df_{t}=f_{t}D$ for each $t>0$. Recall $f=\bigvee_{t}f_{t}$ is the union of the projections $f_{t}$, then clearly $\overline{D}=Df=fD$; on the other hand

[TABLE]

and by Lemma 2.2.3 it is only possible if $(1-f)D=0$, or equivalently $D=\overline{D}$.

Claim: $a$ commutes with $b$, $b$ commutes with all $f_{t}$ and $b^{q}f=b^{q}$. Since $a^{p}$ commutes with all $f_{s}$, then $a^{p}$ commutes with

[TABLE]

Then $a^{p}$ commutes with $b^{q}$ or equivalently, $a$ commutes with $b$. Note that since $f=p_{a}$, then

[TABLE]

therefore $b^{q}f=b^{q}$. Since $a^{p}$ commutes with all $f_{t}$ and $\frac{1}{q}b^{q}=D-\frac{1}{p}a^{p}$, then $b^{q}$ commutes with all $f_{t}$.

Claim: $f_{t}=e_{t}$ for all $t$. Recall that for all $t$, $bf_{t}=e_{t}bf_{t}$, therefore $f_{t}be_{t}=f_{t}b$. Since $f_{t}$ commutes with $b$, $bf_{t}e_{t}=bf_{t}$; since $b$ is injective, $f_{t}e_{t}=f_{t}$. Therefore $f_{t}=f_{t}e_{t}=e_{t}f_{t}e_{t}\leq e_{t}$, and since $\tau(f_{t})=\tau(e_{t})=t$, it must be $f_{t}=e_{t}$ for all $t$.

Finally,

[TABLE]

Let $a_{t}=af_{t},b_{t}=bf_{t}$, then $\frac{1}{p}\mu_{s}(a_{t})^{p}\leq\mu_{s}(D_{t})=\lambda_{s}\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ for $0<s<t$ and likewise with $b$. This means that $a_{t}^{p},b_{t}^{q}\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ and

[TABLE]

and all the operators involved have finite trace. Farenick and Manjegani proved that in that case (see [9, Theorem 3.1] or [14, Theorem 2.1]), it must be $a^{p}f_{t}=a_{t}^{p}=b_{t}^{q}=b^{q}f_{t}$. We give here an alternative argument: taking traces

[TABLE]

by the operator Hölder inequality (applied to $\|a_{t}b_{t}\|_{1}$) and Young’s numeric inequality (applied to $\|a_{t}\|_{p},\|b_{t}\|_{p}$). This implies $\|a_{t}b_{t}\|_{1}=\|a_{t}\|_{p}\|b_{t}\|_{q}$, and this is only possible if $a_{t}^{p}=b_{t}^{q}$ [4, 13]. Since this holds for all $t>0$, $a^{p}=a^{p}f=b^{q}f=b^{q}$ as we claimed.

Case $1<p<4/3$. This implies that $q>4$, but since the ranges of $a$ and $b$ still match by Proposition 4.1, we can assume that $b$ is injective with dense range, and the computation goes through the same lines, modifying the step regarding the commutative operator Young inequality (4) according to [1, Theorem 2] or [7, Proposition 2.3].

Case $p=q=2$. First note that

[TABLE]

therefore $\mu_{s}(a)\in{\mathscr{L}}^{2}_{loc}({\mathcal{A}})$ and likewise with $b$. Proposition 4.1 is of no use here, therefore it suffices to assume $\mu_{s}(ab)\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$.

Let $\tilde{a}=(p_{b}a^{2}p_{b})^{1/2}$. Then ${\rm{Ran\,}}(\tilde{a})\subset{\rm{Ran\,}}(b)$ and $b\tilde{a}^{2}b=ba^{2}b$, therefore $|\tilde{a}b|=|ab|$. Hence

[TABLE]

by (5) applied to the pair $\tilde{a},b$. Therefore, for all $s\geq 0$,

[TABLE]

Since ${\rm{Ran\,}}(\tilde{a})\subset\overline{{\rm{Ran\,}}(b)}$, we can assume that $b$ is injective, and argumenting as in the previous cases, arrive to $\tilde{a}^{2}=b^{2}$, that is $p_{b}a^{2}p_{b}=b^{2}$. In particular $\mu_{s}(b)^{2}\leq\mu_{s}(a)^{2}$ for all $s>0$. Reversing the argument, we also get $p_{a}b^{2}p_{a}=a^{2}$, therefore $\mu_{s}(a)=\mu_{s}(b)$ for all $s>0$.

Let $\{b_{s}\}_{s\geq 0}$ be a complete flag for $b=\int_{0}^{\infty}\mu_{s}(b)db(s)$ with $\tau(b_{t})=t$. Then for all $t\geq 0$, $b_{t}$ commutes with $b$, we have $b_{t}b=\int_{0}^{t}\mu_{s}(b)db(s)$ and since $b_{t}\leq p_{b}$, $b_{t}p_{b}=b_{t}$. Therefore from $p_{b}a^{2}p_{b}=b^{2}$ we obtain $b_{t}a^{2}b_{t}=b_{t}b^{2}$, which implies that

[TABLE]

By Lemma 4.2, this is only possible if $a$ commutes with $b_{t}$. Therefore, $a$ commutes with $b$, then from $p_{a}b^{2}p_{a}=a^{2}$ we have $bp_{a}=p_{a}b=a$. But

[TABLE]

implies (Lemma 2.2.6) $b=p_{a}b=a$. ∎

Remark 4.5**.**

As the proof goes, it suffices to consider $ab\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ if either

[TABLE]

Corollary 4.6**.**

Let $0\leq a,b\in\mathcal{A}\cap{\mathscr{K}}({\widetilde{\mathcal{A}}})$ and assume

[TABLE]

Then $a^{p}=b^{q}$.

4.1 Symmetric norms

We close the paper putting this result in context with the theory of symmetric norms on $\widetilde{\mathcal{A}}$, see for instance [5] and the references therein.

We say that a symmetric norm $\|\cdot\|_{E}$ is strictly increasing if $x,y\in E\subset\widetilde{\mathcal{A}}$, $\mu_{s}(x)\leq\mu_{s}(y)$ for all $s>0$ and $\|x\|_{E}=\|y\|_{E}$ implies $\mu_{s}(x)=\mu_{s}(y)$ for all $s>0$. All ${\mathscr{L}}^{p}$-norms are strictly increasing for $1\leq p<\infty$, while the uniform norm or the Ky-Fan norms $\|x\|_{(t)}=\int_{0}^{t}\mu_{s}(x)ds$ are not.

Theorem 4.7**.**

Let $a,b\in{\mathscr{K}}({\widetilde{\mathcal{A}}})\cap{\mathscr{L}}^{2}_{loc}({\mathcal{A}})$. If $p>1$ and $1/p+1/q=1$, then the following are equivalent:

$|a|^{p}=|b|^{q}$. 2. 2.

$z|ab^{*}|z^{*}=\frac{1}{p}|a|^{p}+\frac{1}{q}|b|^{q}$* for some contraction $z\in\mathcal{A}$* 3. 3.

$\|z|ab^{*}|w\|_{E}=\|\frac{1}{p}|a|^{p}+\frac{1}{q}|b|^{q}\|_{E}$* for a pair of contractions $z,w\in\mathcal{A}$ and $\|\cdot\|_{E}$ a strictly increasing symmetric norm.* 4. 4.

$\mu_{s}(ab^{*})=\mu_{s}\left(\frac{1}{p}|a|^{p}+\frac{1}{q}|b|^{q}\right)$* for all $s>0$.*

Proof.

The proof is much like as in [12, Theorem 2.13], therefore it is omitted. ∎

As in Theorem 4.4, Remark 4.5 or Corollary 4.6, the hypothesis $ab\in{\mathscr{L}}^{2}_{loc}({\mathcal{A}})$ is unnecessary when $ab$ is bounded, and can be relaxed to $ab\in{\mathscr{L}}^{1}_{loc}({\mathcal{A}})$ if $p=q=2$ or if there is an inclusion of ranges.