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A generalized Hölder-type inequalities for measurable operators
Journal of Inequalities and Applications volume 2020, Article number: 200 (2020)
Abstract
We prove a generalized Hölder-type inequality for measurable operators associated with a semi-finite von Neumann algebra which is a generalization of the result shown by Bekjan (Positivity 21:113–126, 2017). This also provides a generalization of the unitarily invariant norm inequalities for matrix due to Bhatia–Kittaneh, Horn–Mathisa, Horn–Zhan and Zou under a cohyponormal condition.
1 Introduction
Let \(\mathbb{M}_{n}\) be the space of \(n\times n\) complex matrices. A norm \(|\!|\!|\cdot |\!|\!|\) on \(\mathbb{M}_{n}\) is called unitarily invariant if \(|\!|\!|UAV|\!|\!|=|\!|\!|A|\!|\!|\) for all \(A\in \mathbb{M}_{n}\) and all unitary matrices \(U, V\in \mathbb{M}_{n}\). Let \(A, B\in \mathbb{M}_{n}\). In 1990, Bhatia and Kittaneh [6] established an arithmetic–geometric mean inequality for unitarily invariant norms, i.e.,
Using tensor algebra techniques, a strengthening inequality of (1.1) was presented by Bhatia and Davis [5]
for \(A, B, X\in \mathbb{M}_{n}\). On the other hand, let \(A,B \in M_{n}\) and \(r>0\), Horn and Mathisa proved in [15] the following Cauchy–Schwarz inequality for unitarily invariant norms
Let \(A, B\in \mathbb{M}_{n}\) and \(\frac{1}{p}+\frac{1}{q}=1\), \(p, q>1\), \(r\geq 0\). With the properties of C-S semi-norms in hand, Horn and Zhan [16] established a stronger version of inequality (1.3) as follows:
which is the Hölder inequality for unitarily invariant norms. In particular, these authors also showed in [16] that
Subsequently, a considerable different proofs, equivalent statements, along with some generalizations, refinements, and applications of inequalities (1.1)–(1.4) were discussed by many authors. We refer to [1–3, 5, 15, 20] for more information on this topic and historical references.
Let \(A, B\in \mathbb{M}_{n}\) and \(\frac{1}{p}+\frac{1}{q}=1\), \(p, q>1\), \(\alpha \in [0, 1]\), \(r\geq 0\) and let \(T_{X}(\alpha )=\alpha AA^{*}X+(1-\alpha )XBB^{*}\). In 2015, by majorization techniques, Audenaert [2] prove an inequality that interpolates between the arithmetic–geometric mean and Cauchy–Schwarz matrix norm inequalities
Recently, Zou [20] presented the inequality for unitarily invariant norms
which is a unified version of inequalities (1.1) and (1.6).
By the concept of uniform Hardy–Littlewood majorization Bekjan [8] gave a Hölder-type inequality (1.4) for τ-measurable operators associated with a semi-finite von Neumann algebra and for symmetric Banach spaces norm. In this paper, we will give a generalized Hölder-type inequality (1.7) for τ-measurable operators under a cohyponormal condition by adopting a technique similar to the one used by Bekjan and Zou. This is a generalization of Bekjan’s result in [8].
2 Preliminaries
Let \(L_{0}\) be the set of all Lebesgue measurable functions on \((0,\infty )\). A Banach space \(E\subseteq L_{0}\) with the norm \(\|\cdot \|_{E}\) satisfying the condition that \(f\in E\) and \(\|f\|_{E}\leq \|g\|_{E}\) whenever \(0\leq f\leq g\), \(f\in L_{0}\) and \(g\in F\), is said to be a Banach function space. A Banach function space \(E\subseteq L_{0}\) is called a symmetric Banach function space if it follows from \(f\in L_{0}\), \(g\in E\) and \(f^{*}\leq g^{*}\) that \(f\in E\) and \(\|f\|_{E}\leq \|g\|_{E}\), where
and m denotes the Lebesgue measure on \((0,\infty )\). The symmetric Banach function space E is called fully if and only if \(f\in E\), \(g\in L_{0}\) and \(\int _{0}^{t}f^{*}(s)\,ds\geq \int _{0}^{t}g^{*}(s)\,ds\) give us that \(g\in E\) and \(\|f\|_{E}\geq \|g\|_{E}\). We say that E has order continuous norm if for every net \(\{f_{i}\}_{i\in \varLambda }\subseteq E\) such that \(f_{i}\downarrow 0\) we have \(\|f_{i}\|_{E}\downarrow 0\). In particular, a symmetric Banach function space which has order continuous norm is automatically fully symmetric. For \(0< r<\infty \), \(E^{(r)}\) will denote the quasi-Banach spaces defined by
For \(r>0\), we know from [17] that if E is a symmetric Banach function space, then \(E^{(r)}\) is a symmetric quasi-Banach space, and if E has order continuous norm, then \(E^{(r)}\) has order continuous norm.
We suppose that \(\mathcal{M}\) is a semi-finite von Neumann algebra, namely a von Neumann algebra equipped with a semi-finite, faithful and normal trace τ. We will denote by 1 the identity in \(\mathcal{M}\) and \(P(\mathcal{M})\) the projection lattice of \(\mathcal{M}\). A closed densely defined linear operator x in \(\mathcal{H}\) with domain \(D(x)\subseteq \mathcal{H}\) is said to be affiliated with \(\mathcal{M}\) if \(u^{*}xu=x\) for all unitary operators u which belong to the commutant \(\mathcal{M^{\prime }}\) of \(\mathcal{M}\). Let \(e^{\bot }_{s}(|x|)=e_{(s, \infty )}(|x|)\) be the spectral projection of \(|x|\) associated with the interval \((s, \infty )\). If x is affiliated with \(\mathcal{M}\), x will be called τ-measurable if and only if \(\tau (e^{\bot }_{s}(|x|))<\infty \) for some \(s>0\). The set of all τ-measurable operators will be denoted by \(L_{0}(\mathcal{M})\).
Definition 2.1
Let \(x\in L_{0}(\mathcal{M})\) and \(t>0\). The “generalized singular number of x” \(\mu _{t}(x)\) is defined by
We will denote simply by \(\lambda (x)\) and \(\mu (x)\) the functions \(t\rightarrow \lambda _{t}(x)\) and \(t\rightarrow \mu _{t}(x)\), respectively. The generalized singular number function \(t\rightarrow \mu _{t}(x)\) is decreasing right-continuous. For \(x, y\in L_{0}(\mathcal{M})\) and \(u, v\in \mathcal{M}\), we obtain
Moreover, let f be a continuous increasing function on \([0, \infty )\) with \(f(0)=0\). It follows from [11, Lemma 2.5, Lemma 2.6 and Corollary 2.8] that
and
See [11] for basic properties and detailed information on generalized singular number of x.
Let E be a symmetric Banach function space on \((0,\infty )\). We define
Then \((E(\mathcal{M}), \|\cdot \|_{E(\mathcal{M})})\) is a noncommutative symmetric Banach function space. If \(E=L^{p}\), then \((E(\mathcal{M}), \|\cdot \|_{E(\mathcal{M})})\) is the usual noncommutative \(L_{p}\) spaces \((L^{p}(\mathcal{M}), \|\cdot \|_{p})\). For \(0< r<\infty \), we define
As is shown in [10, Proposition 3.1], if E is a symmetric Banach function space, then \(E^{(r)}(\mathcal{M})=E(\mathcal{M})^{(r)} \), where
and \(\|x\|_{E^{(r)}(\mathcal{M})}=\|\mu (x)\|_{E^{(r)}}\). It is well known that \(E(\mathcal{M})^{(r)}\) is also a noncommutative fully symmetric Banach function space when \(r\geq 1\) and E is fully (cf. [19]).
In the following, unless stated otherwise, we will keep all previous notations throughout the paper, and we always assume that E is a symmetric Banach function space on \((0,\infty )\) with order continuous norm.
3 Main results
We start this section with several lemmas which will be used in our proof. From [12, Theorem 3.3] and [13, Lemma 3.4] we have the following two results.
Lemma 3.1
Let\(x, y\in \mathcal{M}\)and\(\alpha \in [0, 1]\). Then
Lemma 3.2
Let\(x, y\in \mathcal{M}\)such thatxyis a self-adjoint operator. For every\(r>0\), we obtain
Remark 3.3
If x, y are normal operators in \(L_{0}(\mathcal{M})\), then \(\mu _{s}(xy)=\mu _{s}(yx)\), \(s>0 \). Indeed, we conclude from (2.1) and (2.2) (see also [11, Lemma 2.5]) that
Recall that an operator \(x \in L_{0}(\mathcal{M})\) is said to be hyponormal if \(x^{*}x\geq xx^{*}\), cohyponormal if \(x^{*}\) is hyponormal.
Lemma 3.4
Let\(x, y\in \mathcal{M}\)and\(r\geq 0\). If\(\alpha \in [0, 1]\)and\(xx^{*}(yy^{*})^{\alpha }\)is cohyponormal, then
Proof
By (2.2) and Lemma 3.2 we have
Since \(xx^{*}(yy^{*})^{\alpha }\) is cohyponormal, [8, Corollary 4.5] yields
and hence, [11, Theorem 4.2(iii)] and Lemma 3.1 tell us that
This completes the proof. □
Remark 3.5
Let \(x, y\in \mathcal{M}\) and \(r\geq 0\), \(\alpha \in [0, 1]\). (2.1) now yields \(\mu _{t}(xx^{*}yy^{*})=\mu _{t}(yy^{*}xx^{*})\) for all \(t>0\). If \(yy^{*}(xx^{*})^{\alpha }\) is hyponormal, then from Lemma 3.4 we have
Proposition 3.6
Let\(\alpha \in [0, 1]\), \(r\geq 0\), \(1< p, q<\infty \)with\(\frac{1}{p}+\frac{1}{q}=1\)and let\(x, y\in E(\mathcal{M})^{(2r)}\). If\(xx^{*}(yy^{*})^{\alpha }\)is cohyponormal, then\(x^{*}y\in E(\mathcal{M})^{(r)}\),
where\(T(\alpha )=\alpha xx^{*}+(1-\alpha )yy^{*}\).
Proof
If
or
then the inequality (3.1) is obvious, and so we always suppose that
and
First we assume that \(x, y\in E(\mathcal{M})^{(2r)}\cap \mathcal{M}\). According to [4, Theorem 3] and Lemma 3.4, we have \(x^{*}y\in E(\mathcal{M})^{(r)}\) and
In the general case, for \(y, x\in L_{0}(\mathcal{M})\), let \(x=u|x|\) and \(y=v|y|\) be the polar decomposition of x and y, respectively. We assume also that \(|y|=\int _{0}^{\infty }\lambda \,de_{\lambda }(|y|)\) and \(|x|=\int _{0}^{\infty }\lambda \,de_{\lambda }(|x|)\) are the spectral decomposition of \(|y|\) and \(|x|\), respectively. Set \(y_{n}=v\int _{0}^{n} \lambda \,de_{\lambda }(|y|)\) and \(x_{n}=u\int _{0}^{n} \lambda \,de_{\lambda }(|x|)\). Then
From [18, Proposition 21 of Chapter I] and [11, Lemma 3.1] we conclude that \(\tau (e_{[n, \infty )}(|x|))\rightarrow 0 \) and \(\mu _{t}(x-x_{n})\downarrow 0\) as \(n\rightarrow \infty \). Similarly, \(\mu _{t}(y-y_{n})\downarrow 0\) as \(n\rightarrow \infty \). Since E has order continuous norm, we see that
as \(n\rightarrow \infty \). Thus, [4, Theorem 3] gives
where the constant C from the triangle inequality in \(E(\mathcal{M})^{(r)}\). Therefore, the fact \(\|\mu _{t}(x_{n})^{2r}\|^{\frac{1}{2}}_{E}\leq \|\mu _{t}(x)^{2r}\|^{ \frac{1}{2}}_{E}\) and (3.2) imply that \(\Vert \vert x_{n}^{*}y_{n}-x^{*}y \vert ^{r} \Vert _{E(\mathcal{M})}\rightarrow 0\) as \(n\rightarrow \infty \). Moreover, \(\Vert \vert x_{n}^{*}y_{n} \vert ^{r} \Vert _{E(\mathcal{M})}\rightarrow \||x^{*}y|^{r} \|_{E(\mathcal{M})}\) as \(n\rightarrow \infty \). In the same manner we can see that
and
This completes the proof. □
Remark 3.7
Let \(1< p, q<\infty \) with \(\frac{1}{p}+\frac{1}{q}=1\). If \(\alpha =0\), then \(xx^{*}(yy^{*})^{\alpha }=xx^{*}\) is cohyponormal. Therefore, Proposition 3.6 yields \(x^{*}y\in E(\mathcal{M})^{(r)}\) and
which is a main result of [4].
Remark 3.8
It is necessary for us to remark here that, it can be observed in [7, Lemma 2] without a proof that \(\mu (ab)=\mu (ba)\) when \(ab, ba\in L^{1}(\mathcal{M})\). However, we are not able to give it a proof at this moment. On the other hand, the authors were informed by an anonymous referee that \(\mu (ab)=\mu (ba)\) does not hold even in the matrix case. On account of this, there could be a gap in the proof of [13, Theorem 3.6] and we give a corresponding illustration as follows: Set \(r\geq 1\), \(\alpha \in [0, 1]\) and let \(xx^{*}(yy^{*})^{\alpha }\) be cohyponormal. Using Proposition 3.6 to the case \(E=L_{1}\) and \(p=q=2\), we have
i.e.,
which is the result of [14, Theorem 3.6] under a cohyponormal condition.
Theorem 3.9
Let\(\alpha \in [0, 1]\)and\(1< p, q<\infty \)with\(\frac{1}{p}+\frac{1}{q}=1\). Assume also that\(r\geq \max \{\frac{2}{p}, \frac{2}{q}\}\), \(x, y\in E(\mathcal{M})^{(2r)}\)and\(z\in P(\mathcal{M})\). If\(zxx^{*}z(zyy^{*}z)^{\alpha }\)is cohyponormal, then\(x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}\),
where\(T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}\).
Proof
Let \(T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}\). Then \(z\in P(\mathcal{M})\) and Proposition 3.6 force that
and
and hence
Similarly,
Therefore,
A simple computation shows
According to [11, Theorem 4.4(ii)] and (2.1), we have
Since \(\frac{rp}{2}\geq 1\), from [9, Theorem 2.1] and (2.2) we can assert that
Consequently,
In the same way as used above, we can also prove that
Therefore, inequalities (3.4), (3.5) and (3.6) give
□
Remark 3.10
Let \(\alpha \in [0, 1]\) and \(1< p, q<\infty \) with \(\frac{1}{p}+\frac{1}{q}=1\). Assume also that \(r\geq \max \{\frac{2}{p}, \frac{2}{q}\}\), \(x, y\in E(\mathcal{M})^{(2r)}\) and \(z\in \mathcal{M}\). We write \(T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}\) and we wish to prove
However, we do not succeed in proving it at this moment.
Theorem 3.11
Let\(r>0\)and\(x, y \in E(\mathcal{M})^{(2r)}\), \(0\leq z\in \mathcal{M}\). Assume also that\(\alpha \in [0, 1]\)and\(1< p, q<\infty \)with\(\frac{1}{p}+\frac{1}{q}=1\). If\(z^{\frac{1}{2}}xx^{*}z^{\frac{1}{2}}(z^{\frac{1}{2}}yy^{*}z^{ \frac{1}{2}})^{\alpha }\)is cohyponormal, then\(x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}\)and
where\(T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}\).
Proof
First it follows from [4, Theorem 3] that \(x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}\). Since z is positive, Proposition 3.6 gives
and hence Lemma 3.2 leads to
Similarly,
Therefore,
This completes the proof. □
Remark 3.12
(1) Let \(\alpha \in [0, 1]\) and \(1< p, q<\infty \) with \(\frac{1}{p}+\frac{1}{q}=1\) and let \(r\geq \max \{\frac{2}{p}, \frac{2}{q}\}\). For \(x, y\in E(\mathcal{M})^{(2r)}\) and \(z\in P(\mathcal{M})\), write \(T_{z}(\alpha )= \alpha xx^{*}z+(1-\alpha )zyy^{*}\) and \(T(\alpha )=\alpha xx^{*}+(1-\alpha )yy^{*}\). Assume also that \(zxx^{*}z(zyy^{*}z)^{\alpha }\) is cohyponormal. Combining Theorem 3.11 with Theorem 3.9 we have
where
and
(2) Let \(r>0\), \(x, y\in E(\mathcal{M})^{(2r)}\), \(0\leq z\in \mathcal{M}\) and \(\alpha \in [0, 1]\), \(1< p, q<\infty \) with \(\frac{1}{p}+\frac{1}{q}=1\). If \(z^{\frac{1}{2}}yy^{*}z^{\frac{1}{2}}(z^{\frac{1}{2}}xx^{*}z^{ \frac{1}{2}})^{\alpha }\) is cohyponormal, then \(x^{*}\mathit{zy}\in E(\mathcal{M})^{(r)}\). Moreover, the fact \(\mu _{t}(|x^{*}\mathit{zy}|^{r})=\mu _{t}(x^{*}\mathit{zy})^{r}=\mu _{t}(y^{*}zx)^{r}= \mu _{t}(|y^{*}zx|^{r})\) and Theorem 3.11 yields
where \(T(\alpha )= \alpha xx^{*}+(1-\alpha )yy^{*}\).
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The authors is grateful to the referee for very useful comments.
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This research is partially supported by Tianshan youth project No. 2018Q012 and the National Natural Science Foundation of China No. 11761067 and No. 11701255.
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Han, Y., Shao, J. A generalized Hölder-type inequalities for measurable operators. J Inequal Appl 2020, 200 (2020). https://doi.org/10.1186/s13660-020-02467-w
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DOI: https://doi.org/10.1186/s13660-020-02467-w