Some convexity inequalities in noncommutative $L_p$-spaces

Abstract

In this paper, we prove some convexity inequalities in noncommutative $L_p$ spaces generalizing the previous result of Hiai and Zhan. Moreover, we generalize a variational inequality for positive definite matrices due to Hansen to the case of noncommutative $L_p$ spaces.

MSC:47A30, 47L05, 47L50.

1 Introduction

Let $\parallel \cdot \parallel$ be a unitarily invariant norm on matrices. For matrices A, B, X in $M_n(\mathbb{C})$ with A, B being positive semidefinite and X arbitrary, Bhatia and Davis [1] presented the matrix Cauchy-Schwarz inequality ${\parallel {|A^{\frac{1}{2}}X{B}^{\frac{1}{2}}|}^r\parallel }^2\le \parallel {|AX|}^r\parallel \cdot \parallel {|XB|}^r\parallel$. In 2002, Hiai and Zhan [2] proved that under the same assumption for matrices A, B, and X, the function $t\mapsto \parallel {|A^{t}X{B}^{1-t}|}^r\parallel \cdot \parallel {|A^{1-t}X{B}^t|}^r\parallel$ is convex on $[0,1]$ for each $r>0$. Among other things, this convexity result interpolated the above matrix Cauchy-Schwarz inequality showed by Bhatia and Davis. Another result of interest is that, in 2014, Hansen [3], among other things, gave new and simplified proofs of the Carlen-Lieb theorem concerning concavity of certain trace functions by applying the theory of operator monotone functions. Let ℋ be a Hilbert space and $\mathcal{B}(\mathcal{H})$ denote the $C^{\ast }$-algebra of all bounded linear operators on ℋ. Equipped with the usual adjoint as involution $\mathcal{B}(\mathcal{H})$ becomes a unital $C^{\ast }$-algebra. We say a function $⦀\cdot ⦀:\mathcal{B}(\mathcal{H})\to \mathcal{R}$ is a unitary invariant norm (or symmetric norm) if it is a norm satisfying the invariance property $⦀uxv⦀=⦀x⦀$ for all x and all unitary operators u and v in $\mathcal{B}(\mathcal{H})$.

In this paper, we consider the noncommutative $L_p$ spaces of τ-measurable operators affiliated with a semi-finite von Neumann algebra equipped with a normal faithful semi-finite trace τ. On one hand, we use the method of Hiai and Zhan, via the notion of generalized singular value studied by Fack and Kosaki [4], to prove some convexity inequalities in noncommutative $L_p$ spaces generalizing the previous result of Hiai and Zhan. On the other hand, by making use of the joint convexity and concavity of trace functions obtained by Bekjan [5] we generalize a variational inequality for positive definite matrices due to Hansen to the case of noncommutative $L_p$ spaces.

2 Preliminaries

Throughout the paper, unless specified, we always denote by ℳ a semi-finite von Neumann algebra acting on the Hilbert space ℋ, with a normal faithful semi-finite trace τ. We denote the identity in ℳ by 1 and let denote the projection lattice of ℳ. A closed densely defined linear operator x in ℋ with domain $D(x)\subseteq \mathcal{H}$ is said to be affiliated with ℳ if $u^{\ast }xu=x$ for all unitary u which belong to the commutant ${\mathcal{M}}^{\prime }$ of ℳ. If x is affiliated with ℳ, then x is said to be τ-measurable if for every $ϵ>0$ there exists a projection $e\in \mathcal{M}$ such that $e(\mathcal{H})\subseteq D(x)$ and $\tau (1-e)<ϵ$. The set of all τ-measurable operators will be denoted by $L_0(\mathcal{M},\tau )$, or simply $L_0(\mathcal{M})$. The set $L_0(\mathcal{M})$ is a ∗-algebra with sum and product being the respective closures of the algebraic sum and product. A closed densely defined linear operator x admits a unique polar decomposition $x=u|x|$, where u is a partial isometry such that $u^{\ast }u={(kerx)}^{\perp }$ and $u{u}^{\ast }=\overline{imx}$ (with $imx=x(D(x))$). We call $r(x)={(kerx)}^{\perp }$ and $l(x)=\overline{imx}$ the left and right supports of x, respectively. Thus $l(x)\sim r(x)$. Moreover, if x is self-adjoint, we let $s(x)=r(x)$, the support of x.

Let ${\mathcal{M}}_{+}$ be the positive part of ℳ. Set ${\mathcal{S}}_{+}(\mathcal{M})=\{x\in {\mathcal{M}}_{+}:\tau (s(x))<\mathrm{\infty }\}$ and let $\mathcal{S}(\mathcal{M})$ be the linear span of ${\mathcal{S}}_{+}(\mathcal{M})$, we will often abbreviate ${\mathcal{S}}_{+}(\mathcal{M})$ and $\mathcal{S}(\mathcal{M})$, respectively, as ${\mathcal{S}}_{+}$ and . Let $0<p<\mathrm{\infty }$, the noncommutative $L_p$-space $L_p(\mathcal{M},\tau )$ is the completion of $(\mathcal{S},{\parallel \cdot \parallel }_p)$, where ${\parallel x\parallel }_p=\tau {({|x|}^p)}^{\frac{1}{p}}<\mathrm{\infty }$, $\mathrm{\forall }x\in L_p(\mathcal{M},\tau )$. In addition, we put $L^{\mathrm{\infty }}(\mathcal{M},\tau )=\mathcal{M}$ and denote by ${\parallel \cdot \parallel }_{\mathrm{\infty }}$ ($=\parallel \cdot \parallel$) the usual operator norm. It is well known that $L_p(\mathcal{M},\tau )$ are Banach spaces under ${\parallel \cdot \parallel }_p$ for $1\le p<\mathrm{\infty }$ and they have a lot of expected properties of classical $L_p$-spaces (see [6] or [7]).

Let x be a τ-measurable operator and $t>0$. The ‘t th singular number (or generalized s-number) of x’ is defined by

$${\mu }_t(x)=inf\{\parallel xe\parallel :e\in \mathcal{P},\tau (1-e)\le t\}.$$

See [4] for basic properties and detailed information on the generalized s-numbers.

To achieve one of our main results, we state for easy reference the following fact, obtained from [8], which will be applied below.

Lemma 2.1 If $0<r<1$, then the function $f(z,A)=\tau {(z^r+A^r)}^{\frac{1}{r}}$ is jointly concave in strictly positive operators $(z,A)\in {\mathcal{S}}_{+}\times {\mathcal{S}}_{+}$.

3 Main results

Lemma 3.1 Let $x,y\in \mathcal{S}$ such that xy is a self-adjoint τ-measurable operator and let $1\le p<\mathrm{\infty }$, then

$${\parallel xy\parallel }_p\le {\parallel yx\parallel }_p.$$

Proof Notice that xy is a self-adjoint τ-measurable operator; then

$$\begin{array}{rl}{\mu }_s{(xy)}^{2n}& ={\mu }_s(xyxy\cdots )={\mu }_s({(xy)}^{\ast }xy\cdots )\\ ={\mu }_s\left({|xy|}^{2n}\right)={\mu }_s^{2n}(|xy|)\\ ={\mu }_s^{2n}(xy).\end{array}$$

Let f be an increasing function on ${\mathcal{R}}_{+}$ satisfying $f(0)=0$ and where $t\to f(e^t)$ is convex, then applying Lemma 2 of [9] we have

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}f\left({\mu }_s{(xy)}^{2n}\right)ds& ={\int }_0^{\mathrm{\infty }}f({\mu }_s(xyxy\cdots ))ds\\ \le {\int }_0^{\mathrm{\infty }}f({\mu }_s(yxyx\cdots )\parallel x\parallel \parallel y\parallel )ds\\ ={\int }_0^{\mathrm{\infty }}f\left({\mu }_s(\parallel x\parallel \parallel y\parallel yxyx\cdots )\right)ds\\ \le {\int }_0^{\mathrm{\infty }}f({\mu }_s^{2n-1}(yx)\parallel x\parallel \parallel y\parallel )ds.\end{array}$$

In particular, if $f(s)=s^{\frac{p}{2n-1}}$, then

$${\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{2n}{2n-1}p}(xy)ds\le {\int }_0^{\mathrm{\infty }}{\mu }_s^p(yx){(\parallel x\parallel \parallel y\parallel )}^{\frac{p}{2n-1}}ds.$$

Taking the ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$, by the usual Fatou lemma we get

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}{\mu }_s^p(xy)ds& ={\int }_0^{\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\mu }_s^{\frac{2n}{2n-1}p}(xy)ds\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{2n}{2n-1}p}(xy)ds\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_0^{\mathrm{\infty }}{\mu }_s^p(yx){(\parallel x\parallel \parallel y\parallel )}^{\frac{p}{2n-1}}ds\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{(\parallel x\parallel \parallel y\parallel )}^{\frac{p}{2n-1}}{\int }_0^{\mathrm{\infty }}{\mu }_s^p(yx)ds\\ ={\int }_0^{\mathrm{\infty }}{\mu }_s^p(yx)ds.\end{array}$$

 □

Lemma 3.2 Let $x,y\in \mathcal{S}$ and $z\in \mathcal{M}$. Let $1\le p<\mathrm{\infty }$, then

$${\parallel |x^{\ast }zy{|}^r\parallel }_p^2\le {\parallel |x{x}^{\ast }z{|}^r\parallel }_p\cdot {\parallel |zy{y}^{\ast }{|}^r\parallel }_p,r>0.$$

(1)

Proof It suffices to prove that

$${\left({\int }_0^{\mathrm{\infty }}{\mu }_s{\left(\right|x^{\ast }zy{|}^r)}^{p}ds\right)}^2\le \left({\int }_0^{\mathrm{\infty }}{\mu }_s{\left(\right|x{x}^{\ast }z{|}^r)}^{p}ds\right)\left({\int }_0^{\mathrm{\infty }}{\mu }_s{\left(\right|zy{y}^{\ast }{|}^r)}^{p}ds\right).$$

Since

$${\int }_0^{\mathrm{\infty }}{\mu }_s{\left(\right|x^{\ast }zy{|}^r)}^{p}ds={\int }_0^{\mathrm{\infty }}{\mu }_s^p{\left({\left(x^{\ast }zy\right)}^{\ast }{x}^{\ast }zy\right)}^{\frac{r}{2}}ds={\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{r}{2}p}\left({\left(x^{\ast }zy\right)}^{\ast }{x}^{\ast }zy\right)ds.$$

Applying Lemma 2 of [9] together with the Hölder inequality we obtain

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{r}{2}p}\left(y^{\ast }{z}^{\ast }x{x}^{\ast }zy\right)ds& \le {\int }_0^{\mathrm{\infty }}{\mu }_s^p\left({\left|y{y}^{\ast }{z}^{\ast }x{x}^{\ast }z\right|}^{\frac{r}{2}}\right)ds\\ ={\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{r}{2}p}\left(y{y}^{\ast }{z}^{\ast }x{x}^{\ast }z\right)ds\\ \le {\int }_0^{\mathrm{\infty }}{\mu }_s^{\frac{r}{2}p}\left(zy{y}^{\ast }\right){\mu }_s^{\frac{r}{2}p}\left(x{x}^{\ast }z\right)ds\\ \le {\left({\int }_0^{\mathrm{\infty }}{\mu }_s^{rp}\left(zy{y}^{\ast }\right)ds\right)}^{\frac{1}{2}}{\left({\int }_0^{\mathrm{\infty }}{\mu }_s^{rp}\left(x{x}^{\ast }z\right)ds\right)}^{\frac{1}{2}},\end{array}$$

which implies the lemma. □

Theorem 3.3 Let $x,y\in L_p(\mathcal{M})$ be positive operators with $1\le p<\mathrm{\infty }$ and let $z\in \mathcal{M}$, then for every real number $r>0$, the function

$$\varphi (t)={\parallel |x^{t}z{y}^{1-t}{|}^r\parallel }_p\cdot {\parallel |x^{1-t}z{y}^t{|}^r\parallel }_p$$

is convex on the interval $[0,1]$ and attains its minimum at $t=\frac{1}{2}$.

Proof (i) First we assume that τ is finite. By the density of in $L_p(\mathcal{M})$, we first consider the case $x,y\in \mathcal{S}$. Since ϕ is continuous and symmetric with respect to $t=\frac{1}{2}$, all the conclusions will follow after we show that

$$\varphi (t)\le \frac{1}{2}\{\varphi (t+s)+\varphi (t-s)\}$$

(2)

for $t\pm s\in [0,1]$. By (1) we have

$$\begin{array}{rl}{\parallel |x^{t}z{y}^{1-t}{|}^r\parallel }_p& ={\parallel |x^s\left(x^{t-s}z{y}^{1-t-s}\right)y^s{|}^r\parallel }_p\\ \le {\{{\parallel |x^{t+s}z{y}^{1-(t+s)}{|}^r\parallel }_p\cdot {\parallel |x^{t-s}z{y}^{1-(t-s)}{|}^r\parallel }_p\}}^{\frac{1}{2}}\end{array}$$

and

$$\begin{array}{rl}{\parallel |x^{1-t}z{y}^t{|}^r\parallel }_p& ={\parallel |x^s\left(x^{1-t-s}z{y}^{t-s}\right)y^s{|}^r\parallel }_p\\ \le {\{{\parallel |x^{1-(t-s)}z{y}^{t-s}{|}^r\parallel }_p\cdot {\parallel |x^{1-(t+s)}z{y}^{t+s}{|}^r\parallel }_p\}}^{\frac{1}{2}}.\end{array}$$

Multiplying the above two inequalities we obtain

$${\parallel |x^{t}z{y}^{1-t}{|}^r\parallel }_p\cdot {\parallel |x^{1-t}z{y}^t{|}^r\parallel }_p\le \frac{1}{2}\{\varphi (t+s)+\varphi (t-s)\}.$$

(3)

For the general case, namely, for any $x,y\in L_p(\mathcal{M})$, there exist $x_n,y_n\in {\mathcal{M}}_{+}$ such that $x_n$, $y_n$ are invertible and $x_n\to x$, $y_n\to y$ in $L_p(\mathcal{M})$. Moreover, we have ${\varphi }_n(t)={\parallel {|x_n^{t}z{y}_n^{1-t}|}^r\parallel }_p\cdot {\parallel {|x_n^{1-t}z{y}_n^t|}^r\parallel }_p$ is convex for all $t\in [0,1]$ and attains its minimum at $t=\frac{1}{2}$. Applying Theorem 3.7 of [4] and using the method to prove Lemma 3.3 in [10], we get $x_n^{t}z{y}_n^{1-t}\to x^{t}z{y}^{1-t}$, $x_n^{1-t}z{y}_n^t\to x^{1-t}z{y}^t$ in $L_p(\mathcal{M})$. Hence, we obtain ${\varphi }_n(t)\to \varphi (t)$, $n\to \mathrm{\infty }$. Therefore, $\varphi (t)$ is convex on $[0,1]$ and attains its minimum at $t=\frac{1}{2}$.

(ii) In the general case when τ is semi-finite, there exists an increasing family ${(e_i)}_{i\in \mathcal{I}}\in \mathcal{P}$ such that $\tau (e_i)<\mathrm{\infty }$ for every $i\in \mathcal{I}$ and such that $e_i$ converges to 1 in the strong operator topology (see [6] or [7]). Thus, $e_i\mathcal{M}{e}_i$ is finite for each $i\in \mathcal{I}$. Let $x,y\in L_2{(\mathcal{M})}_{+}$, then $e_{i}x{e}_i,e_{i}y{e}_i\in L_2{(e_i\mathcal{M}{e}_i)}_{+}$. Write $x_i=e_{i}x{e}_i$, $y_i=e_{i}y{e}_i$, it follows from the case (i) that the function $f_i(t)={\parallel {|x_i^{t}z{y}_i^{1-t}|}^r\parallel }_p\cdot {\parallel {|x_i^{1-t}z{y}_i^t|}^r\parallel }_p$ is convex on $[0,1]$ and attains its minimum at $t=\frac{1}{2}$. In view of the fact that $x_i\to x$, $y_i\to y$ in $L_p(\mathcal{M})$, by a similar computation we derive ${lim}_i{f}_i(t)=\varphi (t)$. Therefore, $\varphi (t)$ is convex on $[0,1]$ and attains its minimum at $t=\frac{1}{2}$. □

An immediate consequence of Theorem 3.3 interpolates the inequality (1) as follows.

Corollary 3.4 Let x, y, z be τ-measurable operators as in Theorem  3.3. For every $r>0$,

$$\begin{array}{rl}{\parallel |x^{\frac{1}{2}}z{y}^{\frac{1}{2}}{|}^r\parallel }_p^2& \le {\parallel |x^{t}z{y}^{1-t}{|}^r\parallel }_p\cdot {\parallel |x^{1-t}z{y}^t{|}^r\parallel }_p\\ \le \parallel |xz{|}^r\parallel \cdot {\parallel |zy{|}^r\parallel }_p\end{array}$$

holds for $0\le t\le 1$.

The following is another example of convex functions involving the noncommutative $L_p$-norm.

Theorem 3.5 Let $x_i\in L_p(\mathcal{M})$ ($i=1,2,\dots ,k$) be positive operators and let $1\le p<\mathrm{\infty }$. Then, for every positive real number r, the function $t\mapsto {\parallel {({\sum }_{i=1}^k{x}_i^t)}^r\parallel }_p$ is convex on $(0,\mathrm{\infty })$.

Proof The same proof as of Theorem 4 of Hiai and Zhan [2] works. □

If $\tau (1)=1$, that is to say, we consider the von Neumann algebra ℳ to be equipped with normal faithful finite unital trace, then we can also get the following noncommutative analog of the variational inequality and we refer the readers to Theorem 2.5 of [3] for more details of the Carlen-Lieb theorems concerning the concavity of certain trace functions.

Theorem 3.6 Let $x,y\in L_r(\mathcal{M})$ be positive operators and let $0<r<1$. Then

$$\tau {(x^r+y^r)}^{\frac{1}{r}}\le \tau (z^{\frac{r-1}{r}}x+{(1-z)}^{\frac{r-1}{r}}y)$$

for every $z\in {\mathcal{S}}_{+}$ such that z and $1-z$ are invertible. If $xy=yx$, take

$$z=x^r{(x^r+y^r)}^{-1}$$

and we get the equality.

Proof The same proof as of Theorem 2.5 of Hansen [3] works. □