Norm inequalities related to the Heinz means

Abstract

Let $(I,|\!|\!|\cdot|\!|\!|)$ be a two-sided ideal of operators equipped with a unitarily invariant norm $|\!|\!| \cdot|\!|\!|$. We generalize the results of Kapil’s, using a new contractive map in I to obtain a norm inequality. And we give a new inequality, which is a comparison between the Heinz means and other related inequalities; moreover, we will obtain some correlative conclusions.

1 Introduction and preliminaries

In this paper, let $B(H)$ denote the algebra of all bounded linear operators on a complex separable Hilbert space, $B(H)_{+}$ denote the cone of positive operators and $K(H)$ denote the ideal of compact operators in $B(H)$. And $(I,|\!|\!|\cdot|\!|\!|)$ is a two-sided ideal of $B(H)$ equipped with a unitarily invariant norm $|\!|\!|\cdot|\!|\!|$. We shall denote this by I instead of $(I,|\!|\!|\cdot|\!|\!|)$ for convenience.

For any compact operator $A\in K(H)$, let $S_{1}(A), S_{2}(A), \ldots$ be the eigenvalues of $|A|=(A^{*}A)^{\frac{1}{2}}$ arranged in decreasing order. If $A\in M_{n}$, which is the algebra of all $n\times n$ matrices over C, we take $S_{k}(A)=0$ for $k>n$. A unitarily invariant norm in $K(H)$ is a map $|\!|\!|\cdot|\!|\!|: K(H)\longrightarrow[0,\infty]$ given by $|\!|\!| A|\!|\!| =g(S(A))$, $A\in K(H)$, where g is a symmetric gauge function, cf. [6, 9]. The Schatten p-norms $\| A\|_{p}=(\sum_{j}S_{j}^{p}(A))^{\frac{1}{p}}$ for $p\geq1$ are significant examples of the unitarily invariant norm, and $\|\cdot\|_{2}$ is a special unitarily invariant norm which is the Hilbert–Schmidt norm defined, for $A\in M_{n}$, as follows:

$$\Vert A \Vert _{2}^{2}=\sum _{i,j} \vert a_{ij} \vert ^{2}= \operatorname{tr}\bigl(A^{*}A\bigr). $$

It is well known that the arithmetic–geometric mean inequality

$$\sqrt{ab}\leq\frac{a+b}{2} $$

for positive numbers a and b has been generalized in various directions.

Bhatia et al. [2] have obtained the result that if A, B, and X are $n\times n$ matrices with A and B are positive definite, then

$$\bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\leq\frac {1}{2} |\!|\!| AX+XB|\!|\!|. $$

The Heinz means

$$H_{v}(a,b)=\frac{a^{1-v}b^{v}+a^{v}b^{1-v}}{2} $$

is an interpolation between the arithmetic and geometric means for $a,b\geq0$ and $v\in[0,1]$. It is easy to see that $H_{v}$ is symmetric and convex function for $v\in[0,1]$ and attains its minimum at $v=\frac {1}{2}$; thus we have

$$\sqrt{ab}\leq H_{v}(a,b)\leq\frac{a+b}{2}. $$

The matrix version has been proved in [2]: if A, B and X are positive definite, then for every unitarily invariant norm the function

$$g(v)=\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|$$

is convex on $[0,1]$, and attains its minimum at $v=\frac{1}{2}$. Thus we have

$$2\bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\leq \bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v}\bigr|\!\bigr|\!\bigr|\leq |\!|\!| AX+XB|\!|\!|. $$

The Heron means is defined by

$$F_{\alpha}(a,b)=(1-\alpha)\sqrt{ab}+\alpha\frac{a+b}{2} $$

for $0\leq\alpha\leq1$, see [1]. This family is the linear interpolation between the geometric and the arithmetic mean. Clearly, $F_{\alpha}\leq F_{\beta}$ whenever $\alpha \leq\beta$.

Bhatia and Davis have proved that the inequality

$$\biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha\biggl( \frac {AX+XB}{2}\biggr)\biggr|\!\biggr|\!\biggr|\leq \biggl|\!\biggl|\!\biggl|(1-\beta)A^{\frac{1}{2}}XB^{\frac {1}{2}}+ \beta\biggl(\frac{AX+XB}{2}\biggr)\biggr|\!\biggr|\!\biggr|$$

is always true for $0\leq\alpha\leq\beta\leq1$, $\beta\geq\frac{1}{2}$, and this restriction on β is necessary in [2].

The Heinz means and the Heron means satisfy the inequality

$$H_{v}(a,b)\leq F_{\alpha(v)}(a,b) $$

for $0\leq v\leq1$. And $\alpha(v)=1-4(v-v^{2})$, this is a convex function, its minimum value is $\alpha(\frac{1}{2})=0$, and its maximum value is $\alpha(0)=\alpha(1)=1$.

In [10] the authors have presented the result that

$$\frac{1}{2}\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|\leq \biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \biggl(\frac {AX+XB}{2}\biggr) \biggr|\!\biggr|\!\biggr|$$

for $\frac{1}{4}\leq v\leq\frac{3}{4}$ and $\alpha\in[\frac{1}{2},\infty ]$ and they used the properties of contractive map on I to prove it.

A different version of the Heinz inequality,

$$\bigl|\!\bigl|\!\bigl|A^{\alpha}XB^{1-\alpha}-A^{1-\alpha}XB^{\alpha} \bigr|\!\bigr|\!\bigr|\leq|2\alpha-1||\!|\!| AX-XB|\!|\!|, $$

for $\alpha\in[0,1]$ was proved by Bhatia and Davis [3] in 1995.

A further generalization, namely

$$\frac{2+t}{2}\bigl|\!\bigl|\!\bigl|A^{\alpha}XB^{1-\alpha}+A^{1-\alpha}XB^{\alpha } \bigr|\!\bigr|\!\bigr|\leq \bigl|\!\bigl|\!\bigl|AX+XB+tA^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|, $$

was proved for $t\in[-2,2]$, and $\alpha\in[\frac{1}{4},\frac{3}{4}]$. For more details refer to [13] and [14].

Contractive maps on I play a key role to prove the inequalities, more details refer to [4, 5, 7, 8, 11]. This in turn proves the corresponding Schur multiplier maps to be contractive. We use the ideals to establish the fact that some special maps on I are contractive.

Our paper consists of three parts. In the second part, we will give a new norm inequality which is proved by the properties of contractive maps on I. In the third part, we will present a inequality related to the Heinz means and we notice that this inequality is a comparison between the Heinz means and the geometric mean. Our results are extensions of some previous conclusions about norm inequalities.

2 Norm inequalities with contractive maps

Let $L_{X}$, $R_{Y}$ denote the left and right multiplication maps on $B(H)$, respectively, that is, $L_{X}(T)=XT$ and $R_{Y}(T)=TY$ and we have

$$ e^{L_{X}+R_{Y}}(T)=e^{X}Te^{Y}. $$

(1)

Neeb [12] has proved that $(L_{X}\pm R_{Y})^{-1}\sin(L_{X}\pm R_{Y})$ is an expansive map on I by using the Weierstrass factorization theorem. We use $X_{1}$ and $Y_{1}$ to denote two selfadjoint operators in $B(H)$ and $D=L_{X_{1}}-R_{Y_{1}}$. The following proposition is a result for a contractive map in I; for more details refer to [10].

Proposition 2.1

([10, Proposition 2.4])

Let $0\leq r\leq s$ or $s\leq r\leq0$, then each of the following operator maps is a contraction on I.

• (1) $\frac{(1+t)\cosh(\frac{r}{2}D)}{t+\cosh(sD)}$ for $|t|\leq1$.

• (2) $\frac{t+\cosh(rD)}{t+\cosh(sD)}$ for $|t|\leq1$.

• (3) $\frac{\cosh(rD)}{\cosh(sD)}$. In particular $\frac{1}{r}\int\frac {\cosh(rD)}{\cosh(sD)}\,dr=\frac{\sinh(rD)}{rD\cosh(sD)}$ is a contraction.

• (4) $\frac{s\sinh(rD)}{r\sinh(sD)}$. The case $r=0$ would become $\frac {sD}{\sinh(sD)}$.

Corollary 2.2

([10, Corollary 2.5])

The following maps are contractive on I.

• (1) $\frac{(t+1)\cosh((2v-1)D)}{t+\cosh D}$ for $\frac{1}{4}\leq v\leq \frac{3}{4}$, and $| t|\leq1$.

• (2) $\frac{2\cosh(rD)}{\cosh(s_{1}D)+\cosh(s_{2}D)}$ for $0\leq r\leq \max\{ \frac{s_{1}+s_{2}}{2},\frac{s_{1}-s_{2}}{2}\}$ or $\min \{ \frac {s_{1}+s_{2}}{2},\frac{s_{1}-s_{2}}{2}\}\leq r\leq0$.

• (3) $\frac{(s_{1}+s_{2})\sinh(rD)}{r(\sinh(s_{1}D)+\sinh(s_{2}D))}$ for $0\leq r\leq\frac{s_{1}+s_{2}}{2}$ or $\frac{s_{1}+s_{2}}{2}\leq r\leq 0$.

Corollary 2.3

Let $0\leq v\leq1$ and $|t|\leq1$. Then the map $\frac{(t+1)\cosh ((2v-1)D)}{t+\cosh(2D)}$ is contractive on I.

Proof

This map is a special case of $(1)$ of Proposition 2.1 when $|s|=2$. From the condition $0\leq v\leq1$, we can obtain $| 2v-1|\leq1$. Letting $\frac{r}{2}=2v-1$, we get the desired result. □

With these above results about contractive maps in I, we obtain a new norm inequality which derives from the Heinz means and other related inequalities. Let R be an invertible operator in $B(H)_{+}$, then there exists a selfadjoint operator $S\in B(H)$ such that $R=e^{S}$. To avoid repetitions, we denote the two invertible operators A and B in $B(H)_{+}$ by $e^{2X_{1}}$ and $e^{2Y_{1}}$, respectively, where $X_{1}$ and $Y_{1}$ in $B(H)$ are selfadjoint. The corresponding operator map $L_{X_{1}}-R_{Y_{1}}$ is denoted by D.

Theorem 2.4

Let $v\in[0,1]$ and $\alpha\in[\frac{1}{2},\infty)$. Supposed that A and B are any two invertible operators in $B(H)_{+}$, $X\in I$, then

$$ \frac{1}{2}\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|\leq \biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \frac{A^{\frac {3}{2}}XB^{-\frac{1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2}\biggr|\!\biggr|\!\biggr|. $$

(2)

Proof

Put $A^{\frac{1}{2}}XB^{\frac{1}{2}}=T$ and use (1) we can notice that

$$\begin{aligned} &\frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \\ &\quad =\frac{A^{\frac{2v-1}{2}}A^{\frac{1}{2}}XB^{\frac{1}{2}}B^{\frac {1-2v}{2}}+A^{\frac{1-2v}{2}}A^{\frac{1}{2}}XB^{\frac{1}{2}}B^{\frac {2v-1}{2}}}{2} \\ &\quad =\frac{A^{\frac{2v-1}{2}}TB^{\frac{1-2v}{2}}+A^{\frac {1-2v}{2}}TB^{\frac{2v-1}{2}}}{2} \\ &\quad =\frac{e^{(2v-1)(L_{X_{1}}-R_{Y_{1}})}T+e^{-(2v-1)(L_{X_{1}}-R_{Y_{1}})}T}{2} \\ &\quad =\cosh\bigl((2v-1)D\bigr)T. \end{aligned}$$

(3)

And we also have

$$\begin{aligned} &(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \frac{A^{\frac {3}{2}}XB^{-\frac{1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2} \\ &\quad =(1-\alpha)T+\alpha\frac{AA^{\frac{1}{2}}XB^{\frac {1}{2}}B^{-1}+A^{-1}A^{\frac{1}{2}}XB^{\frac{1}{2}}B}{2} \\ &\quad =(1-\alpha)T+\alpha\frac{ATB^{-1}+A^{-1}TB}{2} \\ &\quad =(1-\alpha)T+\alpha\frac {e^{2(L_{X_{1}}-R_{Y_{1}})}T+e^{-2(L_{X_{1}}-R_{Y_{1}})}T}{2} \\ &\quad =\bigl[(1-\alpha)+\alpha\cosh(2D)\bigr]T. \end{aligned}$$

(4)

Setting $\frac{1}{1+t}=\alpha$ in (4) and applying Corollary 2.3, we can obtain (2) from (3) and (4). □

3 Norm inequality related to the Heinz means

According to the above results, we set

$$\begin{aligned} &H_{v}(a,b)=\frac{a^{v}b^{1-v}+a^{1-v}b^{v}}{2}, \\ &G_{v}(a,b)=(1-v)a^{\frac{1}{2}}b^{\frac{1}{2}}+v\frac{a^{\frac {3}{2}}b^{-\frac{1}{2}}+a^{-\frac{1}{2}}b^{\frac{3}{2}}}{2}. \end{aligned}$$

Then we have the following result, which is a interpolation of $H_{v}$ and $G_{v}$.

Theorem 3.1

Let $a,b>0$ and $v\in[\frac{1}{2},1]$. Then

$$ H_{v}(a,b)\leq\biggl(\frac{H_{\frac{1}{2}}(a,b)}{H_{1}(a,b)}\biggr)^{1-v}G_{v}(a,b). $$

(5)

Proof

Without loss of generality, we can suppose that $a=1$. Then the inequality (5) can be simplified to

$$\frac{b^{v}+b^{1-v}}{2}\leq\biggl(\frac{2\sqrt{b}}{1+b}\biggr)^{1-v}\biggl[(1-v) \sqrt {b}+v\frac{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}{2}\biggr]. $$

To prove this inequality, let

$$f(v)=\log\bigl(b^{v}+b^{1-v}\bigr)-(1-v)\log \frac{2\sqrt{b}}{1+b}-\log\bigl[2(1-v)\sqrt {b}+v\bigl(b^{-\frac{1}{2}}+b^{\frac{3}{2}} \bigr)\bigr]. $$

Calculations show that

$$f^{\prime\prime}(v)=\frac{4b^{v}b^{1-v}\log ^{2}b}{(b^{v}+b^{1-v})^{2}}+\frac{(b^{-\frac{1}{2}}+b^{\frac {3}{2}}-2b^{\frac{1}{2}})^{2}}{[(1-v)b^{\frac{1}{2}}+v(b^{-\frac {1}{2}}+b^{\frac{3}{2}})]^{2}}. $$

We can notice that $f^{\prime\prime}(v)\geq0$ for $\frac{1}{2}\leq v\leq1$, thus f is convex on $[\frac{1}{2},1]$. Then we have $f(v)\leq\max\{f(\frac{1}{2}),f(1)\}$. By calculation, we have

$$f\biggl(\frac{1}{2}\biggr)=\log\biggl(\frac{\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac {1}{2}}}{b^{\frac{1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac {3}{2}}}\biggr)\leq0 $$

because of $\frac{\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac{1}{2}}}{b^{\frac {1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac{3}{2}}}\leq 1$ ($b>0$), which equals $\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac{1}{2}}\leq b^{\frac{1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac {3}{2}}$, that is, $8(1+b)b^{\frac{3}{2}}\leq6b^{2}+b^{4}+4b^{3}+4b+1$.

Setting $b=x^{2}$ ($x>0$), thus we have $x^{8}+4x^{6}-8x^{5}+6x^{4}-8x^{3}+4x^{2}+1\geq0$, that is, $(x-1)^{2}(x^{6}+2x^{5}+7x^{4}+4x^{3}+7x^{2}+2x+1)\geq0$, which is always true when $x>0$, then we get the desired results.

In the same way,

$$f(1)=\log\biggl(\frac{1+b}{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}\biggr)\leq0 $$

because of $\frac{1+b}{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}\leq1$ ($b>0$), which equals $1+b\leq b^{-\frac{1}{2}} +b^{\frac{3}{2}}$, that is $(b^{\frac {1}{2}}-1)(b^{\frac{3}{2}}-1)\geq0$, which is always true when $b>0$. Hence the values of $f(\frac{1}{2})$ and $f(1)$ are both less than 0, that is, $f(v)\leq0$. Then we complete the proof. □

Remark 3.2

The inequality which we obtained from Theorem 3.1 can also be written as

$$H_{v}(a,b)\leq\biggl(\frac{2\sqrt{ab}}{a+b}\biggr)^{1-v}G_{v}, $$

and we notice that it is a new comparison between the Heinz means and the geometric means.

Corollary 3.3

Let $A, B\in M_{n}^{+}$, $X\in M_{n}$ and $\frac{1}{2}\leq v\leq1$. If there are two positive numbers m, M such that $m\leq A, B\leq M$, then

$$\biggl\Vert \frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \biggr\Vert _{2}\leq\biggl( \frac {m+M}{2\sqrt{mM}}\biggr)^{v-1} \biggl\Vert (1-v)A^{\frac{1}{2}}XB^{\frac {1}{2}}+v \frac{A^{\frac{3}{2}}XB^{-\frac{1}{2}}+A^{-\frac {1}{2}}XB^{\frac{3}{2}}}{2} \biggr\Vert _{2}. $$

Proof

Let $A=U\operatorname{diag}(\lambda_{i})U^{*}$ and $B=V\operatorname {diag}(\mu_{j})V^{*}$ be the spectral decompositions of A and B, respectively. Letting $U^{*}XV=Y$, we have

$$\begin{aligned} &\frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \\ &\quad =U\frac{\operatorname{diag}(\lambda_{i}^{v})Y\operatorname{diag}(\mu _{j}^{1-v})+\operatorname{diag}(\lambda_{i}^{1-v})Y\operatorname {diag}(\mu_{j}^{v})}{2}V^{*} \\ &\quad =U\frac{[\lambda_{i}^{v}\mu_{j}^{1-v}+\lambda_{i}^{1-v}\mu _{j}^{v}]\circ[y_{ij}]}{2}V^{*}, \end{aligned}$$

(6)

where ∘ denotes the Schur product. Applying (5) we have

$$\begin{aligned} & \biggl\Vert \frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \biggr\Vert _{2}^{2} \\ &\quad =\sum_{i,j}\biggl(\frac{\lambda_{i}^{v}\mu_{j}^{1-v}+\lambda_{i}^{1-v}\mu _{j}^{v}}{2} \biggr)^{2}| y_{ij}|^{2} \\ &\quad \leq \sum_{i,j}\biggl(\frac{\lambda_{i}+\mu_{j}}{2\sqrt{\lambda_{i}\lambda _{j}}} \biggr)^{2(v-1)}\biggl((1-v)\lambda_{i}^{\frac{1}{2}} \mu_{j}^{\frac {1}{2}}+v\frac{\lambda_{i}^{\frac{3}{2}}\mu_{j}^{-\frac{1}{2}}+\lambda _{i}^{-\frac{1}{2}}\mu_{j}^{\frac{3}{2}}}{2}\biggr)| y_{ij} |^{2} \\ &\quad \leq \biggl(\frac{m+M}{2\sqrt{mM}}\biggr)^{2(v-1)} \biggl\Vert (1-v)A^{\frac {1}{2}}XB^{\frac{1}{2}}+v\frac{A^{\frac{3}{2}}XB^{-\frac {1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2} \biggr\Vert _{2}^{2}, \end{aligned}$$

(7)

where we have used the fact that $m\leq\lambda_{i}$, $\mu_{j}\leq M$ to obtain the last inequality. □