# 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 Kapilâ€™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 Hilbertâ€“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 arithmeticâ€“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 Î² 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. (1)

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

2. (2)

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

3. (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. (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. (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. (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. (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.â€ƒâ–¡

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).â€ƒâ–¡

## 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.â€ƒâ–¡

### 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 âˆ˜ 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.â€ƒâ–¡

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## Funding

This research is supported by the National Natural Science Foundation of China (11701154), the Natural Science Foundation of the Department of Education, Henan Province (16A110003), (19A110020), and the Program for Graduate Innovative Research of Henan Normal University (No.YL201701).

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Correspondence to Fugen Gao.

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Gao, F., Ma, X. Norm inequalities related to the Heinz means. J Inequal Appl 2018, 303 (2018). https://doi.org/10.1186/s13660-018-1892-7