Unexpected relations which characterize operator means

Abstract

We give some characterizations of self-adjointness and symmetricity of operator monotone functions by using the Barbour transform $f \mapsto \frac{t+f}{1+f}$ and show that there are many non-symmetric operator means between the harmonic mean ! and the arithmetic mean $\nabla$. Indeed, we show that there exists a non-symmetric operator mean between any two symmetric operator means.

1. Introduction

A bounded operator $A$ acting on a Hilbert space $H$ is said to be positive if $(Ax, x) \geq 0$ for all $x \in H$. We denote this by $A \geq 0$. Let $B(H)^+$ be the set of all positive operators on $H$ and let $B(H)^{++}$ be the set of all positive invertible operators on $H$.

A real-valued function $f$ on $(0, \infty )$ is operator monotone if whenever bounded operators $A$, $B$ satisfy $0 < A \leq B$, $f(A) \leq f(B)$. Functions $f(t) = t^s$ $(s \in [0, 1])$ and $f(t) = \log t$ are typical operator monotone functions. Let $OM_+$ be the set of positive operator monotone functions on $(0, \infty )$ and $OM_+^1 = \{f \in OM_+ \mid f(1) = 1\}$.

In 6, Kubo and Ando developed an axiomatic theory for operator connections and operator means for pairs of positive operators. That is, a binary operation $\sigma$ on the class of positive operators, $(A, B) \mapsto A\sigma B$, is called a connection if the following requirements are fulfilled:

(I)

$A \leq C$ and $B \leq D$ imply $A \sigma B \leq C \sigma D$.

(II)

$C(A \sigma B)C \leq (CAC) \sigma (CBC)$.

(III)

If $A_n \searrow A$ and $B_n \searrow B$, then $A_n \sigma B_n \searrow A \sigma B$.

A mean is a connection with normalization condition

(IV) $1 \sigma 1 = 1$.

Kubo and Ando showed that there exists an affine order-isomorphism from the class of operator connections onto the class of positive operator monotone functions, $\sigma \mapsto f(t) = 1\sigma (t1)$.

This theory has found a number of applications in operator theory. In particular, Petz 9 connected the theory of monotone metrics with Kubo and Ando’s operator connections. He proved that an operator monotone function $f: (0, \infty ) \longrightarrow {\mathbf{R}}$ satisfying the functional equation

$$f(t)=tf(t^{-1}), \quad t > 0,$$

is related to a Morozova–Chentsov function which gives a monotone metric on the Riemannian manifold of invertible $n \times n$ density matrices.

It is well known that if $f:(0, \infty ) \rightarrow (0, \infty )$ is operator monotone, the transpose $f'(t) = tf(\frac{1}{t})$, the adjoint $f^*(t) = \frac{1}{f(\frac{1}{t})}$, and the dual $f^\perp = \frac{t}{f(t)}$ are also operator monotone (6) and we call $f$ symmetric if $f = f'$ and self-adjoint if $f = f^*$. It is shown in 6 that if $f$ is symmetric with $f(1) = 1$, then the corresponding operator mean exists between the harmonic mean ! and the arithmetic mean $\nabla$, that is, $! \leq \sigma _f \leq \nabla$.

In this note, we characterize symmetric functions and self-adjoint functions using Barbour transform $\widehat{\quad } : OM_+ \rightarrow OM_+^1$ defined by $\hat{f} = \frac{t + f}{1 + f}$, and characterize the class of non-symmetric operator functions and the class of operator connections $\sigma$ which satisfy $\sigma \leq !$ or $\sigma \geq \nabla$. We show the existence of non-symmetric operator means between any pair of symmetric means.

2. Barbour transform

In 7, for any strictly positive continuous functions $\alpha , \beta , \gamma$ on $(0, \infty )$, the Barbour path $\phi _{\alpha , \beta , \gamma }:[0,1] \rightarrow OM_+^1$ was introduced by

$$\phi _{\alpha , \beta , \gamma }(x) = \frac{x \alpha + (1 - x)\beta }{x + (1 - x) \gamma }$$

and its basic properties are elucidated. In 1, Barbour examined a function $\phi _{t, \sqrt {t}, \sqrt {t}}(x)$ which is an approximation of the power function $t^x$. We will denote a Barbour path $\phi _{\alpha , \beta , \gamma } (= \phi )$ such that $\phi (0) = f$, $\phi (\frac{1}{2}) = g$, $\phi (1) = h$ by the triple $[f, g, h]$.

Proposition 2.1 (7).

For $f \in OM_+$, the Barbour path $\phi _{t,f,f}=[1, \frac{t+f}{1 + f}, t]$ exists in $OM_+^1$.

The transform $\widehat{\quad } : OM_+ \rightarrow OM_+^1$ defined by $f \mapsto \phi _{t,f,f}(\frac{1}{2}) = \frac{t+f}{1 + f}$ plays an important role in the analysis of $OM_+$ and we call this transform the Barbour transform.

Proposition 2.2 (cf. 7).

(1)

For $\lambda \in (0,1)$ the transform $B_\lambda : OM_+ \rightarrow OM_+^1$ defined by $f \mapsto \phi _{t,f,f}(\lambda )$ is injective and $B_\lambda ({OM_+}) = OM_+^1\backslash \{1,t\}$.

(2)

For $\lambda \in (0, 1)$, $\{f \in OM_+^1 \mid !_\lambda \leq \sigma _f \leq \nabla _\lambda \} = B_\lambda (OM_+^1)$, where $A \nabla _\lambda B = (1 - \lambda )A + \lambda B$ and $A!_\lambda B = ((1 - \lambda )A^{-1} + \lambda B^{-1})^{-1}$, and notation $!_\lambda \leq \sigma _f \leq \nabla _\lambda$ means that $A!_\lambda B \leq A\sigma _f B \leq A \nabla _\lambda B$ for any positive operators $A$ and $B$.

Note that for $f \in OM_+$, $B_{\frac{1}{2}}(f) = \hat{f}$.

For $g \in OM_+^1$, we can define the inverse map $\check{}$ of the Barbour transform by

$$\check{g}(t) = \frac{t-g}{g-1};$$

then $\check{g} \in OM_+$. Moreover, for $\lambda \in (0, 1)$ and $g \in OM_+^1$, we can define the inverse map $B_\lambda ^{-1}(g)$ of the transform $B_\lambda (g)$ by $B_\lambda ^{-1}(g) = \frac{\lambda }{1 - \lambda }\cdot \check{g}$.

By a simple calculation, we have the following relations.

Lemma 2.3.

For $\lambda \in (0, 1)$ and $f \in OM_+$, we have

$$B_\lambda (f^\bot ) = B_{1 - \lambda }(f)^\bot , B_\lambda (f^{'}) = B_\lambda (f)^*, B_\lambda (f^*) = B_{1 - \lambda }(f)^{'}.$$

In particular, $\widehat{f^\bot } = (\hat{f})^\bot , \widehat{f^*} = (\hat{f})^{'}$, and $\widehat{f^{'}} = (\hat{f})^*$.

We next show some properties of the Barbour transform concerning the usual order relation for operator means.

Lemma 2.4.

For $f, g \in OM_+$ and $\lambda \in (0, 1)$, the following are equivalent:

(1)

$f(t) \leq g(t)$ for all $t > 0$;

(2)

$B_\lambda (B_\lambda (f))(t) \leq B_\lambda (B_\lambda (g))(t)$ for all $t > 0$;

(3)

$B_\lambda (f)(t) \leq B_\lambda (g)(t)$ for all $t < 1$ and $B_\lambda (f)(t) \geq B_\lambda (g)(t)$ for all $t > 1$.

Proof.

The proof follows from the equation

$$B_\lambda (f) - B_\lambda (g) = \dfrac{\lambda (1-\lambda )(g - f)(t -1)}{\{\lambda + (1 - \lambda )f\}\{\lambda + (1 - \lambda )g\}}.$$ ■

In the following section, we show that the Barbour transform is a powerful tool for finding examples of non-self-adjoint operator means.

3. Self-adjoint means

In 6, Kubo and Ando asked whether there exist self-adjoint operator means besides the trivial means $w_l$, $w_r$, the weighted geometric means corresponding to the operator monotone functions $x^p$ $(0 < p < 1)$. Later, Hansen (3) gave an integral form for a strictly positive self-adjoint operator monotone function based on an exponential map.

Using the Barbour transform, we characterize the self-adjointness in $OM_+$ and give concrete examples in this section.

Proposition 3.1.

Let $f$ be a positive continuous function on $(0, \infty )$ and let $\mathcal{A} = \{c, ct\mid c > 0\}$. The following are equivalent:

(1)

$f \in OM_+^1\backslash \{1,t\}$ and $f = f^*$;

(2)

there exists an operator monotone function $g \in OM_+\backslash \mathcal{A}$ such that $f = \sqrt {gg^*}$;

(3)

there exists an operator monotone function $g \in OM_+$ such that$$f = \frac{t + g + g'}{1 + g + g'}.$$

Proof.

$(1) \leftrightarrow (2)$: Set $g(t) = \sqrt {f}(t)$. Then $g \in OM_+\backslash \mathcal{A}$, $g = g^*$, and $f = \sqrt {gg^*}$.

Conversely, if $f = \sqrt {gg^*}$ for some $g \in OM_+\backslash \mathcal{A}$, it is obvious that $f \in OM_+^1$.

$(1) \rightarrow (3)$: Since $f \in OM_+^1$, there exists a $g \in OM_+$ such that $f = \hat{g}$. Since $\hat{g^{'}} = (\hat{g})^* = f^* = f = \hat{g}$ and the Barbour transform $\hat{\quad }$ is injective, we have $g = g^{'}$; that is, $g$ is symmetric.

Hence,

$$f = \frac{t + \frac{g}{2} + \left(\frac{g}{2}\right)'}{1 + \frac{g}{2} + \left(\frac{g}{2}\right)'}.$$

$(3) \rightarrow (1)$: Set $h = g + g'$. Then $h$ is symmetric operator monotone and $\hat{h} = f$. Furthermore,

$$\begin{align*} f^* &= \left(\hat{h}\right)^*\\ &= \hat{h'}\\ &= \hat{h} = f. \end{align*}$$ ■

The above result shows that we can construct a self-adjoint mean by using a symmetric mean.

Remark 3.2.

Using Proposition 3.1, we can construct many examples of self-adjoint means. For example, if $g(t) = \log (t + 1)$, then the corresponding operator means of functions $\sqrt {\log (t + 1)/\log (t^{-1} + 1)}$ and $\dfrac{t+ \log (t + 1) + t\log (t^{-1} +1)}{1 + \log (t + 1) + t\log (t^{-1} + 1)}$ are self-adjoint.

4. Symmetric means and non-symmetric means

4.1. Symmetric means

Symmetric means have been discussed many times in the literature 4568. In contrast to self-adjoint means, many examples of symmetric means are known and appear in the quantum information literature 9.

Proposition 4.1.

Let $f$ be a positive continuous function on $(0, \infty )$. The following are equivalent:

(1)

$f \in OM_+$ and $f = f'$;

(2)

there exists an operator monotone function $g \in OM_+$ such that$$f = g + g';$$

(3)

there exists an operator monotone function $g \in OM_+\backslash \mathcal{A}$ such that$$f = \frac{t - \sqrt {gg^*}}{\sqrt {gg^*} - 1}.$$

Proof.

$(1) \leftrightarrow (2)$: This is obvious.

$(1) \rightarrow (3)$: Set $h = \hat{f}$. Then $h \in OM_+\backslash \{1, t\}$. Since $f = f'$, $h^* = {(\hat{f})}^* = \widehat{f'} = \hat{f} = h$. Hence by Proposition 3.1 there exists $g \in OM_+\backslash \mathcal{A}$ such that $h = \sqrt {gg^*}$. Therefore

$$\begin{align*} f &= {\check{\hat{f}}}\\ &= \check{h}\\ &= \frac{t - h}{h -1}\\ &= \frac{t - \sqrt {gg^*}}{\sqrt {gg^*} - 1}. \end{align*}$$

$(3) \rightarrow (1)$: Set $h = \sqrt {gg^*}$. Since $h^* = h$ implies that $(\check{h})' = \check{h}$, $f' = (\check{h})' = \check{h} = f$.

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Proposition 4.2.

Let $f$ be a positive continuous function on $(0, \infty )$. The following are equivalent:

(1)

$f \in OM_+^1\backslash \{1,t\}$ and $f = f'$,

(2)

there exists an operator monotone function $g \in OM_+$ such that$$f = \frac{t + \sqrt {gg^*}}{1 + \sqrt {gg^*}}.$$

Proof.

This follows from the same argument as for Proposition 3.1 using the formula $(\hat{h})' = \widehat{h^*}$ for $h \in OM_+$.

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4.2. Non-symmetric means between ! and $\nabla$

It is well known that a symmetric operator mean must be between ! and $\nabla$. To show that the converse is not true, we present an algorithm for constructing a non-symmetric mean $\sigma$ such that $! \leq \sigma \leq \nabla$.

Lemma 4.3.

Let $f$ be a positive operator monotone function on $(0, \infty )$ with $f(1) = 1$. The following are equivalent:

(1)

$\sigma _{\hat{f}}$ is non-symmetric mean;

(2)

$f$ is non-self-adjoint.

Proof.

$(2) \rightarrow (1)$: Since $\left(\widehat{f}\right)^{'} = \widehat{f^*}$, if $f$ is non-self-adjoint operator monotone and $f(1)=1$, $\hat{f}$ is non-symmetric; that is, $\sigma _{\hat{f}}$ is non-symmetric.

$(1) \rightarrow (2)$: If $\sigma _{\hat{f}}$ is a non-symmetric mean, then $\hat{f}\not = \left( \hat{f} \right)' = \widehat{f^*}$ , which implies $f\not = f^*$.

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Lemma 4.4.

If a symmetric operator mean $\sigma$ is self-adjoint, then $\sigma = \sharp$.

Proof.

Let $f$ be an operator monotone function corresponding to $\sigma$. Then

$$f(t) = tf(\frac{1}{t}) = \dfrac{1}{f(\dfrac{1}{t})}.$$

Hence, $f(t) = \sqrt {t}$, and $\sigma = \sharp$.

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Remark 4.5.

From Lemma 4.4, we know that all the following operator means are non-self-adjoint: arithmetic mean, logarithmic mean, harmonic mean, Heinz mean, and Lehmer mean 8.

Hence, we have the following result.

Proposition 4.6.

$$\begin{align*} &\left\{f\mid f:\ \text{non-symmetric}, \ f_{!} \leq f \leq f_\nabla \right\}\\ &= \left\{\hat{f}\mid f:\ \text{non-self-adjoint, $f(1)=1$}\right\}\\ &= \left\{\hat{\hat{f}}\mid f:\ \text{non-symmetric}\right\}\\ &\supset \left\{\hat{f}\mid f:\ \text{symmetric, $f(1)=1$}\right\}\backslash \left\{f_\sharp \right\}. \end{align*}$$

Proof.

The first equality is clear from Lemma 4.3.

It follows from $\widehat{\widehat{f'}}=\left( \hat{\hat{f}} \right)'$ and Proposition 2.2 (2) that

$$\left\{f\mid f:\ \text{non-symmetric}, \ f_{!} \leq f \leq f_\nabla \right\} = \left\{\hat{\hat{f}}\mid f:\ \text{non-symmetric}\right\}.$$

Thus, we have the second equality.

From Lemma 4.4, we have

$$\{ \hat{f}\nobreakspace |\nobreakspace f\not =f^*, f(1)=1\} \supset \{ \hat{f}\nobreakspace |\nobreakspace f =f', f(1)=1\} \backslash \left\{f_\sharp \right\},$$

which means the last inclusion holds.

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Remark 4.7.

From Proposition 4.6, a non-self-adjoint positive operator monotone function $f$ with $f(1) = 1$ gives a non-symmetric operator mean $\sigma _{\hat{f}}$ such that $! \leq \sigma _{\hat{f}} \leq \nabla$. For example, let $p \in [- 1, \frac{1}{2}) \cup (\frac{1}{2}, 2]$ and $\mathrm{ALG}_p$ be the function corresponding to the power difference mean defined by

$$\mathrm{ALG}_p (t) = \left\{\begin{array}{ll} \dfrac{p-1}{p}\dfrac{1-t^p}{1 - t^{p-1}}&t \not = 1,\\1 & t = 1. \end{array} \right.$$

Then $\mathrm{ALG}_p$ is symmetric and non-self-adjoint by Lemma 4.4. Hence, $\sigma _{\widehat{\mathrm{ALG_p}}}$ is non-symmetric. Moreover, the Petz-Hasegawa function $f_p$ which is defined by

$$f_p(t) = p(p - 1)\dfrac{(t - 1)^2}{(t^p - 1)(t^{1-p} - 1)}$$

is non-self-adjoint. Hence, $\sigma _{\widehat{f_p}}$ is a non-symmetric operator mean between ! and $\nabla$.

Later, in Section 5, we shall give an algorithm for constructing non-symmetric operator means between ! and $\nabla$ (see Lemmas 5.1 and 5.2).

4.3. Non-symmetric means between symmetric means

In this section, we show the existence of non-symmetric operator means between any pair of symmetric means.

Theorem 4.8.

Let $\sigma _1, \sigma _2$ be symmetric operator means. If $\sigma _1 \leq \sigma _2$ and $\sigma _1\not =\sigma _2$, then there exists a non-symmetric operator mean $\sigma$ such that $\sigma _1 \leq \sigma \leq \sigma _2$.

To prove the above theorem, we need the following lemma.

Lemma 4.9.

Let $h_1$ and $h_2$ be self-adjoint positive operator monotone functions on $(0, \infty )$ with $h_1 \not = h_2$ and $h_1(1) = h_2(1) = 1$. If $h_1(t) \leq h_2(t)$ for all $t < 1$ and $h_1(t) \geq h_2(t)$ for all $t > 1$, then there exists a non-self-adjoint positive operator monotone function $h$ such that

$$\begin{array}{ll} h_1(t) \leq h(t) \leq h_2(t) &\text{for all} \ t < 1,\\h_1(t) \geq h(t) \geq h_2(t) &\text{for all}\ t > 1. \end{array}$$

Proof.

Fix a non-self-adjoint mean $g$. If $h$ is an operator monotone function corresponding to $\sigma _{h_1}(\sigma _g)\sigma _{h_2}$ (see 6, (2.9)), then $h$ satisfies the above inequalities.

It follows from the assumption $h_1\not =h_2$ that there exists $\delta >0$ such that

$$(1-\delta , 1+\delta ) \subset \{ h_1(t) h_2(t)^{-1}\nobreakspace |\nobreakspace t\in (0,\infty ) \},$$

and

$$g(t)\not = g^*(t) \text{ on } (1 - \delta , 1 + \delta )\backslash \{1\}.$$

So there exists $t_0 \in (0, \infty )\backslash \{1\}$ such that $g\left( h_1(t_0) h_2(t_0)^{-1} \right) \not = g^*\left( h_1(t_0) h_2(t_0)^{-1} \right) .$ Thus,

$$\begin{align*} h\left({1\over {t_0}}\right) &=h_1\left({1\over {t_0}}\right) g \left( {{h_2\left({1\over {t_0}}\right)} \over {h_1 \left({1\over {t_0}}\right) } } \right) \\ &= {1\over {h_1(t_0)}} g \left( {{h_1(t_0)} \over {h_2(t_0)}} \right)\\ &\not = {1\over {h_1(t_0)}} g^* \left( {{h_1(t_0)} \over {h_2(t_0)}} \right)\\ &= {1\over {h_1(t_0)}} {1 \over {g \left( {{h_2(t_0)} \over {h_1(t_0)}} \right)}} \\ &= {1 \over {h(t_0)}}, \end{align*}$$

which implies $h\not =h^*$.

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Proof of Theorem 4.8.

Let $f_1, f_2$ be positive operator monotone functions which correspond to $\sigma _1, \sigma _2$, respectively. If we define $h_1:=\check{f_1}, h_2:=\check{f_2}$, then $h_1,h_2$ satisfy the conditions appearing in Lemma 4.9. Thus, there exists a non-self-adjoint positive operator monotone function $h$ such that

$$h_1(t) \le h(t) \le h_2(t)\quad \text{ for all }t <1$$

and

$$h_1(t) \ge h(t) \ge h_2(t) \quad \text{ for all } t >1 .$$

By a simple calculation, we have

$$f_1=\widehat{h_1} \le \hat{h} \le \widehat{h_2}=f_2$$

and

$$\hat{h}\not = (\widehat{h^*}) = (\hat{h})',$$

which means that $\sigma _{\hat{h}}$ is a desired mean.

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If we define $g(t):={{1+t} \over 2}$ in the proof of Lemma 4.9, the function $h$ which is an operator monotone function corresponding to $\sigma _{\check{f_1}}(\sigma _g)\sigma _{\check{f_2}}$ can be written as

$$h=\left(\dfrac{\check{f_1} + \check{f_2}}{2}\right) .$$

Corollary 4.10.

Let $f_1$, $f_2$ be symmetric positive operator monotone functions on $(0, \infty )$ with $f_1(1) = f_2(1) = 1$. If $f_1 \leq f_2$ and $f_1 \not = f_2$, then

$$\widehat{\left(\dfrac{\check{f_1} + \check{f_2}}{2}\right)} = {{2t(1-f_1)(1-f_2) + (f_1-t)(1-f_2) + (f_2- t)(1-f_1)} \over {2(1-f_1)(1-f_2) + (f_1-t)(1-f_2) + (f_2- t)(1-f_1)}}$$

is a non-symmetric positive operator monotone function between $f_1$ and $f_2$.

Example 4.11.

(1)

Since $f_! = \hat{t}$ and $f_\nabla = \hat{1}$, $\widehat{\frac{t+1}{2}}={{3t+1}\over {t+3}}$ is a non-symmetric positive operator monotone function between $f_!$ and $f_\nabla$.

(2)

Since $f_\sharp = \hat{f_\sharp }$ and $f_\nabla = \hat{1}$, $\widehat{\frac{\sqrt {t} + 1}{2}}={{2t+\sqrt {t} +1 } \over { \sqrt {t} +3}}$ is a non-symmetric positive operator monotone function between $f_\sharp$ and $f_\nabla$.

5. Application

It is well known that a symmetric operator mean exists between ! and $\nabla$. To construct non-symmetric means between ! and $\nabla$, we introduced the Barbour transform in Section 2 to characterize self-adjoint means and symmetric means.

Therefore, we can systematically construct non-symmetric operator means between ! and $\nabla$.

Lemma 5.1.

Let $f \colon (0, \infty ) \rightarrow (0, \infty )$ be a continuous function. The following are equivalent:

(1)

$f \in OM_+$ and $f \geq f_\nabla$, that is, $f(t) \geq \frac{1+t}{2}$ for $t \in (0, \infty )$;

(2)

there exists an operator monotone function $g \in OM_+ \cup \{0\}$ and non-negative real numbers $a, b \geq \frac{1}{2}$ such that $\lim _{t\rightarrow 0}g(t) = 0$, $\lim _{t \rightarrow \infty }\frac{g(t)}{t} = 0$, and$$f(t) = a + bt + g(t) \ (t \in (0, \infty )).$$

Proof.

The implication from $(1)$ to $(2)$: From Löwner’s Theorem (for example, see 2, Exercise V.4.9), there exists a positive Radon measure $\rho$ on $[0, \infty ]$ such that

$$f(t) = a + bt + \int _0^\infty \frac{t(1 + \lambda )}{t + \lambda } d\rho (\lambda ).$$

We have then

$$\begin{align*} a = \lim _{t\rightarrow 0}f(0) &\geq f_{\nabla }(0) = \frac{1}{2}, \\ b = \lim _{t\rightarrow \infty }\frac{f(t)}{t} &\geq \lim _{t\rightarrow \infty }\frac{f_{\nabla }(t)}{t} = \frac{1}{2}. \end{align*}$$

Define a function $g$ on $(0, \infty )$ by $g(t) = \int _0^\infty \frac{t(1 + \lambda )}{t + \lambda } d\rho (\lambda )$. Then we have $g \in OM_+$, $\lim _{t\rightarrow 0}g(t) = 0$, and $\lim _{t\rightarrow \infty }\frac{g(t)}{t} = 0$.

The implication from $(2)$ to $(1)$ is obvious.

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Lemma 5.2.

Let $f \colon (0, \infty ) \rightarrow (0, \infty )$ be a continuous function. The following are equivalent:

(1)

$f \in OM_+$ and $f \leq f_!$, that is, $f(t) \leq \frac{2t}{1+t}$ $(t \in (0, \infty ))$;

(2)

there exists an operator monotone function $g \in OM_+ \cup \{0\}$ and non-negative real numbers $a, b \geq \frac{1}{2}$ such that $\lim _{t\rightarrow 0}g(t) = 0$, $\lim _{t\rightarrow \infty }\frac{g(t)}{t} = 0$, and$$f(t) = \frac{t}{a + bt + g(t)} \ (t \in (0, \infty )).$$

Proof.

The lemma follows from the observation that $f \leq f_!$ if and only if $\frac{t}{f} \geq f_\nabla$.

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Remark 5.3.

The followings should be well known:

(1)

If $f \in OM_+^1$ and $f \leq f_!$, then $f = f_!$.

(2)

If $f \in OM_+^1$ and $f \geq f_\nabla$, then $f = f_\nabla$.

Proposition 5.4.

Suppose that $f \in OM_+$ and $f \le f_{!}$. Then $f_!\le \hat{\hat{\hat{f}}} \le f_\nabla$ and $\hat{\hat{\hat{f}}}$ is not symmetric.

Proof.

We first consider the case $f(1)<1$. Suppose that $\hat{f}$ is symmetric. Then $(\hat{f} )'= (\widehat{f^*} )=(\hat{f} )$. Since $\hat{ }$ is injective, $f^*=f$; that is, $f$ is self-adjoint. This is a contradiction and therefore $\hat{f}$ is not symmetric (so is $\hat{\hat{\hat{f}}}$). If $f(1)=1$, then, by Remark 5.3, we have $f=f_!$ and $\hat{\hat{\hat{f}}}$ is not symmetric.

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Corollary 5.5.

Let $a,b$ be non-negative real numbers greater than or equal to $1\over 2$ and assume $g \in OM_+\cup \{0\}$ satisfies condition (2) in Lemma 5.2. Define a function $f : (0,\infty ) \rightarrow (0,\infty )$ by $f(t)={t \over {a+bt+g(t)}}\nobreakspace (t\in (0,\infty ))$. Then $f\in OM_+, f_!\le \hat{\hat{\hat{f}}} \le f_\nabla$ and $\hat{\hat{\hat{f}}}$ is not symmmetric.

Proof.

It is obvious that $f\in OM_+$ and $f\le f_!$. By Proposition 5.4, $f_!\le \hat{\hat{\hat{f}}} \le f_\nabla$ and $\hat{\hat{\hat{f}}}$ is not symmetric.

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