数理解析研究所講究録
第 1996 巻 2016 年 54-65

54

RESEARCH OF WEIGHTED 0PERATOR MEANS
FROM TWO POINTS OF VIEW
YOICHI UDAGAWA, TAKEAKI YAMAZAKI, AND MASAHIRO YANAGIDAABSTRACT. In the recent year, P\’alfia and Petz have given to make a weighted operator mean from an arbitrary operator mean. In this report, we shall give concrete
formulae of the dual, orthogonal and adjoint of weighted operator means. Then thecharacterization of operator interpolational means is obtained. We shall show thatthe operator interpolational means is only the weighted power means.

1. INTRODUCTIONLet
be a complex Hilbert space with inner productandbe the setof all bounded linear operators on . An operator
is said to be positive$x\rangle>0$
definite (resp. positive semi-definite) if and only if
resp.
for all non-zero vectors. Let
be the set of all positive definite operatorsin. If an operator
is positive semi-definite, then we write $A\geq$ O. For selfadjoint operators $A,$
, $A\leq B$ means $B-A$ is positive semi-definite. A map
is called an operator mean [6] if the operator
satisfies
$A,$
$C,$
$B,$
$D\geq0,$
the following four conditions; for(i) $A\leq C$ and $B\leq D$ implies,$X(\mathfrak{M}(A,B))X\leq\mathfrak{M}(XAX,
XBX)$
(ii)
for all self-adjoint
,
(iii)
imply
and
in the strong topology,$\mathcal{H}$$\mathcal{B}(\mathcal{H})$$\mathcal{H}$$A\in \mathcal{B}(\mathcal{H})$$\langle Ax,$$x\in \mathcal{H}$$($$\langle Ax, x\rangle\geq 0)$$\mathcal{B}(\mathcal{H})_{+}$$A$$\mathcal{B}(\mathcal{H})$$B\in \mathcal{B}(\mathcal{H})$$\mathfrak{M}:\mathcal{B}(\mathcal{H})_{+}^{2}arrow \mathcal{B}(\mathcal{H})_{+}$$\mathfrak{M}(A, B)$$\mathfrak{M}(A, B)\leq \mathfrak{M}(C, D)$$X\in \mathcal{B}(\mathcal{H})$$A_{n}\searrow A$(iv)$B_{n}\searrow B$$\mathfrak{M}(A_{n}, B_{n})\searrow \mathfrak{M}(A, B)$$\mathfrak{M}(I, I)=I.$

We remark that by the above condition (iii), we may assume $A,$
. It isknown many examples of operator means, for instance, the weighted geometric mean,
the weighted power mean and the logarithmic mean. In particular, the weighted power
mean has been studied by many researchers (cf. [3, 5, 9]) because of its usefulness,for instance, the weighted power means interpolate weighted arithmetic, geometricand harmonic means. One of the fact is that weighted power means derive power
diﬀerence means by integrating their weight [9]. However, we do not know any explicit formula of the weighted operator means except the weighted power means. For
the problem, P\’alfia-Petz [7] has given an algorithm to get weighted operator meansfrom arbitrary operator means. On the other hand, J.I. Fujii-Kamei have consideredanother algorithm to get weighted operator means from arbitrary symmetric operator
means, and they have considered about the operator interpolational means [2]. It is$B\in \mathcal{B}(\mathcal{H})_{+}$

2010 Mathematics Subject Classification. Primary $47A64$ . Secondary $47A63.$
Key words and phrases. Positive definite operator; operator mean; operator monotone function;interpolational mean; power mean.

55
a family of weighted operator means$\{\mathfrak{M}_{t}\}_{t\in[0,1]}$

with the weight$t$such that$\mathfrak{M}_{(1-\lambda)\alpha+\lambda\beta}(A, B)=\mathfrak{M}_{\lambda}(\mathfrak{M}_{\alpha}(A, B), \mathfrak{M}_{\beta}(A, B))$

. In [1], a characterization of the, 1 and $A,$
interpolational
operatormeans have been obtained. But it has not been given any
concrete example of the operator interpolational means.
In this report, we shall study about weighted operator means. In Section 2, weshall introduce the algorithm to get weighted opel.ator means due to P\’alfia-Petz [7],and introduce some properties of weighted operator means. In Section 3, we will givethe formulae of the dual, adjoint and orthogonal of weighted operator means, they
have very intuitive forms. In Section 4, we shall give another characterization of theoperator interpolational means. It says that the operator interpolational means are
just only the weighted power means.
holds for all$\alpha,$$\beta,$$\lambda\in[0$$]$$B\in \mathcal{B}(\mathcal{H})_{+}$

2. WEIGHTED OPERATOR MEANS
is called an operator monotone
A function $f(x)$ defined on an interval$A\leqB$
implies $f(A)\leq f(B)$ for every seif-adjoint operators
function, provided for$A,$lie inandwhose spectral
The next theorem is so important to study operator means.
$I\subseteq \mathbb{R}$$\sigma(A)$$B\in \mathcal{B}(\mathcal{H})$$\sigma(B)$$I.$

Theorem $A$ ([6]). For any operator mean , there uniquely exists an operator monotone function onwith $f(1)=1$ such that$\mathfrak{M}$$f$$(0, \infty)$(2.1)$f(x)I=\mathfrak{M}(I, xI) , x>0.$

The function satisfying (2.1) is called the representing function of . The followinghold:(i) The map
is one to-one onto aﬃne mapping from the set of all
operator means to the set of all non-negative operator monotone functions onandpreserves the order, i.e., letwith $f(1)=1$ . Moreover,respectively,
thenand
be operator means with representing functions$\mathfrak{M}$$f$$\mathfrak{M}\mapsto f$$a$$(0, \infty)$$\mathfrak{M}$$\mathfrak{M}\mapsto f$$f$$\mathfrak{R}$$g_{f}$$\mathfrak{M}(A, B)\leq \mathfrak{M}(A, B)(A, B\geq 0)\Leftrightarrow f(x)\leq g(x)(x>0)$(ii) If$A\in \mathcal{B}(\mathcal{H})_{+}$.

, then$\mathfrak{M}(A, B)=A^{\frac{1}{2}}f(A^{\frac{-1}{2}BA^{\frac{-1}{2}}})A^{\frac{1}{2}}.$

denote the sets of all operator means andIn this paper, The symbol
and, we use the symbol
representing functions, respectively. Especially, foras the representing function of , i.e., tn is an operator monotone function, s.t.,
on ,with$\mathcal{O}\mathcal{M}$$\mathcal{R}\mathcal{F}$$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$\mathfrak{M}$$\mathfrak{m}\in \mathcal{R}\mathcal{F}$$(\zeta)$$\infty)$$\uparrow n(1)=\lambda$$\mathfrak{M}(A, B)=A^{\frac{1}{2}}\uparrow \mathfrak{n}(A^{\frac{-1}{2}BA^{\frac{-1}{2})A^{\frac{1}{2}}}}$

holds for all $A,$
with the representing functions
For operator means, respectively. We can obtain a representing function of a composition mean
$B\in \mathcal{B}(\mathcal{H})_{+}.$$\mathfrak{M},$$\mathcal{R}\mathcal{F}$$\mathfrak{R}_{1},$$\mathfrak{N}_{2}\in \mathcal{O}\mathcal{M}$$\mathfrak{m},$$n_{1}\mathfrak{n}_{2}\rangle\in$56$\mathfrak{M}(\mathfrak{N}_{1},\mathfrak{N}_{2})$as follows: For $x>0,$$\mathfrak{M}(\mathfrak{N}_{1}(I,xI), \mathfrak{N}_{2}(I, xI))=\mathfrak{M}(\mathfrak{n}_{1}(x)I, \mathfrak{n}_{2}(x)I)$$=\mathfrak{n}_{1}(x)\mathfrak{m}(\mathfrak{n}_{1}(x)^{-1}\mathfrak{n}_{2}(x))I.$In what follows, we will use the symbolby the representing function, i.e.,
of(2.2).
For the following discussion, we shall define the -weighted operator means as follows.
$\mathfrak{M}(\mathfrak{n}_{1}(x), n_{2}(x))$$\mathfrak{M}(\mathfrak{N}_{1}, \mathfrak{N}_{2})$$\mathfrak{M}(\mathfrak{n}_{1}(x), \mathfrak{n}_{2}(x))=\mathfrak{n}_{1}(x)\mathfrak{m}(\mathfrak{n}_{1}(x)^{-l}n_{2}(x))$$t$Definition 1. Let. Thenand only if its representing function$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$\mathfrak{M}$is said to be a -weighted operator mean ifsatisfies$t$$\mathfrak{m}\in \mathcal{R}\mathcal{F}$$\mathfrak{m}’(1)=t.$

We remark that if, then, 1 by [7].
P\’alfia-Petz [7] suggested an algorithm for making a -weighted operator mean fromgiven an operator mean, recently. It can be regarded as a kind of binary searchalgorithm:$\mathfrak{m}\in \mathcal{R}\mathcal{F}$$\mathfrak{m}’(1)\in[0$

$]$

$t$

Definition 2 ([7]). Letwith the representing function $m(x)$ . For $A,$ $B\in$
and $t\in[0$ , 1 , let $a_{0}=0and^{\backslash }b_{0}=1,$ $A_{0}=A$ and $B_{0}=B$ . Defineandrecursively by the following procedure defined inductively for all $n=0$ , 1, 2,(i) If $a_{n}=t$ , then $a_{n+1}:=a_{n}$ and $b_{n+1}:=a_{n},$ $A_{n+1}:=A_{n}$ and $B_{n+1}:=A_{\eta},$$:=b_{n}$ and$:=b_{n},$$:=B_{n},$$:=B_{n}$ and(ii) if $b_{n}=t$ , then(iii) if $(1-\mathfrak{m}’(1))a_{n}+\mathfrak{m}’(1)b_{n}\leq t$ , then $a_{n+1}:=(1-\mathfrak{m}’(1))a_{n}+\mathfrak{m}’(1)b_{n}$ and$:=b_{n},$$:=B_{n},$and(iv) if $(1-\mathfrak{m}’(1))a_{n}+\mathfrak{m}’(1)b_{n}>t$ , then $b_{n+1}:=(1-\mathfrak{m}’(1))a_{n}+\mathfrak{m}’(1)b_{n}$ and$a_{n+1}:=a_{n},$and$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$]$$\mathcal{B}(\mathcal{H})_{+}$$A_{n},$$a_{n},$$b_{n}$$B_{n}$$b_{n+1}$$A_{n+1}$$:=\mathfrak{M}(A_{n}, B_{n})$$B_{n+1}:=\mathfrak{M}(A_{n\rangle}B_{n})$For $A,$$B\in \mathcal{B}(\mathcal{H})_{+}$$A_{n+1}$$b_{n+1}$$a_{n+1}$$B_{n+1}$$B_{n+1}$$\mathcal{A}_{n+1}:=A_{n}.$, the Thompson metric $d(A, B)$ is defined by$d(A, B)= \max\{\log M(A/B), \log M(B/A)\},$

where $M(A/B)= \sup\{\alpha>0|\alpha A\leq B\}$ . It is known that
respected to the Thompson metric [8].

$\mathcal{B}(\mathcal{H})_{+}$

is complete

Theorem $B$ ([7]). The operator sequences
anddefined in Definition
2 converge to the same limit point in the Thompson metric, In what follows, we shalldenoteby the limit point of
and$\{A_{n}\}_{n=0}^{\infty}$$\mathfrak{M}_{t}(A, B)$$\{A_{n}\}_{n=0}^{\infty}$$\{B_{n}\}_{n=0}^{\infty}$$\{B_{n}\}_{n=0}^{\infty}.$$A,$
Proposition $C$ ([7]). For
and $t\in[0$ , 1
fulfill the following properties:(i) ifthen,$\mathfrak{M}_{m’(1)}(A,B)=\mathfrak{M}(A,
B)$
(ii)
,
(iii)
is continuous in on the norm topology.
$\mathfrak{N}\in \mathcal{O}\mathcal{M},$$\mathfrak{M},$$B\in \mathcal{B}(\mathcal{H})_{+}$$],$$\mathfrak{M}_{t}(A, B)$and$\mathfrak{N}_{t}(A, B)$$\mathfrak{N}(A, B)\leq \mathfrak{M}(A, B)$$\mathfrak{N}_{t}(A, B)\leq \mathfrak{M}_{t}(A, B)$$\mathfrak{B}b(A, B)$$t$

Corollary $D$ ([7]). For a nontrivial operator mean , there is a corresponding one
parameter family of weighted means. Let $m(x)$ be the representing function. Then similarly we have one parameter family of operator monotone functions
corresponding to. The family
is continuous in$\mathfrak{M}$$\{\mathfrak{M}_{t}\}_{t\in[0,1]}$$of\mathfrak{M}$$\{\mathfrak{m}_{t}(x)\}_{t\in[0,1]}$$a$$\{\mathfrak{M}_{t}\}_{t\in[0,1]}$$\{\mathfrak{m}_{t}(x)\}_{t\in[0,1]}$

57, and $m_{0}(x)=1$ andtrivial means, so actually$\mathfrak{m}_{1}(x)=x$$t$

It is easy to see thatare two extremal points which correspond to the twointerpolates between these two points.

$\mathfrak{m}_{t}(x)$$\frac{d}{dx}\mathfrak{M}_{t}(x)_{x=1}=t.$

3. THE DUAL, ADJOINT AND ORrHOGONAL OF WEIGHTED OPERATOR MEANS
In this section, we shall give concrete formulae of the dual, adjoint and orthogonalof weighted operator means.
and $m(x)$ be the representing function ofDefinition 3. Let
dual, adjoint and orthogonal of are defined by the representing functions, respectively.
and$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$\mathfrak{M}$$\mathfrak{M}$. The$\cdot$$Xt\mathfrak{n}(x)^{-1},$$x\mathfrak{m}(x^{-1})$$\uparrow \mathfrak{n}(x^{-1})^{-1}$

We remark that if $m’(1)=t$ , then$\frac{d}{dx}\mathfrak{m}(x^{-1})^{-1}|_{x=1}=t,$$\frac{d}{dx}xm(x)^{-1}|_{x=1}=1-t$

and$\frac{d}{dx}x\uparrow \mathfrak{n}(x^{-1})|_{x=1}=1-t.$

Proposition 1. Let

and its representing function. $Thenfort\in[O$ , 1$\mathfrak{M}\in 0\mathcal{M}$$m(x^{s})^{\frac{1}{s}}(s\in[-1,1]\backslash \{O\})$$\mathfrak{m}\in \mathcal{R}\mathcal{F}$. Let $g(x)=$$],$$\mathfrak{m}_{t}(x^{s})^{\frac{1}{s}}=g_{t}(x) (\ddagger \mathfrak{n}(x^{8}\rangle^{\frac{1}{s}})_{t})$holds for $altx>0,$
Proposition 1 says that the maps
tative like the following diagram.

$\mathfrak{m}(x)\mapsto t\mathfrak{n}(x^{s})^{\frac{1}{\delta}}$and$\mathfrak{m}(x)\mapsto \mathfrak{m}_{t}(x)$are $commuarrow$$m(x)\frac{g}{r}g(x)=\mathfrak{m}(x^{s})^{\frac{1}{s}}$$t\downarrow t\downarrow$$m_{t}(x)arrow^{g}g_{t}(x)=m_{t}(x^{s})^{\frac{1}{s}}$in Proposition 1 is an operator monotone function
Remark 1. The function
since $m(x)$ is an operator monotone function. Especially, by putting $s=-1$ , we get$g(x)=m(x^{-1})^{-1}$ and, namely, we obtain a relation between.
and the adjoint of$m(x^{s})^{\frac{1}{s}}$$\mathfrak{n}4(x)$$\backslash \mathfrak{n}_{t}(x^{-1})^{-1}=g_{t}(x)$$m_{i}(x)$Before proving Proposition 1, we would like to define some notations which will bebe the representing function of, andused in the proof. Let
as in Definition
For $t\in[0$ , 1 , we define the sequences, we define the corresponding operator sequences
2. For $A,$
respectively, by Definition 2. We remark
andand-weighted-weighted operator mean (resp.
) is an(resp.
that each), and(resp.
operator mean). Then we give a representing function of.
resp.
denote$\mathfrak{M}.$$(\mathfrak{n}\in \mathcal{R}\mathcal{F}$$\mathfrak{M}\in O\mathcal{M}$$\{a_{\mathfrak{m},n}^{(t)}\}_{n=0}^{\infty},$$\{b_{\mathfrak{m},n}^{(t)}\}_{n=0}^{\infty}\subset[0_{\}}1]$$]$$\{A_{r\mathfrak{n},n}^{(t)}\}_{n=0}^{\infty}$$B\in \mathcal{B}(\mathcal{H})_{+}$$\{B_{\mathfrak{n}\iota,n}^{(t)}\}_{n=0}^{\infty}t\circ\{a_{\mathfrak{m},n}^{(t)}\}_{n=0}^{\infty}$$\{b_{\mathfrak{m},n}^{(t)}\}_{n=0)}^{\infty}$$b_{\tau}^{(}$$A_{1\mathfrak{n},n}^{(t)}$$B_{m,n}^{(t)}$$a_{r\mathfrak{n},n}^{(t)}$$A_{t\mathfrak{n},n}^{(t)}$$\mathfrak{m}_{L,n}^{(t)}(x)\in \mathcal{R}\mathcal{F}$$($$m_{R,r\iota}^{(t)}(x)\in \mathcal{R}\mathcal{F})$$B_{tti,n}^{(t)}$58
Proof of Proposition 1. For a given
and $t\in[0$ , 1 , letThen, since $m’(1)=g’(1)$ , we have$]$$\{a_{m,n}^{(t)}\},$$a_{\mathfrak{m},n}^{(t)}=a_{g,n}^{(t)}$andwith the representing function,, 1 be sequences constructed by Definition 2.

$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$\{b_{\mathfrak{m},n}^{(t)}\}\subset[0$$\mathfrak{m}(x)$$]$$(n=0,1,2,$$b_{\tau \mathfrak{n},n}^{(t)}=b_{g,n}^{(t)}$Then we shall show
(3.1)$\mathfrak{m}_{L,n}^{(t)}(x^{s})^{\frac{1}{\theta}}=g_{L,n}^{(t)}(x)$and$\mathfrak{m}_{R,n}^{(t)}(x^{s})^{\frac{1}{s}}=g_{R,n}^{(t)}(x)$hold for $n=0$ , 1, 2, by mathematical induction on
Assume that (3.1) holds in the case $n=k$ . If
alently $(1-g’(1))a_{g,k}^{(t)}+g’(1)b_{g,k}^{(t)}\leq t$ ), then
and
. So

$n$

. The case $n=0$ is clear.

$(1-\mathfrak{m}’(1))a_{\mathfrak{m},k}^{(t)}+\mathfrak{m}’(1)b_{\mathfrak{m},k}^{(t)}\leq t$and$b_{\mathfrak{m},k+1}^{(t)}=b_{m,k}^{(t)}$$\mathfrak{m}_{R,k+1}^{(t)}(x)=\mathfrak{m}_{R,k}^{(t)}(x)$(equiv, i.e.,$b_{g,k+1}^{(t)}=b_{g,k}^{(t)}$$g_{R,k+1}^{(t)}(x)=g_{R,k}^{(t)}(x)$$\mathfrak{m}_{R,k+1}^{(t)}(x^{s})^{\frac{1}{s}}=\mathfrak{m}_{R,k}^{(t)}(x^{s})^{\frac{1}{\delta}}=g_{R,k}^{(t)}(x)=9_{R,k+1}^{(t)}(x)$hold from the assumption. On the other hand, by (2.2), we have$\mathfrak{m}_{L,k+1}^{(t)}(x)=\mathfrak{M}(\mathfrak{m}_{L,k}^{(t)}(x), \mathfrak{m}_{R,k}^{(t)}(x))$(3.2)$=\mathfrak{m}_{L,k}^{(t)}(x)\mathfrak{m}(\mathfrak{m}_{L,k}^{(t)}(x)^{-1}\mathfrak{m}_{R,k}^{(t)}(x))$$g_{L,k+1}^{(t)}(x)=9_{L,k}^{(t)}(x)g(g_{L,k}^{(t)}(x)^{-1}g_{R,k}^{(t)}(x))$,.

By (3.2), we have$g_{L,k+1}^{(t)}(x)=g_{L,k}^{(t)}(x)\mathfrak{m}((g_{L,k}^{(t)}(x)^{-1}g_{R,k}^{(t)}(x))^{s})^{\frac{1}{s}}$$=\mathfrak{m}_{L,k}^{(t)}(x^{s})^{\frac{1}{s}}\mathfrak{m}((\mathfrak{m}_{L,k}^{(t)}(x^{s})^{\frac{1}{s}})^{-s}(\mathfrak{m}_{R,k}^{(t)}(x^{s})^{\frac{1}{s}})^{s})^{\frac{1}{s}}$$=\{\mathfrak{m}_{L,k}^{(t)}(x^{s})\mathfrak{m}(\mathfrak{m}_{L,k}^{(t)}(x^{s})^{-1}\mathfrak{m}_{R_{)}k}^{(t)}(x^{s}))\}^{\frac{1}{s}}=\mathfrak{m}_{L,k+1}^{(t)}(x^{s})^{\frac{1}{s}}.$Similarly, we can also showin (3.1) for the case $n=k+1$ . If
(equivalently $(1-g’(1))a_{g,k}^{(t)}+9’(1)b_{g,k}^{(t)}>t$ ). From the$\mathfrak{M}_{R,n}^{(t)}(x^{s})^{\frac{1}{s}}=g_{R,n}^{(t)}(x)$$(1-\mathfrak{m}’(1))a_{\mathfrak{m},k}^{(t)}+\mathfrak{m}’(1)b_{\mathfrak{m},k}^{(t)}>t$

above, we get$\mathfrak{m}_{L,n}^{(t)}(x^{s})^{\frac{1}{s}}=g_{L,n}^{(t)}(x) , \mathfrak{m}_{R_{)}n}^{(t)}(x^{s})^{\frac{1}{s}}=g_{R,n}^{(t)}(x)$hold for $n=0$ , 1, 2, Since, respectively, on$\mathfrak{n}4(x^{8})^{\frac{1}{s}}$$g_{L,n}^{(t)}(x)$$(0, \infty)$

and$\mathfrak{m}_{L,n}^{(\ell)}(x^{s})^{\frac{1}{s}}$converge point wise to$g_{t}(x)$and,$0\leq|g_{l}(x)-\mathfrak{m}_{t}(x^{8})^{\frac{1}{s}}|$$=|g_{t}(x)-g_{L,n}^{(t)}(x)+\mathfrak{m}_{L,n}^{(t)}(x^{s})^{\frac{1}{s}}-\mathfrak{m}_{t}(x^{s})^{\frac{1}{s}}|$$\leq|g_{t}(x)-g_{L,n}^{(t)}(x)|+|\mathfrak{m}_{L,n}^{(t)}(x^{s})^{\frac{1}{s}}-\mathfrak{m}_{t}(x^{s})^{\frac{1}{s}}|arrow 0 (as narrow\infty)$.
$\square$59
In the next theorem, we obtain intriguing results for the relations among the dual
and the orthogonal of weighted operator means. It complements our intuitive understanding of the weighted operator means.
Theorem 2. Let

Define $k(x)$be the representing function of
and $l(x)$ $:=xm(x^{-1})$ . Then for $t\in[0$ , 1$\mathfrak{M}\in \mathcal{O}\mathcal{M}$$:=x(\mathfrak{n}(x)^{-1}$, and let$\iota \mathfrak{n}\in \mathcal{R}\mathcal{F}$$\mathfrak{M},$$],$$k_{1-t}(x)=x\mathfrak{m}_{t}(x)^{-1}$and$t_{1-t}(x)=x\mathfrak{m}_{t}(x^{-1})$.

Theorem 2 gives similar consequences to Proposition 1 as the following diagram.
$m(x)$$arrow^{k}$$m(x)$$k(x)=x\iota \mathfrak{n}(x^{-1})$$1-t\downarrow$$1-t\downarrow$$arrow^{l}$$l(x)=xm(x)^{-1}$$1-t\downarrow$$1-t\downarrow$$\mathfrak{m}_{\lambda-t}(x)arrow^{k}k_{1-t}(x)=xm_{t}(x^{-1})$$m_{1-t}(x)arrow^{l}t_{1\sim t}(x)=x\mathfrak{m}_{t}(x)^{-1}$

Proof. First we shall show $l_{1-t}(x)=xm_{t}(x^{-1})$ . Let,1be the sequences constructed by Definition 2 for a given function $l(x)$ and a constantbe so for $t\in[0$ , 1 . Then, since$1-t\in[0$ , 1 , and let$1-l’(1)$ , we have$\{a_{t,n}^{(1-t)}\}_{n=0}^{\infty},$$\{a_{t\mathfrak{n},n}^{(t)}\}_{n=0\}}^{\infty}\{b_{m,n}^{(t)}\}_{n=0}^{\infty}$$]$$a_{ttt,n}^{\langle t)}=1-b_{i,n}^{(1-t)}$and$\{b_{l_{\}}n}^{(1-t)}\}_{n=0}^{\infty}\subseteq[\zeta\}$$]$$\mathfrak{m}’(1)=$$]$$(n=0$ , 1, 2,$b_{m,n}^{(t)}=1-a_{l,n}^{(1-t)}$

To prove $l_{1-t}(x)=xm_{t}(x^{-1}\rangle$ , it is enough to show$(3.3\rangle$$xm_{L,n}^{(t\rangle}(x^{-1})=t_{R,n}^{(1-t)}(x)$and$x\dagger \mathfrak{n}_{R,n}^{(t)}(x^{-1})=l_{L,n}^{(1-t)}(x)$hold for $n=0$ , 1, 2, by mathematical induction on as in the proof of Proposition
1. The case $n=0$ is clear. Assume that (3.3) holds in the case $n=k$ . If (1(equivalently $(1-l’(1))a_{i,k}^{(1-t)}+t’(1)b_{l,k}^{(1-t)}\geq 1-t$ ), then
. Therefore
andholds from the assumption. On the other hand, by (2.2), we have$n$$m^{;}(1))a_{n\backslash ,k}^{(t)}+m’(1)b_{rr\iota,k}^{(t)}\leq t$$m_{R,k+1}^{(t\rangle}(x)=\mathfrak{n}\iota_{R,k}^{(t)}(x)$$l_{L,k+1}^{(1-t)}(x)=l_{L,k}^{(1-t)}(x)$$xm_{R,k+1}^{(t)}(x^{-1})=l_{L,k+1}^{(1-t)}(x)$$\mathfrak{m}_{L,k+1}^{(t)}(x)=\mathfrak{M}(m_{L,k}^{(t)}(x), \mathfrak{m}_{R,k}^{(t)}(x))$$=\mathfrak{m}_{L,k}^{(t)}(x)\mathfrak{m}(m_{L_{\}}k}^{(t)}(x)^{-\lambda}\mathfrak{m}_{R,k}^{\{t)}(x))$,$l_{R,k+1}^{(1-i)}(x)=l_{L,k}^{(1-t)}(x)t(t_{L,k}^{(1-t\rangle}(x)^{-1}t_{R,k}^{(\lambda-t\rangle}(x))$.

Ikom the assumption, we get$l_{R,k+1}^{(1-t)}(x)=l_{R,k}^{(1-t)}(x)l(l_{R,k}^{(1-t)}(x)^{-1}l_{L,k}^{(1-t)}(x))$$=x\mathfrak{n}\tau_{R,k}^{(t)}(x^{-1})t(m_{R,k}^{(t)}(x^{-1})^{-1}m_{L,k}^{(t\rangle}(x^{-1}))$$=x\mathfrak{m}_{R,k}^{(t)}(x^{-1})(m_{R_{)}k}^{(t)}(x^{-1})^{-1}m_{l_{J},k}^{(t)}(x^{-1}))\mathfrak{m}(m_{R,k}^{(t)}(x^{-1})\mathfrak{m}_{L,k}^{(t)}(x^{-1})^{-1})$$=x\mathfrak{m}_{L,k}^{(t)}(x^{-1})\mathfrak{m}(m_{R,k}^{(t)}(x^{-1}){\}\mathfrak{n}_{L,k}^{(t)}(x^{-1})^{--1})$$=xm_{L,k+\lambda}^{(t)}(x^{-1})$.

60
Likewise, we can also show the case(equivalently $(1-l’(1))a_{l,k}^{(1-t)}+t’(1)b_{l,k}^{(1-l)}<1-t$ ).

$(1-\mathfrak{m}’(1))a_{\mathfrak{m},k}^{(t)}+m’(1)b_{\mathfrak{m},k}^{(t)}>t$

From the above, we obtain$x\mathfrak{m}_{L,n}^{(t)}(x^{-1})=l_{R,n}^{(1-t)}(x)$We can show

and$k_{1-t}(x)=x\mathfrak{n}v(x)^{-1}$$x\mathfrak{m}_{R,n}^{(t)}(x^{-1})=l_{L,n}^{(1-t)}(x)$$(n=0,1,2,$

by the same way.

$\square$Example 1. Let $f(x)= \frac{1+x}{2}$ (Arithmetic mean) andDefinition 2 implies$t= \frac{1}{4}$. Then applying the$f_{\frac{\lambda}{4}}(x)= \frac{3}{4}+\frac{1}{4}x$and we have$xf_{\frac{1}{4}}(x)^{-1}=[ \frac{1}{4}+\frac{3}{4}x^{-1}]^{-1}$

On the other hand, $k(x)=xf(x)^{-1}= \frac{2x}{1+x}$ (Harmonic mean) and$k_{\frac{s}{4}}(x)=[ \frac{1}{4}+\frac{3}{4}x^{-1}]^{-1}$from the algorithm of Definition 2. So we obtain$k_{1-\frac{1}{4}}(x)=xf_{\frac{1}{4}}(x)^{-1}.$

4. INTERPOLATIONAL MEANS

In this section, characterizations of interpolatinal means will be obtained. We shallconsider them in the cases of numerical and operator interpolatinal means, separately.

Definition 4 (Interpolational mean, [2]).
: $(0, \infty)^{2}arrow(0, \infty)$ be a continuous function.
(i) For each $t\in[0$ , 1 , let
Assumeis point wise continuous on $t\in[0$ , 1 . The family of continuous
is said to be an interpolatinal mean if and only if thefunctionsfollowing condition is satisfied;$]$$m_{t}$$]$$m_{t}$$\{m_{t}\}_{t\in[0,1]}$$m_{\delta}(m_{\alpha}(a, b), m_{\beta}(a, b))=m_{(1-\delta)\alpha+\delta\beta}(a, b)$, 1 and $b\in(O, \infty)$ .
for all(ii) Letbe a family of weighted operator means. If, 1 and satisfies the following condition, thenon
be an operator interpolational mean;$\alpha,$$\beta,$$\delta\in[0$

$]$

$a,$

$\{\mathfrak{M}_{\alpha}\}_{\alpha\in[0,1]}$$\alpha\in[0$$]$$\alpha,$$\beta,$$\delta\in[0$is continuous
is said to$\{\mathfrak{M}_{\alpha}\}_{\alpha\in[0,1]}$$\mathfrak{M}_{\delta}(\mathfrak{M}_{\alpha}(A, B), \mathfrak{M}_{\beta}(A, B))=\mathfrak{M}_{(1-\delta)\alpha+\delta\beta}(A, B)$

for all$\mathfrak{M}_{\alpha}$, 1 and all $A,$$]$

$B\in \mathcal{B}(\mathcal{H})_{+}.$

61
A typical example of operator interpolational mean is the weighted power mean
whose representing function is$P_{s,\alpha}(x)=[(1-\alpha)+\alpha x^{s}]^{\frac{1}{s}} (s\in[-\lambda, 1]\backslash \{0\})$(The case $s=0$ is considered as
Firstly, we think about the numerical case.

$\lim_{sarrow 0}P_{s,\alpha}(x)=x^{\alpha}.$.

): $(0, \infty)^{2}arrow(0, \infty)$ be a continuous function.
Theorem 3. For each $t\in[0$ , 1 , let
Assume thatis point wise continuous on $t\in[O$ , 1 , and is satisfying the following
conditions) and $t\in[O$ , 1(i) $m_{0}(a, b)=a,$ $m_{1}(a, b)=b$ and $m_{t}(a, a)=a$ for all(ii) if
or , then $a=b$ for all $b\in(O, oo)$ ,Then the following assertions are equivalent:(1)
is an interpolational mean,(2) there exists a real-valuel function such that$]$$m_{t}$$]$$m_{t}$$a,$ $b\in\langle O,$$b$$m_{\frac{1}{2}}((x, b)=a$$],$$\infty$$a,$$\{m_{t}\}_{t\in[0,1]}$$f$$m_{t}(a, b)=f^{-1}[(1-t)f(a)+tf(b)]$for each $t\in[0$ , 1 and$]$$a,$ $b\in(O, \infty)$.

. For fixed), letis clear. .We shall prove (1)$M_{a,b}(t)$$a\neqb$$m_{t}(a, b)$ $:=M_{a,b}(t)$ . We may assume
is a one-to-one. First we shall prove$\beta\in[0$$t\in[0$,that1,mapping
1
.
such
on
onto
Assume that there exists$Fo1$, 1 such: any
there uniquely exists
and $M_{a,b}(\alpha)=M_{a,b}(\beta)=\mu.$,that
so we have

Proof. (2)$a,$ $b\in\langle 0,$$\Rightarrow(2)$$\Rightarrow(1)$$]$$\alpha<\beta$$]$$\alpha,$$\delta\in[0$$\gamma\in[\alpha, \beta]$$\infty$$]$$\gamma=(1-\delta)\alpha+\delta\beta$$M_{a,b}(\gamma)=M_{a,b}((1-\delta)\alpha+\delta\beta)$$=m_{(1-\delta)\alpha+\delta\beta}(a, b)$$=m_{\delta}(m_{\alpha}(a, b), m_{\beta}(a, b))$$=m_{\delta}(\mu, \mu)=\mu$,(by (1))(by (i) )namely,(4.1)

holds for allhave$M_{a,b}(\gamma)=\mu$$\gamma\in[\alpha, \beta]$. Moreover, for each$\epsilon>0$such that$[\alpha-\epsilon, \alpha+\epsilon]\subseteq[0, \beta]$, we$\mu=M_{a,b}(\alpha)$(by (1))$=m_{\frac{1}{2}}(M_{a,b}(\alpha-e), M_{a,b}(\alpha+\epsilon))$$=m_{\frac{1}{2}}(M_{a,b}(\alpha-6), \mu)$(by (4.1)).

By (ii), it implies $M_{a,b}(\alpha-\epsilon)=\mu$ , so we have $M_{tt,b}(\gamma)=\mu$ for all. Thususing this way several times, we get $M_{a,b}(\gamma)=\mu$ for all$\gamma\in[\alpha-\epsilon, \beta]$. By$\gamma\in[0, \beta]$$\mu=M_{a,b}(0)=(x.$Likewise we can show $\mu=M_{a,b}(1)=b$ , and hence we get $a=b$ . It is a contradiction
to $a\neq b$ . Therefore $M_{a,b}(t)$ is a one to-one mapping on $t\in[0$ , 1 . Also we can show
$M_{a,b}(t)$ is an onto mapping because $M_{a,b}(0)=a,$ $M_{a,b}(1)=b$ and $M_{a,b}(t)$ is continuous
$]$

62
. on
$a,$

$t\in[0_{\}}1]$

is a one-to-one onto mapping. Hence, for fixed, there exists a function
defined on $[a, b]$ such that

. From the above,$b\in(O, \infty)$

such that(4.2)$a\neq b$$M_{a,b}(t)$$f_{a,b}$$f_{a,b}(m_{t}(a, b))=(1-t)f_{a,b}(a)+tf_{a,b}(b)$.

Here we may assume $a<b$ . Next we shall prove that this function
is independentof the interval $[a, b]$ and unique up to aﬃne transformations of. Because for any$M,$ $N\in \mathbb{R}(M\neq 0)$ , let $g(x)=Mf_{a,b}(x)+N$ . Then we can easily obtain$f_{a,b}$$f_{a,b}$$f_{a,b}^{-1}[(1-t)f(a)+tf(b)]=g^{-1}[(1-t)g(a)+tg(b)].$Case 1. $[a, b]\subset[a’,$: $[a, b]arrow[O$ , 1 be the inverse function of $M_{a,b}(t)(=m_{t}(a,$Let$[a’,$, 1 satisfying
it is clear that there exists. Sincemean,
we haveis an interpolational
$b$

$]$

$M_{a,b}^{-1}$

$b$

$b$

$\delta_{1},$$\delta_{2}\in[0$

$b$

From

$[a, b]\subset$$m_{\delta_{1}}(a’, b’)=a,$ $m_{\delta_{2}}(a’, b’)=$$]$$\{m_{t}\}_{t\in[0,1]}$$m_{t}(a, b)=m_{t}(m_{\delta_{1}}(a’, b m_{\delta_{2}}(a’, b$$=m_{(1-t)\delta_{1}+t\delta_{2}}(a’, b’)$$=M_{a’,b’}((1-t)\delta_{1}+t\delta_{2})$.

It is equivalent to$M_{a,b}^{-1}, (m_{t}(a, b))=(1-t)\delta_{1}+t\delta_{2}.$Put $x=m_{t}(a, b)\in[a, b]$ , then$M_{a_{)}b}^{-1}(x)=t$. We have$M_{a,b}^{-1}, (x)=(1-M_{a,b}^{-1}(x))\delta_{1}+M_{a,b}^{-1}(x)\delta_{2},$hence we have$M_{a,b}^{-1}(x)= \frac{1}{M_{a’,b’}^{-1}(b)-M_{a’,b’}^{-1}(a)}M_{a’,b’}^{-1}(x)-\frac{M_{a’,b’}^{-1}(a)}{M_{a’,b’}^{-1}(b)-M_{a’,b’}^{-1}(a)}.$Here by putting$\omega_{1}=\frac{1}{M_{a,b}^{-1},(b)-M_{a,b}^{-1},(a)}$and$\omega_{2}=\frac{M_{a’,b’}^{-1}(a)}{M_{\alpha,b}^{-1},(b)-M_{a,b}^{-1},(a)}$we have$M_{a,b}^{-1}(x)=\omega_{1}M_{a,b}^{-1},(x)+\omega_{2} (x\in[a, b$For$M_{a,b}^{-1}(x)$and$M_{a_{)}b}^{-1},(x)$, let$k(x)=\{\begin{array}{ll}M_{a,b}^{-1}(x) (x\in[a, b])\omega_{1}M_{a,b}^{-1},(x)+\omega_{2} (x\in[a’, b’]\backslash [a, b])\end{array}$

Then $k(x)=\omega_{1}M_{a,b}^{-1},(x)+\omega_{2}$ holds for $x\in[a’,$This result, (4.2) and putting$x=m_{t}(a, b)$ $(or m_{t}(a’, b$
imply $t=M_{a,b}^{-1}(x)=k(x)$ ,$b$$f_{a,b}(x)=k(x)(f_{a,b}(b)-f_{a,b}(a))+f_{a,b}(a)$and$f_{a’,b’}(x)= \frac{k(x)-\omega_{1}}{\omega_{2}}(f_{a’,b’}(b’)-f_{a’,b’}(a^{l}))+f_{a’,b’}(a’)$.

63
From the above, we can find that there exists$\omega_{1}’,$$\omega_{2}’\epsilon \mathbb{R}$such that$f_{c\iota’,b^{t}}(x)=\omega_{1}’f_{a,b}(x)+\omega_{2}’.$

Case 2. $[a, b],$ $[c, d](a<b<c<d)$ ;It’s enough to think about the case []$(\iota, b$$\subseteq[a, d]$and $[c, d]c[a,$$d\rfloor$.

ロCorollary 4. For $t\in[O$ , 1
be a real-valued continuous function
on each variables satisfying the follontng condition
$].$

(4.3)

for all$ietm_{t}:(0, \infty)^{2}arrow \mathbb{R}$$[(1-t)a^{-1}+tb^{-1}]^{-1}\leq m_{t}(a, b)\leq(1-t)a+tb$

and $t\in[O$ ,Assume that
, 1 . Then the following assertions are equivalent:(1)
is an interpolational mean,(2) there exists a real-valued function such that$t\in[0$$a,$ $b\in(O, \infty)$$\{m_{t}\}_{t\in[0,1]}$is point wise continuous on$]$$\{m_{t}\}_{t\in[0,1]}$$f$$m_{t}((j\iota, b)=f^{-1}[(1-t)f(a)+tf(b)]$for all$a,$ $b\in(O, \infty)$

and $t\in[O, 1].$

Proof. It is enough to show that the condition (4.3) satisfies the conditions (i) and(ii) of Theorem 3. Since (i) is easy, here we only show that (4.3) implies the condition(ii) of Theorem 3. If
satisfies, then$m_{\frac{1}{2}}(x, y)=x$$( \frac{x^{-1}+y^{-1}}{2})^{1}\leq x\leq\frac{x+y}{2}$by (4.3). By the first inequality of the above, we get $y\leq x$ , and also we obtain $x\leq y$from the second inequality of the above. Therefore $x=y$ holds and condition (ii) is[3satisfied.
Lastly we derive a characterization of operator interpolational means from the above
results. The characterization gives us the fact that the weighted power mean is the
only operator interplational mean.

Theorem 5. For

functions$m_{\alpha}(x)$$\alpha\in[0$. If, 1 , let$]$be a weighted operator mean with the representing, 1 andis point wise continuous on$\mathfrak{M}_{\alpha}$$\{\iota\iota\iota_{\alpha}(x)\}_{\alpha\epsilon[0,1]}$$\alpha\in[0$$]$$[(1-\alpha)+\alpha x^{-1}]^{-1}\leq m_{\alpha}(x)\leq(1-\alpha)+\alpha x$

holds for all$\alpha\in[O$, 1 and $x>0$ , then they are mutually equivalent:
is an operator interpolational mean,$]$(1)(2) there exists $r\in[-1_{\}}1],$
In (2), we consider the case $r=0$ as$\{\mathfrak{M}_{\alpha}\}_{\alpha\in[0,1]}$$m_{\alpha}(x)=[(1-\alpha)+\alpha x^{r}]^{\frac{1}{r}}.$$x^{\alpha}.$To prove Theorem 5, we prepare the next

Lemma

$1emma_{\}}.$, Theorem 84
For a real-valued continuous function such that its$m_{\alpha}(a,b)=f^{-1}[(1-\alpha)f(a)+\alpha f(b)]$inverse function, let
for $b>0$ and, 1 . If$\mathfrak{B}([4$$f$$exist\fbox{Error::0x0000}$$\alpha\in[0$

$a,$

$]$

$m_{\alpha}(ka, kb)=km_{\alpha}(a, b)$holds for all $k>0$ andbe determined as$a,$ $b\in(O, \infty)$, then there exists$r\in \mathbb{R}$$m_{\alpha}(a_{:}b)=[(1-\alpha)a^{r}+\alpha b^{r}]^{\frac{1}{r}}.$such that$m_{\alpha}(a, b)$

can

64
Proof of Theorem 5. (2)$\alpha\in[0$

,1

$],$

$\Rightarrow(1)$$\mathfrak{M}_{\alpha}(aI, bI)=a\mathfrak{m}_{\alpha}(\frac{b}{a})I$

is clear. We only show (1)holds from Theorem A and$\Rightarrow(2)$

. For

$a,$

$b>0$ and$a[(1- \alpha)+\alpha(\frac{b}{a})^{-1}]^{-1}\leq a\mathfrak{m}_{\alpha}(\frac{b}{a})\leq a[(1-\alpha)+\alpha\frac{b}{a}]$follows from the assumption of Theorem 5. This relation is equivalent to the following
inequality;$[(1-\alpha)a^{-1}+\alpha b^{-1}]^{-1}\leq \mathfrak{M}_{\alpha}(a, b)\leq(1-\alpha)a+\alpha b,$

and $cI$ , respectively. Hence by theand byhere we identify
assumption and Corollary 4, there exists a real-valued function such that$\mathfrak{M}_{\alpha}(a, b)$$c$$\mathfrak{M}_{\alpha}(aI, bI)$$f$$\mathfrak{M}_{\alpha}(a, b)=f^{-1}[(1-\alpha)f(a)+\alpha f(b)].$Moreover,$\mathfrak{M}_{\alpha}$satisfies the transformer equality
for $c>0$$\mathfrak{M}_{\alpha}(cA, cB)=c\mathfrak{M}_{\alpha}(A, B)$

because$\mathfrak{M}_{\alpha}$

is an operator mean. These facts and Lemma$E$

implies$\mathfrak{M}_{\alpha}(a, b)=[(1-\alpha)a^{r}+\alpha b^{r}]^{\frac{1}{f}}, r\in \mathbb{R}.$Moreover since it is increasing on
Therefore we obtain$r\in \mathbb{R}$, we have $r\in[-1, 1]$ by the assumption.

$\mathfrak{m}_{\alpha}(xI)=\mathfrak{M}_{\alpha}(I, xI)=[(1-\alpha)+\alpha x^{r}]^{\frac{1}{f}}I.$$\square$

REFERENCES[1] J.I. Fujii, Interpolationality for symmetric operator means, Sci. Math. Jpn., 75 (2012), 267-274.
[2] J.I. Fujii and E. Kamei, Uhlmann’s interpolational method for operator means, Math. Japon.,
34 (1989), 541-547.
[3] T. Fnruta, Concrete examptes of operator monotone functions obtained by an elementary method
without appealing to L wner integral representation, Linear Algebra Appl., 429 (2008), 972-980.
[4] G. H. Hardy, J. E. Littlewood and G. P\’olya, Inequalities, 2d ed. Cambridge, at the University
Press, 1952. x\"u 324 pp.
[5] F. Hiai and D. Petz, Introduction to matrix analysis and applications, Universitext. Springer,
Cham; Hindustan Book Agency, New Delhi, 2014. $viii+332$ pp.
[6 F. Kubo and T. Ando, Means of positive hnear operators, Math. Ann., 246 (1979/80), 205-224.
[7 M. P\’alfia and D. Petz, Weighted multivariable operator means of positive definite operators,
Linear Algebra Appl., 463 (2014), 134-153.
[8 A.C. Thompson, On certain contraction mappings in a partially ordered vector space, Proc.
. Math. Soc., 14 (1963), 438-443.
[9] Y. Udagawa, S. Wada, T. Yamazaki and M. Yanagida, On a family of operator means involving
the power diﬀerence means, Linear Algebra Appl., 485 2015), 124-131.
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DEPARTMENT OF MATHEMATICAL SCIENCE FOR INFORMATION SCIENCES, GRADUATE SCHOOLOF SCIENCE, TOKyO UNIVERSITY OF SCIENCE, TOKYO, 162-8601, JAPAN.
-mail address: $1414701\Phi ed$ . tus. ac. jp$E$

DEPARTMENT OF ELECTRICAL, ELECTRONIC AND COMPUTER ENGINEERING, TOYO UNIVERSITY, $KAWAGO8-SHI,$ $SAi^{r}$lAMA, 350-85S5, JAPAN.
-mail address: t-yarnazaki@toyo.jp$E$

DEPARTMENT OF MATHEMATICAL INFORMA’rION SCIENCE, FACULTY
OF SCIENCE, TOKYO, 162-8601, JAPAN.
-mail address: yanag\’ida@rs.tus.ac.jp$VERS\ddagger TY$$E$$O\ddagger^{r}$

SCIENCE, TOKYO UN1-