数理解析研究所講究録
第 1839 巻 2013 年 49-55

49

Elementary proofs of Petz-Hasegawa theorem and Hansen resultsby only using L\"owner-Heinz inequality
弘前大学 (名誉教授) 古田孝之 (TAKAYUKI FURUTA)

Hirosaki University (Emeritus)

To the memory of Professor Tsuneo Kanno passed away by tsunami disasterat Tohoku district on 2011.3.11 with deep sorrow
2011.3.11 東日本大津波で亡くなられた菅野恒雄東京工業大学名誉教授の霊に捧げますWe show elementary proofs of useful and important operator monotone functions byPetz-Hasegawa and Frank Hansen. Also we consider some extensions of results by Hansen.
A capital letter means a bounded linear operator on a Hilbert space $H$ . An operator
is said to be positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H$ and also an operator
is said to be strictly positive (denoted by $T>0$ ) if is positive and invertible.
realvalued continuous function $f(t)$ on
is said to be operator monotone if $f(A)\geq f(B)$holds for $A\geq B.$

$T$

$T$

$T$

$A$

$(0, \infty)$\S 1 An elementary proof of Theorem$B$by Petz-HasegawaTheorem is the essential factor of the operator monotone function giving the famousand important Wigner- Yanase-Dyson skew information and it turns out that WignerYanase-Dyson skew information is closely ralated to the special case of quantum Fisher$B$information.
We cite the following almost obvious Proposition A which is an immediate consequenceof the well known celebrated result as ( $LH$ ) (abbreviation of L\"owner-Heinz) thatis anoperator monotone for any $\alpha\in[0,1]$ to give a proof of Theorem B.
$t^{\alpha}$

Proposition $A$ ([2]).
( $LH$ )$t^{\alpha}$

is an operator monotone for any $\alpha\in[0,1].$( $LH$ -1) Let$\alpha_{j},$$\beta_{j},$$\gamma_{j}\ldots\in[0,1]$$forj=1,2,$$\ldots,$$n$. Then$( \frac{1}{t^{\alpha}1+t^{\alpha}2+\ldots+t^{\alpha_{n}}}+\frac{1}{t^{\beta_{1}}+t^{\beta_{2}}+\ldots+t^{\beta_{n}}}+\frac{1}{t^{\gamma_{1}}+t^{\gamma_{2}}+\ldots+t\gamma_{n}}+\ldots)^{-1}$in particular.

$(t^{-\alpha}1+t^{-\alpha 2}+\ldots+t^{-\alpha_{n}})^{-1}$is operator monotone,

rs opemtor monotone.

First of all, recall the following obvious Lemma 1.

50Lemma 2. Let natuml number and such that $n-1\geq k\geq 1$ . Then the following
inequality holds for any positive real number :
$k$

$n$

$t$

$(t^{n-1}+t^{n-2}+\ldots\ldots t^{2}+t+1)^{2}$$=(t^{n-k-1}+t^{n-k-2}+\ldots..+t^{2}+t+1)(t^{n+k-1}+t^{n+k-2}+\ldots..+t^{2}+t+1)$(1.1)$+t^{n-k}(t^{k-1}+t^{k-2}+\ldots..+t^{2}+t+1)^{2}$ .

Proof. We show (1.1) by mathematical inductioin on$k$such that $n-1\geq k\geq 1.$(i) In case $k=n-1$ . In faet (1.1) holds for $k=n-1$ because (1.1) putting $k=n-1,$which just coincides with (1.0) of Lemma 1.
(ii) Assume (1.1) for some$k$such that $n-1\geq k\geq 1$ . We show that (1.1) holds for $k-1.$$(t^{n-1}+t^{n-2}+\ldots\ldots t^{2}+t+1)^{2}$$=(t^{n-k-1}+t^{n-k-2}+\ldots\ldots+t+1)\{(t^{n+k-1}+t^{n+k-2}+\ldots t^{2k-1})+(t^{2k-2}+t^{2k-3}\ldots+t+1)\}$$+t^{n-k}\{(t^{2k-2}+t^{2k-3}+\ldots\ldots+t+1)+t(t^{k-2}+t^{k-3}+\ldots\ldots+t+1)^{2}\}$by (1.0)$=(t^{n-k-1}+t^{n-k-2}+\ldots\ldots+t+1)(t^{n-k}+t^{n-k-1}+\ldots\ldots+t+1)t^{2k-1}$$+(t^{n-k}+t^{n-k-1}+\ldots\ldots t+1)(t^{2k-2}+t^{2k-3}+\ldots\ldots+t+1)+t^{n-k+1}(t^{k-2}+t^{k-3}\ldots+t+1)^{2}$$=(t^{n-k}+t^{n-k-1}+\ldots\ldots t+1)\{(t^{n+k-2}+t^{n+k-3}+\ldots+t^{2k-1})+(t^{2k-2}+t^{2k-3}\ldots+t+1)\}$(1.2)$+t^{n-k+1}(t^{k-2}+t^{k-3}+\ldots\ldots t^{2}+t+1)^{2}$and (1.2) shows that (1.1) holds for any
induction on by (i) and (ii).

$k$such that $n-1\geq k\geq 1$ by mathematical$\square$$k$

Next we state an elementary proof of the following Theorem

Theorem$B$$B$(see Remark 1.1).

([6], [1]). $f_{p}(t)=p(1-p) \frac{(t-1)^{2}}{(t^{p}-1)(t^{1-p}-1)}$ is an opemtor monotone function$for-1\leq p\leq 2.$

51so that$g_{p}(t)$is operator monotone by ( $LH$ -1) of Proposition A ans so is $f_{p}(t)=p(1-p)g_{p}(t)$ .

(ii) In case $1<p<2$ . Since $p(1-p)<0$ we have only to prove that$h_{p}(t)=- \frac{(t-1)^{2}}{(t^{p}-1)(t^{1-p}-1)}=\frac{t^{\frac{k}{n}}(t-1)^{2}}{(t^{\frac{k}{n}}-1)(t^{1+\frac{k}{n}}-1)}$

is an operator monotone function for natural number
continuity of an operator. Then we have by (1.6)$n$and$k$(1.6)

such that $n-1\geq k\geq 1$ by52$h_{p}(t)= \frac{t^{\frac{k}{n}}(t-1)^{2}}{(t^{\underline{n}\pm}n-1)(t^{\frac{k}{n}}-1)\underline{k}}$$=t^{\frac{k}{n}}[ \frac{(t^{\frac{n.-1}{n}}+t^{\frac{n-2}{n}}+..+t^{\frac{k}{n}}+t^{\frac{k-1}{n}}\ldots t^{\frac{2}{n(}}+t^{\frac{1}{n1}}+1)^{2}}{(t^{\underline{n}\mapsto A_{\frac{k-2}{n}}}nk-1+t^{n}+..+t^{\frac{n}{n}}+t^{\frac{n-1}{n}}.+t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)t^{\frac{k-}{n}}+t^{\frac{k-2}{n}}+\ldots..t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)}]$

By applying modified Lemma 2 replacing by
the numerator in (1.7) can be rewritten as follows:$t^{\frac{1}{n}}$$t$(1.7).

to (1.7), so that the part of$[$of$]$$(t^{\frac{n-1}{n}}+t^{\frac{n-2}{n}}+..+t^{\frac{k}{n}}+t^{\frac{k-1}{n}\ldots t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)^{2}}$$=(t^{\frac{n-k-1}{n}}+t^{\frac{n-k-2}{n}}+..+t^{\frac{1}{n}}+1)(t\mapsto nk-1n\mapsto k-2n+t^{-}n+\ldots..+t^{\frac{1}{n}}+1)+t^{\frac{n-k}{n}}(t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}\ldots t^{\frac{2}{n}}}+t^{\frac{1}{n}}+1)^{2}$

and$h_{p}(t)$in (1.7) can be rewitten as follows$(1.8)$$h_{p}(t)=t^{\frac{k}{n}}[ \frac{(t^{\frac{n-k-1}{n}}+t^{\frac{n-k-2}{n}}+..+t^{1}n+1)(.t^{\lrcorner_{\frac{k-1}{n}}}n+t^{-}n..nk\mapsto-2+\ldots..+t^{12\perp}n+1+t^{\frac{n-k}{n}}(t^{\frac{k-.1}{n}}.+t^{\frac{k-2}{n}}\ldots tn+tn+1)^{2}}{(t^{nk-1\mapsto_{n+..+t^{\frac{n}{n}}+t^{\frac{n-1}{n}}+t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)(t^{\frac{k-1)}{n}}+t^{\frac{k-2}{n}}+...t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)}}\mapsto_{n+t^{nk-2}}}].$By (1.8) we have$h_{p}(t)$$=t^{\frac{k}{n}}[ \frac{t^{\frac{n-k-1}{n}}+t^{\frac{n-k-.2}{n}}.+..+t^{\frac{1}{n}}+1}{t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}+...t^{\frac{2}{n}}+t^{\frac{1}{n}+1}}}]+[\frac{t.(t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}\ldots.t^{\frac{2}{n}}+t^{\frac{1}{n}}+1)}}{t(t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}+..t^{\frac{1}{n}}+1)+(t^{\frac{n-1}{n}}.+tn)+(t^{\frac{k-1}{n}}+..tn+t^{\frac{1}{n}}+1)}A2}]$$=(t^{\frac{-(n-k)}{n}}+\ldots+t^{\frac{-(n-2)}{n}}+t^{\frac{-(n-1)}{n})^{-1}}+(t^{\frac{-(n-k-1)}{n}}+\ldots+t^{\frac{-(n-2)}{n})^{-1}}+\ldots..+(t^{\frac{-1}{n}}+\ldots+t^{\frac{-k}{n}})^{-1}$$+[1+ \frac{1}{t}+\frac{1}{t^{\frac{k}{n}}+t^{\frac{k-1}{n}}+\ldots t^{\frac{1}{n}}}+\frac{1}{t^{\sim}n^{\underline{1}}+t^{\frac{k}{n}}+\ldots t^{\frac{2}{n}}k\pm}+\ldots.+\frac{1}{t^{\frac{n-1}{n}}+t^{\frac{n-2}{n}+\ldots t^{\frac{n-\kappa}{n}}}}]^{-1}$is operator monotone by ( $LH$-1).
(iii). In case-l $<p<0.$$1<p<2$ by symmetry.
(iv)$f_{p}(t)$is operator monotone because the result reduces to the case$\lim_{parrow 0}f_{p}(t)=\lim_{parrow 1}f_{p}(t)=\frac{t-1}{\log t}$

and $f_{-1}(t)=f_{2}(t)= \frac{2t}{t+1}$ and these functions are both

operator monotone.

Whence the proof is complete by (i), (ii), (iii) and (iv).

$\square$Remark 1.1. The proof of the case $0<p<1$ in Theorem was obtained in [6], who
also conjectured the case $1<p<2$ . Imcomplete proofs of this statement have appeared in
the literature, but it seems that the first correct proof was obtained by Cai And Hansen [1,Theorem 5.2] (see the footnote on page 11 of [1]). The proof of Theorem in this paper is$B$$B$based on [3].

53
\S 2 Elementary proofs of the results by Hansen and related ones
Theorem$C$([4]).

$f(t)= \frac{t^{q}-1}{t^{p}-1}$

is an opemtor monotone function for $1\geq q\geq p>0$ and $t\geq 0.$(ii)$f^{*}(t)= \frac{t}{f(t)}$is also an opemtor monotone, where $f(t)$ is in (i)(iii)$f^{\#}(t)=tf(t^{-1})$(i)

is also an opemtor monotone, where $f(t)$ is in (i).

Proof. (i). We have only to prove the result for
and$m,$
such that $n\geq m\geq k\geq 1$ by continuity of an operator.
$p= \frac{k}{n}$$n,$$q= \frac{m}{n}$for natural numbers$k$$f(t)= \frac{t^{\frac{m}{n}}-1}{t^{\frac{k}{n}}-1}=\frac{(t^{\frac{1}{n}}-1)(t^{\frac{m-1}{n}}+t^{\frac{m-2}{n}}+\ldots+t^{\frac{k}{n}}..+t^{\frac{k-1}{n}}+\ldots t^{\frac{1}{n}}+1)}{(t^{\frac{1}{n}}-1)(t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}}+.+t^{\frac{1}{n}}+1)}$(2.1)$= \frac{t^{\underline{m}}+t^{\underline{m}}+t^{\frac{k}{n}}}{t^{\frac{k-1}{n}\frac{k-2}{n}\frac{1}{n}}+1}+1$$=(t^{\frac{-(m-k)}{n}}+t^{\frac{-(m-k+1)}{n}}+\ldots+t^{\frac{-(m-1)}{n})^{-1}+}(t^{\frac{-(m-k-1)}{n}}+t^{\frac{-(m-k)}{n}}+\ldots+t^{\frac{-(m-2)}{n}})^{-1}$$+\ldots\ldots.+(t^{\frac{-1}{n}}+t^{\frac{-2}{n}}+\ldots+t^{\frac{-k}{n}})^{-1}+1$

so that $f(t)$ is an operator monotone function by ( $LH$ -1) of Proposition A.
(ii).

by (2.1)$f^{*}(t)= \frac{t}{f(t)}=\frac{t}{1+\frac{t^{\frac{m-1}{n}}+t^{\frac{m-2}{n}}+\ldots+t^{\frac{k}{n}}}{t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}}+\ldots+t^{\frac{1}{n}}+1}}$$= \{\frac{1}{t}+\frac{1}{t^{\frac{n-m+k}{n}}+t^{\frac{n-m+k-1}{n}}+..+t^{\frac{n-m+1}{n}}}+\frac{1}{t^{\frac{n-m+k+1}{n}}+t^{\frac{n-m+k}{n}}+..+t^{\frac{n-m+2}{n}}}+..+\frac{1}{t^{\frac{n-1}{n}}+t^{\frac{n-2}{n}}+..+t^{\frac{n-k}{n}}}\}^{-1}$is operator monotone by ( $LH$ -1) of Proposition A.
(iii).

$f^{\#}(t)=tf(t^{-1})=( \frac{t}{t}\frac{m}{\frac{\pi^{n}}{n}}\frac{-1}{-1})t^{1-\frac{m}{n}+\frac{k}{n}}$$= \frac{(t^{\frac{1}{n}}-1)(t^{\frac{m-1}{n}}+t^{\frac{m-2}{n}}+\ldots t^{\frac{m-k}{1n}}+t^{\frac{m-k-1}{n}}+\ldots+t^{\frac{1}{n}}+1)t^{1-\frac{m}{n}+\frac{k}{n}}}{(t^{\frac{1}{n}}-1)(t^{\frac{k-}{n}}+t^{\frac{k-2}{n}}+\ldots+t^{\frac{1}{n}}+1)}$$= \frac{(t^{\frac{m-1}{n}}+t^{\frac{m-2}{n}}+\ldots t^{\frac{m-k}{1n}}+t^{\frac{m-k-1}{2n}}+\ldots+t^{\frac{1}{n}}+1)t^{1-\frac{m}{n}+\frac{k}{n}}}{t^{\frac{k-}{n}}+t^{\frac{k-}{n}}+\ldots+t^{\frac{1}{n}}+1}$$= \frac{t(t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}}+t^{\frac{1}{n}}+1)+t^{\frac{n-1}{n}}+t^{\frac{n-2}{n}}+\ldots+t^{\frac{n-m+k+1}{n}}+t^{\frac{n-m+k}{n}}}{t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}}+\ldots+t^{\frac{1}{n}}+1}$$=t+ \frac{t^{\frac{n-1}{n}}+t^{\frac{n-2}{n}}+\ldots+t^{\frac{n-m.+k+1}{n}}+t^{\frac{n-m+k}{n}}}{t^{\frac{k-1}{n}}+t^{\frac{k-2}{n}}+..+t^{\frac{1}{n}}+1}$

54$=t+(t^{\frac{-(n-k)}{n}}+t^{\frac{-(n-k+1)}{n}}+\ldots+t^{\frac{-(n-1)}{n}})^{-1}+(t^{\frac{-(n-k-1)}{n}}+t^{\frac{-(n-k)}{n}}+\ldots+t^{\frac{-(n-2)}{n}})^{-1}$$+t^{\frac{-(n-m+k)}{n}})^{-1}$$+\ldots\ldots+(t^{\frac{-(n-m+1)}{n}}+t^{\frac{-(n-m+2)}{n}}+$$\cdots$so that$f^{\#}(t)$is operator monotone by ( $LH$ -1) of Proposition A.

$\square$Proposition 2.1. If $f(t)$ and $g(t)$ are both positive opemtor monotone on $t\geq 0$ , then(i)is positive operator monotone for any $\alpha\in[0,1]$ on $t\geq 0$ .

(2.2)

is positive opemtor monotone for any $\alpha\in[0,1]$ on $t\geq 0$ .

(2.3)

$f(t)^{1-\alpha}g(t)^{\alpha}$

In particular,(ii)(iii)$f(t)^{1-2\alpha}t^{\alpha}$$f(t)^{1-\alpha}f(t^{-1})^{\alpha}t^{\alpha}$

is positive opemtor monotone for any $\alpha\in[0,1]$ on $t\geq 0$ .

(2.4)

Proof. Let $A\geq B>0$ . Since $f(A)\#_{\alpha}g(A)\geq f(B)\#_{\alpha}g(B)$ for $\alpha\in[0,1]$ , so we have (2.2).
Since $g(t)=f^{*}(t)= \frac{t}{f(t)}$ is operator monotone, (2.3) follows by (2.2).

Also since $g(t)=f^{\#}(t)=tf(t^{-1})$ is operator monotone, (2.4) follows by (2.2).

$\square$Proposition 2.2. Let $1\geq q\geq p>0$ and $\alpha\in[0,1]$ . Then(i)$F_{p,q,\alpha}= \frac{t^{q}-1}{t^{p}-1}t^{(1-q+p)\alpha}$

is opemtor monotone on $t\geq 0$and(ii)Proof.

$G_{p,q,\alpha}=( \frac{t^{q}-1}{t^{\rho}-1})^{1-2\alpha}t^{\alpha}$(i).

Put$tf(t^{-1})= \frac{t^{q}-1}{t^{p}-1}t^{1-q+p}$

is opemtor monotone on $t\geq 0.$

for $1\geq q\geq p>0$ on $t\geq 0$ and $g(t)=f^{\#}(T)=$on $t\geq 0$ . Since $f(t)$ and $g(t)$ are operator monotone shown in \S 1,$f(t)= \frac{t^{q}-1}{t^{p}-1}$$f(t)^{1-\alpha}g(t)^{\alpha}= \frac{t^{q}-1}{t^{p}-1}t^{(1-q+p)\alpha}$

is operator monotone by (2.2). And (ii) follows by (2.3).

$\square$

Theorem $D$ ([5]). Let the exponent $\gamma\in[0,1]$ . The functions$f_{\gamma}(t)= \frac{1}{2}(1+t)(\frac{4t}{(t+1)^{2}})^{\gamma}=t^{\gamma}(\frac{1+t}{2})^{1-2\gamma}$

are opemtor monotone, normalized in the sense $f_{\gamma}(1)=1$ and$f_{\gamma}(t)=tf_{\gamma}(t^{-1})$

for$t\geq 0.$

Proof. Since $f(t)= \frac{1+t}{2}$ is operator monotone andis operator monoyone by (2.3) of Proposition 2.1. Others are immediate consequences of$f_{\gamma}(t)=f(t)^{1-2\gamma}t^{\gamma}=t^{\gamma}( \frac{1+t}{2})^{1-2\gamma}$

calculations.

$\square$55
Remark 2.1.
is operator monotone on $t\geq 0$ in Proposition
2.2 contains the following useful operator monotone functions as follows;$F_{p,q,\alpha}(t)= \frac{t^{q}-1}{t^{p}-1}t^{(1-q+p)\alpha}$$F_{p,p,\alpha}(t)=t^{\alpha}$

for $\alpha\in[0,1]$

Although it is well known thatand$f^{*}(t)= \frac{t}{f(t)}$$F_{p,q,0}(t)= \frac{t^{q}-1}{t^{p}-1}$for $1\geq q\geq p>0.$and $f^{\#}(t)=tf(t^{-1})$ in Theorem$C$in \S 2 areboth operator monotone if $f(t)$ is operator monotone, we give elementary direct proofs of(ii) and (iii) in Theorem without use of the operator monotonicity of $f(t)$ in (i).
$C$

Both Proposition 2.1 and Proposition 2.2 forin \S 2 are shown in [4] and Theorem
is shown in [5] by using a canonical representation and these results are considered in[4][5] closely associated with useful Morozova-Chentsov function.
$\alpha=\frac{1}{2}$$D$

Acknowledgment. We would like to express our cordial thanks to Professor Frank
Hansen since he kindly informs us of the content of Remark 1.1 and also the results in \S 2
are obtained via useful and instructive discussions with him.

REFERENCES[1] L.Cai and F.Hansen, Metric-adjusted skew information: convexity and restricted forms ofsuperadditivity, Lett Math Phys.,93(2010), 1-13.
[2] T.Furuta, Concrete examples of operator monotone functions obtained by an elementary

method without applealing to L\"owner integral representaion, Linear Algebra Appl., 429(2008),972-980.
[3] T.Furuta, Elementary proof of Petz-Haswgawa Theorem, Lett Math Phys (2012) 101: 355-359,

DOI 10.1007/sll005-Ol2-0568-3.
[4] F.Hansen, Some operator monotone functions, Linear Algebra Appl., $430(2009),795-799.$[5] F.Hansen, Characterization of symmetric monotone metrics on the space of quantum systems,
Quant. Inform. Comput., 6 $(2006),597-605.$[6] D.Petz and H.Hasegawa, On the Riemannian metric of -entropies of density matrices,
Lett Math. Phys.,38 (1996), 221-225.
$\alpha$T. Furuta
1-4-19 Kitayamachou,
Fuchu city
Tokyo 183-0041
Japan