数理解析研究所講究録
第 1893 巻 2014 年 143-150

143

AN OPERATOR INEQUALITY FOR OPERATOR MONOTONE

FUNCTIONS

Masatoshi Fujii
Osaka Kyoiku University

Young Ok Kim
Suwon University
mfujii@cc.osaka-kyoiku.ac.jp evergreen1317@gmail.com

Ritsuo Nakamoto
Ibaraki University
r-naka@net1.jway.ne.jp

ABSTRACT. We give a characterization of convex functions in terms of difference among
values of a function. As an application, we propose an estimation of operator monotonefunctions: If $A>B\geq 0$ and is operator monotone on, then$f$$(0, \infty)$$f(A)-f(B)\geq f(\Vert B\Vert+\epsilon)-f(\Vert B\Vert)>0,$

where. As a consequance, we give a refined estimation of L\"ownerHeinz inequality under the assumption $A>B\geq 0$ . Moreover it gives a simple proof to
Furuta’s theorem: If $\log A>\log B$ for $A,$ $B>0$ and is operator monotone on,then there exists a $\beta>0$ such thatfor all
Finally we discuss strict positivity of Furuta inequality which is a beautiful extension of$\epsilon=\Vert(A-B)^{-1}\Vert^{-1}$$f$$f(A^{\alpha})>f(B^{\alpha})$$(0, \infty)$$0<\alpha\leq\beta.$

L\"owner-Heinz inequality.

1. INTRODUCTION
For a twice differentiable real-valued function , its convexity is characterized by $f”\geq 0.$
Since there are many non-differentiable convex functions, we consider a characterizationof general convex functions. We cannot use the differentiation, but the average rateof change is available. Roughly speaking, we claim that the convexity of a function ischaracterized by the non-decreasingness of average rate of change. It seems to be natural
as a generalization of the condition $f”\geq 0$ . Actually it will be formulated as Lemma 1
in the next section.
To explain operator monotone functions, we introduce the operator order $A\geq B$ among
selfadjoint operators $A,$ $B$ on a Hilbert space $H$ by $(Ax, x)\geq(Bx, x)$ for all $x\in H$ . In
particular, is positive if $A\geq 0$ , i.e., $(Ax, x)\geq 0$ for all $x\in H$ . Next, a positive operator
is said to be strictly positive, denoted by $A>0$ , if $A\geq c$ for some constant $c>0$ . So
$A>B$ means that $A-B>0.$
A real-valued continuous function defined on
is called operator monotone
if it preserves the operator order, i.e., $f(A)\geq f(B)$ for $A\geq B\geq 0$ . One of the mostimportant examples is the power function
for $0\leq p\leq 1$ (L\"owner-Heinz inequality).
In general, is called operator monotone on an interval if $f(A)\geq f(B)$ for $A\geq B$
whose spectra contained in . For this, we pose $\log t$ as a fundamental example of an
operator monotone function on.
Very recently, Moslehian and Naj afi [13] proposed an excellent extension of the L\"ownerHeinz inequality as follows:
$f$

$A$

$A$

$f$

$[0, \infty)$$t\mapsto t^{p}$

$f$

$J$

$J$

$(0, \infty)$

2010 Mathematics Subject aassification. Primary $47A63$ ; Secondary $47B10,47BA30.$
Key words and phrases. convex function, operator monotone function, L\’owner-Heinz inequality, Furuta
inequality and chaotic order.

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Theorem $MN$ . If $A>B\geq 0$ and $0<r\leq 1$ , then $A^{r}-B^{r}\geq\Vert A\Vert^{r}-(\Vert A\Vert-\epsilon)^{r}>0,$
and $\log A-\log B\geq\log\Vert A\Vert-\log(\Vert A\Vert-\epsilon)>0$ , where $\epsilon=\Vert(A-B)^{-1}\Vert^{-1}.$
In this note, we apply our characterization of concave functions and give an improvement and a generalization of Theorem $MN$ (Theorem 5). As another application, we cangive a short proof to a recent result due to Furuta [9, Theorem 2.1], which is an operator inequality related to operator monotone functions and chaotic order, i.e., the orderdefined by $\log A\geq\log B$ among positive invertible operators.
Incidentally, this note is based on our paper [5].

2. A CHARACTERIZATION OF CONVEX FUNCTIONSIn this section, we propose an elementary characterization of convex functions. We
essentially use average rate of change.

on an interval $J=[a, b)$ with $b\in$real valued continuous function
convex $(resp.$ concave) if and only if, for each $0<\epsilon<b-a,$$f(t+\epsilon)-f(t)$ is non-decreasing (resp. non-increasing) on $[a, b-\epsilon)$ .

Lemma 2.1.

$A$

$f$

$(-\infty, +\infty] is$$D_{\epsilon}(t)=$is convex on . Take $s,t\in J$ with $s<t$ and $t+\epsilon\in J$ . We
may assume that $t-s<\epsilon$ . Let $y=L(t)$ be the linear function through $(s, f(s))$ and. Then we have
$L(t)\geq f(t)$ and $L(t+\epsilon)\leq f(t+\epsilon)$by the convexity of . Hence it implies that

Proof. Suppose that

$J$

$f$

$(s+\epsilon, f(s+\epsilon))$$f$$D_{\epsilon}(t)=f(t+\epsilon)-f(t)$$\geq L(t+\epsilon)-L(t)$$=L(s+\epsilon)-L(s)$by the linearity of$L$$=f(s+\epsilon)-f(s)$$=D_{\epsilon}(s)$,as desired.

is non-decreasing. Take
Conversely suppose that
, we haveSince$D_{\epsilon}(t)$$t,$$s\in J$

with $s<t=s+2\epsilon.$$D_{\epsilon}(s)\leq D_{\epsilon}(s+\epsilon)$$2f( \frac{s+t}{2})=2f(s+\epsilon)\leq f(s+2\epsilon)+f(s)=f(t)+f(s)$So.

is convex.

$f$$\square$forfor all $t\in[a, b].$Remark 2.3. Analogous argument on convexity of functions as above has been done in
Corollary 2.2. If

some$\delta>0$$f$

is strictly increasing and concave on an interval $[a, b+\delta]$ in, then for each$0<\epsilon\leq\delta,$$\mathbb{R}$$D_{\epsilon}(t)\geq D_{\epsilon}(b)>0$[12, page 2],

3. APPLICATIONS To OPERATOR MONOTONE FUNCTIONSAs an application of Corollary 2.2, we give an estimation of operator monotone functions.

Lemma 3.1. If is non-constant and operator monotone on the interval
then is strictly increasing.
$f$

$f$

$\mathbb{R}_{+}=[0, \infty)$,

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Proof. First of all, we note that is non-decreasing. Next we suppose that $f’(c)=0$ forsome $c>0$ . Noting that the L\"owner matrix$f$$(_{f^{[1]}(d,c)}f’(c) f^{[1]}(c,d)f’(d))$

is positive semidefinite for any $d>0$ by the operator monotonicity of , where $f^{[1]}(c, d)=$is the devided difference.
Therefore its determinant is nonnegative, so that $f^{[1]}(c, d)=0$ for any $d>0$ . Thismeans that is constant, which is a contradiction. Consequently we have $f’>0.$$f$$\frac{f(-f(d)}{-d}$$f$$\square$Lemma 3.2. If $C\geq 0$ and is a concave and strictly increasing function on an interval$[a, d)$ containing the spectrumof , then for each $0<\epsilon<d-\Vert C\Vert,$ $f(C+\epsilon)\geq f(C)+$
.
$f$

$C$

$D_{\epsilon}(\Vert C\Vert)$

Proof. We first note that for a given $0<\epsilon<d-\Vert C\Vert$ , we can take $c>0$ satisfying$0<c<d$ and$\epsilon<c-\Vert C\Vert$follows that

. Applying Corollary 2.2 to$b=\Vert C\Vert$$f(C+\epsilon)-f(C)\geq D_{\epsilon}(\Vert C\Vert)$

and $\delta=c-\Vert C\Vert$ , it.
$\square$

We here give a precise estimation of [9, Theorem 2.1] and [12, Proposition 2.2], cf. [13].
Theorem 3.3. If $A>B\geq 0$ and$f$is non-constant operator monotone on, where $\epsilon=\Vert(A-B)^{-1}\Vert^{-1}.$$[0, \infty)$, then$f(A)-f(B)\geq f(\Vert B\Vert+\epsilon)-f(\Vert B\Vert)>0$

Proof. Since$A\geq B+\epsilon$

for $\epsilon=\Vert(A-B)^{-1}\Vert^{-1}>0$ , we have$f(A)\geq f(B+\epsilon)$.

Furthermore Lemmas 3.1 and 3.2 imply that$f(B+\epsilon)\geq f(B)+D_{\epsilon}(\Vert B\Vert)$.

Hence we have$f(A)-f(B)\geq D_{\epsilon}(\Vert B\Vert)=f(\Vert B\Vert+\epsilon)-f(\Vert B\Vert)>0.$$\square$

As a consequence, we have an improvement of the estimation due to Moslehian andNajafi [13]:Corollary 3.4. If $A>B\geq 0$ and $0<r\leq 1$ , then $A^{r}-B^{r}\geq(\Vert B\Vert+\epsilon)^{r}-(\Vert B\Vert)^{r}>0,$
and $\log A-\log B\geq\log(\Vert B\Vert+\epsilon)-\log\Vert B\Vert>0$ , where$\epsilon=\Vert(A-B)^{-1}\Vert^{-1}.$

Remark 3.5. We note that Corollary 3.4 actually improves Theorem $MN$. Since
and the function
is strictly concave, it follows
that$\Vert A\Vert-$$(\Vert A\Vert-\epsilon)=\epsilon=(\Vert B\Vert+\epsilon)-\Vert B\Vert$$t\mapsto t^{r}$$\Vert A\Vert^{r}-(\Vert A\Vert-\epsilon)^{r}\leq(\Vert B\Vert+\epsilon)^{r}-\Vert B\Vert^{r}.$We here pose an example:$A=(\begin{array}{ll}4 00 2\end{array})$and$B=(\begin{array}{ll}2 00 1\end{array}).$

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Then

$A-B=(\begin{array}{ll}2 00 l\end{array})\geq 1$

and so$\epsilon=1$. Hence we have$\Vert A\Vert^{r}-(\Vert A\Vert-\epsilon)^{r}=4^{r}-3^{r}<(\Vert B\Vert+\epsilon)^{r}-\Vert B\Vert^{r}=3^{r}-2^{r}.$Now Theorem 3.3 can be regarded as a difference version. So we give a ratio version of
it. It is obtained by Theorem 3.3 itself:
Corollary 3.6. If $A>B>0$ and$f$is non-constant operator monotone on$(0, \infty)$, then$f(B)^{-\frac{1}{2}}f(A)f(B)^{-\frac{1}{2}}\geq 1+(f(\Vert B\Vert+\epsilon)-f(\Vert B\Vert))\Vert f(B)\Vert^{-1},$

where $\epsilon=\Vert(A-B)^{-1}\Vert^{-1}.$

Proof. Put$\delta=f(\Vert B\Vert+\epsilon)-f(\Vert B\Vert)$. It follows from Theorem 3.3 that$f(B)^{-\frac{1}{2}}f(A)f(B)^{-\frac{1}{2}}\geq f(B)^{-\frac{1}{2}}(f(B)+\delta)f(B)^{-\frac{1}{2}}$$=1+\delta f(B)^{-1}\geq 1+\delta\Vert f(B)\Vert^{-1}.$$\square$As another application of Theorem 3.3, we need the chaotic order: For $A>0$ , we candefine the selfadjoint operator $\log A$ . So a weaker order than the operator order appears
by $\log A\geq\log B$ for $A,$ $B>0$ . We call it the chaotic order. The chaotic order plays an
substantial role in operator inequalities. Among others, it brightens the Furuta inequality[7], [3], [4], [1], [6], [10] and recent development of Karcher mean theory [16].
Now we give a simple and elementary proof to the following recent theorem [9, Theorem
2.1] due to Furuta, in which we don’t use any integral representation of operator monotonefunctions.
Theorem 3.7. If $\log A>\log B$ for $A,$ $B>0$ andthen there exists $\beta>0$ such that$f(A^{\alpha})>f(B^{\alpha})$for all$f$

for all$(0, \infty)$,$0<\alpha\leq\beta.$

Proof. Since $\log A>\log B$ , it is known that there exists$A^{\alpha}>B^{\alpha}$

is operator monotone on$\beta>0$such that$0<\alpha\leq\beta.$

Therefore it follows from Theorem 3.3 that, for each fixed$f(A^{\alpha})>f(B^{\alpha})$$\alpha\in(0, \beta],$,as desired.

$\square$

4. FURUTA INEQUALITY.
First of all, we cite the Furuta inequality ( $FI$ ) in [7], see also [2], [8], [11] and [14] forthe best possibility of it.
The Furuta inequality. If $A\geq B\geq 0$ , then for each $r\geq 0,$$A^{\epsilon_{\frac{+r}{q}}}\geq(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2}})^{\frac{1}{q}}$holds for $p\geq 0,$ $q\geq 1$ with$(1+r)q\geq p+r.$

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To extend Corollary 3.4, we remark that the case $r=0$ in ( $FI$ ) is just the L\"owner-Heinz
inequality. Now we introduce a constant $k(b, m,p, q, r)$ for $m,p,$ $r\geq 0$ by
$b,$

$q,$

$k(b, m,p, q, r)=(b+m)^{a+\underline{r}}q^{-r}-b^{L+\underline{r}}q^{-r}$As a matter of fact, we have an extension of Corollary 3.4 in the form of Furuta inequality
as follows:
Theorem 4.1. Let

for

$A$

and

$B$

be invertible positive operators with $A-B\geq m>0$ . Then

$0<r\leq 1,$$A^{\epsilon_{\frac{+r}{q}}}-(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2}})^{\frac{1}{q}}\geq k(\Vert B\Vert, m,p, q, r)(\Vert B^{-1}\Vert^{-1}+m)^{r}$holds for $p\geq 0,$ $q\geq 1$ with $(1+r)q\geq p+r\geq qr.$and $(1+r)q\geq p+r\geq qr$ assure the exponent
belongs to $[0,1]$ . Since $0\leq r\leq 1$ , it follows from Theorem that

Proof. We note that

constant$k$$q\geq 1$$\frac{p+r}{q}-r$

in the$B$$A^{L+\underline{r}}q-(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2}})^{\frac{1}{q}}=A^{e_{\frac{+r}{q}}}-A^{\frac{r}{2}}B^{fi}2(B^{\epsilon 2\frac{1}{q}-1\xi i}2A^{r}B2)B2A^{\frac{r}{2}}$$=A^{\frac{p+r}{q}}-A^{\frac{r}{2}}B^{\frac{p}{2}}(B^{\frac{-p}{2}A^{-r^{-s\rho}}}B^{-}2)^{1-\frac{1}{q}B2}A^{\frac{r}{2}}$$\geq A^{a_{\frac{+r}{q}}}-A^{\frac{r}{2}}B^{2}2(B^{-}2\epsilon B^{-r}B^{-}\epsilon 2)^{1-\frac{1}{q}}B2A^{\frac{r}{2}}e$$=A^{\rho_{\frac{+r}{q}}}-A^{\frac{r}{2}}B^{p-(J^{3+r)(1-\frac{1}{q})}}A^{\frac{r}{2}}$$=A^{\frac{r}{2}}(A^{p_{\frac{+r}{q}-r}}-B^{g_{\frac{+r}{q}-r}})A^{\frac{r}{2}}$$\geq k(\Vert B\Vert, m,p, q, r)A^{r}$$\geq k(\Vert B\Vert, m,p, q, r)(B+m)^{r}$$\geq k(\Vert B\Vert, m,p, q, r)(\Vert B^{-1}\Vert^{-1}+m)^{r}.$$\square$For a general case on , we have the following estimation of Furuta inequality byrepeating method as in a proof of Furuta inequality.
$r$

Theorem 4.2. Let andbe invertible positive operators with $A-B\geq m>0$ andand $0<s\leq 1$ . Thensomenaturalnumber$r=n+s$ for
$A$

$B$

$n$

$A^{\rho_{\frac{+r}{q}}}-(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2}})^{\frac{1}{q}}\geqk(\Vert B_{n}\Vert^{\frac{1}{q}}, m_{n-1},p, q, r)(\Vert B^{-1}\Vert^{-1}+m)^{s}$holds for $p\geq 1,$ $q\geq 1$ with $p+1 \geq q\geq\frac{p+1}{2}$ , where$B_{n}=A^{\frac{n}{2}}B^{p}A^{\frac{n}{2}},$$m_{n}=k(\Vert B_{n}\Vert^{\frac{1}{q}}, m_{n-1},p, q, 1)(\Vert B^{-1}\Vert^{-1}+m)$for$n\geq 1$

and $m_{0}=k(\Vert B\Vert, m,p, q, s)(\Vert B^{-1}\Vert^{-1}+m)^{s}.$Proof. Taking $r=1$ in the above theorem, we have$A^{a+\underline{1}}q-(A^{\frac{1}{2}}B^{p}A^{\frac{1}{2}})^{\frac{1}{q}}\geq k(\Vert B\Vert, m,p, q, 1)(\Vert B^{-1}\Vert^{-1}+m) :=m_{1}.$Next we put$C=A^{\frac{1}{2}}B^{p}A^{\frac{1}{2}}$. Since$A\geq C^{\frac{1}{p+1}}$

and $0\leq s\leq 1$ , it follows that$(A^{\frac{1+s}{2}B^{p}A^{\frac{1+s}{2}})^{\frac{1}{q}}}=(A^{\frac{8}{2}}CA^{\frac{8}{2}})^{\frac{1}{q}}$$=A^{\frac{S}{2}}C^{\frac{1}{2}}(C^{\frac{1}{2}}A^{s}C^{\frac{1}{2}})^{\underline{1}-z}qC^{\frac{1}{2}}A^{\frac{s}{2}}$148$\leq A^{\frac{S}{2}}C^{\frac{1}{2}}(c^{-}\frac{1}{2}c\frac{-\epsilon}{p+1}C^{-\frac{1}{2}})^{L^{-\underline{1}}}qC^{\frac{1}{2}}A^{\frac{s}{2}}$$=A^{\frac{\epsilon}{2}}(C^{\frac{1}{q}})^{\frac{p+1-(q-1)s}{p+1}A^{\frac{\epsilon}{2}}}.$

Consequently we have$A^{\epsilon_{\frac{+1+\epsilon}{Q}}}-(A^{\frac{1+\iota}{2}B^{p}A^{\frac{1+s}{2}})^{\frac{1}{q}}}$$\geq A^{\frac{s}{2}}((A^{L+\underline{1}p+1-(q-1)\epsilon}q)p+1-(C^{\frac{1}{q}})^{\frac{p+1-(q-1)*}{p+1}})A^{\frac{\epsilon}{2}}$$\geq((\Vert C^{\frac{1}{q}}\Vert+m_{1})^{\frac{p+1-(q-1)s}{p+1}}-\Vert C^{\frac{1}{q}}\Vert^{\frac{p+1-(q-1)\epsilon}{p+1}})(\Vert B^{-1}\Vert^{-1}+m)^{s}.$Taking $s=1$ in the above, we have$A^{\epsilon_{\frac{+2}{q}}}-(AB^{p}A)^{\frac{1}{q}}\geq((\Vert C^{\frac{1}{Q}}\Vert+m_{1})^{\frac{+2}{p+1}}-\Vert C^{\frac{1}{q}}\Vert^{\frac{+2}{p+1}})(\VertB^{-1}\Vert^{-1}+m)^{s}:=m_{2}.$Inductively we put $D=AB^{p}A$ and then we have$(A^{\frac{2+\epsilon}{2}B^{p}A^{\frac{2+\epsilon}{2})^{\frac{1}{q}}}}=(A^{\frac{\delta}{2}}DA^{\frac{s}{2}})^{\frac{1}{q}}\leq A^{\frac{s}{2}}(D^{\frac{1}{q}})^{\frac{p+2-(q-1)s}{p+2}A}$and so$A^{\mapsto+2\underline{+s}}q-(A^{\frac{2+\epsilon}{2}B^{p}A^{\frac{2+s}{2}})^{\frac{1}{g}}}$$\geq A^{\frac{*}{2}}((A^{\epsilon_{\frac{+2}{q}}})^{p+2-(q-1)\iota}p+2-(D^{\frac{1}{q}})^{\frac{p+2-(q-1)s}{p+2}})A^{\frac{\epsilon}{2}}$$\geq((\Vert D^{\frac{1}{q}}\Vert+m_{2})^{\frac{p+2-(q-1)}{p+2}}-\Vert D^{\frac{1}{q}}\Vert^{\frac{p+2-(q-1)s}{p+2}})(\Vert B^{-1}\Vert^{-1}+m)^{s}.$Repeating this, we obtain the conclusion.

$\square$In the Furuta inequality, the optimal case where $p\geq 1$ and $(1+r)q=p+r$ is themost important by virtue of the L\"owner-Heinz inequality. So we would like to mention
the following result:Corollary 4.3. Let

$A$

and

$B$

be invertible positive operators with $A-B\geq m>0$ . Then

$A^{1+r}-(A^{\frac{r}{2}}B^{p}A^{\frac{f}{2}})^{\frac{1+r}{p+r}}\geq m(\Vert B^{-1}\Vert^{-1}+m)^{r}$holds for $p\geq 1$ and $r\geq 0.$

Proof. First of all, we note that if

for $p\geq 1$ and $r\geq 0$ , then for each $M>0,$$k(b, M,p, q, r)=M$ for arbitrary $b>0$ . Hence we have the conclusion for $0<r\leq 1$ by
Theorem 2.1.
Next, if $r>1$ , that is, $r=n+s$ for some natural number and $0<s\leq 1$ , thenTheorem 2.2 implies that$q= \frac{p+r}{1+r}$$n$$A^{1+r}-(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2}})^{\frac{1+r}{p+r}}\geq m_{n-1}(\Vert B^{-1}\Vert^{-1}+m)^{s},$where $m_{n-1}$ is the constant defined in Theorem 2.2. On the other hand, since$m_{n-1}=m_{n-2}(\Vert B^{-1}\Vert^{-1}+m)=m_{n-2}(\Vert B^{-1}\Vert^{-1}+m)=\cdots$$=m_{0}(\Vert B^{-1}\Vert^{-1}+m)^{n-1}=m(\Vert B^{-1}\Vert^{-1}+m)^{n},$we get the desired lower bound.

$\square$149
5. CONCLUDING REMARKS.
We now pose a proof of Theorem 3.3 by the use of integral representation for operator
monotone functions.

Proof of Theorem 3.3. We first prepare the basic tool: If $A>B>0$ and $m=\Vert(A-$$B)^{-1}\Vert^{-1}$, then(5.1)$B^{-1}-A^{-1} \geq\frac{m}{(\Vert B\Vert+m)\Vert B\Vert}.$

It is shown by$B^{-1}-A^{-1} \geq B^{-1}-(B+m)^{-1}=mB^{-1}(B+m)^{-1}\geq\frac{m}{\Vert B\Vert(\Vert B\Vert+m)}$

because of $A-B\geq m$ . Note that$f$admits the integral representation:$f(t)=a+bt+ \int_{-\infty}^{0}\frac{1+ts}{s-t}dm(s)=a+bt+\int_{-\infty}^{0}(-s-\frac{1+s^{2}}{t-s})dm(s)$where $b\geq 0$ and $m(s)$ is a positive measure. Hence it follows that$f(A)-f(B)=b(A-B)+ \int_{-\infty}^{0}(1+s^{2})((B-s)^{-1}-(A-s)^{-1})dm(s)$$\geq bm+\int_{-\infty}^{0}(1+s^{2})(\frac{1}{\Vert B||-s}-\frac{1}{\Vert B\Vert-s+m})dm(s)$$=f(\Vert B\Vert+m)-f(\Vert B\Vert)(>0)$.
$\square$Finally we discuss an operator extension of Lemma 2.1. Namely we may expect thefollowing conjecture:
A real valued function on an interval $J=(a, b)$ with $b\in(-\infty, +\infty]$ is operator convex
is operator monotone on $(a, b-\epsilon)$ .
if and only if, for each $0<\epsilon<b-a,$
Unfortunately we have a negative answer as follows: We choose the functionon. It is a typical example of operator convex functions. Nevertheless,is not operator monotone. As a matter of fact, we take two
matrices and :$f$$D_{\epsilon}(t)$$f(t)= \frac{1}{t}$$(0, \infty)$$D_{1}(t)=- \frac{1}{t(t+1)}$$2\cross 2$$A=(\begin{array}{ll}3 11 2\end{array})$

and

$A$

$B$

$B=(\begin{array}{ll}2 00 1\end{array}).$Note that $D_{1}(A)\geq D_{1}(B)$ if and only if $A(A+1)\geq B(B+1)$ . Clearly $A\geq B$ , but$A(A+1)-B(B+1)=(\begin{array}{ll}13 66 7\end{array})-(\begin{array}{ll}6 00 2\end{array})=(\begin{array}{ll}7 66 5\end{array})\not\geq 0.$

This is a counterexample.
Incidentally, the operator convexity of the function
enough to prove the inequality

1 is easily shown as follows: It is$( \frac{A+B}{2})^{-1}\leq\frac{1}{2}(A^{-1}+B^{-1})$

And it is simplified by putting$C=A^{\frac{1}{2}}B^{-1}A^{\frac{1}{2}}$that$4(1+C^{-1})^{-1}\leq 1+C,$.

150
which follows from the numerical inequality $4\leq(1+x^{-1})(1+x)$ .

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