数理解析研究所講究録 1452 巻 2005 年 30-37

30SEVERAL REVERSE INEQUALITIES OF OPERATORS
大阪教育大学藤井 正俊 (Masatoshi Fujii)

Osaka Kyoiku University

瀬尾祐貴 (Yuki Seo)
大阪教育大学附属高等学校天王寺校舎
Tennoji Branch, Senior Highschool, Osaka Kyoiku University

ABSTRACT In this report, we show reverse inequalities to Araki’s inequality and investigate the equivalence among reverse inequalities of Araki, Cordes and L\"owner-Heinz
inequalities. Am ong others we show that if and are positive operators on a Hilbert
space $H$ such that $0<mI\leq \mathrm{A}\leq MI$ for some scalars $m<M$ , then$K(m, M,p)||BAB||^{p}\leq||B^{p}A^{p}B^{\mathrm{P}}||$
for all $0<p<1$ ,where $K(m, M, p)$ is a generalized Kantorovich constant by Furuta.
$A$

$B$

1. INTRODUCTION
be positive operators on a Hilbert space $H$ . The equivalence among Cordes
and L\"owner-Heinz inequalities was discussed by many authors. In [8], Furuta showed that
the Cordes inequality

Let

$A$

and

(1)

$B$

$||A^{p}B^{p}||\leq||AB||^{p}$for $0<p<1$

is equivalent to the L\"owner-Heinz inequality (cf.[14])(2)$A\geq B\geq 0$(cf. [5]). In [1], Araki showed a(3)

implies$\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}$$A^{p}\geq B^{p}$for $0<p<1$

inequality which entails the following inequality:$||B^{p}A^{p}B^{p}||\leq||BAB||^{p}$for $0<p<1$ .

Moreover, it was shown in $[6, 2]$ that the Cordes inequality (1) is equivalent to Araki’s
inequality (3).
On the other hand, Furuta [9] showed the following Kantorovich type inequalities: If
and are positive operators with $0<mI\leq A\leq MI$ for some scalars $m<M$ , then

$A$

(4)

$B$

implies$A\geq B\geq 0$($K$ $m,$ $M$, p)$A^{p}\geq B^{p}$

for

$p$

$>1$

,

where a generalized Kantorovich constant $K(m, M,p)[3,7,11]$ is defined as(5)$K(m, M,p)= \frac{mM^{p}-Mm^{p}}{(p-1)(M-m)}(\frac{p-1}{p}\frac{M^{p}-m^{p}}{mM^{p}-MmP})^{p}$for all real numbers .
$p$

We here cite Furuta’s textbook [10] as a pertinent reference to Kantorovich inequalities.
Also, Yamazaki [16] showed the following difference type reverse inequalities of the
L\"owner-Heinz inequality: If andare positive operators with $0<mI\leq B\leq MI$ forsome scalars $m<M$ , then$A\geq B\geq 0$(6)implies $C(m, M_{j}p)+A^{p}\geq B^{p}$ for $p>1$ ,
$A$

$B$

31
where the constant $C(m, M,p)[12,16]$ is defined as(7)$C(m, M,p)=(p-1)( \frac{M^{p}-m^{p}}{p(M-m)})^{\overline{p}-\overline{1}}\Delta+\frac{Mm^{p}-mM^{p}}{M-m}$

for all real numbers .
$p$

In this report, we show reverse inequalities to Araki’s inequality (3) and the Cordes
inequality (1): If and are positive operators with $0<mI\leq A\leq MI$ for some scalars$m<M$ , then the following inequalities hold
$A$

$B$

(8)

$K(m, M,p)||BAB||^{p}\leq||B^{p}A^{p}B^{p}||$for $0<p<1$ ,(9)$K(m_{i}^{2}M^{2},p)^{1/2}||AB||^{p}\leq||A^{p}B^{p}||$for $0<p<1$ ,respectively. We moreover show that reverse inequalities (4), (8) and (9) are mutuallyequivalent.

2. MAIN RESULTSFirst of all, we present our main theorem which is a reverse inequality to Araki’s
inequality (3).

are positive operators on $H$ such that $0<mI\leq A\leq MI$ for
Theorem 1. If andsome scalars $m<M$ , then for each $\alpha>0$
$B$

$A$

(10)

$||BAB||^{p} \leq\alpha||B^{p}A^{p}B^{p}||+\beta(m^{p}, M^{p},\frac{1}{p}, \alpha)||B||^{2p}$for all $0<p<1$ ,or equivalently(11)$||B^{p}A^{p}B^{p}||^{\frac{1}{p}}\leq\alpha||BAB||+\beta(m, M,p, \alpha)||B||^{2}$for all $p>1$ ,

where
(12)

if
if

$\frac{p-1}{p}(\frac{M^{p}-m^{p}}{\alpha p(M-m)})^{\frac{1}{p-1}}+$$\beta(m, M,p, \alpha)=\{$$(1-\alpha)M$

if$(1-\alpha)m$$\frac{M^{p}-m^{p}}{pM^{p-1}(M-m)}\leq\alpha\leq\frac{M^{p}-m^{p}}{pm^{p-1}(M-m)}$$\underline{M^{p}-\underline{m^{p}}}$$0<\alpha\leq pM^{\mathrm{p}-1}(M-m$$\alpha$$\geq pm^{\mathrm{p}}\ovalbox{\tt\small REJECT}-(M-m)M^{p}-m^{\mathrm{p}}$

]

,

,

.

If we choose a satisfying $\beta(m, M,p, \alpha)=0$ in Theorem 1, then we have the following
ratio type reverse inequalities.
Corollary 2. If and aresome scalars $m<M$ , then
$A$

(13)

$B$

$pos\mathrm{i}t\acute{\iota}ve$

operators on $H$ such that $0<mI\leq A\leq MI$ for$K(m, M,p)||BAB||^{p}\leq||B^{p}A^{p}B^{p}||$for $0<p<1$ ,or equivalently(11)$||BAB||^{p}\leq K(m, M,p)||B^{p}A^{p}B^{p}||$where $K(m, M,p)$ is defined as (5) in \S 1.

for $p>1$ ,32
If we put $\alpha=1$ in Theorem 1, then we have the following difference type reverse
inequalities.
andare positive operators on $H$ such that $0<mI\leq A\leq MI$ forsome scalars $m<M_{f}$ thenCorollary 3. If

(15)

$A$

$B$

for $0<p<1$ ,$||BAB||^{p}-||B^{p}A^{p}B^{p}||\leq-C(m, M,p)||B||^{2p}$or equivalently(16)

for $p>1$ ,$||B^{p}A^{p}B^{p}||^{\frac{1}{p}}-||BAB|| \leq-C(m^{p}, M^{p}\frac{1}{p}\})||B||^{2}$

where $C(m, M,p)$ is defined as (7) in \S 1.

As special cases of Corollary 2 and Corollary 3, we have the following corollary
Corollary 4. If and are positive operators on $H$ such that $0<mI\leq A\leq MI$ forsome scalars $m<M$ , then
$A$

$B$

(17)

$||B^{2}A^{2}B^{2}|| \leq\frac{(M+m)^{2}}{4Mm}||BAB||^{2}$(18)$||B^{2}A^{2}B^{2}||^{\frac{1}{2}}-||BAB|| \leq\frac{(M-m)^{2}}{4(M+m)}||B||^{2}$(19)$\frac{2\sqrt[4]{Mm}}{\sqrt{M}+\sqrt{m}}||BAB||^{\frac{1}{2}}\leq||B^{\frac{1}{2}}A^{\frac{1}{2}}B^{\frac{1}{2}}||$

.

.

.
.

(20)

$||BAB||^{\frac{1}{2}}-||B^{\frac{1}{2}}A^{\frac{1}{2}}B^{\frac{1}{2}}|| \leq\frac{(\sqrt{M}-\sqrt{m})^{2}}{4(\sqrt{M}+\sqrt{m})}||B||$

Since $||X^{*}X||=||X||^{2}$ for an operator $X$ , we obtain the following reverse inequality to
the Cordes inequality by Corollary 2.
Theorem 5. If and are positive operators on $H$ such that $0<mI\leq A\leq MI$ forsome scalars $m<M$ , then
$A$

(21)

$B$

$K(m^{2}, M^{2},p)^{\frac{1}{2}}||AB||^{p}\leq||A^{p}B^{p}||$for all $0<p<1$ ,or equivalently(22)$||A^{p}B^{p}||\leq K(m^{2}, M^{2},p)^{\frac{1}{2}}||AB||^{p}$for all $p>1$ .

In particular,(23)$\leq||A^{\frac{1}{2}}B^{\frac{1}{2}}||$

.

and

(24)

$||A^{2}B^{2}|| \leq\frac{M^{2}+m^{2}}{2Mm}||AB||^{2}$

The equivalence among the reverse inequalities of Araki, Cordes and L\"owner-Heinz
inequalities is now given as follows

33
Theorem 6. For a given $p>1$ , the following are mutually equivalent: For all
and $0<mI\leq A\leq MI$

(A)
(B)
(C)
$(\mathrm{B}$$(\mathrm{C}$’implies$A\geq B\geq 0$$K(m, M,p)A^{p}\geq B^{p}$$||A^{p}B^{p}||\leq K(m^{2}, M^{2},p)^{1/2}||AB||^{p}$.

$||B^{p}A^{p}B^{p}||\leq K(m, M_{1}p)||BAB||^{p}$.

$A$, $B\geq 0$.

.
$K(m, M, 1/p)||BAB||^{p}\leq||B^{p}A^{p}B^{p}||$ .
$K(m^{2}, M^{2},1/p)^{1/2}||AB||^{p}\leq||A^{p}B^{p}||$

$)$

’ $)$

3. LEMMAS
We start with the following three lemmas before we give proofs of the results in 52.
Let be a positive operator on a Hilbert space $H$ and a unit vector in $H$ . Then it
follows from H\"older-McCarthy inequality that
$A$

$x$

(Ax,

(25)
By using the(25) $[15, 4]$ :$\mathrm{M}\mathrm{o}\mathrm{n}\mathrm{d}- \mathrm{P}\mathrm{e}\check{\mathrm{c}}\mathrm{a}\mathrm{r}\mathrm{i}\acute{\mathrm{c}}$$x1,$$\leq(A^{p}x, x)^{\frac{1}{\mathrm{p}}}$for all $p.>1$ .

method $[12, 13]$ , we have the following reverse inequality ofLemma 7. if is a positive operator on $H$ such that $0<mI\leq A\leq MI$ for sorne
scalars $0<m<M$ , then for each $\alpha>0$
$A$

(26)

$(A^{p}x,x)^{\frac{1}{p}}\leq\alpha(Ax, x)+\beta(m, M,p, \alpha)$

holds for every unit vector $x\in H$ , where$\beta(m, M,p_{\backslash }\alpha)$

for all $p>1$
is defined as (12) in Theorem 1.

Proof For the sake of reader’s convenience, we give a proof. Put$\beta=\beta(m, M, p, \alpha)$, then we have
. It follows that the equation $f’(t)=0$ has exactly one solution$m\leq t_{0}\leq M$ , then we have $\beta=\max_{m\leq t\leq M}f(t)=f(t_{0})$ sinceand the condition $m\leq t_{0}\leq M$ is equivalent to the condition$f(t)=(at+b)^{\frac{1}{p}}-\alpha t$for$a= \frac{M^{p}-m^{p}}{M-m}$and$b= \frac{Mm^{;p}-mM^{1^{\mathrm{J}}}}{M-m}$and$f’(t)= \frac{a}{p}(at+b)^{\frac{1}{p}-1}-$$t_{0}= \frac{1}{a}(\frac{\alpha p}{a})^{\frac{p}{1-p}}-\frac{b}{a}$$\alpha$. If$f’(t)= \frac{a^{2}(1-p]}{R,p},(at+b)^{\frac{1}{p}-2}<$$0$$\frac{M^{p}-m^{p}}{pM^{p-1}(M-m)}\leq ce$ $\leq\frac{M^{p}-m^{p}}{pm^{p-1}(M-m)}$.

, then $f(t)$ is increasing on $[m, M]$ and hence we have $\beta=\max_{m\leq t\leq M}f(t)=$$f(t_{0})=(1-\alpha)M$ for $t_{0}=M$ . Similarly, we have $\beta=\max_{m\leq t\leq M}f(t)=f(t_{0})=(1-\alpha)m$. Hence it follows thatfor $t_{0}=m$ if
for all $t\in[m, M]$ .
Since is convex for $p>1$ , it follows that $t^{p}\leq at+b$ for $t\in[m, M]$ . By the spectral
theorem, we have $A^{p}\leq aA+b$ and hence $(A^{p}x, x)\leq$ $(Ax, x)+b$ for every unit vector$x\in H$ . Therefore we have$(a(Ax, x)+b)^{\frac{1}{p}}-\alpha(Ax, x)$$-$ a $(Ax, x)$
If$M\leq t\mathit{0}$$t_{0}\leq m$$(at+b)^{\frac{1}{p}}$–$\alpha t\leq\beta$$t^{p}$$(A^{p}x, x)^{\frac{1}{\mathrm{p}}}$$\leq$$\leq$$\max_{m\leq t\leq M}f(t)=\beta(m, M,p, \alpha)$.
$\square$34
By Lemma 7, we have the following estimates of both the difference and the ratio in
the inequality (25),Lemma 8. ij A is a positive operator on H such that $0<mI\leq A\leq MI$ for somescalars $0<m<M_{f}$ then for each p $>1$
(27)$(A^{p}x, x)^{\frac{1}{p}}\leq K(m, M,p)^{\frac{1}{p}}(Ax, x)$

and

(28)

$(A^{p}x, x)^{\frac{1}{p}}-(Ax, x) \leq-C(m_{)}^{p}M^{p},\frac{1}{p})$hold for every unit vector $x\in H$ , where $K(m, M,p)$ is defined as (5) in \S 1 and $C(m, M,p)$is defined as (7) in \S 1.
Proof. If we choose satisfying $\beta(m, M, p, \alpha)=0$ in Lemma 7, then we have
. If we put $=1$ in Lemma 7, then we have $\beta(m, M,p, 1)=-C(m^{p}, M^{p},\frac{1}{p})\square$.
$\alpha=$$\alpha$$K(m, M,p)^{\frac{1}{p}}$$\alpha$We remark that $K(m, M, 2)$ coincides with the Kantorovich constant$\frac{(M+m)^{2}}{4Mm}$

if $p=2$ .

We summarize some important properties of a generalized Kantorovich constant $[3, 11]$ .
Lemma 9. Let $m<M$ be given. Then a generalized Kantorovich constant $K(m, M,p)$
has the following properties.
.
(i) $K(m, M,p)=K(M, m,p)$ for all.
(ii) $K(m_{)}M,p)=K(m, M, 1-p)$ for all.
(ii) $K(m, M, 0)$ $=K(m, M, 1)=1$ for all.
and decreasing for(iv) $K(m, M,p)$ is increasing for(v)
for $pr\neq 0$ .
$p\in \mathbb{R}$$p\in \mathbb{R}$$p\in \mathbb{R}$$p< \frac{1}{2}$$p> \frac{1}{2}$$K(m^{r}, M^{r}, \frac{p}{r})^{\frac{1}{p}}=K(m^{p}, M^{p}, \frac{r}{p})^{-\frac{1}{r}}$4. PROOF OF RESULTS
Based on Lemmas in the preceding section, we give proofs of the results mentioned in
the second section.
Proof of Theorem 1.
For every unit vector $x\in H$ , it follows that$((BAB)^{p}x,x)$$\leq$$=$$\leq$

$($

BABx,

(
(

by H\"older-McCarthy inequality and $0<p<1$$x)^{p}$$(A^{p})^{\frac{1}{p}} \frac{Bx}{||Bx||}$,$\frac{Bx}{||Bx||}$)$\alpha(A^{p}\frac{Bx}{||Bx||}, \frac{Bx}{||Bx||})$$||Bx||^{2p}$$+ \beta(m^{p},M^{p},\frac{1}{p}, \alpha)$)$||Bx||^{2p}$by Lemma 7$= \alpha(A^{p}Bx, Bx)||Bx||^{2p-2}+\beta(m^{p}, M^{p},\frac{1}{p},\alpha)||Bx||^{2p}$$=\alpha$($B^{p}A^{p}B^{p} \frac{B^{1-p}x}{||B^{1-p}x||}$,$\frac{B^{1-p}x}{||B^{1-p}x||}$)$||Bx||^{2p-2}||B^{1-p}x||^{2}+ \mathrm{X}(m^{p}, M^{p},\frac{1}{p}, \alpha)||Bx||^{2p}$35and$||Bx||^{2p-2}||B^{1-p}x||^{2}$$=$$(B^{2}x, x)^{p-1}(B^{2-2p}x, x)$$\leq$$(B^{2}x, x)^{p-1}(B^{2}x, x)^{1-p}=1$

by $0<1-p<1$ .

By combining two inequalities above, we have$||BAB||^{p}$$=$$||(BAB)^{p}||$$\leq$$\alpha||B^{p}A^{p}B^{p}||+\beta(m^{p}, M^{p},\frac{1}{p}, \alpha)||B||^{2p}$and hence we have the desired inequality (10).
Next, we show (10)\Rightarrow (11). For $p>1$ , since $0< \frac{1}{p}<1$ , it follows from (10) that$||BAB||^{\frac{1}{p}}\leq\alpha||B^{\frac{1}{p}}A^{\frac{1}{p}}B^{\frac{1}{p}}||+\beta(m^{\frac{1}{p}}, M^{\frac{1}{p}},p_{)}\alpha)||B||^{\frac{2}{p}}$By replacing

$A$

by

$A^{p}$

and

$B$

by

$B^{p}$.

in the above inequality respectively, we have,$||B^{p}A^{p}B^{p}-||^{\frac{1}{p}}\leq\alpha||BAB||+\beta(m, M,p, \alpha)||B^{p}||^{\frac{2}{\mathrm{p}}}$and so we have the desired inequality (11). Similarly we can show (11)\supset (10). Therefore(10) is equivalent to (11).
$\square$

Proof of Corollary 2.
For $p>1$ , if we put $\beta(m, M,p, a)$ $=0$ in Theoren 1, then it follows that$\frac{p-1}{p}(\frac{M^{p}-m^{p}}{p(M-m)})^{\frac{1}{p-1}}+\alpha^{\frac{p}{p-1}}\frac{(Mm^{p}-mM^{p})}{M^{p}-m^{p}}=0$

and hence.

$\alpha^{\frac{p}{\mathrm{p}-1}}=-\frac{p-1}{p}(\frac{M^{p}-m^{p}}{p(M-m)})^{\frac{1}{p-1}}\frac{M^{p}-m^{p}}{MmP-mM^{p}}$Therefore , we have$\alpha^{p}=\frac{M^{p}-m^{p}}{p(M-m)}(\frac{p-1}{p}\frac{M^{p}-m^{p}}{mM^{p}-MmP})^{p-1}$$=K(m, M,p)$and we obtain the desired inequality (14).
For $0<p<1$ , since $1/p>1$ , it follows from (14) that$||BAB||^{\frac{1}{p}} \leq K(m, M,\frac{1}{p})||B^{\frac{1}{p}}A^{\frac{1}{p}}B^{\frac{1}{\mathrm{p}}}||$

By replacing

$A$

and

$B$

by

$A^{p}$

and$B^{p}$.

respectively, then we have.

$||B^{p}A^{p}B^{p}||^{\frac{1}{p}} \leq K(m^{p}, M^{p},\frac{1}{p})||BAB||$

Hence it follows from Lemma 9 that$||B^{p}A^{p}B^{p}|| \leq K(m^{p}, M^{p},\frac{1}{p})^{p}||BAB||^{p}$$\leq K(m, M,p)^{-1}||BAB||^{p}$,

and we have the desired inequality (13). Similarly we have the implication (13)\Rightarrow (14).
Cl38
Proof of Corollary 3.
If we put $\alpha=1$ in Theorem 1, then it follows that$\beta(m^{p}, M^{p},\frac{1}{p},1)=\frac{\frac{1}{p}-1}{\frac{1}{p}}(\frac{M-m}{\frac{1}{p}(M^{p}-m^{p})})\frac{1}{\frac{1}{p}-1}+\frac{M^{p}m-m^{p}M}{M-m}$$=(1-p)( \frac{p(M-m)}{M^{p}-m^{p}})^{\frac{p}{1-p}}+\frac{M^{p}m-m^{p}M}{M-m}$$=-C(m, M,p)$ .

Similarly it follows that(15)$\beta(m, \mathrm{M},\mathrm{p}, 1)=-C(m^{p}, M^{p},\frac{1}{p})$. Hence we have the equivalence$\square$$\Leftrightarrow(16)$Proof of Corollary 4.
In Corollary 2 and 3, we have only to put $p=2$ and $p=1/2$ .

cl

Proof of Theorem 5
By Corollary 2, it follows that.

$K(m, M,p)||A^{\frac{1}{2}}B||^{2p}\leq||A^{\frac{p}{2}}B^{p}||^{2}$

By replacing

$A$

by

$A^{2}$, we have.
Therefore we have (21). Similarly, we have the equivalence (21)\Leftrightarrow (22).
$K(m^{2}, M^{2},p)||AB||^{2p}\leq||A^{p}B^{p}||^{2}$$\square$

Proof of Theorem 6
The proof is divided into three parts, namely the equivalence $(A)\Rightarrow(B)\Rightarrow(C)\supset$(A), $(B)\Leftrightarrow(B’)$ and $(C)\Leftrightarrow(C’)$ .
$(A)\Rightarrow(B)$ . It follows that$(A)\Leftrightarrow||A^{-\frac{1}{2}}B^{1}\yen||\leq 1arrow||A^{-\frac{p}{2}}B^{\frac{\mathrm{p}}{2}}||^{2}\leq K(m, M,p)$$\Leftrightarrow||A^{\frac{1}{2}}B^{\frac{1}{2}}||\leq 1arrow||A^{\frac{p}{2}}B^{\frac{p}{2}}||^{2}\leq K(M^{-1}, m^{-1},p)=K(m, M,p)$$\Leftarrow\neq||AB||\leq 1arrow||A^{p}B^{p}||\leq K(m^{2}, M^{2},p)$.

If we put $B_{1}=B/||AB||$ , then it follows from $||AB_{1}||=1$ that$||A^{p}B\mathrm{i}1$$(B)\Rightarrow(C)$$\leq K(m^{2}, M^{2},p)^{\frac{1}{2}}\Leftrightarrow||A^{p}B^{p}||\leq K(m^{2}, M_{1}^{2}p)^{\frac{1}{2}}||AB||^{p}$. If we replace

$A$

by

$A^{\frac{1}{2}}$.

in (A), then it follows that$||A^{\frac{p}{2}}B^{p}||\leq K(m, M,p)^{\frac{1}{2}}||A^{\frac{\dot{1}}{2}}B||^{p}$.

Square both sides, we have$||B^{p}A^{p}B^{p}||\leq K(m, M,p)||BAB||^{p}$$(C)\supset(A)$. If we replace

$B$

by

$B^{\frac{1}{2}}$

and

$A$

by

$A^{-1}$.

in (C), then it follows that.

$||B^{\frac{p}{2}}A^{-p}B^{E}2||\leq K(M^{-1},m^{-1},p)||B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}||^{p}$

By rearranging it, we have$||A^{-_{2}}B^{p}A^{-_{2}}||\mathrm{z}\mathrm{g}\leq K(m, M,p)||A^{-\frac{1}{2}}BA^{-\frac{1}{2}}||^{p}$.

37
Since $A\geq B\geq 0$ , it follows from$A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\leq 1$

that$||A^{-\frac{p}{2}}B^{p}A^{-\mathrm{E}}2||\leq K(m, M,p)$and hence$B^{p}\leq K(m, M,p)A^{p}$$(B)\Leftrightarrow(B’)$: If we replace A and

$B$

by

$A^{\frac{1}{p}}$and$B^{\frac{1}{p}}$.
in (B) respectively, then it follows

that$(B)\Leftrightarrow||AB||\leq K(m^{\frac{2}{p}}, M^{\frac{2}{p}},p)^{\frac{1}{2}}||A^{\frac{1}{p}}B^{\frac{1}{p}}||^{p}$$\Leftrightarrow||AB||^{\frac{1}{p}}\leq K(m^{\frac{2}{p}}, M^{\frac{2}{p}},p)^{\frac{1}{2p}}||A^{\frac{1}{p}}B^{\frac{1}{p}}||$$=$$K(m^{2}, M^{2},p)^{\frac{1}{2}}||AB||^{\frac{1}{p}}\leq||A^{\frac{1}{p}}B^{\frac{1}{p}}||$by Lemma 9$\Leftrightarrow(B’)$

Similarly we have $(C)\Leftrightarrow(C’)$ and so the proof is complete.

口

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