数理解析研究所講究録 1427 巻 2005 年 98-102

98
A MODIFIED HEINZ’S INEQUALITY[東北大学名誉教授]

崇吉野Concerning the Heinz’s inequality, we got the following result.
Theorem.

The region of$\gamma$such that the operator inequality$(A^{\gamma}A^{\alpha}A^{\wedge\gamma}’)^{\beta}\geq\backslash /A^{\prime\gamma}B^{\alpha}A^{\gamma})^{\beta}$

holds for any operators and $B$ such as $A\geq B\geq bI$ (some $b>0$ ) and for any given
is as follows ;andand such as a$A$$\alpha$$\backslash ,>0$$\beta$(1)(2)$\beta>\mathrm{t}1$, $0<\beta\leq 1$ ,$0<\alpha\leq 1$ ,$0<\alpha\leq 1$$1<\beta\leq 2_{j}$

znax
(3)
$(4_{J}^{\backslash }$,$\{-\underline{\frac{1}{9}}$$\frac{-\alpha\beta}{2_{\backslash }^{(\beta-11}},\int 1\leq’\gamma\underline{\backslash }\min/\{0$$0<\alpha\underline{\backslash /}1,2$$< \beta\leq\frac{1}{\alpha}$$0<\alpha\leq 1,2<\beta$$(^{\mathrm{K}}.)$$1<\alpha$,,$\gamma=0$,$\frac{-\alpha\beta}{2(\beta-1)}\}\leq\gamma\leq\min\{..\frac{\underline{9}\cong-\alpha\beta}{2(\beta-1)}$,$\frac{1-\alpha\beta}{2(\beta-1)}\}$, $0<\beta<1$ ,$\max\{0$,$\frac{\alpha\beta-1}{2(1-\beta)}\})1\leq\gamma$.

In this lecture, we constituted counter examples of$0<ce$ $\leq 1$ ,$\frac{1-\alpha\beta}{2(\beta-1)}\}$,$\max\{\frac{2\alpha-1-\alpha\beta}{2(\beta-1)}$and$-\infty<\gamma<_{\backslash }’+\infty$$1<\beta$,$\min\int_{[}\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{r}_{0}\backslash ’\frac{2\alpha-\alpha\beta}{2\acute{(}\beta-1)})$$A$,

and $B$ for the cases where$\frac{1-\alpha\beta}{2\acute{(}\beta-1)}\}./\backslash \gamma(|\neq 0)$

and$0<\alpha\leq 1$

as follows,$1<\beta$,$\gamma<\mathrm{l}\mathrm{I}\mathrm{l}\mathrm{a}\mathrm{x}$$\{\max$ $(- \frac{1}{2}$,$\frac{2\alpha-1-\alpha\beta}{2(\beta-1)}$),$\frac{-\alpha\beta}{2(\beta-1)}\}$

$\mathfrak{g}$For any ,$b$$a$and$\mu$such as $1<a$$\leq\frac{1T}{\mathrm{x}\epsilon}$, $0<b$$\leq\frac{1}{2}$and $0<\mu$$\leq$$=a$ $+ \frac{\{a-1\rangle b^{1-\mu}}{a-1+b\langle 1rightarrow b\}}-\frac{\langle a-1\}(b^{2-\mu}+b^{1-\mu})}{b\langle 1-b\}}-\{1$$x$mirx{o,$1-\alpha$$-b\rangle b^{2-\alpha}$a}, let.

$\{\beta_{1})$Then$a$$-x$ $=$ $(1 -b)b^{2-\alpha}+ \frac{\{a-1\rangle[\mathrm{P}^{-\mu}\{a-1+b(\mathrm{I}-b)\}+\{a-\}\}b^{1-\mu}]}{b\{1-b)\{a-1+\dot{b}(1-b)\}}$

and$(\beta_{2}\}$.

$1- \frac{dx}{\ }=\frac{(aarrow 1)b^{1-\mu}\{a-1+2b(1-b)\}}{b(1-b\}\{a-1+b\{1-b)\}^{2}}+\frac{b^{2-\mu}}{b\zeta 1arrow b)}$$( \int_{3}\}$Since $\mu+2>2-\alpha$ , $b^{\mu+2}<b^{2-}$ ” and, by taking ( $1-b\}y\iota+2(<a-1)$ sufficiently)
and hence we can hoose such as
1, we have $a-1<\{1$near by- $b$$a$ -$b^{2-\alpha}$$b$$\mathrm{c}$$(1 -b \}b^{\mu+2}<a -1<(1 -b)b^{2-\alpha}.

By {$\beta_{2})$and {$a$$\beta_{4})$\{\oint_{4})$, we have$-1$ $<(1$ $-b\}b^{2-\alpha}<a$ $-x$$<(1-b)b^{2-\alpha}+ \frac{(a-1)b^{1-\mu}}{1-b}+\frac{\{a-1)^{2}b^{-1-\mu}}{(1-b)^{\mathit{2}}}$$<$$(1 -b)b^{2-\alpha}+b^{@-\{\mu+\alpha\}}+b^{3-(\mu+2\alpha)}-+\mathrm{O}$(as $aarrow 1$ )$3\alpha$

because$b$$arrow\theta$

And, by(as$(\mathfrak{g}_{3})$$a$ $arrow 1$) by {and by$(\# 4)$$\beta_{4})$and$0$$<(\mu+2\alpha$}$( \int_{5}\}$$(0 <\alpha \leq_{\tilde{2}}1)\leq 2$$\leq\{$$1+\alpha$.

$( \frac{1}{2}\leq\alpha<1)$, we have$\frac{b^{1-\mu}}{1-b}<1arrow\frac{dx}{da}<\frac{2\{a-1\rangle b^{\mathrm{B}-\mu}}{b\{1-b)\{_{ll}-1+b(1-b)\}}+\frac{b^{1-\mu}}{1-b}$$b^{1-\mu}$$2\langle a-1)b^{-1-\mu}$$<+\overline{\langle 1-b\}^{2}}\overline{1-b}$(\S \epsilon )$< \frac{2b^{1-\{\mu+\alpha\rangle}}{1-b}+\frac{b^{\mathrm{I}-\mu}}{1-b}$

For any such as }And leta -b$\zeta$$\epsilon$$-\delta$$<\epsilon<$

let$\delta$$= \frac{b\{\mathrm{I}}{a}-\frac{-b}{-1}\mathrm{L}^{\epsilon}$. Then$0<\delta$$<a$ -b and$>\mathfrak{g}_{\iota}$

A

and B

$=$

$=(\begin{array}{ll}x 0\mathrm{O} y\end{array})$

and clearly B. Then A and B are
where y $=b$
, we may ass ume that $\min\{\#,$y\rangle(as a $-*1$ } bySince x$\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{k}\mathrm{a}\mathrm{d}\tilde{\mathrm{g}}\dot{\mathrm{o}}\mathrm{M}$$- \frac{\epsilon\langle a-b-\delta\}}{\overline{a}-1}$$arrow \mathrm{I}$$(\#\mathrm{s})$,.

$\geq \mathrm{H}\mathrm{i}\mathrm{n}(x_{2}y)I$$> \frac{b}{2}\ddagger\}\mathrm{a}\mathrm{e}\mathrm{a}\iota \mathrm{l}\mathrm{S}\mathrm{B}$we shal100
1. Since the proper polynomial of$\lambda^{2}-(a$and sittceIf$a$$-x$ $+\epsilon+$$\mathit{5}+\frac{\epsilon\{a-b-\delta\}}{a-1}$$-x$ $>a$ $-1$ by$( \int_{\mathrm{S}})$,$\lambda$$+(a-x)\{\epsilon+\mathit{5}\}$ $+( \frac{a-x}{a-1}-\mathrm{f})$ $\epsilon(a-b -\delta)$$\{a-x\}(\not\in+\delta\}+\{\frac{\mathrm{e}\mathrm{r}-x}{a-1}-1)$$\zeta A^{\gamma}A^{a}A^{\gamma}\}^{\beta}\geq\{A^{\gamma}B^{\alpha}A^{\gamma}\}^{\beta}$$a^{\mathit{2}\gamma}b^{2\gamma}\{x^{\alpha}-y^{\alpha}\}^{2}\zeta a-b)$ $)\epsilon\{2$), then$\infty$$\epsilon(\ -b-\delta)>0$

and$A$ $\geq B$.

have$+ \frac{\triangleleft\epsilon\}}{\epsilon}\}$‘$\ovalbox{\tt\small REJECT}\{a^{2\gamma\beta}x^{\alpha\beta}-a^{\{\alpha+2\gamma)\beta}\}$$\mathrm{x}$$+a^{2\gamma\beta}\{$ $\frac{2\gamma\{x^{\alpha\beta}-a^{\alpha\beta})}{a}+\frac{x^{\alpha\beta}(y^{\alpha}-x^{\approx})}{P(a-b\}}$$+ \frac{x^{\alpha\beta}b^{2\gamma}\{x^{\alpha}-y^{\alpha})^{2}}{(a-b)_{\mathrm{J}\mathrm{i}^{\alpha}}(a^{2\gamma}x^{a}-\Psi b^{2}?\}}-\alpha a^{\alpha\beta-1\}\mathcal{B}\epsilon+\mathit{0}\{\epsilon)\ovalbox{\tt\small REJECT}}$$b^{\{\alpha+2\gamma\}\beta}\ovalbox{\tt\small REJECT}\{$$\mathrm{x}$$\frac{\alpha(1-b)}{a-1}-\frac{x^{\alpha}-y^{\alpha}}{\{a-b)y^{a}}+\frac{a(a-b)}{b(a-1\}}$$a^{2\gamma}(x^{\alpha}-y^{a})^{2}$$+_{\overline{\zeta a-b)y^{\alpha}(a^{2\gamma}x^{\alpha}-\wp\wp_{\gamma}\}}}\}\beta\epsilon+\theta(\epsilon\}\ovalbox{\tt\small REJECT}$$\leq\frac{a^{4\gamma}b^{4\gamma}(x^{\alpha}-y^{\alpha})^{4}\epsilon^{2}}{\langle a^{2\gamma}\Phi-y^{\alpha}b^{2\gamma})^{2}}\{1$$+ \frac{o(\mathrm{e})}{\epsilon}\}$$\rangle\zeta\ovalbox{\tt\small REJECT}\{\mathrm{s}\mathrm{z}^{\{\infty+2\gamma\}\beta}-y^{\alpha\beta}b^{\mathfrak{R}\beta}\}+\{$$\{\alpha+2\gamma)a^{\{\alpha+2\gamma]\beta-1}-\{$ $\frac{2\gamma(1-b\}}{a-1}$$+ \frac{x^{\alpha}-y^{\alpha}}{(a-b\}y^{\alpha}}-\frac{a^{2\gamma}(x^{\mathrm{c}\mathrm{r}}-y^{\alpha})^{2}}{(a-b)y^{\alpha}(a^{2\gamma}x^{\alpha}-y^{\alpha}b^{2}?\}})y^{\alpha\beta}b^{2\gamma\beta\}\mathcal{B}\epsilon+a(\epsilon)\ovalbox{\tt\small REJECT}}$$\ovalbox{\tt\small REJECT}\{b^{(a+2\gamma)\beta}-a^{2\gamma\beta}x^{\alpha\beta}\}+$$\mathrm{x}$$\{\frac{(\alpha+2\gamma\}(l-b\}b^{\{a+2\gamma\}\beta}}{a-1}-\{\frac{2\gamma}{a}$(i)$+ \frac{y^{\alpha}-x^{\Phi}}{x^{\alpha}(a-b)}+\frac{b^{2\gamma}(x^{\alpha}-y^{\alpha}\rangle^{2}}{\zeta a-b\}x^{\alpha}(\mathrm{s}\epsilon^{2\gamma}x^{\alpha}-y^{\alpha}b^{2}?\}}\# ax2’\gamma\beta"\beta\}\beta\epsilon+d_{\backslash }\epsilon\rangle\ovalbox{\tt\small REJECT}$

where$a(\epsilon)$

is a function of such that$\epsilon$

By multiplying, we haveand by putting$\epsilon.\prec \mathrm{O}\mathrm{h}\ln\frac{q(\epsilon)}{\not\in}=0$$\{a^{2\gamma(1+\beta\}}b^{4\gamma}\xi x^{\alpha}-y^{\alpha})^{4}\epsilon^{2}\}^{\wedge 1}$$\epsilon$.

to the both side of the inequality (i)$rightarrow\succ\theta$$\frac{\beta(a^{\alpha\beta}-x^{*\phi})b^{(a+2\gamma)\beta\sim 2\gamma}}{\{x^{\alpha}-P)^{2}}\{\frac{x^{\alpha}-b^{\alpha}}{\wp}-\frac{\alpha\zeta a-\theta\}\{a\sim b)}{b\langle\alpha-1\}}-\frac{a^{2\gamma}(x^{\alpha}-\Psi\}^{2}}{b^{\alpha}(a^{2\gamma}x^{a}-b^{\alpha+2\gamma})}\}$$\leq\frac{a^{2\gamma\{1-\beta\}}}{(b^{\alpha+2\gamma}-a^{2\gamma}x^{a})^{2}}\{a\}\beta\sim \mathrm{t}\infty+2\gamma b\{\infty+2\gamma\}\mathcal{B}\}\{b^{\langle\alpha+2\gamma)\beta}-a^{2\gamma\theta}x^{a\beta\}}$$\{\dot{\tilde{\mathrm{u}}})$

101and, by multiplying$b^{-2(\alpha+2\gamma)(\beta-1)}$

to the both side of (ii), we have$\frac{\beta(a^{\alpha\beta}-x^{\alpha}}{(ae}\frac{\beta\}b^{2\gamma\zeta 1-\beta\}-\alpha\beta+2\alpha}}{\alpha_{-\#)^{2}}}\{\frac{x^{\alpha}-b^{\alpha}}{b^{\alpha}}-\frac{\alpha(a-b^{2})(a-b)}{b\{_{\mathrm{c}}a-1)}$$arrow\frac{a^{2\gamma}(x^{\alpha}-b^{\alpha})^{2}}{b^{a}(a^{2\gamma}x^{\alpha}-b+\mathrm{z}_{\gamma}\rangle}"\}$$\leq\frac{a^{2\gamma(}}{\xi \mathrm{I}-a^{2\gamma}x^{\Phi}}\frac{2-\beta\}}{b^{-\{\alpha+2\gamma\}}\}^{2}}\{a^{\{+2\gamma\rangle\beta}" b^{-(\alpha+2\gamma\rangle\beta}-\mathrm{I}\}\{1$

Since $0<a$ -x $<3b$ by$|a^{\alpha\beta-1}$$(\#\mathrm{s})$(case 1) Let $0<\alpha\leq 1$ ,.

$\{\mathrm{i}\ddot{\alpha})$, we have$x^{a\beta-t}$–$-a^{2\gamma\beta}ae^{\alpha\beta}b^{-\{\alpha+2\gamma)\beta}\}$

I$\leq Kb$$1<\beta$,

for some constant .K $>0$

mun,$\{\mathrm{m}\grave{\mathrm{l}}\mathrm{n}\{9$$\frac{2\alpha-\alpha\beta}{2(\beta-1\}}\}$,$( \int_{1})$$\frac{1--\alpha\beta}{2(\beta-15}\}<\gamma(\neq 0\}$.

In this case$\{\mathrm{a}\mathrm{n}\mathrm{d}‘.(\alpha+2\gamma\}(\beta-1)=(\alpha+2\gamma)\beta 2\gamma-\alpha \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\alpha+2\gamma>0\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e}\beta>1(\alpha+_{\mathrm{A}^{1}}\gamma\}\beta-2\gamma-\alpha-\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{n}\{\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{n}\{\alpha(\beta-1\rangle,\alpha),1-\alpha\}=2\gamma(\beta-1\grave{f}+\alpha\beta-\alpha-\min\{\min_{-}\{\alpha\langle\beta-1\},\alpha),-\alpha\}>0>\mathrm{m}\mathrm{i}\alpha\{\dot{\mathrm{m}}\mathrm{n}(\alpha\{\beta-1),\alpha),1-:\}\geq 0 \{\mathrm{b}_{1})$Let$a$ $arrow 1$. Then$barrow \mathrm{O}$$\xi$}$\mathrm{y}\xi \mathfrak{g}_{4}$) and $xarrow 1$ by {$\oint_{5})$and hence we have$\ell\iota.arrow \mathrm{I}\mathrm{h}\mathrm{f}\mathrm{f}\mathrm{l}\frac{a^{2\gamma\langle 1-\beta)}}{\{b^{\alpha+2\gamma}-a^{2\gamma}x^{\alpha}\}^{2}}\{a^{(\alpha+2\gamma)\beta}-b^{(a+2\gamma)\beta}\}\{b^{\{\alpha+2\gamma\rangle\beta}-a^{2\gamma\beta}x^{\alpha\beta}\}=\sim 1$

because$\alpha$Since$+2\gamma>$(} by$\langle \mathrm{b}_{1}\}$.

$0< \min\{\dot{\mathrm{m}}\mathrm{n}\{\alpha\{\beta-1\}, \alpha), 1-\alpha\}\leq \mathrm{I}\mathrm{n}\mathrm{i}\mathrm{n}\{\alpha, 1 -\alpha.\}$$\mu=\mathrm{l}\mathrm{n}\mathrm{i}\mathrm{n}\{\min(\alpha(\beta-1\},$Then$xarrow$ I (as$(\alpha+ 2\gamma)\beta$ -$a$ $arrow 1$$2\gamma$-$\alpha$-$\mu$$>0$ by ($\mathrm{b}_{1}\}$$\alpha)$,$\mathrm{k}\mathrm{t}$, $1-\alpha\}$ .

and since $barrow 9$ (as$\mathrm{a}arrow 1$) by (I4) and} by (ii), $K$ have$a \prec 11^{*}\mathrm{m}\frac{a^{\alpha\beta}-x^{\alpha\beta}}{a-1}b^{\langle\alpha+2\gamma\rangle\beta-2\gamma\sim 1}\overline{\sim}\mathrm{h}.\mathrm{n}\alpha\beta aarrow 1(a^{\alpha\beta\sim 1}-x^{a\beta\sim 1}\frac{dx}{da})b^{\{\alpha+2\gamma\}\beta-2’\gamma-1}$$=\alpha\beta \mathrm{l}\mathrm{i}\mathrm{n}\mathrm{b}\prec 0$$(1$ $\sim\frac{dx}{da}\}b^{(\alpha+2\gamma\}\beta-2\gamma-1}=0$by$(\mathfrak{b}\mathrm{z}\}$102
because$\frac{dx}{da}<\frac{2b^{1-\{\mu+\alpha)}}{1-b}+\frac{b^{1\sim\mu}}{1-b}$$\frac{b^{\mathrm{t}-\mu}}{\mathrm{I}-b}<1\sim$by$(\beta_{6})$and hence we have$a \prec 11\mathrm{i}\Re\frac{\beta(a^{\alpha\beta}-x^{\alpha\beta}\}b^{(\alpha+2\gamma\}\beta-2\gamma}}{(\theta-b^{\alpha}\}^{2}}\{\frac{x^{\alpha}-b^{\alpha}}{b^{\alpha}}-\frac{\alpha(a-\mathfrak{X})(a-b)}{b(a-1)}$.

$- \frac{a^{2\gamma}(x^{a}\sim b^{\alpha})^{2}}{b^{\alpha}(a^{2\gamma}x^{\alpha}-b^{\alpha+2\gamma})}\}=0$This contradicts {\’ii),(case 2) Let $8<\alpha$$\leq 1$,$1<\beta$,$\gamma<\mathscr{L}\{\max(-\frac{1}{2}f$}),$\underline{2\alpha_{\acute{1}}-\mathrm{I}-\underline{\alpha\beta}}2\beta-1$$\frac{-\alpha\beta}{\overline{2}\Gamma\beta-1)}\}$.

In this case$\{\mathrm{b}_{2})$$\mathrm{a}\mathrm{J}\}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{h}e\mathrm{n}\mathrm{c}\mathrm{e}\alpha+2\gamma<0\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e}\beta>12\gamma(\beta-\mathrm{l})+\alpha\beta+1-\alpha-\mathrm{m}\mathrm{f}\mathrm{f}\mathrm{i}\{\mathrm{n}\mathrm{a}\mathrm{x}(\alpha\beta+2-\beta-\alpha,\alpha\},1-\alpha\}<9\mathrm{d}(\alpha+2\gamma)(\beta-1\}<-\mathrm{l}+\Pi \mathrm{l}\mathrm{m}\{x\mathrm{n}\mathrm{a}\mathrm{x}\{\alpha\beta+2-\sqrt-\alpha,\alpha\},1\sim\alpha\}=-\mathrm{x}\mathrm{n}\mathrm{i}\mathrm{n}\{(\mathrm{l}\sim\alpha\rangle \mathrm{z}\mathrm{l}\acute{\mathrm{I}}\mathrm{h}(\beta-l,1\rangle,\alpha\}\leq 0$$\ovalbox{\tt\small REJECT}$

Let$0$ $arrow 1$. Then$barrow \mathrm{O}$

by$(\# 4)$

and $xarrow 1$ by {$\mathfrak{g}_{5})$and hence we have$\lim_{a\prec 1}\frac{a^{2\gamma\acute{1}1-\beta]}}{\{1-a^{2\gamma}x^{a}b^{-\{\alpha+2\gamma)}\}^{2}}\{a^{(\alpha+2\gamma)\beta}b^{-\{a+2\gamma)\beta}-1\}$ $\{1-a^{2\gamma\beta}x^{\alpha\beta}b^{-(\alpha+2\gamma)\beta}\}$$=-1$

because$\alpha+2\eta$

Since$<0$ by$(\mathrm{b}_{2}\}$.

$0< \min\{(1--\alpha\rangle\dot{\mathrm{m}}\mathrm{n}(\sqrt-1_{1}1), a\}$$\leq \mathrm{m}\mathrm{i}_{\mathrm{X}1}\{\alpha, 1-\alpha\}$$\mu=\min\{(1-\alpha?\mathrm{n}\mathrm{i}\mathrm{n}(\beta -1, 1), \alpha\}$Then$x$ $arrow 1$(as$2\gamma(1-\beta)$$a$ $arrow 1$) by– $\alpha\beta+\alpha-\mu$$( \oint_{5})$, we have$>0$ by{$\mathrm{k}_{2})$and si ce,$1\mathrm{e}1$.

$barrow \mathrm{O}$(as$a$ $arrow 1$) by$\langle$$\mathrm{h}.\mathrm{n}\underline{a^{<t\beta}-x^{\alpha\beta}}_{b^{2\gamma(1-\beta\}-\alpha\beta+2\alpha-1}}$$a\prec 1$$a-1$$=1 \acute{1}\mathrm{m}\alpha\beta\epsilon-*1(a^{\alpha\beta-1}-x^{\alpha\beta-1}\frac{\ }{da})b^{2\gamma(2-\beta\rangle-\alpha\beta+Z\alpha-1}$$=\alpha\beta \mathrm{h}.\mathrm{m}_{\mathfrak{g}}b\prec(1$because$\frac{b^{1-\mu}}{1arrow b}<1$$- \frac{dx}{da}\ovalbox{\tt\small REJECT} b^{2\gamma(1-\beta)-<\approx\beta+2\alpha-1}=0$$- \frac{dx}{da}<\frac{2b^{1-(\mu+\infty)}}{1-b}+\frac{b^{1-\mu}}{\mathrm{I}-b}$

by {$\int_{S})$by$\langle \mathfrak{y}_{1}\}$and hence we have$a. \prec 1\mathrm{h}\mathrm{m}\frac{\beta(a^{\alpha\beta}-x^{\alpha\beta})b^{2\gamma[1-\beta)-\alpha\beta+2\alpha}}{(x^{\alpha}-b^{\alpha})^{2}\backslash }\{\frac{x^{a}-b^{\alpha}}{b^{\alpha}}\sim\frac{\alpha(a-b^{2})(a-b)}{b(a\sim 1\rangle}$$- \frac{a^{2\gamma}(x^{\alpha}\sim b^{\alpha})^{\underline{\mathrm{o}}}}{b^{\alpha}(a^{2\gamma}x^{\alpha}-b^{\alpha+2\gamma})}\}-\sim 0$

This contradicts (ii),.

$\oint_{4})$and