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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">\chapter[پیوستگی خودکار   ‎$n$-‎هم‌ریختی‌های پوشا]{\huge پیوستگی خودکار ‎$n$-‎هم‌ریختی‌های پوشا}</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">مفهوم ‎$n$-‎هم‌ریختی‌ها</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{$n$-هم‌ریختی}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‌برای جبرهای باناخ مختلط توسط حجازیان</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Hejazian}‎،</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">میرزاوزیری</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Mirzavaziri}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> و مصلحیان</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎\LTRfootnote{Moslehian}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">  مطالعه شد</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{6}‎.</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">براچیچ</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Bara$\check{c}$i$\check{c}$}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> و مصلحیان در </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎	\cite{1}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">در مورد ‎$3$-‎هم‌ریختی‌ها روی جبر‌‌‌‌های باناخ با واحد تقریبی کراندار تحقیق کردند و نشان دادند هر ‎$3$-‎هم‌ریختی حافظ برگشت بین </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$C^\ast$-‎جبرها پیوسته و نرم کاهنده است.</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">این نتیجه بعدا در سال ‎$2009$‎ توسط پارک</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Park}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> و تروت</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Trout}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">   برای هر ‎$n$-‎هم‌ریختی  حافظ برگشت بین ‎$C^\ast$-‎جبرها تعمیم داده شد</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{7}.\\‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> پیوستگی خودکار  ‎$n$-‎هم‌ریختی‌ها برای جبرهای باناخ تجزیه‌پذیر</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎\index{تجزیه‌پذیر}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">  در</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{4}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">بیان شده است.</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">مشابه همین مسئله برای جبرهای توپولوژیک در </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{5}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">مطالعه شده است. در این فصل ابتدا مفاهیم اولیه برای مطالعه این موضوع آورده شده است و پس از آن به قضایای اصلی می‌پردازیم.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">\section{مفهوم ‎$n$-هم‌ریختی}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$A$‎ و ‎$B$‎ جبرهای مختلط باشند. نگاشت خطی      </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\theta:A\to B$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">را یک ‎$n$-‎هم‌ریختی</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{$n$-هم‌ریختی}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> نامیم اگر به‌ازای هر     </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a_1,‎\l‎dots,a_n\in A $‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">داشته باشیم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$$\theta(a_1\cdots a_n)=\theta(a_1)\cdots\theta(a_n).$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> %-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">  نگاشت خطی    </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\theta:A\to B$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">را یک ‎$n$-‎پادهم‌ریختی</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{$n$-پادهم‌ریختی}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> نامیم هرگاه به‌ازای هر </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a_1,\ldots,a_n\in A$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">داشته باشیم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\theta(a_1\cdots a_n)=\theta(a_n)\cdots\theta(a_1).$$‎ </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">جبر ‎$B$‎را تجزیه‌پذیر</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{تجزیه‌پذیر}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> نامیم هرگاه به‌ازای هر </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a\in B$‎عضوهای  ‎$b,c\in B$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">موجود باشند به‌طوری‌که ‎$a=bc$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">نگاشت خطی </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\theta:A\to B$‎ </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">را ‎$n$-‎هم‌ریختی ژوردان</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{$n$-‎هم‌ریختی ژوردان}</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> نامیم اگر:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\theta(a^n)=\theta(a)^n\quad(a\in A).$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$(G,\circ)$‎ و ‎$(G',\ast)$‎ گروه باشند. نگاشت ‎$h:G\to G'$‎ را هم‌ریختی گروهی گوییم هرگاه:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$h(a \circ b)=h(a) \ast h(b) \quad (a,b \in G).$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">برخی نتایج در مورد پیوستگی خودکار ‎$n$-‎هم‌ریختی‌های ژوردان روی جبرهای باناخ و ‎$C^\ast$-‎جبرها در</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{3.2}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">بررسی شده است.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">\section{طیف و شعاع طیفی جبر باناخ}</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">در این بخش طیف و شعاع طیفی را برای جبرهای یکدار و غیر یکدار تعریف و بررسی می‌کنیم.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$A$‎ یک جبر مختلط یکدار با یکه‌ی ‎$e_A$‎ باشد. اگر ‎$InvA$‎ نمادی برای اعضای وارون‌پذیر ‎$A$‎ باشد، طیف</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{طیف}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> و شعاع طیفی</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> \index{شعاع طیفی}</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">  عضو ‎$a\in A$‎ را به‌صورت زیر تعریف می‌کنیم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$sp_A(a)=\lbrace\lambda\in \mathbb{C}‎ :‎\lambda e_A-a\notin InvA\rbrace,$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\rho_A(a)=\sup\lbrace\vert\lambda\vert‎: ‎\lambda\in sp_A(a)\rbrace.$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">در حالتی که ‎$A$‎ جبر باناخ باشد بنابه </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\emph{\lr{\cite[Theorem 2.3.8]{5.2}}}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">داریم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\rho_A(a)=\lim_{n \to \infty} \|a^n\|^{\frac{1}{n}}=\inf_{n\in \mathbb{N}}\|a^n\|^{\frac{1}{n}}.$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">اگر ‎$A$‎یک جبر باناخ دلخواه باشد شبه‎‏‌ضرب  ‎$ \diamond$‎ روی ‎$A$‎ را به‌صورت زیر داریم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{center}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$a\diamond b=a+b-ab.$‎ </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{center}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">عضو  ‎$a\in A$‎را شبه-وارون‌پذیر چپ ‎(راست)‎ گوییم هرگاه عضوی مانند ‎$b\in A$‎ یافت شود به‌طوری‌که:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$b\diamond a=0\hspace{2mm}(a\diamond b=0).$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">عضو  </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$a\in A$‎ را شبه-وارون‌پذیر گوییم اگر هم شبه-وارون‌پذیر چپ  باشد و هم  شبه-وارون‌پذیر راست باشد. مجموعه‌ی تمام اعضای شبه-وارون‌پذیر را با ‎$q-InvA$‎ نشان می‌دهیم.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> %-----------------------------------------------------------------------------------------------------------</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$A$‎ جبر باناخ ‎(مختلط)‎ غیر‌یکدار باشد. د‎‏راین‌صورت به وضوح به‌ازای هر ‎$a,b \in A$‎، </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$a\diamond b=0$‎ اگر و تنها اگر </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$(e_{A^{\#}}-a)(e_{A^{\#}}-b)=e_{A^{\#}}$‎، که در اینجا ‎$.e_{A^{\#}}=‎(0,‎1)$‎</p>
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<p style="-qt-paragraph-type:empty; margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px;"><br /></p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{proof}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$a\diamond b=0$‎ پس  ‎$a+b=ab$‎ و لذا </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$(e_{A^{\#}}-a)(e_{A^{\#}}-b)=\big((0,1)-(a,0)\big)\big((0,1)-(b,0)\big)=(-a,1)(-b,1)=(ab-a-b,1)=e_{A^{\#}}.$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">برعکس از تساوی بالا پیداست که اگر تساوی ‎$(e_{A^{\#}}-a)(e_{A^{\#}}-b)=e_{A^{\#}}$‎ برقرار باشد آنگاه </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$a\diamond b=0$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{proof}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> طیف</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎\index{طیف}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">  یک عضو در جبر‌های غیریکدار با حالت یکدار متفاوت است و به‌صورت زیر تعریف می‌شود:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$sp_{A}(a)= \lbrace 0\rbrace \cup \lbrace\lambda\in\mathbb{C}\setminus \lbrace0\rbrace‎ :‎\frac{1}{\lambda}a \notin q-InvA\rbrace.$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{thm}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">به‌ازای هر ‎$ a$‎ از جبر ‎$ A$‎ داریم  ‎$sp_{A}(a)=sp_{A^{\#}}\big((a,0)\big)$‎ و  </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\rho_{A}(a)=\rho_{A^{\#}}\big((a,0)\big)$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{thm}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">%-----------------------------------------------------------------------------------------------------------</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{proof}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">ابتدا نشان می‌دهیم </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$.sp_{A}(a)\subseteq sp_{A^{\#}}\big((a,0)\big) $‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{center}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\lambda \in sp_{A}(a)=\lbrace0\rbrace\cup \lbrace\lambda\in\mathbb{C}\setminus \lbrace0\rbrace‎ :‎\frac{1}{\lambda}a \notin q-InvA\rbrace.$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{center}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">اگر ‎$ \lambda =0$‎  به‌سادگی ثابت می‌شود که ‎$ 0\in sp_{A^{\#}}\big((a,0)\big) $‎.  پس فرض می‌کنیم که ‎$\lambda \neq 0$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> به برهان خلف فرض کنیم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\lambda\notin sp_{A^{\#}}\big((a,0)\big)=\lbrace\lambda\in\mathbb{C}:\lambda e_{A^{\#}}-(a,0)\notin Inv A^{\#}\rbrace =\lbrace \lambda\in\mathbb{C}‎: ‎(-a,\lambda)\notin Inv A^{\#}\rbrace.$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">پس ‎$(b,\lambda^{\prime})\in A^{\#}$‎ هست به‌طوری‌که </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$(b,\lambda')(-a,\lambda)=(-a,\lambda)(b,\lambda^{\prime})=(0,1)$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">بنابراین </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{align*}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎(-ba-\lambda' a+\lambda b,\lambda \lambda')=(-ab-a\lambda^{\prime}+\lambda b,\lambda \lambda^{\prime})=(0,1)‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎&amp;\Rightarrow‎  ‎-ab-\frac{a}{\lambda}+‎ ‎\lambda b=0\\‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎&amp;\Rightarrow \frac{a}{\lambda}\diamond (-\lambda b)=0=-ba-\frac{a}{\lambda}+\lambda b‎,</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{align*}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">که تناقض است، زیرا ‎$\frac{1}{\lambda} a \notin q-InvA$.\\‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">حال برعکس  نشان می‌دهیم  </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$sp_{A^{\#}}\big((a,0)\big)\subseteq sp_{A}(a) $‎. </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم  ‎$\lambda\neq0$‎  عضوی از </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$sp_{A^{\#}}\big((a,0)\big)$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">باشد‎.\\‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> به برهان خلف فرض کنیم  	</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\frac{a}{\lambda}\in q-InvA$‎ </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">باشد پس ‎$b\in A$‎ هست به‌طوری‌که ‎$\frac{1}{\lambda} a \diamond b=0$‎. لذا خواهیم داشت:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\frac{1}{\lambda}a+b-\frac{1}{\lambda}ba=b\diamond \frac{a}{\lambda}=\frac{1}{\lambda}a+b-\frac{1}{\lambda}ab=0\Rightarrow \frac{1}{\lambda}a+b=\frac{1}{\lambda}ab=\frac{1}{\lambda}ba,$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">از طرفی بنابه فرض ‎$(0,\lambda)-(a,0)=(-a,\lambda)\notin InvA^{\#}$‎. کافی است عضوی در ‎$A^{\#}$‎ پیدا کنیم که وارون ‎$(-a,\lambda)$‎ باشد و به تناقض برسیم. از تساوی </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$(\frac{-b}{\lambda},\frac{1}{\lambda})(-a,\lambda)=(-a,\lambda)(\frac{-b}{\lambda},\frac{1}{\lambda})=(0,1)$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">پیداست که ‎$(-a,\lambda)$‎ وارون‌پذیر است و لذا تناقض حاصل می‌شود. حال اگر ‎$\lambda=0$‎‎‏‏، بدیهی است که بنابه تعریف </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$sp_A(a)$‎، ‎$0\in sp_A(a)$‎. همچنین داریم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$\rho_A(a)=\sup\lbrace |\lambda|‎: ‎\lambda \in  sp_A(a)\rbrace=\sup\lbrace |\lambda|‎: ‎\lambda \in  sp_‎{‎A^{\#}}(a,0)‎\rbrace‎ =\rho_{A^{\#}}(a,0)‎. ‎$$‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{proof}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">\section{پیوستگی خودکار}</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">در این بخش روش رنسفورد</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\LTRfootnote{Ransford}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{8}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> برای ‎$n$-‎هم‌ریختی‌ها</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎\index{$n$-هم‌ریختی}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">روی جبر‌های باناخ غیریکدار تعمیم داده می‌شود. نتایج این بخش با نتایج مرجع‌های </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{4}‎، </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{5}‎، </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{7}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">و </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\cite{8}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">متفاوت است.</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{lem}\label{2.1}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\emph{\lr{\cite[Lemma 1]{8}}}‎ </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$A$‎ یک جبر باناخ یکدار با یکه‌ی ‎$e$‎ باشد و ‎$a\in A$‎. اگر به‌ازای هر ‎$a_{1}\in A$‎ داشته باشیم ‎$\rho_{A}(a_1 a)=0$‎ آنگاه ‎$a\in Rad(A)$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{lem}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{proof}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">به برهان خلف فرض کنیم ‎$a\in A$‎ در فرض صدق کند ولی ‎$a\notin Rad(A)$‎. لذا یک ایده‌آل چپ ماکسیمال مانند ‎$L$‎ یافت می‌شود به‌طوری‌که</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a\notin L$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">و در نتیجه داریم:</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$$L\subset Aa+L\subseteq A\Rightarrow Aa+L=A.$$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">چون ‎$e\in A$‎ پس ‎$a_1\in A$‎  موجود است به‌طوری‌که ‎$e-a_1a\in L$‎. لذا ‎$e-a_1a$‎ نمی‌تواند وارون‌پذیر باشد، یعنی</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$1\in sp_A(a_1a)$‎.</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">از طرفی بنابه فرض به‌ازای هر ‎$a_1\in A$‎ داریم </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$\rho_{A}(a_1 a)=\sup \big\lbrace |z|‎: ‎z\in  sp_A(a_1 a)\big\rbrace= 0$‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">که یک تناقض است، لذا ‎$a\in Rad(A)$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{proof}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{rem}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">می‌توانیم این نتیجه را برای جبر‌های باناخ غیریکدار هم تعمیم دهیم. به طور کلی این مطلب برای هر جبری برقرار است</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\emph{\lr{\cite[Theorem 2.3.3]{palmer}}}‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{rem}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">عضو ‎$x$‎ در جبر نرم‌دار ‎$A$‎ پوچ‌توان</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\index{پوچ‌توان}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> توپولوژیکی نامیده می‌شود درصورتی‌که ‎$\rho_A(x)=0$‎. مجموعه‌ی تمام اعضای پوچ‌توان توپولوژیکی را با ‎$Q(A)$‎ نشان می‌دهیم. البته این‌گونه اعضا به شبه‌ پوچ‌توان هم معروفند.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{dfn}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">ایده‌آل ‎$I$‎ در جبر نرم‌دار ‎$A$‎ را ایده‌آل پوچ گوییم درصورتی‌که مشمول در ‎$Q(A)$‎ باشد.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{dfn}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{lem}\label{2.2}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم ‎$A$‎ یک جبر باناخ باشد و ‎$a\in A$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{itemize}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\item[1.]‎ </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">اگر به‌ازای هر تعداد متناهی ‎$a_{1},\ldots,a_{n-1}\in A$‎ تساوی ‎$\rho_{A}(a_1 a_2\ldots a_{n-1} a)=0$‎  برقرار باشد آنگاه‎\\‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">$a\in Rad(A)$.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\item[2.]‎ </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">اگر به‌ازای هر تعداد متناهی ‎$a_{1},\ldots,a_{n-1}\in A$‎ تساوی ‎$\rho_{A}(a a_1 a_2\ldots a_{n-1})=0$‎  برقرار باشد آنگاه\\‎‎‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">$a\in Rad(A)$.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{itemize}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{lem}‎</p>
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<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{proof}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\begin{itemize}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\item[1.]‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم به‌ازای هر ‎$a_{1},\ldots,a_{n-1}\in A$‎ رابطه  ‎$\rho_{A}(a_1 a_2\cdots a_{n-1} a)=0$‎ برقرار باشد. بنابه لم</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\ref{2.1}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">نتیجه می‌شود  </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a_2\cdots a_{n-1} a\in Rad(A) $‎. </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">چون رادیکال هر جبر نرم‌دار یک ایده‌آل پوچ توپولوژیکی است، یعنی به‌ازای هر ‎$a\in A$‎، ‎$Rad(A)\subseteq Q(A)$‎ که در آن </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$Q(A)=\left\lbrace a\in A‎: ‎\rho_A(a)=0\right\rbrace$‏.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\lr{\cite[Theorem 2.3.4]{9}}‎،</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">نتیجه می‌شود ‎$\rho_A( a_2a_3\cdots a_{n-1} a)=0$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎%\newpage‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">با ادامه این روند به‌ازای هر ‎$a_{n-1}\in A$‎ خواهیم داشت ‎$\rho_A(a_{n-1} a)=0$‎، که بنابه لم </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\ref{2.1}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">نتیجه می‌دهد ‎$a\in Rad(A)$‎. </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\item[2.]‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">فرض کنیم به‌ازای هر ‎$a_{1},\ldots,a_{n-1}\in A$‎ رابطه  ‎$\rho_{A}(a a_1 a_2\cdots a_{n-1})=0$‎ برقرار باشد. بنابه لم</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\ref{2.1}‎ نتیجه می‌شود  </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;"> ‎$a a_1\cdots a_{n-2} \in Rad(A) $‎. </p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">چون رادیکال هر جبر نرم‌دار یک ایده‌آل پوچ توپولوژیکی است، یعنی به‌ازای هر ‎$a\in A$‎، ‎$Rad(A)\subseteq Q(A)$‎ که در آن </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎$Q(A)=\left\lbrace a\in A‎: ‎\rho_A(a)=0\right\rbrace$.‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\lr{\cite[Theorem 2.3.4]{9}}‎،</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">پس نتیجه می‌شود ‎$\rho_A(a a_1a_2\ldots a_{n-2})=0$‎. با ادامه این روند به‌ازای هر ‎$a_1\in A$‎ خواهیم داشت ‎$\rho_A(a a_1)=0$‎، که بنابه لم </p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\ref{2.1}‎</p>
<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">نتیجه می‌دهد ‎$a\in Rad(A)$‎.</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{itemize}‎</p>
<p style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">‎\end{proof}‎</p>
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<p dir='rtl' style=" margin-top:0px; margin-bottom:0px; margin-left:0px; margin-right:0px; -qt-block-indent:0; text-indent:0px; -qt-user-state:0;">... (ادامه مطالب حذف شده است)</p></body></html>