When does the spectrum of an element in a Banach algebra with involution lie in the open right half-plane?












2












$begingroup$


Let $A$ be a Banach algebra with involution, $xin A$ and $tin
{mathbb R}$
such that $t>rho(xx^*)$. Show that $sigma(te-xx^*)$
lies in the open right half-plane.



I have no idea! It's obvious that $t-rho(xx^*)>0$. Maybe we should use this to
conclude $rho(te-xx^*)>0$ and then getting the statement! Would you
please help me with that?










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$endgroup$








  • 1




    $begingroup$
    Is $e$ the identity element of $A$?
    $endgroup$
    – Math1000
    Dec 25 '18 at 22:02
















2












$begingroup$


Let $A$ be a Banach algebra with involution, $xin A$ and $tin
{mathbb R}$
such that $t>rho(xx^*)$. Show that $sigma(te-xx^*)$
lies in the open right half-plane.



I have no idea! It's obvious that $t-rho(xx^*)>0$. Maybe we should use this to
conclude $rho(te-xx^*)>0$ and then getting the statement! Would you
please help me with that?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Is $e$ the identity element of $A$?
    $endgroup$
    – Math1000
    Dec 25 '18 at 22:02














2












2








2





$begingroup$


Let $A$ be a Banach algebra with involution, $xin A$ and $tin
{mathbb R}$
such that $t>rho(xx^*)$. Show that $sigma(te-xx^*)$
lies in the open right half-plane.



I have no idea! It's obvious that $t-rho(xx^*)>0$. Maybe we should use this to
conclude $rho(te-xx^*)>0$ and then getting the statement! Would you
please help me with that?










share|cite|improve this question









$endgroup$




Let $A$ be a Banach algebra with involution, $xin A$ and $tin
{mathbb R}$
such that $t>rho(xx^*)$. Show that $sigma(te-xx^*)$
lies in the open right half-plane.



I have no idea! It's obvious that $t-rho(xx^*)>0$. Maybe we should use this to
conclude $rho(te-xx^*)>0$ and then getting the statement! Would you
please help me with that?







spectral-theory banach-algebras






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asked Dec 25 '18 at 21:34









SsFfSsFf

485




485








  • 1




    $begingroup$
    Is $e$ the identity element of $A$?
    $endgroup$
    – Math1000
    Dec 25 '18 at 22:02














  • 1




    $begingroup$
    Is $e$ the identity element of $A$?
    $endgroup$
    – Math1000
    Dec 25 '18 at 22:02








1




1




$begingroup$
Is $e$ the identity element of $A$?
$endgroup$
– Math1000
Dec 25 '18 at 22:02




$begingroup$
Is $e$ the identity element of $A$?
$endgroup$
– Math1000
Dec 25 '18 at 22:02










1 Answer
1






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2












$begingroup$

I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.



Note that,
begin{align*}
&s in sigma(y) \
iff , &(y - se)^{-1} text{ doesn't exist} \
iff , &(se - y)^{-1} text{ doesn't exist} \
iff , &(te - y - (t - s)e)^{-1} text{ doesn't exist} \
iff , &t - s in sigma(te - y).
end{align*}

If $s in sigma(y)$, then
$$Re(t - s) = t - Re(s) ge t - |s| > 0.$$






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    1 Answer
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    1 Answer
    1






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    active

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    2












    $begingroup$

    I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.



    Note that,
    begin{align*}
    &s in sigma(y) \
    iff , &(y - se)^{-1} text{ doesn't exist} \
    iff , &(se - y)^{-1} text{ doesn't exist} \
    iff , &(te - y - (t - s)e)^{-1} text{ doesn't exist} \
    iff , &t - s in sigma(te - y).
    end{align*}

    If $s in sigma(y)$, then
    $$Re(t - s) = t - Re(s) ge t - |s| > 0.$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.



      Note that,
      begin{align*}
      &s in sigma(y) \
      iff , &(y - se)^{-1} text{ doesn't exist} \
      iff , &(se - y)^{-1} text{ doesn't exist} \
      iff , &(te - y - (t - s)e)^{-1} text{ doesn't exist} \
      iff , &t - s in sigma(te - y).
      end{align*}

      If $s in sigma(y)$, then
      $$Re(t - s) = t - Re(s) ge t - |s| > 0.$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.



        Note that,
        begin{align*}
        &s in sigma(y) \
        iff , &(y - se)^{-1} text{ doesn't exist} \
        iff , &(se - y)^{-1} text{ doesn't exist} \
        iff , &(te - y - (t - s)e)^{-1} text{ doesn't exist} \
        iff , &t - s in sigma(te - y).
        end{align*}

        If $s in sigma(y)$, then
        $$Re(t - s) = t - Re(s) ge t - |s| > 0.$$






        share|cite|improve this answer









        $endgroup$



        I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.



        Note that,
        begin{align*}
        &s in sigma(y) \
        iff , &(y - se)^{-1} text{ doesn't exist} \
        iff , &(se - y)^{-1} text{ doesn't exist} \
        iff , &(te - y - (t - s)e)^{-1} text{ doesn't exist} \
        iff , &t - s in sigma(te - y).
        end{align*}

        If $s in sigma(y)$, then
        $$Re(t - s) = t - Re(s) ge t - |s| > 0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 23:23









        Theo BenditTheo Bendit

        19.4k12353




        19.4k12353






























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