$a=cc^*c$ for some $c$. $a in A$ a $C^*$ algebra. The Next CEO of Stack OverflowSome examples...

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$a=cc^*c$ for some $c$. $a in A$ a $C^*$ algebra.



The Next CEO of Stack OverflowSome examples in C* algebras and Banach * algebrasMap to multiplier algebra for C*-subalgebraIntegration with values in a $C^*$-algebraShow the algebra of all bouned operators is a $W^*$-algebraInfinite dimensional C*-algebra contains infinite dimensional commutitive subalgebraExistence of state on a C*-algebra satisfying $|tau(ab)|=|ab|$Nonunital C*-Algebra: Proper IdealsAn equivalent condition for a cyclic von Neumann algebraA question on positive elements bigger than $1_A$ in a unital C*-algebra $A$Increasing continuous function preserves inequality for elements of $C^*$-algebra












1












$begingroup$



Let $A$ be a $C^*$ algebra. Let $a in A$, then there exists $c in A$ such that $a=cc^*c$.




This fact is used from example (1) of Prop 4.25. How does one show this?






functional-analysis operator-algebras adjoint-operators







share|cite|improve this question









$endgroup$












  • $begingroup$
    The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
    $endgroup$
    – MaoWao
    Mar 16 at 18:17










  • $begingroup$
    Ah, ok, I did not read the proof, and jumped to the example for some intuition.
    $endgroup$
    – CL.
    Mar 16 at 18:36
















1












$begingroup$



Let $A$ be a $C^*$ algebra. Let $a in A$, then there exists $c in A$ such that $a=cc^*c$.




This fact is used from example (1) of Prop 4.25. How does one show this?






functional-analysis operator-algebras adjoint-operators







share|cite|improve this question









$endgroup$












  • $begingroup$
    The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
    $endgroup$
    – MaoWao
    Mar 16 at 18:17










  • $begingroup$
    Ah, ok, I did not read the proof, and jumped to the example for some intuition.
    $endgroup$
    – CL.
    Mar 16 at 18:36














1












1








1





$begingroup$



Let $A$ be a $C^*$ algebra. Let $a in A$, then there exists $c in A$ such that $a=cc^*c$.




This fact is used from example (1) of Prop 4.25. How does one show this?






functional-analysis operator-algebras adjoint-operators







share|cite|improve this question









$endgroup$





Let $A$ be a $C^*$ algebra. Let $a in A$, then there exists $c in A$ such that $a=cc^*c$.




This fact is used from example (1) of Prop 4.25. How does one show this?






functional-analysis operator-algebras adjoint-operators




functional-analysis operator-algebras adjoint-operators






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 16 at 17:31









CL.CL.

2,3452925




2,3452925












  • $begingroup$
    The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
    $endgroup$
    – MaoWao
    Mar 16 at 18:17










  • $begingroup$
    Ah, ok, I did not read the proof, and jumped to the example for some intuition.
    $endgroup$
    – CL.
    Mar 16 at 18:36


















  • $begingroup$
    The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
    $endgroup$
    – MaoWao
    Mar 16 at 18:17










  • $begingroup$
    Ah, ok, I did not read the proof, and jumped to the example for some intuition.
    $endgroup$
    – CL.
    Mar 16 at 18:36
















$begingroup$
The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
$endgroup$
– MaoWao
Mar 16 at 18:17




$begingroup$
The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems?
$endgroup$
– MaoWao
Mar 16 at 18:17












$begingroup$
Ah, ok, I did not read the proof, and jumped to the example for some intuition.
$endgroup$
– CL.
Mar 16 at 18:36




$begingroup$
Ah, ok, I did not read the proof, and jumped to the example for some intuition.
$endgroup$
– CL.
Mar 16 at 18:36










1 Answer
1






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oldest

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1












$begingroup$

$a^*a ge 0$ so the function $t mapsto frac{t^{1/6}}{sqrt{t+frac1n}}$ is continuous on $sigma(a^*a) subseteq [0,infty]$ and satisfies $0mapsto 0$.



Thus it makes sense to define a sequence $(c_n)_n$ as $$c_n = aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6} in A$$



Denote $d_{mn} = left(a^*a + frac1mright)^{-1/2} - left(a^*a + frac1nright)^{-1/2}$ for $m,ninmathbb{N}$. We have
begin{align}
|c_m - c_n|^2 &= |a ,d_{mn}(a^*a)^{1/6}|^2 \
&= |(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}|\
&= |{d_{mn}}^2(a^*a)^{4/3}|\
&= |{d_{mn}}(a^*a)^{2/3}|^2\
&= left|left(a^*a + frac1mright)^{-1/2}(a^*a)^{2/3} - left(a^*a + frac1nright)^{-1/2}(a^*a)^{2/3}right|^2\
&xrightarrow{m,ntoinfty} |(a^*a)^{1/6}-(a^*a)^{1/6}|^2 \
&= 0
end{align}



because $$f_n(t) = frac{t^{2/3}}{sqrt{t+frac1n}} xrightarrow{ntoinfty} t^{1/6}$$ uniformly on $sigma(a^*a)$.



Hence $(c_n)_n$ converges to an element $c in A$. We claim that $c$ is the desired element.



begin{align}
cc^*c &= lim_{ntoinfty} c_nc_n^*c_n \
&= lim_{ntoinfty} aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}left(a^*a + frac1nright)^{-1/2}(a^*a)left(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}\
&= lim_{ntoinfty} acdot (a^*a)^{3/2}left(a^*a + frac1nright)^{-3/2}\
&= a cdot 1\
&= a
end{align}
because $$g_n(t) = frac{t^{3/2}}{left(t+frac1nright)^{3/2}} xrightarrow{ntoinfty} 1$$ uniformly on $sigma(a^*a)$. The limit function $1$ does not satisfy $0 mapsto 0$ but we can adjoin a unit to the algebra $A$ so that it makes sense.



Therefore $a = cc^*c$.



This proof was mostly from Pedersen's $C^*$-algebras and their Automorphism Groups, Lemma $1.4.4.$ and $1.4.5.$






share|cite|improve this answer









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    $begingroup$

    $a^*a ge 0$ so the function $t mapsto frac{t^{1/6}}{sqrt{t+frac1n}}$ is continuous on $sigma(a^*a) subseteq [0,infty]$ and satisfies $0mapsto 0$.



    Thus it makes sense to define a sequence $(c_n)_n$ as $$c_n = aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6} in A$$



    Denote $d_{mn} = left(a^*a + frac1mright)^{-1/2} - left(a^*a + frac1nright)^{-1/2}$ for $m,ninmathbb{N}$. We have
    begin{align}
    |c_m - c_n|^2 &= |a ,d_{mn}(a^*a)^{1/6}|^2 \
    &= |(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}|\
    &= |{d_{mn}}^2(a^*a)^{4/3}|\
    &= |{d_{mn}}(a^*a)^{2/3}|^2\
    &= left|left(a^*a + frac1mright)^{-1/2}(a^*a)^{2/3} - left(a^*a + frac1nright)^{-1/2}(a^*a)^{2/3}right|^2\
    &xrightarrow{m,ntoinfty} |(a^*a)^{1/6}-(a^*a)^{1/6}|^2 \
    &= 0
    end{align}



    because $$f_n(t) = frac{t^{2/3}}{sqrt{t+frac1n}} xrightarrow{ntoinfty} t^{1/6}$$ uniformly on $sigma(a^*a)$.



    Hence $(c_n)_n$ converges to an element $c in A$. We claim that $c$ is the desired element.



    begin{align}
    cc^*c &= lim_{ntoinfty} c_nc_n^*c_n \
    &= lim_{ntoinfty} aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}left(a^*a + frac1nright)^{-1/2}(a^*a)left(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}\
    &= lim_{ntoinfty} acdot (a^*a)^{3/2}left(a^*a + frac1nright)^{-3/2}\
    &= a cdot 1\
    &= a
    end{align}
    because $$g_n(t) = frac{t^{3/2}}{left(t+frac1nright)^{3/2}} xrightarrow{ntoinfty} 1$$ uniformly on $sigma(a^*a)$. The limit function $1$ does not satisfy $0 mapsto 0$ but we can adjoin a unit to the algebra $A$ so that it makes sense.



    Therefore $a = cc^*c$.



    This proof was mostly from Pedersen's $C^*$-algebras and their Automorphism Groups, Lemma $1.4.4.$ and $1.4.5.$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $a^*a ge 0$ so the function $t mapsto frac{t^{1/6}}{sqrt{t+frac1n}}$ is continuous on $sigma(a^*a) subseteq [0,infty]$ and satisfies $0mapsto 0$.



      Thus it makes sense to define a sequence $(c_n)_n$ as $$c_n = aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6} in A$$



      Denote $d_{mn} = left(a^*a + frac1mright)^{-1/2} - left(a^*a + frac1nright)^{-1/2}$ for $m,ninmathbb{N}$. We have
      begin{align}
      |c_m - c_n|^2 &= |a ,d_{mn}(a^*a)^{1/6}|^2 \
      &= |(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}|\
      &= |{d_{mn}}^2(a^*a)^{4/3}|\
      &= |{d_{mn}}(a^*a)^{2/3}|^2\
      &= left|left(a^*a + frac1mright)^{-1/2}(a^*a)^{2/3} - left(a^*a + frac1nright)^{-1/2}(a^*a)^{2/3}right|^2\
      &xrightarrow{m,ntoinfty} |(a^*a)^{1/6}-(a^*a)^{1/6}|^2 \
      &= 0
      end{align}



      because $$f_n(t) = frac{t^{2/3}}{sqrt{t+frac1n}} xrightarrow{ntoinfty} t^{1/6}$$ uniformly on $sigma(a^*a)$.



      Hence $(c_n)_n$ converges to an element $c in A$. We claim that $c$ is the desired element.



      begin{align}
      cc^*c &= lim_{ntoinfty} c_nc_n^*c_n \
      &= lim_{ntoinfty} aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}left(a^*a + frac1nright)^{-1/2}(a^*a)left(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}\
      &= lim_{ntoinfty} acdot (a^*a)^{3/2}left(a^*a + frac1nright)^{-3/2}\
      &= a cdot 1\
      &= a
      end{align}
      because $$g_n(t) = frac{t^{3/2}}{left(t+frac1nright)^{3/2}} xrightarrow{ntoinfty} 1$$ uniformly on $sigma(a^*a)$. The limit function $1$ does not satisfy $0 mapsto 0$ but we can adjoin a unit to the algebra $A$ so that it makes sense.



      Therefore $a = cc^*c$.



      This proof was mostly from Pedersen's $C^*$-algebras and their Automorphism Groups, Lemma $1.4.4.$ and $1.4.5.$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $a^*a ge 0$ so the function $t mapsto frac{t^{1/6}}{sqrt{t+frac1n}}$ is continuous on $sigma(a^*a) subseteq [0,infty]$ and satisfies $0mapsto 0$.



        Thus it makes sense to define a sequence $(c_n)_n$ as $$c_n = aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6} in A$$



        Denote $d_{mn} = left(a^*a + frac1mright)^{-1/2} - left(a^*a + frac1nright)^{-1/2}$ for $m,ninmathbb{N}$. We have
        begin{align}
        |c_m - c_n|^2 &= |a ,d_{mn}(a^*a)^{1/6}|^2 \
        &= |(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}|\
        &= |{d_{mn}}^2(a^*a)^{4/3}|\
        &= |{d_{mn}}(a^*a)^{2/3}|^2\
        &= left|left(a^*a + frac1mright)^{-1/2}(a^*a)^{2/3} - left(a^*a + frac1nright)^{-1/2}(a^*a)^{2/3}right|^2\
        &xrightarrow{m,ntoinfty} |(a^*a)^{1/6}-(a^*a)^{1/6}|^2 \
        &= 0
        end{align}



        because $$f_n(t) = frac{t^{2/3}}{sqrt{t+frac1n}} xrightarrow{ntoinfty} t^{1/6}$$ uniformly on $sigma(a^*a)$.



        Hence $(c_n)_n$ converges to an element $c in A$. We claim that $c$ is the desired element.



        begin{align}
        cc^*c &= lim_{ntoinfty} c_nc_n^*c_n \
        &= lim_{ntoinfty} aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}left(a^*a + frac1nright)^{-1/2}(a^*a)left(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}\
        &= lim_{ntoinfty} acdot (a^*a)^{3/2}left(a^*a + frac1nright)^{-3/2}\
        &= a cdot 1\
        &= a
        end{align}
        because $$g_n(t) = frac{t^{3/2}}{left(t+frac1nright)^{3/2}} xrightarrow{ntoinfty} 1$$ uniformly on $sigma(a^*a)$. The limit function $1$ does not satisfy $0 mapsto 0$ but we can adjoin a unit to the algebra $A$ so that it makes sense.



        Therefore $a = cc^*c$.



        This proof was mostly from Pedersen's $C^*$-algebras and their Automorphism Groups, Lemma $1.4.4.$ and $1.4.5.$






        share|cite|improve this answer









        $endgroup$



        $a^*a ge 0$ so the function $t mapsto frac{t^{1/6}}{sqrt{t+frac1n}}$ is continuous on $sigma(a^*a) subseteq [0,infty]$ and satisfies $0mapsto 0$.



        Thus it makes sense to define a sequence $(c_n)_n$ as $$c_n = aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6} in A$$



        Denote $d_{mn} = left(a^*a + frac1mright)^{-1/2} - left(a^*a + frac1nright)^{-1/2}$ for $m,ninmathbb{N}$. We have
        begin{align}
        |c_m - c_n|^2 &= |a ,d_{mn}(a^*a)^{1/6}|^2 \
        &= |(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}|\
        &= |{d_{mn}}^2(a^*a)^{4/3}|\
        &= |{d_{mn}}(a^*a)^{2/3}|^2\
        &= left|left(a^*a + frac1mright)^{-1/2}(a^*a)^{2/3} - left(a^*a + frac1nright)^{-1/2}(a^*a)^{2/3}right|^2\
        &xrightarrow{m,ntoinfty} |(a^*a)^{1/6}-(a^*a)^{1/6}|^2 \
        &= 0
        end{align}



        because $$f_n(t) = frac{t^{2/3}}{sqrt{t+frac1n}} xrightarrow{ntoinfty} t^{1/6}$$ uniformly on $sigma(a^*a)$.



        Hence $(c_n)_n$ converges to an element $c in A$. We claim that $c$ is the desired element.



        begin{align}
        cc^*c &= lim_{ntoinfty} c_nc_n^*c_n \
        &= lim_{ntoinfty} aleft(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}left(a^*a + frac1nright)^{-1/2}(a^*a)left(a^*a + frac1nright)^{-1/2}(a^*a)^{1/6}\
        &= lim_{ntoinfty} acdot (a^*a)^{3/2}left(a^*a + frac1nright)^{-3/2}\
        &= a cdot 1\
        &= a
        end{align}
        because $$g_n(t) = frac{t^{3/2}}{left(t+frac1nright)^{3/2}} xrightarrow{ntoinfty} 1$$ uniformly on $sigma(a^*a)$. The limit function $1$ does not satisfy $0 mapsto 0$ but we can adjoin a unit to the algebra $A$ so that it makes sense.



        Therefore $a = cc^*c$.



        This proof was mostly from Pedersen's $C^*$-algebras and their Automorphism Groups, Lemma $1.4.4.$ and $1.4.5.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 16 at 21:58









        mechanodroidmechanodroid

        29k62648




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            Why did Rep. Omar conclude her criticism of US troops with the phrase "NotTodaySatan"? Contradiction proof for inequality of P and NP? What is meant by electrons moving to lower potential energy? Negative Resistance Zonal Statistics is returning null values in ArcGIS Desktop? Why do distances seem to matter in the Foundation world? Injection into a proper class and choice without regularity A faster way to compute the largest prime factor "Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English? Do I need to watch Ant-Man and the Wasp and Captain Marvel before watching Avengers: Endgame? How bug prioritization works in agile projects vs non agile Is this a typo in Section 1.8.1 Mathematics for Computer Science? Drawing a german abacus as in the books of Adam Ries How to have a sharp product image? Creating a chemical industry from a medieval tech level without petroleum Partitioning va...
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            Joseph Lister

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            Joseph Lister Fødd 5. april 1827 Upton House, Newham Død 10. februar 1912 Kent Nasjonalitet Det sameinte kongeriket Storbritannia og Irland Område kirurgi Yrke lege, universitetslærar, kirurg Institusjonar King’s College London University of Edinburgh University of Glasgow Alma mater University College London Medlem Royal Society Kungliga Vetenskapsakademien Det ungarske vitskapsakademiet Det greske filologiske selskap i Konstantinopel grunnlagt 1861 American Academy of Arts and Sciences Kungliga Vetenskaps- och Vitterhetssamhället i Göteborg Deutsche Akademie der Naturforscher Leopoldina Académie nationale de médecine Royal Academy of Medicine of Belgium Joseph Lister (fødd 5. april 1827 i Upton i Essex, død 10. februar 1912 i Walmer) var ein britisk kirurg. Han er kjend for å ha oppdaga at karbolsyre påverkar bakteriane. Ved å rense sår og å leggje på desifiserte banda...
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