Is $|a+b|leqsqrt{|a|^2+|b|^2}+2varepsilon$ when $|ae-a|leqvarepsilon$ and $|be|leqvarepsilon$ in a...












2












$begingroup$


Let $mathcal{A}$ be a $C^*$-algebra with $a,b,einmathcal{A}$ such that $egeq0$ and $|e|leq1$. If $|ae-a|leqvarepsilon$ and $|be|leqvarepsilon$, then is $|a+b|leqsqrt{|a|^2+|b|^2}+2varepsilon$?



This is a follow up on Can we bound $|a+b|$ if $|ae-a|$ and $|be|$ are small, in a $C^*$-algebra? where we showed an even stronger result holds if $A$ is abelian. Here is what I have so far.



Assume $mathcal{A}=mathcal{B}(mathcal{H})$ for some Hilbert space $mathcal{H}$. For convenience, we denote $A=a$, $B=b$, $E=e$ and $I=mbox{id}$. For all $hinmathcal{H}$ with $|h|leq1$ we have
begin{align}
|(A+B)h|
&=|(A-AE)h+AEh+(B-BE)h+BEh|
\&leq|A||Eh|+|B||(I-E)h|+|A-AE||h|+|BE||h|
end{align}
Here we have $|A-AE||h|+|BE||h|leq2varepsilon$, so we only need to show that $|A||Eh|+|B||(I-E)h|leqsqrt{|A|^2+|B|^2}$. Let us denote $h_1=Eh$ and $h_2=(I-E)h$. Note $h_1+h_2=h$.



(*) I believe that you should be able to use $Egeq0$ and $|E|leq1$ to show that $langle h_1,h_2ranglegeq0$. With this, we get $$|h_1|^2+|h_2|^2leqlangle h_1,h_1rangle+langle h_1,h_2rangle+langle h_2,h_1rangle+langle h_2,h_2rangle=langle h,hrangleleq1.$$ By Cauchy Schwartz, we then get $$|A||h_1|+|B||h_2|=langle(|A|,|B|),(|h_1|,|h_2|)rangleleqsqrt{|A|^2+|B|^2}.$$



Apart from a detail that I hope is fixable (*), this solves the problem for operator spaces on Hilbert spaces. Since an even stronger result turns out to be true for abelian $C^*$-algebras, basically every naturally occuring $C^*$-algebras has been handled. However, I would like to be able to generalize this proof to all $C^*$-algebras. I have the idea that the Gelfand-Naimark-Segal construction might be useful to reduce the general case to the $mathcal{B}(mathcal{H})$ case. However, I have never done anything with this theorem before, so I'm not very confident with my skills in applying it.



In summary: Please help me fix the detail (*) and generalize the proof. Or give a counterexample.






inequality operator-algebras c-star-algebras







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $mathcal{A}$ be a $C^*$-algebra with $a,b,einmathcal{A}$ such that $egeq0$ and $|e|leq1$. If $|ae-a|leqvarepsilon$ and $|be|leqvarepsilon$, then is $|a+b|leqsqrt{|a|^2+|b|^2}+2varepsilon$?



    This is a follow up on Can we bound $|a+b|$ if $|ae-a|$ and $|be|$ are small, in a $C^*$-algebra? where we showed an even stronger result holds if $A$ is abelian. Here is what I have so far.



    Assume $mathcal{A}=mathcal{B}(mathcal{H})$ for some Hilbert space $mathcal{H}$. For convenience, we denote $A=a$, $B=b$, $E=e$ and $I=mbox{id}$. For all $hinmathcal{H}$ with $|h|leq1$ we have
    begin{align}
    |(A+B)h|
    &=|(A-AE)h+AEh+(B-BE)h+BEh|
    \&leq|A||Eh|+|B||(I-E)h|+|A-AE||h|+|BE||h|
    end{align}
    Here we have $|A-AE||h|+|BE||h|leq2varepsilon$, so we only need to show that $|A||Eh|+|B||(I-E)h|leqsqrt{|A|^2+|B|^2}$. Let us denote $h_1=Eh$ and $h_2=(I-E)h$. Note $h_1+h_2=h$.



    (*) I believe that you should be able to use $Egeq0$ and $|E|leq1$ to show that $langle h_1,h_2ranglegeq0$. With this, we get $$|h_1|^2+|h_2|^2leqlangle h_1,h_1rangle+langle h_1,h_2rangle+langle h_2,h_1rangle+langle h_2,h_2rangle=langle h,hrangleleq1.$$ By Cauchy Schwartz, we then get $$|A||h_1|+|B||h_2|=langle(|A|,|B|),(|h_1|,|h_2|)rangleleqsqrt{|A|^2+|B|^2}.$$



    Apart from a detail that I hope is fixable (*), this solves the problem for operator spaces on Hilbert spaces. Since an even stronger result turns out to be true for abelian $C^*$-algebras, basically every naturally occuring $C^*$-algebras has been handled. However, I would like to be able to generalize this proof to all $C^*$-algebras. I have the idea that the Gelfand-Naimark-Segal construction might be useful to reduce the general case to the $mathcal{B}(mathcal{H})$ case. However, I have never done anything with this theorem before, so I'm not very confident with my skills in applying it.



    In summary: Please help me fix the detail (*) and generalize the proof. Or give a counterexample.






    inequality operator-algebras c-star-algebras







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $mathcal{A}$ be a $C^*$-algebra with $a,b,einmathcal{A}$ such that $egeq0$ and $|e|leq1$. If $|ae-a|leqvarepsilon$ and $|be|leqvarepsilon$, then is $|a+b|leqsqrt{|a|^2+|b|^2}+2varepsilon$?



      This is a follow up on Can we bound $|a+b|$ if $|ae-a|$ and $|be|$ are small, in a $C^*$-algebra? where we showed an even stronger result holds if $A$ is abelian. Here is what I have so far.



      Assume $mathcal{A}=mathcal{B}(mathcal{H})$ for some Hilbert space $mathcal{H}$. For convenience, we denote $A=a$, $B=b$, $E=e$ and $I=mbox{id}$. For all $hinmathcal{H}$ with $|h|leq1$ we have
      begin{align}
      |(A+B)h|
      &=|(A-AE)h+AEh+(B-BE)h+BEh|
      \&leq|A||Eh|+|B||(I-E)h|+|A-AE||h|+|BE||h|
      end{align}
      Here we have $|A-AE||h|+|BE||h|leq2varepsilon$, so we only need to show that $|A||Eh|+|B||(I-E)h|leqsqrt{|A|^2+|B|^2}$. Let us denote $h_1=Eh$ and $h_2=(I-E)h$. Note $h_1+h_2=h$.



      (*) I believe that you should be able to use $Egeq0$ and $|E|leq1$ to show that $langle h_1,h_2ranglegeq0$. With this, we get $$|h_1|^2+|h_2|^2leqlangle h_1,h_1rangle+langle h_1,h_2rangle+langle h_2,h_1rangle+langle h_2,h_2rangle=langle h,hrangleleq1.$$ By Cauchy Schwartz, we then get $$|A||h_1|+|B||h_2|=langle(|A|,|B|),(|h_1|,|h_2|)rangleleqsqrt{|A|^2+|B|^2}.$$



      Apart from a detail that I hope is fixable (*), this solves the problem for operator spaces on Hilbert spaces. Since an even stronger result turns out to be true for abelian $C^*$-algebras, basically every naturally occuring $C^*$-algebras has been handled. However, I would like to be able to generalize this proof to all $C^*$-algebras. I have the idea that the Gelfand-Naimark-Segal construction might be useful to reduce the general case to the $mathcal{B}(mathcal{H})$ case. However, I have never done anything with this theorem before, so I'm not very confident with my skills in applying it.



      In summary: Please help me fix the detail (*) and generalize the proof. Or give a counterexample.






      inequality operator-algebras c-star-algebras







      share|cite|improve this question











      $endgroup$




      Let $mathcal{A}$ be a $C^*$-algebra with $a,b,einmathcal{A}$ such that $egeq0$ and $|e|leq1$. If $|ae-a|leqvarepsilon$ and $|be|leqvarepsilon$, then is $|a+b|leqsqrt{|a|^2+|b|^2}+2varepsilon$?



      This is a follow up on Can we bound $|a+b|$ if $|ae-a|$ and $|be|$ are small, in a $C^*$-algebra? where we showed an even stronger result holds if $A$ is abelian. Here is what I have so far.



      Assume $mathcal{A}=mathcal{B}(mathcal{H})$ for some Hilbert space $mathcal{H}$. For convenience, we denote $A=a$, $B=b$, $E=e$ and $I=mbox{id}$. For all $hinmathcal{H}$ with $|h|leq1$ we have
      begin{align}
      |(A+B)h|
      &=|(A-AE)h+AEh+(B-BE)h+BEh|
      \&leq|A||Eh|+|B||(I-E)h|+|A-AE||h|+|BE||h|
      end{align}
      Here we have $|A-AE||h|+|BE||h|leq2varepsilon$, so we only need to show that $|A||Eh|+|B||(I-E)h|leqsqrt{|A|^2+|B|^2}$. Let us denote $h_1=Eh$ and $h_2=(I-E)h$. Note $h_1+h_2=h$.



      (*) I believe that you should be able to use $Egeq0$ and $|E|leq1$ to show that $langle h_1,h_2ranglegeq0$. With this, we get $$|h_1|^2+|h_2|^2leqlangle h_1,h_1rangle+langle h_1,h_2rangle+langle h_2,h_1rangle+langle h_2,h_2rangle=langle h,hrangleleq1.$$ By Cauchy Schwartz, we then get $$|A||h_1|+|B||h_2|=langle(|A|,|B|),(|h_1|,|h_2|)rangleleqsqrt{|A|^2+|B|^2}.$$



      Apart from a detail that I hope is fixable (*), this solves the problem for operator spaces on Hilbert spaces. Since an even stronger result turns out to be true for abelian $C^*$-algebras, basically every naturally occuring $C^*$-algebras has been handled. However, I would like to be able to generalize this proof to all $C^*$-algebras. I have the idea that the Gelfand-Naimark-Segal construction might be useful to reduce the general case to the $mathcal{B}(mathcal{H})$ case. However, I have never done anything with this theorem before, so I'm not very confident with my skills in applying it.



      In summary: Please help me fix the detail (*) and generalize the proof. Or give a counterexample.






      inequality operator-algebras c-star-algebras




      inequality operator-algebras c-star-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 7:20









      Martin Argerami

      128k1184184




      128k1184184










      asked Dec 27 '18 at 5:18









      SmileyCraftSmileyCraft

      3,749519




      3,749519






















          1 Answer
          1






          active

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          2












          $begingroup$

          Yes, I think this works. You have
          $$
          langle h_1,h_2rangle=langle Eh,(I-E)hrangle=langle (I-E)Eh,hrangle=langle (I-E)^{1/2}E(I-E)^{1/2}h,hranglegeq0.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:33






          • 1




            $begingroup$
            It's more general: if $bgeq0$, then $aba^*geq0$.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:37










          • $begingroup$
            Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:40






          • 1




            $begingroup$
            It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:43






          • 1




            $begingroup$
            Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:47











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          $begingroup$

          Yes, I think this works. You have
          $$
          langle h_1,h_2rangle=langle Eh,(I-E)hrangle=langle (I-E)Eh,hrangle=langle (I-E)^{1/2}E(I-E)^{1/2}h,hranglegeq0.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:33






          • 1




            $begingroup$
            It's more general: if $bgeq0$, then $aba^*geq0$.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:37










          • $begingroup$
            Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:40






          • 1




            $begingroup$
            It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:43






          • 1




            $begingroup$
            Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:47
















          2












          $begingroup$

          Yes, I think this works. You have
          $$
          langle h_1,h_2rangle=langle Eh,(I-E)hrangle=langle (I-E)Eh,hrangle=langle (I-E)^{1/2}E(I-E)^{1/2}h,hranglegeq0.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:33






          • 1




            $begingroup$
            It's more general: if $bgeq0$, then $aba^*geq0$.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:37










          • $begingroup$
            Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:40






          • 1




            $begingroup$
            It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:43






          • 1




            $begingroup$
            Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:47














          2












          2








          2





          $begingroup$

          Yes, I think this works. You have
          $$
          langle h_1,h_2rangle=langle Eh,(I-E)hrangle=langle (I-E)Eh,hrangle=langle (I-E)^{1/2}E(I-E)^{1/2}h,hranglegeq0.
          $$






          share|cite|improve this answer









          $endgroup$



          Yes, I think this works. You have
          $$
          langle h_1,h_2rangle=langle Eh,(I-E)hrangle=langle (I-E)Eh,hrangle=langle (I-E)^{1/2}E(I-E)^{1/2}h,hranglegeq0.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 6:23









          Martin ArgeramiMartin Argerami

          128k1184184




          128k1184184












          • $begingroup$
            Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:33






          • 1




            $begingroup$
            It's more general: if $bgeq0$, then $aba^*geq0$.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:37










          • $begingroup$
            Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:40






          • 1




            $begingroup$
            It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:43






          • 1




            $begingroup$
            Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:47


















          • $begingroup$
            Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:33






          • 1




            $begingroup$
            It's more general: if $bgeq0$, then $aba^*geq0$.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:37










          • $begingroup$
            Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
            $endgroup$
            – SmileyCraft
            Dec 27 '18 at 6:40






          • 1




            $begingroup$
            It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:43






          • 1




            $begingroup$
            Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
            $endgroup$
            – Martin Argerami
            Dec 27 '18 at 6:47
















          $begingroup$
          Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
          $endgroup$
          – SmileyCraft
          Dec 27 '18 at 6:33




          $begingroup$
          Ah of course, I had to use that $I-E$ is hermitian. Only one question, why is $a^{1/2}ba^{1/2}$ positive if $a$ and $b$ commute and are positive? I tried to look it up and found math.stackexchange.com/questions/1339851/…, but exactly you gave the exact same argument :P
          $endgroup$
          – SmileyCraft
          Dec 27 '18 at 6:33




          1




          1




          $begingroup$
          It's more general: if $bgeq0$, then $aba^*geq0$.
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:37




          $begingroup$
          It's more general: if $bgeq0$, then $aba^*geq0$.
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:37












          $begingroup$
          Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
          $endgroup$
          – SmileyCraft
          Dec 27 '18 at 6:40




          $begingroup$
          Oh because then $b=xx^*$ so $aba^*=axx^*a^*=(ax)(ax)^*$, right. By the way, even though this is very helpful, I hope you understand I only want to accept an answer when the statement is resolved for arbitrary $C^*$-algebras.
          $endgroup$
          – SmileyCraft
          Dec 27 '18 at 6:40




          1




          1




          $begingroup$
          It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:43




          $begingroup$
          It's your question to accept or not accept something, and I don't really mind. But your argument works on any C$^*$-algebra. You are unnecessarily assuming $A=B(H)$, when you could have assumed $Asubset B(H)$, which is true for any C$^*$-algebra.
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:43




          1




          1




          $begingroup$
          Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:47




          $begingroup$
          Look "universal representation". You first prove that for any positive $ain A$ there exists a state with $varphi(a)>0$. Then you do GNS for all states and take a representation $bigoplus_{varphiin S(A)}pi_varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points).
          $endgroup$
          – Martin Argerami
          Dec 27 '18 at 6:47


















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