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Let $\mathcal{A}$ be a $C^*$-algebra with $a,b,e\in\mathcal{A}$ such that $e\geq0$ and $\|e\|\leq1$. If $\|ae-a\|\leq\varepsilon$ and $\|be\|\leq\varepsilon$, then is $\|a+b\|\leq\sqrt{\|a\|^2+\|b\|^2}+2\varepsilon$?

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 $h\in\mathcal{H}$ 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 $\|A-AE\|\|h\|+\|BE\|\|h\|\leq2\varepsilon$, so we only need to show that $\|A\|\|Eh\|+\|B\|\|(I-E)h\|\leq\sqrt{\|A\|^2+\|B\|^2}$. 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 $E\geq0$ and $\|E\|\leq1$ to show that $\langle h_1,h_2\rangle\geq0$. With this, we get $$\|h_1\|^2+\|h_2\|^2\leq\langle h_1,h_1\rangle+\langle h_1,h_2\rangle+\langle h_2,h_1\rangle+\langle h_2,h_2\rangle=\langle h,h\rangle\leq1.$$ By Cauchy Schwartz, we then get $$\|A\|\|h_1\|+\|B\|\|h_2\|=\langle(\|A\|,\|B\|),(\|h_1\|,\|h_2\|)\rangle\leq\sqrt{\|A\|^2+\|B\|^2}.$$

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.

Martin Argerami
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SmileyCraft
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Yes, I think this works. You have $$ \langle h_1,h_2\rangle=\langle Eh,(I-E)h\rangle=\langle (I-E)Eh,h\rangle=\langle (I-E)^{1/2}E(I-E)^{1/2}h,h\rangle\geq0. $$

Martin Argerami
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  • 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 https://math.stackexchange.com/questions/1339851/product-ab-of-postive-elements-a-b-is-again-positive-if-ab-ba, but exactly you gave the exact same argument :P – SmileyCraft Dec 27 '18 at 06:33
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    It's more general: if $b\geq0$, then $aba^*\geq0$. – Martin Argerami Dec 27 '18 at 06:37
  • 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. – SmileyCraft Dec 27 '18 at 06:40
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    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 $A\subset B(H)$, which is true for any C$^$-algebra. – Martin Argerami Dec 27 '18 at 06:43
  • Right, I was hoping something like that would be true. Was I right that this has something to do with the Gelfand-Naimark-Segal construction? I think it is a bit confusing as the theorem requires you to define a state. If this has nothing to do with it, do you have a reference / proof why $A\subset B(H)$? – SmileyCraft Dec 27 '18 at 06:45
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    Look "universal representation". You first prove that for any positive $a\in A$ there exists a state with $\varphi(a)>0$. Then you do GNS for all states and take a representation $\bigoplus_{\varphi\in S(A)}\pi_\varphi$, which is faithful by the aforementioned property for states (that they separate poisitive points). – Martin Argerami Dec 27 '18 at 06:47