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

The reason I believe this is true, is because this holds for all homomorphisms $h$ on $A$. We get $|h(a-ae)|\leq\varepsilon$, $|h(be)|\leq\varepsilon$ and $0\leq h(e)\leq1$. This gives \begin{align} |h(a+b)| &=|h(a-ae)+h(ae)+h(b-be)+h(be)| \\&\leq|h(e)h(a)+(1-h(e))h(b)|+|h(a-ae)|+|h(be)| \\&\leq\max\{|h(a)|,|h(b)|\}+2\varepsilon. \end{align} If I remember correctly, only in abelian $C^*$-algebras we have the equality $\|a\|=\sup h(a)$ where $h$ is any homomorphism. This means that we have solved the question for abelian $C^*$-algebras, but not for others.

I tried to solve this with only norm inequalities, but I only found one in the wrong direction: \begin{align} \|a+b\| &\geq\|a+b\|\|e\| \\&\geq\|ae+be\| \\&\geq\|a\|-\|ae-a\|-\|be\| \\&\geq\|a\|-2\varepsilon. \end{align} I could not even find an inequality of the form $\|a+b\|\geq\|b\|-\phi(\varepsilon)$ where $\phi(\varepsilon)\to0$ as $\varepsilon\to0$.

SmileyCraft
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It's not true. Let $A=M_2(\mathbb C)$, and $$ a=\begin{bmatrix} 1&0\\0&0\end{bmatrix}, \ b=\begin{bmatrix} 0&1\\0&0\end{bmatrix},\ e=\begin{bmatrix} 1-\varepsilon&0\\0&0\end{bmatrix} . $$ Then $$ ae-a=\begin{bmatrix} -\varepsilon&0\\0&0\end{bmatrix}, $$ so $\|ae-a\|=\varepsilon$. Also $be=0$, so $\|be\|<\varepsilon$. And $$ a+b=\begin{bmatrix} 1&1\\ 0&0\end{bmatrix}, $$ so $$ \|a\|=1,\ \|b\|=1,\ \ \|a+b\|=\sqrt2. $$

Martin Argerami
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  • Very nice, thank you! Do you think we can reduce it to $|a+b|\leq\sqrt{|a|^2+|b|^2}+2\varepsilon$? I have a feeling this might be possible with Gelfand-Naimark-Segal construction. – SmileyCraft Dec 26 '18 at 22:56
  • I made a separate thread for this follow up question. https://math.stackexchange.com/questions/3053615/is-ab-leq-sqrt-a-2-b-22-varepsilon-when-ae-a-leq-varepsilo – SmileyCraft Dec 27 '18 at 05:19