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Let $A$ be a $C^*$-algebra. Consider the unitalization $A^+$ with underlying vector space $A^+=A\oplus \mathbb{C}$. The norm on $A^+$ is defined as $||a+\lambda||_{A^+}=\mathrm{sup}_{||b||\leq 1}\ ||ab+\lambda b||_A$. In order to show that this is positive definite I consider $(a,\lambda)\in A^+$ such that $||a+\lambda||_{A^+}=0$. Without loss of generality $||a||\leq 1$. This means that for all $b\in A$ with $||b||\leq 1$ we have $ab+\lambda b=0$ using positive definiteness of the norm on $A$. Now $b=a$ implies $a^2=-\lambda a$ and therefore $|\lambda|=||a||$. Therefore it suffices to show that $a$ or $\lambda$ equals zero.

Can anybody help me with the rest? In every book or paper a read, there is no proof of this.

Roland
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If $\lambda=0$, take $b=a^*$ and you get $aa^*=0$ and hence $a=0$.

If $\lambda\ne0$, then $ab+\lambda b$ can be read as $cb=b$, where $c=-a/\lambda$. Taking adjoints it can also be seen as $bd=d$, where $d=-a^*/\overline\lambda$. This implies that $-a/\lambda=1$, so $a=-\lambda$. This is in contradiction by construction of the unitization, where the only element of $a$ that is a scalar is zero.

Martin Argerami
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  • What do you mean by the only element that is a scalar is zero? For example if $A$ itself is the unitalization of some $C^*$-algebra $B$, i.e. $A$=$B^+$. Then $a$ can be of the form $(0,\lambda)$ and $(0,\lambda)$ acts on $B^+$ the same way as the scalar $\lambda\in \mathbb{C}$ would. – Roland Feb 21 '23 at 09:35
  • If $A$ is unital to begin with, then what you defined is not a norm. Consider the element $(1_A,-1)$. Then $$|1_Ab+(-1)b|=|b-b|=0$$ for every $b$, hence $|(1_A,-1)|=0$. – Martin Argerami Feb 21 '23 at 11:44
  • Ah okay, I see. Maybe can you help me with the following question. If we want to do it uniformly, like here https://mathoverflow.net/questions/210025/unitization-process-of-unital-and-non-unital-c-algebras#comment520923_210025 then I should be able to proof that the operator norm as in the linked question equals the norm above if $A$ is non-unital. This comes down to the question, why there exists an $b\in A, ||b||\leq 1$ such that $||ab+\lambda b||\geq |\lambda|$. Or should i rather open a new question for this or edit this question? – Roland Feb 21 '23 at 12:57
  • I don't immediately see how to prove that inequality directly (not saying there isn't an easy proof). But if you put on $A\oplus\mathbb C$ the norm $$|(a,\lambda)|=\max{|\lambda|,\sup{|ab+\lambda b|:\ |b|≤1},$$ it is not too hard to check that this a C$^$-norm. As the norm on a C$^$-algebra is unique and $ \sup{|ab+\lambda b|:\ |b|≤1}$ is a C$^*$-norm when $A$ is non-unital, they have to agree. Which in particular implies your inequality in the case where $A$ is not unital. – Martin Argerami Feb 21 '23 at 13:53