Norm product inequality for unitisation of a $C^*$-algebra

Question

Let $A$ be a $C^*$-algebra without a unit. Define $\widetilde{A}=\{(\alpha,a):\alpha \in \mathbb{C}, \;a\in A\}$ equipped with componentwise addition and scalar multiplication. Vector multiplication is defined by $(\alpha,a)(\beta,b)=(\alpha \beta, \alpha b +\beta a+ab)$. Then $\widetilde{A}$ forms a unital complex algebra, and we equip it with the norm $\|(\alpha, a)\|=\sup_{b\in A, \;\|b\|=1} \|\alpha b+ab\|$. I could check that this is a norm, and that it has the $C^*$-algebra property, assuming that the norm product inequality $\|(\alpha,a)(\beta,b)\|\leq \|(\alpha,a)\|\|(\beta,b)\|$ holds. However, I have not been able to show that the norm product inequality holds. I suppose one should use the $C^*$-algebra norm property in $A$, and in the RHS part, maybe one should use adjoints somehow, but I have not yet managed. This is from Operator Algebras and Quantum Statistical Mechanics vol. I by Bratteli and Robinson, where this is left as an "easy" verification to the reader on page 22, line 1 of the proof of Prop. 2.1.5 (which basically says that $\widetilde{A}$ is a unital $C^*$-algebra). I would be grateful for any hints! Thank you.

Answer 1

There is a trick to see this quickly. Here is a hint:

Try to think of elements in $\widetilde{A}$ as elements of $B(A)$, the bounded operators on $A$. You know that the submultiplicativity of the norm is satisfied for $B(A)$. If you do not want a complete answer, stop reading here.

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Here are more details. Define the map $$\Phi: \widetilde{A}\to B(A): (\alpha,a)\mapsto L_a+\alpha\operatorname{id}_A$$ where $L_a: A \to A$ is left multiplication: $L_a(b) = ab$.

By definition of the norm on $\widetilde{A}$, $\Phi$ is an isometry. Moreover, it is also easily verified that $\Phi$ preserves the multiplication. Hence, $$\|(\alpha,a)(\beta,b)\|= \|\Phi(\alpha,a)\Phi(\beta,b)\| \le \|\Phi(\alpha,a)\|\|\Phi(\beta,b)\|= \|(\alpha,a)\|\|(\beta,b)\|$$ where we made use of the fact that the norm on $B(A)$ is submultiplicative.