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The following is about a proof in Bratteli Robinson vol 1. Let $\mathcal{A}$ be some C*-algebra. Show that $$\mathcal{B}=\{(A,\alpha)~|~A\in\mathcal{A}, \alpha\in\mathbb{C}\}$$ together with the norm $$\lVert(A,\alpha)\lVert=\sup_{B\in\mathcal{A}, \lVert B\lVert\leq 1} \lVert AB+\alpha B \lVert$$ and the operations $$(A,\alpha)(B,\beta)=(AB+\alpha B+\beta A, \alpha \beta)$$ $$\lambda(A,\alpha)=(\lambda A, \lambda\alpha)$$ $$(A,\alpha)+(B,\beta)=(A+B, \alpha+ \beta)$$ $$(A,\alpha)^*=(A^*,\bar{\alpha})$$ is again a C*-algebra. I understood the proof in general, but I don't see why the norm product inequality ($\lVert (A,\alpha)(B,\beta)\lVert\leq\lVert (A,\alpha)\lVert \lVert(B,\beta)\lVert$) holds. Any hints?

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Hint: Plug $\dfrac{BC + \beta C}{\Vert BC + \beta C\Vert}$ into the norm computation of $(A,\alpha)$ and compare to plugging in $C$ to the norm computation of $(A,\alpha)(B,\beta)$.

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