Proving $||AB|| \le ||A||||B||$ for $\mathbb{C}\oplus \mathcal{U}$ Let $\mathcal{U}$ be a $C^{*}$-algebra without a unit and consider the algebra $\mathbb{C}\oplus \mathcal{U}$ formed by the ordered pairs $(\alpha, A)$, $\alpha \in \mathbb{C}$ and $A \in \mathcal{U}$ with the operations $(\alpha, A) + (\beta, B) := (\alpha+\beta, A+B)$ and $(\alpha,A)(\beta, B) := (\alpha \beta, \alpha B + \beta A + AB)$. I'm trying to prove that $||(\alpha, A)(\beta, B)|| \le ||(\alpha,A)||||(\beta, B)||$, where $||(\alpha, A)|| := \sup_{||B||=1, B \in \mathcal{U}}||\alpha B + AB||$. I know that:
$$||(\alpha, A)(\beta, B)|| = ||(\alpha\beta, \alpha B + \beta A + AB)|| = \sup_{||C||=1}||\alpha\beta C +\alpha BC+\beta AC+ABC||$$
but I'm stuck there. Can anyone help me? Thanks!

We have :
\begin{align} 
\|(\alpha,A)(\beta,B)\| &= \|(\alpha\beta,\alpha B+\beta A + AB)\| \\
&= \sup_{\|C\| = 1 } \| \alpha\beta C + (\alpha B  + \beta A + AB)C \|
\end{align}
For $C\in\mathcal U$ with $\|C\|=1$, we have :
\begin{align}
\alpha\beta C + (\alpha B  + \beta A + AB)C = \alpha (\beta C+ BC) + A(\beta C + BC)
\end{align}
If $\beta C+ BC\neq 0$, we have :
\begin{align}
\| \alpha\beta C + (\alpha B  + \beta A + AB)C \| &= \|\beta C +BC\| \cdot \left\| \alpha \frac{\beta C +BC}{\|\beta C +BC\|} + A \frac{\beta C +BC}{\|\beta C +BC\|}\right\| \\
&\leq \|(\beta,B)\|\|(\alpha,A)\|
\end{align}
This also holds if $\beta C+ BC = 0$. By taking the supremum over $C$, we get :
$$\|(\alpha,A)(\beta,B)\| \leq \|(\alpha,A)\|\|(\beta,B)\|$$
