Let $A$ be a nonunital $C^{\ast}$-algebra and let $B$ be a unital $C^{\ast}$-algebra such that $A \subseteq B.$ Then $A + \mathbb C 1_{B} : = \left \{a + \lambda 1_{B}\ |\ a \in A, \lambda \in \mathbb C \right \}$ is a unital $C^{\ast}$-subalgebra of $B$ and is isometrically isomorphic to the unitization $A^+$ of $A,$ where $A^+ = A \oplus \mathbb C$ equipped with the componentwise addition and scalar multiplication by complex numbers and the product, involution and norm are defined respectively in the following way $:$
For any $(a,\lambda), (b, \mu) \in A^+$
$(1)\ (a,\lambda) \cdot (b,\mu) = (ab + \lambda b + \mu a, \lambda \mu),$
$(2)\ ((a, \lambda))^{\ast} = \left (a^{\ast}, \overline {\lambda} \right ),$
$(3)\ \|(a,\lambda)\|_{1} = \sup\limits_{\substack {b \in A \\ \|b\| \leq 1}} \|ab + \lambda b\|.$
I have managed to prove almost everything except the fact that the map $a + \lambda 1_{B} \longmapsto (a,\lambda)$ is an isometry. For this I need to show that $$\|a + \lambda 1_{B}\| = \sup\limits_{\substack {b \in A \\ \|b\| \leq 1}} \|ab + \lambda b\|.$$ It is clear that RHS $\leq$ LHS. How to prove the other way round? Any help or suggestion in this regard would be warmly appreciated.
Thanks a bunch.
Let's start with a lemma:
Lemma: If $\phi: A \to B$ is a $*$-isomorphism between $C^*$-algebras, then $\phi$ is isometric.
Proof lemma: Define a norm $|||a|||:= \|\phi(a)\|_B$ on $A$. Routine verification shows that this is a $C^*$-norm on $A$. By uniqueness of the $C^*$-norm, we obtain $\|a\| = \|\phi(a)\|$ for all $a \in A$, so $\phi$ is isometric. $\quad \square$
Consider the map $$\Psi: A^+ \to A + \mathbb{C}1_B: (a, \lambda) \mapsto a + \lambda 1_B.$$ This is a $*$-homomorphism, because $$\Psi((x, \lambda)(y, \mu)) = \Psi (xy + \lambda y + \mu x, \lambda \mu) = xy + \lambda y + \mu x + \lambda \mu 1_B$$ and $$\Psi(x,\lambda)\Psi(y,\mu) = (x + \lambda 1_B) (y + \mu 1_B) = xy + \mu x + \lambda y + \lambda \mu 1_B.$$ Clearly, this map is also injective, because if $a+ \lambda 1_B = 0$, then $a= -\lambda 1_B$, which implies $\lambda = 0$ since $1_B \notin A$ (or $A$ would be unital) and thus also $a=0$.
It follows that $\Psi$ is a $*$-isomorphism. In particular, it is isometric, and the desired equality of norms follows from this.