Unitization of a $C^{*}$-algebra - completeness of the constructed norm I have a question about the unitization of $C^{*}$-algebras. More precisely, a question about the proof of the following statement:

If $A$ is a (possibly unital) $C^{*}$-algebra, then there is a unique $C^{*}$-norm on the unitization $A_{1}:=A\oplus\mathbb{C}$.

Here $A_{1}$ is endowed with the component wise linear operations. The multiplication is given by $(a\oplus\lambda)(b\oplus\mu):=ab+\lambda b+\mu a\oplus\lambda\mu$ and the involution by $(a\oplus\lambda)^{*}:=a^{*}\oplus\bar{\lambda}$.
Let us restrict to the common proof of the non-unital case. Consider the bounded linear operator $L_{a}\colon A\to A$ defined by $L_{a}(b):=ab$. Using the fact that $A$ is non-unital, one can prove that $$A_{1}\to B(A),\qquad a\oplus\lambda\mapsto L_{a}+\lambda I$$ is injective. So $\|a\oplus\lambda\|:=\|L_{a}+\lambda I\|$ is a well-defined norm on $A_{1}$. With some effort one can show that it is submultiplicative and that it satisfies the $C^{*}$-identity.
In order to conclude existence (and automatically uniqueness), one also needs that this norm is complete. The proofs I found ignore completeness of this norm.
Is completeness of the above norm meant to be obvious? I really don't see it. Any help would be greatly appreciated!
 A: If you have a Cauchy sequence $(a_n \oplus \lambda_n) \subset A_1$, then you have two Cauchy sequences $(a_n)$ and $(\lambda_n)$ in $A$ and $\mathbb{C}$ respectively.
Indeed,
$$
||a_n - a_m|| = ||a_n \oplus 0 - a_m \oplus 0|| = ||P_A (a_n \oplus \lambda_n - a_m \oplus \lambda_m)|| \leq ||P_A||||a_n \oplus \lambda_n - a_m \oplus \lambda_m||
$$
where $P_A$ is a projector on the first summand. $P_A$ is bounded because it is continuous at zero: if $L_{b_n} + I\beta_n \to 0$, then $L_{b_n} \to 0$ since $A$ is strictly nonunital.
Then by triangle inequality $$ |\lambda_n - \lambda_m| \leq ||a_n - a_m|| + ||a_n \oplus \lambda_n - a_m \oplus \lambda_m||.$$
$A$ and $\mathbb{C}$ are Banach, so you can say $a_n \to a$ and $\lambda_n \to \lambda$. Then $a \oplus \lambda \in A_1$ is the limit of the original sequence:
$$
||a_n \oplus \lambda_n - a \oplus \lambda|| = ||(a_n - a) \oplus (\lambda_n - \lambda)|| \leq ||a_n - a|| + ||\lambda_n - \lambda|| \to 0.
$$
The triangle inequality follows from $||a\oplus\lambda|| = ||L_a + I\lambda||$. Also identities $||L_a|| = ||a||$ and $||I\lambda|| = |\lambda|$ were used.
