Unitisation of an algebra: Is the norm $\text{max}\lbrace\|a\|_A,|\lambda|\rbrace$ submultiplicative? I feel as if I may have missed something really obvious here. It says in my notes that if $A$ is an algebra (over $\mathbb{C}$), then its unitisation $A^1:=A\times\mathbb{C}$ is a unital algebra with respect to the multiplication $$(a,\lambda)(b,\mu):=(ab+\lambda b+\mu a,\lambda\mu)$$ and that if $A$ is a normed algebra (with submultiplicative norm $\|\cdot\|_A$), this becomes a normed algebra with respect to the norm $$\|(a,\lambda)\|_\infty:=\text{max}\lbrace\|a\|_A,|\lambda|\rbrace$$
To prove that $\|\cdot\|_\infty$ is submultiplicative, we have to prove that $$\text{max}\lbrace\|ab+\lambda b+\mu a\|_A,|\lambda|\,|\mu|\rbrace\leq\text{max}\lbrace\|a\|_A,|\lambda|\rbrace\text{max}\lbrace\|b\|_A,|\mu|\rbrace.$$
If $\|ab+\lambda b+\mu a\|_A\leq|\lambda|\,|\mu|$, then this is obvious, but what about if $\|ab+\lambda b+\mu a\|_A>|\lambda|\,|\mu|$?
EDIT: Just to clarify, I'm not 100% sure if this result is true or not, it just seems to implicitly suggest that it is true in notes I am reading.
 A: You norm is not submultiplicative. For example, in the C$^*$-algebra setting if $A=C_0(\mathbb R)$ with the infinity norm, take $f(t)=\frac2{1+t^2}$, $\lambda=1$. Then
$$
\|(f,\lambda)\|_\infty=2,\qquad \|(f,\lambda)^2\|_\infty=\max\{\|f^2+2\lambda f\|,|\lambda|^2\}=8.
$$
So $\|(f,\lambda)^2\|_\infty>\|(f,\lambda)\|^2_\infty$.
To do the unitization and get a C$^*$-algebra, the norm you need to use is
$$
\|(a,\lambda)\|=\sup\{\|ab+\lambda b\|:\ \|b\|=1\}. 
$$
This new norm is a norm if $A$ is non-unital: if $\|(a,\lambda)\|=0$, then $ab=-\lambda b$ for all $b\in A$. If $\lambda=0$, then $a=0$. If $\lambda\ne0$, you have that $ab=-\lambda b$ for all $b\in A$, which may be written as $\big(-\tfrac1\lambda a\big)b=b$ for all $b\in A$. This is a contradiction, as $A$ is non-unital. Then $\lambda=0$ and so $a=0$.
And submultiplicative: if $\|c\|=1$,
\begin{align}
\|(ab+\lambda b+\mu a+\lambda\mu)c\|
&=\|a(bc+\mu c)+\lambda(bc+\mu c)\|\\[0.3cm]
&=\bigg\|a\bigg(\frac{bc+\mu c}{\|bc+\mu c\|}\bigg)+\lambda\bigg(\frac{bc+\mu c}{\|bc+\mu c\|}\bigg)\bigg\|\,\|bc+\mu c\|\\[0.3cm]
&\leq \|(a,\lambda)\|\,\|bc+\mu c\|\\[0.3cm]
&\leq\|(a,\lambda)\|\,\|(b,\mu)\|.
\end{align}
Taking supremum over $c$ with $\|c\|=1$ we get
$$
\|(a,\lambda)(b,\mu)\|\leq\|(a,\lambda)\|\,\|(b,\mu)\|. 
$$
