I recently discussed unitization of a non-unital $C^\ast$-algebra. I proved some facts and so on (which I afterwards found on the internet).
Anyway let's consider the following $C^\ast$-algebra. That is, the
$C^\ast$-algebra $\mathcal{A}$ with norm $\|\cdot\|$. Let $\tilde{\mathcal{A}}=\mathcal{A}\oplus \mathbb{C}$ as a vector space. We endow it with multiplication and involution,
$$(a,\lambda)\cdot (b,\mu):=(ab+\lambda b+\mu a,\lambda \mu)$$
and
$$(a,\lambda)^\ast:=(a^\ast,\bar{\lambda})$$
As mentioned I did some proofs of this and discussed with fellow members of this site. However I can see people discuss norm of this beast, let me clarify this.
We consider $\omega:\mathcal{A}\rightarrow\tilde{\mathcal{A}}$ by the map $\omega(a):=(a,0)$. This is indeed a two sided ideal $\omega(\mathcal{A})$ and of course also a $\ast$-homomorphism. Now we will suppress the inclusion $\omega$ and think of $\mathcal{A}$ as an ideal in $\mathcal{A}$.
I claim: Take $x\in\tilde{\mathcal{A}}$ then we consider the linear map $\tilde{L}_x:\tilde{\mathcal{A}}\rightarrow \tilde{\mathcal{A}}$ by $y\mapsto xy$ restricts to a map, $L_x:\mathcal{A}\rightarrow \mathcal{A}$ which is bounded with $\|L_x\|_\infty \leq \|x\|_1$.
I want to show the above claim. Here we know that the one-norm is given by $x=(a,\lambda)\in \tilde{\mathcal{A}}$ by $\|x\|_1=\|a\|+|\lambda|$ and of course $\|\cdot\|_\infty$ denote the operator norm.
But how can we construct such a proof of this claim? I've tried to read more about this kind of unitization and norms on the following link however I don't think they do the claim that I give.
http://www.math.nagoya-u.ac.jp/~richard/teaching/s2020/Terasawa.pdf
So, how would one do it?
Best Answer
Take $x=(a,\lambda)\in\bar{\mathcal A}.$ For any $b\in\mathcal A,$ $$\|L_x(b)\|=\|ab+\lambda b\|$$ (by definition of $L_x$) and $$\begin{align}\|ab+\lambda b\|&\le\|ab\|+\|\lambda b\|\\&\le\|a\|\|b\|+|\lambda|\|b\|\\ &=\|x\|_1\|b\|\end{align} $$ hence $$\forall b\in\mathcal A\quad\|L_x(b)\|\le\|x\|_1\|b\|,$$ which is exactly the definition of $$\|L_x\|_\infty\le\|x\|_1.$$
Edit: This answers the initial post. Another question has been added meanwhile, which should however be the object of another post.