In the proof of proposition 1.5 of Takesaki's Theory of Operator Algebras: Part 1, after defining defining a norm on the unitisation $\tilde{A}$ of a non-unital C-algebra $A$ making it a Banach algebra (the norm of $x \in \tilde{A}$ is the norm of the map $L_x: A \to A, a \mapsto xa$), he then proves that the C-identity holds for this norm. The proof of this starts by noting that for each $x \in \tilde{A}$ and $\varepsilon > 0$, there exists $y \in A$ such that $||xy|| \geq (1 - \varepsilon) ||x||$ with $||y|| \leq 1$. I can prove that there is a $y \in A$ satisfying the inequality involving $\varepsilon$, but not necessarily with norm not exceeding $1$.
If $x = 0$ or $\varepsilon \geq 1$, we may choose $y = 0$. If $x \neq 0$ and $\epsilon < 1$, then $||x|| = ||L_x|| \neq 0$, so there exists $y \in A$ such that $xy \neq 0$. Given $\lambda \in \mathbb{C}$, we have that
\begin{align} ||x(\lambda y)|| &= |\lambda| ||xy|| \\ &\geq (1 - \varepsilon) ||x|| \end{align} if we choose $\lambda$ satisfying $|\lambda| \geq \frac{(1-\varepsilon)||x||}{||xy||}$. In the spirit of keeping the norm of $\lambda y$ as small as possible, let's choose $\lambda$ such that $|\lambda| = \frac{(1-\varepsilon)||x||}{||xy||}$. From here I can't show that $||\lambda y|| \leq 1$. The only thing I seem to be able to show is that since $||xy|| \leq ||x|| \ ||y||$ we have that \begin{align} ||\lambda y|| = \frac{(1-\varepsilon) ||x|| \ ||y||}{||xy|| } \geq 1- \varepsilon \end{align}
My question is:
How can we find a suitable $y$ with $||y|| \leq 1$?
Takesaki leaves out a lot of details, but filling them in always makes a good exercise.
Now, let $\epsilon > 0$ and fix $x \in \widetilde{A}$. By the equality $$\|x\| = \|L_x\|= \sup_{y \in A, \|y\| \le 1} \|xy\|$$ proven earlier in Takesaki (?), there exists $y \in A$ in the unit ball of $A$ such that $$\|x\| = \|L_x\| \le \|L_x y\| + \epsilon \|x\| = \|xy\| + \epsilon \|x\|.$$ Rearranging, $(1-\epsilon)\|x\| \le \|xy\|.$