Let $\mathcal A$ be a non-unital complex $C^\ast$algebra with norm $\|\cdot\|$.
I'm trying to proof that given $a \in \mathcal A$ and $\lambda \in \mathbb C$ $$ \sup \{ \|ax + \lambda x \|: x \in \mathcal A, \, \|x\| \leq 1 \} \geq |\lambda|.$$
I'm looking for a hint to solve this question.
Help?
You can represent $\mathcal A$ as an algebra of bounded operators on $\mathcal A$ using left multiplication: $\varphi:\mathcal A\to B(\mathcal A)$, $\varphi(a)x=ax$. This is an isometric representation by the C*-identity $\left(\|\varphi(a)\|\leq \|a\|\right.$ would be true in any Banach algebra, and the equality holds because $\|\varphi(a)a^*\|=\|a\|\|a^*\|\left.\right)$. Let $I\in B(\mathcal A)$ be the identity operator, and let $\overset{\sim}{\mathcal{A}}=\varphi(\mathcal A)+\mathbb CI$, which is a Banach algebra unitization of $\varphi(\mathcal A)$, having $\varphi(\mathcal A)$ as a $1$-co-dimensional closed two sided ideal. If $a\in \mathcal A$ and $\lambda\in \mathbb C$, then $$\|\varphi(a)+\lambda I\|=\sup \{ \|ax + \lambda x \|: x \in \mathcal A, \, \|x\| \leq 1 \}. $$
On the other hand, if $\pi: \overset{\sim}{\mathcal{A}}\to \overset{\sim}{\mathcal{A}}/\varphi(\mathcal A)\cong\mathbb C$ is the quotient map, then $\|\pi\|=1$, hence
$$\|\varphi(a)+\lambda I\|\geq \|\pi(\varphi(a)+\lambda I)\|=|\lambda|.$$
OK, here's an easier way if you take the unitization for granted. Note that $\sup \{ \|ax + \lambda x \|: x \in \mathcal A, \, \|x\| \leq 1 \}$ is the norm of $a+\lambda 1$ in the C*-unitization of $A$. For a contradiciton, suppose $\|a+\lambda 1\|<|\lambda|$. Then $\left\|\frac1{\lambda}a+1\right\|<1$, hence $-\frac{1}{\lambda}a$ is invertible, which is absurd because $a$ is in the proper ideal $A$ of the unitization.
The more complicated method above has little advantage for proving this result, but it describes how the C*-unitization gets its norm, which is the context that makes it easier to see the easier way (even though I didn't at first).
