Showing that $\lim \sup\limits_{\gamma} \left \|a - a e_{\gamma} \right \| \leq \|a - b\|$ for all $a \in A$ and $b \in J.$

Let $J$ be a closed ideal in a $C^{\ast}$-algebra $A.$ Let $\left (e_{\gamma} \right )_{\gamma}$ be a (bounded) approximate identity in $J.$ Then for $a \in A,$ one has $\left \|a + J \right \| = \lim\limits_{\gamma} \left \|a - a e_{\gamma} \right \|.$

Proof $:$ For any $b \in J,$ we have $\lim\limits_{\gamma} \left \|b - b e_{\gamma} \right \| = 0.$ Since

$$\left \|a - a e_{\gamma} \right \| \leq \left \|b - b e_{\gamma} \right \| + \|a - b\| + \left \|a e_{\gamma} - b e_{\gamma} \right \| \leq \left \|b - b e_{\gamma} \right \| + \|a - b\| \left (1 + \left \|e_{\gamma} \right \| \right ),$$

it follows that

$$\left \|a + J \right \| \leq \lim \inf\limits_{\gamma} \left \|a - a e_{\gamma} \right \| \leq \color{red} {\lim \sup\limits_{\gamma} \left \|a - a e_{\gamma} \right \| \leq \|a - b\|}$$

for all $b \in J.$ Therefore one has $\left \|a + J \right \| = \lim\limits_{\gamma} \left \|a - a e_{\gamma} \right \|.$

This completes the proof.

I don't understand the inequality in the proof highlighted in red. From the first sequence of inequalities we have

$$\lim \sup\limits_{\gamma} \left \|a - a e_{\gamma} \right \| \leq 2 \|a - b\|$$ for all $b \in J$ since $\lim\limits_{\gamma} \left \|b - b e_{\gamma} \right \| = 0$ for all $b \in J$ and $\left \|e_{\gamma} \right \| \leq 1$ for all $\gamma.$ But I have no idea as to how to bound the limit superior by $\|a - b\|.$ Any suggestion regarding this inequality will be greatly welcomed.

Thanks in advance.

Rabin.

Solution 1:

Yeah, the red inequality is true, but the inequality above it does not seem like a proof of this fact to me.

This argument is around in the literature. The key is to pass to the unitization $\tilde{A}$ of $A$ (if necessary) and make note of the fact that $\|1_{\tilde{A}} - e_\gamma\| \leq 1$ (note its important that we've taken positive contractions in our approximate unit). Now you get \begin{align} \|a - ae_\gamma\| &= \|a - b + b - be_\gamma + be_\gamma- ae_\gamma\| \\ &\leq \|a - b -ae_\gamma + be_\gamma\| + \|b- be_\gamma\| \\ &= \|(1_{\tilde{A}} - e_\gamma)(a-b)\| + \|b - be_\gamma\| \\ &\leq \|a - b\| + \|b- be_\gamma\|. \end{align} Now one can take limsups and you get your inequality.