Consider the Cuntz algebra $\mathcal{O}_2$ in two generators $s_1,s_2$. Let $\lambda:\mathcal{O}_2\to \mathcal{O}_2$ given by $$\lambda(a)=s_1as_1^*+s_2as_2^*.$$ Let $(a_n)_{n\in\mathbb{N}}\subset \mathcal{O}_2$ be a bounded sequence.
In going through a proof in which one uses the following fact:
$\lim\limits_{n\to\infty}\|a_nb-ba_n\|=0$ if and only if $\lim\limits_{n\to\infty}\|\lambda(a_n)-a_n\|=0$ $\;\star$.
it's only calculation in both directions, but I don't get this..
I tried to prove both directions using the following facts: Set $u=\sum\limits_{i=1}^2\lambda(s_i)s_i^*\in\mathcal{O}_2$. $u$ is an unitary element (with $u=u^*)$. Let $\lambda_u:\mathcal{O}_2\to \mathcal{O}_2$ given by $\lambda_u(s_j)=us_j.$ One has $\lambda=\lambda_u$. However, my calculations were pointless.
Why is $\star$ correct? I appreciate your help.
If $a_nb-ba_n\to0$ for all $b\in \mathcal O_2$, then write (using $s_1s_1^*+s_2s_2^*=I$) $$ \lambda(a_n)-a_n=(s_1a_n-a_ns_1)s_1^*+(s_2a_n-a_ns_2)s_2^*\to0. $$ Conversely, if $\lambda(a_n)-a_n\to0$, this is $$ s_1a_ns_1^*+s_2a_ns_2^*-a_n\to0. $$ Using that $s_2^*s_1=0$ we get by multiplying by $s_1$ on the right (recall that $s_1,s_2$ are isometries) that $$s_1a_n-a_ns_1\to0.$$Multiplying by $s_1^*$ on the left, $$a_ns_1^*-s_1^*a_n\to0.$$ And similarly with $s_2$. Now we can use this to show that $$a_nb-ba_n\to0$$ for any $b$ that is a polynomial in $s_1,s_1^*,s_2,s_2^*$. And now we can pass to any $b$ by a density argument.