On a distance formula for elements of $C^*$-algebras and closed left ideals Let $A$ be a unital $C^*$-algebra and let $\{p_n\}$ be an increasing sequence of projections in $A$. Set
$$L:=\overline{\bigcup_{n}Ap_n} $$
and observe that $L$ is a closed left ideal in $A$. Note also that $\{p_n\}$ is an approximate unit for $L\cap L^*$ and a left (resp. right) approximate unit for $L$ (resp. $L^*$).
Through some technicalities, I was able to show that $L+L^*$ is a closed (self-adjoint of course) subspace of $A$ (I showed that  the canonical map $L/(L\cap L^*)\to (L+L^*)/L^* $ is isometric, hence $(L+L^*)/L^*$ is complete and thus $L+L^*$ is complete, this is a 2 out of 3 property).
My question is this: can we show that, for all $x\in A$, we have
$$\|x+(L+L^*)\|=\lim_{n\to\infty}\|(1-p_n)x(1-p_n)\|\;?$$
I have been trying this for days with no success. Of course $\|x+(L+L^*)\|\leq\|(1-p_n)x(1-p_n)\|$ for all $n$. Also, if $x\in L+L^*$ then this limit is indeed zero, this is easily seen. I know that when $J\triangleleft A$ is a closed, two-sided ideal with approximate unit $\{e_\lambda\}$ then $\|x+J\|=\lim_{\lambda}\|x(1-e_\lambda)\|=\lim_\lambda\|(1-e_\lambda)x\|$, but I was not able to apply this here.
It would also be interesting to ask if this question is true even more generally, for any closed left ideal and any left approximate unit (instead of the increasing sequence of projections).
 A: Given $a,b \in L$, $\varepsilon > 0$, we can show that there is some $N \in \mathbb{N}$ such that whenever $n \geq N$, we have
$$ \|(1-p_n)x(1-p_n)\| < \|x - a - b^*\| + 2\varepsilon. $$
This would mean that $\limsup_n \|(1-p_n)x(1-p_n)\| \leq \|x - a - b^*\|$ (2 is there out of laziness for the approximations below). Consequently
$$ \|x + (L + L^*)\| \leq \liminf_n\|(1-p_n)x(1-p_n)\| \leq \limsup_n\|(1-p_n)x(1-p_n)\| \leq \|x - a - b^*\| $$
whenever $a,b \in L$. So taking the infimum over all $a,b \in L$ gives the equality you want (along with the fact that the limit exists, which a priori wasn't clear to me).
So lets do that approximation. We can approximate $a \approx_\varepsilon a'p_N$ and $b^* \approx_\varepsilon p_Nb'$ where $a',b' \in A$ and $N \in \mathbb{N}$. Then whenever $n \geq N$, $(1 - p_n)p_N = p_N(1 - p_n) = 0$ since the projections are increasing and so
$$ (1 - p_n)a'p_N(1-p_n) = (1-p_n)p_Nb'(1-p_n) = 0. $$
Consequently for $n \geq N$,
$$ \|(1-p_n)x(1-p_n)\| = \|(1-p_n)(x - a'p_N - p_Nb')(1 - p_n)\| \leq \|x - a'p_N - p_Nb'\|. \tag{1} $$
By going back to our approximations of $a,b$, we have
$$ \|x - a'p_N - p_Nb'\| \leq \|x - a - b^*\| + \|a - a'p_N\| + \|b^* - p_Nb'\| < \|x - a - b^*\| + 2\varepsilon. \tag{2}$$
Combining (1) and (2), we've got what we wanted at the beginning.
