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This is a follow up of this question. Let $I$ be closed left ideal of $C^*$-algebra $A$.

Assume we are given a sequence of left $A$-module morphisms $R_n:I\to A$ with $\sum_n \Vert R_n\Vert<\infty$ and a sequence $\{b_n\}_{n\in\mathbb{N}}$ in the unit ball of $I$ with the property $$ \sum_n R_n(a)b_n=a\tag{1} $$ for all $a\in I$. Is it true that $I=Ap$ for some $p\in I$.

If the sum in $(1)$ would be finite the result would be true, since all finitely generated left ideals of $C^*$-algebras are principal. Clearly, this is also true for $A$ being commutative, because we can set $p=\sum_n R_n(b_n)$.

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Yes, $I=Ap$ for some projection $p$. More generally, suppose that a closed left ideal $I$ contains a sequence $c_1,c_2,\dots$ such that $ \sum_{i=1}^\infty \|c_i\|<\infty $ and for any $x\in I$ there exist $x_1,x_2,\dots$ such that $\|x_i\|\leq \|x\|$ and $$ \sum_{i=1}^\infty x_ic_i=x. $$ Then $I$ is generated by a projection. (In the question, $c_i=\|R_i\|b_i$ and $x_i=R_i(x)/\|R_i\|$.)

Here is a proof: Let $c=\sum_{i=1}^\infty (c_i^*c_i)^{\frac 1 2}$ (which converges). Since $(c_i^*c_i)^{\frac 1 2}\leq c$, we have $c_i=d_ic$ for some contraction $d_i\in A^{**}$. Choose $\epsilon>0$ small enough ($\epsilon=\frac 1 6$ will suffice). Choose $N$ such that $\sum_{i>N}^\infty \|c_i\|<\epsilon$. Since $c^{1/k}\in I$ for all $k=1,2,\dots$, we must have $$ c^{\frac 1 k} =\sum_{i=1}^N x_ic_i+\sum_{i>N} x_ic_i=Xc^{\frac 1 2}+\Delta, $$ where $X=\sum_{i=1}^Nx_id_i$ and $\|\Delta\|\leq \sum_{i>N} \|x_i\|\|c_i\| \leq \epsilon \|c\|^{1/k}$. Then, \begin{align*} c^{\frac 2 k} &=c^{\frac 1 2}X^*Xc^{\frac 1 2}+ \Delta'\leq \|X\|^2c+\|\Delta'\|\cdot 1. \end{align*} The norms of $X$ and $\Delta'$ can be estimated: $\|X\|\leq \sum_{i=1}^N\|x_id_i\|\leq N\|c\|^{1/k}$ and $\|\Delta'\|\leq 5\epsilon \|c\|^{2/k}$ (probably not a sharp estimate; I'm skipping the details). So $$ c^{\frac 2 k}\leq N^2\|c\|^{\frac 2 k}\cdot c+ 5\epsilon \|c\|^{\frac 2 k}\cdot 1. $$ This inequality must hold in $C^*(c,1)$ for all $k$. For $k$ large enough, $\|c\|^{2/k}$ is close to 1, so the right hand side is bounded by a linear function in $c$ which at 0 is less than $6\epsilon$. So $0$ can only be an isolated point of the spectrum of $c$. Now, with $p\in I$ the support projection of $c$, we have $c_ip=d_icp=d_ic=c_i$ for all $i$, and so $xp=x$ for all $x\in I$.

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  • $\begingroup$ Thank you Leonel! I was totally missing the trick with usage of $k$, and stuck with inequality of type $c^{1/2}\leq Ac+B$ $\endgroup$ – Norbert Sep 6 '14 at 0:37

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