Let $\mathbb{F}\_2$ denotes the free group generated by a,b, denote this group by $G$. Then consider the von Neumann algebra $L(G)$ generated by the family 
$\{L\_{x\_g} : g \in G\}$, here, with $g \in G$, we denote by $x\_g$ the function on $G$ that takes the value 1 at g and 0 at other elements of $G$. Then, note that we have the following relations:


$(L\_{x\_g})^\*=L\_{(x\_g)^{-1}} , L\_{x\_g}L\_{x\_h}=L\_{x\_g\*x\_h}=L\_{x\_{gh}}$,
then, for any $ A \in L(G),$we can set $ A=\sum\_{g \in g}\mu\_g L\_{x\_g},$ 
with $\mu\_g \in \mathbb{C}.$


When we calculate $ ||Ax\_{h}||^2 $, we find that $ \sum\_{g \in G}|\mu\_g|^2 < \infty.$


Then, is this condition sufficient for $ A \in L(G) $? Or some stronger condition is necessary? Like $ \sum\_{g \in G}|\mu\_g| < \infty,$ or something else?



Restriction on the coefficients for an operator in the free group factor $ L(\mathbb{F}_2) $
It is more common to just write $L\_g$ for $L\_{x\_g}$. As $L(G)$ admits a finite trace, there is a natural injective map $L(G)$ into $\ell^2(G)$-- this is your map $A \mapsto (\mu\_g)$. It is absolutely not true that this map surjects (Open Mapping Theorem). It is obviously sufficient that $(\mu\_g)\in\ell^1(G)$ for there to be some $A$ giving rise to $(\mu\_g)$.


With $G=\mathbb F\_2$, one can say a bit more. For example, Haagerup showed in:  

Haagerup, Uffe  

An example of a nonnuclear C∗-algebra, which has the metric approximation property.  

Invent. Math. 50 (1978/79), no. 3, 279–293.   

See Lemma 1.4 that if $f$ is a function of finite support, then denoting $f\_n$ the function which agrees with $f$ on the collection of words of reduced length $n$, and is zero elsewhere, we have that there is $A\in L(G)$ inducing $f$, with $\|A\| \leq \sum\_{n\geq 0} (n+1) \|f\_n\|\_2$. From this, it's easy to construct functions not in $\ell^1(G)$, but which are nonetheless induced by members of $L(G)$.


