Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.



> 
> For $A= (A\_1,\cdots,A\_d)\in\mathcal{L}(E)^d$ (not necessary to be commuting). Why
>  $$\lim\_{n\to+\infty}\bigg(\bigg\|\sum\_{f\in F(n,d)} A\_{f}^\* A\_{f}\bigg\|^{\frac{1}{2n}} \bigg)\;\text{exists}?,$$
>  where $F(n,d)$ denotes the set of all functions from $\{1,\cdots,n\}$ into $\{1,\cdots,d\}$ and $A\_f:=A\_{f(1)}\cdots A\_{f(n)}$, for $f\in F(n,d)$.
> 
> 
> 


I try to apply Fekete's lemma:
For every subadditive sequence $(a\_n)\_{n\in \mathbb{N}^\*}$, the limit $\displaystyle \lim \_{n\to \infty }\frac{a\_n}{n}$ exists and is equal to $\displaystyle \inf\_{n\in \mathbb{N}^\*}\frac{a\_n}{n}$. (The limit may be ${\displaystyle -\infty }$).


But I'm facing difficulties in the choice of the sequence $(a\_n)\_{n\in \mathbb{N}^\*}$.


Thank you for your help.



Why $\lim_{n\to+\infty}\bigg(\bigg\|\sum_{f\in F(n,d)} A_{f}^* A_{f}\bigg\|^{\frac{1}{2n}} \bigg)\;\text{exists}?$
A simple calculation shows that 
$$\sum\_{f\in F(n+m,d)}A\_f^\*A\_f=\sum\_{g\in F(m,d)} A\_g^\*\left(\sum\_{f\in F(n,d)}A\_f^\*A\_f\right)A\_g$$
Let $a\_n:=\left\|\sum\_{f\in F(n,d)}A\_f^\*A\_f\right\|$. By use of the above equality we have 
$$0\leq a\_{m+n}\leq a\_ma\_n$$
so $\lim\_n a\_n^{\frac 1n}$ exists.


