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.


2 Answers 2


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.

  • 1
    $\begingroup$ Could you explain in more detail how you obtain the submultiplicative inequality? $\endgroup$
    – R W
    Jan 24, 2018 at 20:39
  • $\begingroup$ A matrix trick will show that. In fact, for any positive operator $E$ and any collection $D_1,\ldots,D_m$ of operators, one has $$\|\sum_{j=1}^{m}D_j^{*}ED_j \| \leq \|\sum_{j=1}^{m}D_j^{*}D_j\|\cdot\|E\|.$$ $\endgroup$
    – T. Le
    Jan 24, 2018 at 20:56

Short answer: because it's an instance of the spectral radius formula.

Details. Let's start with the following inequality on operator norms. Given a finite family $A:=(A_j)_{j\in J}$ of operators $A_j \in\mathcal{L}(H)$, and a symmetric operator $L\in \mathcal{L}_{sym}(H)$, one has $$\big\|\sum_{j\in J}A_j^*LA_j\big\|\le \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|L\|\ ,$$ with an obvious equality, incidentally, for $L=I$. Indeed, for any $x\in H$, $$ \big(\sum_{j\in J}A_j^*LA_jx, \ x\big) = \sum_{j\in J}(A_j^*LA_jx,x) = \sum_{j\in J}(LA_jx,A_jx)\le \sum_{j\in J}\|L\|(A_jx,A_jx)$$ $$= \|L\| \sum_{j\in J}(A_j^*A_jx,x)\le\|L\|\ \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|x\|^2,$$ and since the same holds for $-L$ the claimed inequality between operator norms follows.

We can interpret the latter in the following simple way: the bounded linear operator ${\bf T} \in\mathcal{L} \big(\mathcal{L}_{sym}(H)\big)$ defined by ${\bf T} L=\sum_{j\in J}A^*_jLA_j\in \mathcal{L}_{sym}(H)$ for any $L\in \mathcal{L}_{sym}(H)$ has operator norm $\|{\bf T} \|=\|\sum_{j\in J}A_j^*A_j\big\|$.

Also note that ${\bf T}^n$ is of the same nature of ${\bf T}$, corresponding to a larger family, namely $(A_f)_{f\in J^n}$:

$${\bf T}^nL=\sum_{f\in J^n} A_f^*LA_f $$ so that in particular $\|{\bf T}^n\|=\|\sum_{f\in J^n} A_f^*A_f\|$, and the limit you wrote is (the square root of) the spectral radius of ${\bf T}$, namely $$\rho({\bf T})=\lim_{n\to \infty}\|{\bf T}^n\|^{1/n}=\inf_{n\in\mathbb{N}}\|{\bf T}^n\|^{1/n}.$$

  • $\begingroup$ I think that in order to get $$\big\|\sum_{j\in J}A_j^*LA_j\big\|\le \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|L\|\ ,$$ the operator $L$ should be positive not $L\in \mathcal{L}_{sym}(H)$. $\endgroup$
    – Student
    Feb 1, 2018 at 16:45
  • $\begingroup$ I added a line of explanation. Is it ok now? $\endgroup$ Feb 1, 2018 at 16:57
  • 1
    $\begingroup$ Yes. Incidentally $\mathbf{T}$ is completely positive. $\endgroup$
    – lcv
    Feb 1, 2018 at 20:47
  • $\begingroup$ @Icv and what is the Stinespring representation of ${\bf T}$? $\endgroup$ Feb 1, 2018 at 22:43

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