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Let $E$ be a complex Hilbert space. Let ${\bf S} = (S_1,...,S_d) \in \mathcal{L}(E)^d$. We recall that $\|{\bf S}\|$ is defined by \begin{eqnarray*} \|{\bf S}\| &:=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^d\|S_kx\|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\}, \end{eqnarray*}

If the operators $S_k$ are commuting, why we have $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2\;?? \;,$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{N}^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d\alpha_j$; ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\diamond{\bf S}\diamond\cdots\diamond{\bf S}$.

Note that ${\bf S}^2 :={\bf S}\diamond{\bf S}= (S_1 S_1,\cdots,S_1 S_d,S_2S_1,\cdots,S_2S_d,S_dS_1\cdots,S_d S_d)$.

Thank you!!

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  • $\begingroup$ you have $n$ operators $\alpha_i$ from each how many ways can you reorder them if order matters? $\endgroup$ – Adam Nov 23 '17 at 19:01
  • $\begingroup$ I think ${n \choose{\alpha_1, \alpha_2, \ldots , \alpha_d }} = \frac{n!}{\alpha_1!\ldots \alpha_d!}$. $\endgroup$ – Schüler Nov 23 '17 at 19:05
  • $\begingroup$ well there is the missing factor $\endgroup$ – Adam Nov 23 '17 at 19:06
  • $\begingroup$ I mean your formula $||{\bf S}^n||^2=\displaystyle\sup_{\|x\|=1}\displaystyle\sum _{k=1}^{d^n}\| S_1 ^{\alpha _1} S_2^{\alpha_2} \cdots S_d^{\alpha_d}x \|^2.$ is false as you already reordered the components (using the commutivity) without this factor $\endgroup$ – Adam Nov 23 '17 at 19:09
  • $\begingroup$ Yes, but how can I get a formula which is equal to the following: $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2 ?$$ $\endgroup$ – Schüler Nov 23 '17 at 19:12
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Recall that the number of ways in which you can permute $ n $ objects where $ \alpha_1 $ are of $ 1 $st type, $ \alpha_2 $ are of $ 2 $nd type,..., and $ \alpha_d $ are of $ d $th type is $$ \frac{n!}{\alpha_1!...\alpha_d!}=\frac{n!}{\alpha!}$$

As you already mentioned, $ S^n $ is $ d^n $ tuple of operators (each operator is product of $ n $ operators from $\{S_1,...,S_d\}$) . Hence $$||S^n||^2=\sup_{||x||=1} \sum_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2.$$

A typical operator in the $ d^n $ tuple of operators will look like $ S_{i_1}S_{i_2}...S_{i_n}$ where each $i_j\in \{1,2,...,d\}.$ Let $\alpha_k=$ number of $ j $ such that $ i_j=k .$ Then the operator $S_{i_1}S_{i_2}...S_{i_n}$ will become $S_1^{\alpha_1}...S_d^{\alpha_d}.$ Thus every operator in the $d^n$ tuple will become some $S_1^{\alpha_1}...S_d^{\alpha_d}$ for some $\alpha_1,...,\alpha_d$ with $\sum \alpha_i=n.$

Now fix any $\alpha_1,...,\alpha_d$ with $\alpha_1+...+\alpha_d=n.$ We have to count number of $ S_{i_1}S_{i_2}...S_{i_n}$ in the $d^n$ tuple which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$. Take $\alpha_1$ copies of $S_1, \alpha_2$ copies of $S_2$,...,$\alpha_d$ copies of $S_d.$ Let $\mathcal A$ be the set of all permutations of this operators. Note that each operator in $\mathcal A$ will appear in the $d^n$ tuple of operators and each operator (in the $d^n$ tuple of operators) which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$ (after using commutative property) must be present in $\mathcal A.$ Hence number of $ S_{i_1}S_{i_2}...S_{i_n}$ in the $d^n$ tuple which will become $S_1^{\alpha_1}...S_d^{\alpha_d}$ is same as the number of ways in which you can permute $ n $ objects where $ \alpha_1 $ are of $ 1 $st type, $ \alpha_2 $ are of $ 2 $nd type,..., and $ \alpha_d $ are of $ d $th type. Since the later number is $\frac{n!}{\alpha!},$ we hav $$||S^n||^2=\sup_{||x||=1} \sum_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2.$$

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  • $\begingroup$ @Thierry please let me know the proof is clear or not. $\endgroup$ – Black-horse Nov 25 '17 at 14:54
  • $\begingroup$ @Thierry I have edited my answer..please check and let me know. $\endgroup$ – Black-horse Nov 25 '17 at 17:28
  • $\begingroup$ @Thierry Okay I will see. $\endgroup$ – Black-horse Dec 1 '17 at 11:05

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