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Let ${\bf S} = (S_1,...,S_d) \in \mathcal{L}(E)^d$. We recall that the norm of $\|{\bf S}\|$ 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*}

I want to show that if the operators $S_k$ are commuting, then $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$ where ${\bf S}^n:={\bf S}\cdot{\bf S}\cdot\cdots\cdot{\bf S}$.

Note that ${\bf S}^2 :={\bf S}\cdot{\bf S}= (S_1 S_1,\cdots,S_1 S_d,S_2S_1,\cdots,S_2S_d,S_dS_1\cdots,S_d S_d)$, and ${\bf S}^n$ is defined by induction.

Thank you!

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    $\begingroup$ Erm... And what is the difference between the LHS and the RHS (except the fancy notation that is there to confuse the reader)? $\endgroup$
    – fedja
    Commented Nov 21, 2017 at 20:57

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As it was mentioned in the comments, this is basically writing down the definition.

Since all the $S_i$'s commute with each other, then each component of $\boldsymbol{S}^n$ is of the form $(S_1 \ldots S_1)(S_2 \ldots S_2) \ldots (S_d \ldots S_d)$, where $S_i$ apprears $\alpha _i$ times and $\sum _i \alpha _i =n$. The number of times the sequence $\alpha = (\alpha _1 , \ldots , \alpha _d)$ apprears, is the number of ways you can choose the first option(which is $S_1$) $\alpha_1$ times, the second option (that is $S_2$) $\alpha_2$ times and so on, and this is given by the binomial expansion ${n \choose{\alpha_1, \alpha_2, \ldots , \alpha_d }} = \frac{n!}{\alpha_1!\ldots \alpha_d!}$.

Hence, we get: $$||S^n||^2=\sup_{||x||=1} \sum_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2,$$

because the 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 $\frac{n!}{\alpha!}$, where each $i_j\in \{1,2,...,d\}$.

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  • $\begingroup$ I think this formula is not true: $$ \Vert \boldsymbol{S}^n \Vert^2 = sup \Big\{ \sum _{k=1}^{d^n} \Vert S_1 ^{\alpha _1} S_2^{\alpha_2} \ldots S_d^{\alpha_d} \Vert^2, x\in E, \Vert x \Vert =1 \Big\} .$$ $\endgroup$
    – Schüler
    Commented Nov 24, 2017 at 17:42
  • $\begingroup$ This is a standard "take $\epsilon >0$ and prove each side is less than the other" argument. Let $F= sup _{x\in A}\{f(x)\}$. Then for each $n\in \mathbb{N}$, there exists $x_n \in A$ such that $f(x_n) > F - 1/n$. Hence assuming that $f(x)$ is non-negative for each $x\in A$ (which is the case in your example.), $sup _{x\in A}\{ f(x)^2\} \geq f(x_n)^2 > (F-1/n)^2$ for each $n$, hence $sup _{x\in A}\{ f(x)^2\} \geq F^2$. Also, for each $x\in A$, $F \geq f(x)$, therefore $F^2 \geq f(x)^2$. Now by taking supremum on both sides, we get $F^2 \geq sup _{x\in A}\{ f(x)^2\}$. This finishes the proof. $\endgroup$
    – Kashayar
    Commented Nov 24, 2017 at 18:32
  • $\begingroup$ I'm sorry, I had a few typos, so I deleted the comment and added a new one. In my comment above, take $f(x) = (\sum _{k=1}^{d^n} \Vert S_1^{\alpha_1} \ldots S_d^{\alpha_d} \Vert ^2)^{1/2}$. Then you can see if $A=\{ x\in E, \Vert x \Vert =1\}$, we get $\Vert \boldsymbol{S}^n \Vert = sup_{x\in A}\{f(x)\}$. $\endgroup$
    – Kashayar
    Commented Nov 24, 2017 at 18:36
  • $\begingroup$ Thank you very much but perhaps you don't understand me very well: I tested your formula for $n=2$ and $d=2$ and I don't find that it is the same as the left hand side. $\endgroup$
    – Schüler
    Commented Nov 24, 2017 at 18:37
  • $\begingroup$ Thanks for your clarification. I see where the confusion is. In my original answer, The sumation is from $k=1$ to $d^n$. What I meant to say was to consider all the $d^n$ possible cases which appear in $\boldsymbol{S}^n$, and the summation in my answer, is NOT on different combinations of $\alpha$ . For example, if $d=2, n=2$, there are three choices for $\alpha$: $(2,0) , (0,2) , (1,1)$, and hence three terms in your summation on the left. However, in the summation on your right, there are four terms, $S_1 S_1 ,S_1 S_2 , S_2 S_1 , S_2 S_2$. $\endgroup$
    – Kashayar
    Commented Nov 24, 2017 at 19:12

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