In this question $E$ stands for a complex Hilbert space and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

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 $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\diamond{\bf S}\diamond\cdots\diamond{\bf S}$.

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

Thank you!

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!

In this question $E$ stands for a complex Hilbert space and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

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 $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\diamond{\bf S}\diamond\cdots\diamond{\bf S}$.

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

Thank you

In this question $E$ stands for a complex Hilbert space and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

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 $$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\diamond{\bf S}\diamond\cdots\diamond{\bf S}$.

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

Thank you!