In this question $F$ stands for a complex Hilbert space. Let ${\bf S} = (S_1,...,S_d) \in \mathcal{B}(F)^d$. We recall that $\|{\bf S}\|$ is defined by \begin{eqnarray*} \|{\bf S}\| &=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^d\|S_kg\|^2\bigg)^{\frac{1}{2}},\;g\in F,\;\|g\|=1\;\right\}. \end{eqnarray*} Moreover, if the operators $S_k$ are commuting, then $r({\bf S})$ is given by $$r({\bf S})=\lim_{n\longrightarrow \infty}\left[\displaystyle\sup_{\|g\|=1}\left(\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}g\|^2\right)^{\frac{1}{2n}}\right],$$ with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{N}^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d\alpha_j$; and ${\bf S}^\alpha:=S_1^{\alpha_1} S_2^{\alpha_2}\cdots S_d^{\alpha_d}$.

I want to prove that if the operators $S_k$ are commuting and each $S_k$ is self-adjoint, then $$r({\bf S})=\|{\bf S}\|.$$

  • $\begingroup$ $S_1,...,S_d$ belong to $\mathcal{B}(F)$ as I understand. Do you consider ${\bf S} = (S_1,...,S_d) \in \mathcal{B}(F)^d$ as an element of $ \mathcal{B}(F^d)$? $\endgroup$ – Yurii Savchuk Dec 1 '17 at 7:37
  • $\begingroup$ Do you mean $\mathcal{B}(F)^d\subseteq\mathcal{B}(F^d)$? $\endgroup$ – Yurii Savchuk Dec 1 '17 at 7:39
  • $\begingroup$ No $\mathcal{B}(F)^d:=\mathcal{B}(F)\times\cdots\times \mathcal{B}(F)$ $\endgroup$ – Schüler Dec 1 '17 at 7:41
  • $\begingroup$ But there is a natual inclusion $\mathcal{B}(F)^d\subseteq\mathcal{B}(F^d)$, so that ${\bf S}$ becomes an operator on a Hilbert space. $\endgroup$ – Yurii Savchuk Dec 1 '17 at 7:44
  • $\begingroup$ What is the importance of this inclusion to solve my question? Thank you $\endgroup$ – Schüler Dec 1 '17 at 7:47

Here is a sketch of the solution. You can indeed use a spectral theorem for tuples of bounded commuting self-adjoint operators on a Hilbert space, similar to what you have mentioned.

Example. Consider the following operators $S_1,\dots,S_d$. Let $F$ be the Hilbert space $L^2(\mathbb R^d,\mu)$, let ${\operatorname{supp}}\,\mu=\Omega\subseteq\mathbb R^d$ be compact and let $S_k$ be the operator of multiplication by $x_k$ i.e. $$(S_kf)(x_1,\dots,x_d)=x_kf(x_1,\dots,x_d)\ \ \ \ \ (*)$$ Then $S_1,\dots,S_d$ is a $d$-tuple of bounded commuting self-adjoint operators.

Theorem. Every $d$-tuple of bounded commuting self-adjoint operators on a Hilbert space is (up to unitary equivalence) equal to a direct sum of tuples, which act like $(*)$.

You can do the following.

1. Consider the case, when ${\bf S}=(S_1,\dots,S_d)$ act like in the Example. Show that the Harte spectrum of a tuple of bounded operators given in the paper you have mentioned coincides with $\Omega$. Show that $r({\bf S})$ and $\|{\bf S}\|$ are both equal to the maximum of the euclidean norm of the elements in $\Omega\subseteq\mathbb R^d$.

2. Show that for the case, when ${\bf S}$ is a direct sum of such ${\bf S}_i,\ i\in I$, which act as $(*)$, then the Harte spectrum of ${\bf S}$ is the union of correponding $\Omega_i$'s, the norm $\|{\bf S}\|$ is the supremum of $\|{\bf S_i}\|$ and hence $r({\bf S})$ and $\|{\bf S}\|$ are still equal.

Hope this helps.

  • $\begingroup$ Thank you for your answer, but I hope to use only the formula of $r(S)$ not a spectral consideration (Harte spectrum) $\endgroup$ – Schüler Dec 2 '17 at 7:54
  • $\begingroup$ @Thierry It is easier to work with the spectrum of ${\bf S}$ in this case $\endgroup$ – Yurii Savchuk Dec 2 '17 at 18:08
  • $\begingroup$ Yes but i hope working without it $\endgroup$ – Schüler Dec 2 '17 at 20:12
  • $\begingroup$ @Thierry Please think on the following exercise: calculate the spectrum, the norm and the spectral raduis of the operator $(Tf)(x)=xf(x)$ on the Hilbert space $L^2(\mathbb R,\mu)$, where $\mu$ is compactly supported. This is the lite-version of your problem. $\endgroup$ – Yurii Savchuk Dec 4 '17 at 7:44
  • $\begingroup$ OK thank you very much for your advice $\endgroup$ – Schüler Dec 4 '17 at 7:55

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