Let $E$ be a complex Hilbert space. Let $A=(A_1,\cdots,A_d)\in \mathcal{L}(E)^d$. Consider \begin{eqnarray*} W_{max}(A) &=&\{\alpha\in \mathbb{C}^d:\;\exists\,(z_n)\subset E\;\;\hbox{such that}\;\|z_n\|=1,\displaystyle\lim_{n\rightarrow+\infty}\langle A_j z_n,z_n\rangle=\alpha_j,\\ &&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}\|A_jz_n\|\rightarrow \|A_j\|,\;\forall j=1,\cdots,d \}. \end{eqnarray*} It is well known if $d=1$, we have $W_{Max}(A)$ is convex. If $d\geq2$, is $W_{Max}(A)$ convex??

Thank you for your help.

  • $\begingroup$ Can you precise what you note $\mathcal{B}(\mathcal{H})$? $\endgroup$ – mathcounterexamples.net Jul 26 '17 at 9:57
  • $\begingroup$ Have you tried an internet search? For example, I found this: sciencedirect.com/science/article/pii/S0096300302009451 (unfortunately it is behind a paywall) $\endgroup$ – Giuseppe Negro Jul 26 '17 at 10:30
  • $\begingroup$ If anyone is still interested in the obtaining the paper mentioned by Giuseppe Negro without the fee, one can use this brilliant website: sci-hub.io (Note: when opening the page, it may say "This site is not secure," or something to that effect, but that appears to be complete rubbish). $\endgroup$ – Eli Bashwinger Jul 26 '17 at 13:15
  • $\begingroup$ What does that $\to$ mean in $\langle T_k x_n\; |\;x_n\rangle\rightarrow \lambda_k$ mean? Is it the same as $=$? It is not clear what qualifies $x_n$ in the definition of $W_{\text{Max}}(T)$. Are the conditions to be satisfied by all $x_n\in \mathcal{H}$ etc. or so long there exists an $x_n$ that $x_n\in \mathcal{H}$ such that $\langle T_k x_n\; |\;x_n\rangle\rightarrow \lambda_k$ (whatever that means)? $\endgroup$ – Hans Oct 27 '17 at 19:11
  • $\begingroup$ That means $$\sum_{k=1}^d|\langle T_k x_n\; |\;x_n\rangle-\lambda_k|^d\longrightarrow 0$$ $\endgroup$ – Student Oct 27 '17 at 19:16

It is not convex and there is an example in the following paper: C. K. Li and Y. T. Poon, The joint essential numerical range of operators: convexity and related result.


This is not an answer but hopes to help the OP and is too long for a comment.

When you don't have the restriction that $||T_kx_n||=||T_k||$, then what you defined boils down to the standard definition of Joint numerical range. In 2-D, it is known to be convex, a result referred to as Toeplitz-Hausdorff theorem. For higher dimensions, counter-examples are indeed known. So essentially, you sample a point $x_n$ from the unit-norm $n-$dimensional sphere then map it to the $d-$dimensional point $(\lambda_1,\dots,\lambda_d)$ such that $\lambda_i=<x_n,T_ix_n>$. Now you do this for points in the sphere and you have joint numerical range. Now, you have put a restriction on the set of points from the sphere. Now you have the additional constraint that these points be such that $||T_ix_n||=||T_i||~,~\forall i$. In some sense, all matrices $T_i$ should preserve the norm. For instance, if you consider the 2-norm. This will mean that $x_n$ should be the eigenvector corresponding to the largest eigenvalue for all $T_i$. Such a vector may-not even exist. I don't know how to proceed if it does exist.

  • $\begingroup$ See example 1.1 in the paper "Convexity of the Joint Numerical range" at pdfs.semanticscholar.org/44db/… . I think this is a counter-example you are looking for. Setting $n=2$ in that example should give your answer. $\endgroup$ – dineshdileep Jul 27 '17 at 13:14

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