Solving the following problem of convexity 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.
 A: 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. 
A: 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.
