Let $(\mathcal{H}, \langle\cdot,\cdot\rangle)$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ is the algebra of all bounded linear operators on $\mathcal{H}$.
For ${\bf T} = (T_1,\cdots,T_n)\in \mathcal{B}(\mathcal{H })^n$, we consider the following subset $\mathbb{C}^n$ \begin{eqnarray*} JtMaxW({\bf T}) &=&\{(\lambda_1,\cdots,\lambda_n)\in \mathbb{C}^n:\;\exists\,(x_i)\subset \mathcal{H}\;\;\hbox{such that}\;\|x_i\|=1,\displaystyle\lim_{i\rightarrow+\infty}\langle T_k x_i,x_i\rangle=\lambda_k,\\ &&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{i\rightarrow+\infty}\|T_kx_i\|\rightarrow \|T_k\|,\; 1\leq k \leq n\;\}. \end{eqnarray*}
Why the following facts which is taken from this reference(theorem 3.8) hold?
Note that Lemma 3.4 indicates that $\pi_k(JtMaxW_A({\bf T}))$ is convex for all $k\in\{1,\cdots,n\}$, where $\pi_k$ is the projection from $\mathbb{C}^n$ to $\mathbb{C}$ i.e. $\pi_k(x_1,\cdots,x_n)=x_k$, for all $k\in\{1,\cdots,n\}$.
If the above claim is false. Is it true if $n=1$?
My attempt: Suppose now that $0\notin JtMaxW({\bf T})$, then $\forall\{x_i\}\in\mathcal{H}$ such that $\|x_i\|=1$ we have, $$\langle T_k x_i,x_i\rangle\nrightarrow 0 \quad \mbox{or}\quad\|T_kx_i\|\nrightarrow \|T_k\|,\,\, 1\leq k \leq n.$$ Let us rotate ${\bf T}$, that is we consider ${\bf \widehat{T}}=(e^{i\alpha_1}T_1,...,e^{i\alpha_n}T_n)$. Moreover, if $\lambda=(\lambda_1,...,\lambda_n)\in JtMaxW({\bf \widehat{T}})$, then there exists a sequence $\{y_i\}$ in $\mathcal{H}$ such that $$\|y_i\|=1\quad \mbox{and}\quad\|\widehat{T}_ky_i\|\rightarrow \|\widehat{T}_k\|,\,\, 1\leq k \leq n.$$ i.e. $$\|y_i\|=1,\,\,e^{-i\alpha_k}\langle T_k y_i,y_i\rangle\rightarrow \lambda_k \quad \mbox{and}\quad\|T_ky_i\|\rightarrow \|T_k\|,\,\, 1\leq k \leq n.$$ Since, $\displaystyle\lim_{i\rightarrow\infty}\langle T_k y_i,y_i\rangle\neq0$, then $\lambda_k\neq0$ for all $1\leq k \leq n.$ Assume that there exists $\delta_k$ such that $\Re e(\lambda_k)\geq\delta_k>0$
Why for all $x\in \mathcal{H},\;\|x\|=1$ we have $\Re e\langle \widehat{T}_k x,x\rangle\geq\delta_k>0$)?
The claim as it stands is false because $JtMaxW(T)$ might be empty. Then clearly, $(0,\ldots,0)\notin JtMaxW(T)$, but also $JtMaxW(T') = \emptyset$ for any "rotation" $T'$ of $T$.
An example for which $JtMaxW(T)$ is empty is given by $$ T_1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}\quad\text{and}\quad T_2 = \begin{pmatrix}0&0\\0&1\end{pmatrix}. $$