Let $T\in\mathscr{B(\mathcal{H})}$ where $\mathcal{H}$ is an infinite dimensional seperable Hilbert Space and $k\in\mathbb{N}\cup\{\infty\}$. Now we define k-rank numerical range of $T$ denoted by $\Lambda_k(T)$ is defined as $$\Lambda_k(T):=\{\lambda\in\mathbb{C}: PTP=\lambda P, \text{ for some orthogonal projection } P \text{ of rank } k\}$$ or equivalently we can write $$\lambda\in\Lambda_k(T) \text{ iff there exists an orthonormal set } \{f_j\}_{j=1}^k \text{ s.t. } \langle Tf_j,f_r\rangle=\lambda\delta_{j,r} \text{ for } j,r\in\{1,2\ldots, k\}$$ where $\delta_{j,r}$ is Kronecker delta. Clearly $\Lambda_1(T)=W(T)$ i.e. $\Lambda_k(T)$ is a genaralization of numerical range $W(T)$.

Question: Let $T\in\mathscr{B(\mathcal{H})}$ be a normal operator. Show that if $\Re(e^{i\theta}\lambda)\in\Lambda_k(\Re(e^{i\theta}T))\text{ for all }\theta\in[0,2\pi)\text{ then }\lambda\in\Lambda_k(T)$

Comments: I could able to prove this using some big result of k-rank numerical range of an operator in terms of intersection of half-planes. But unfortunately I could not able to prove it using just definition of k-rank numerical range (written in the beginning) which I feel one could. This result is easy to see for numerical range $W(T)=\Lambda_1(T)$.

Any Hints or comments is highly appreciated. Instead of showing $\lambda\in\Lambda_k(T)$, if one show $\lambda\in\overline{\Lambda_k(T)}$ is also highly appreciable.


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