Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. Let $T\in \mathcal{L}(E)$.

We recall for $S=(S_1,S_2)\in \mathcal{L}(E)^2$, the numerical range of $S$ is defined as $$W(S)=\{(\langle S_1 x\;,\;x\rangle,\langle S_2 x ,\;x\rangle)\,;\,x \in E,\;\;\|x\|=1\}.$$ We say that $S=(S_1,S_2)$ satisfy the convex property $(^*)$ if $\forall\;\lambda_1,\lambda_2\in W(S)$, and for any point $\lambda_3$ on the line segment joining $\lambda_1$ and $\lambda_2$, $\exists\,\alpha,\,\beta\in \mathbb{C}$ such that $\|\alpha x+\beta y\|=1$ and $(\langle S_1(\alpha x+\beta y)\; ,\;\alpha x+\beta y\rangle,\langle S_2(\alpha x+\beta y)\; ,\;\alpha x+\beta y\rangle)=\lambda_3$.

If $T\in \mathcal{L}(E)$, it is well known by the Toeplitz-Hausdorff Theorem for linear operators that $W(T)$ is convex. One of the shortest proofs I have seen is one page by K Gustafson. Moreover, $T$ satisfy the convex property $(^*)$. When I see the proof by K Gustafson, I remark that for construction of $\alpha,\beta$, it suffices to look at the (possibly complex) plane spanned by unit vectors $x$ and $y$ in order to find for any $0<\lambda<1$ a unit element $z=\alpha x + \beta y$ so that $$(Tz|z) = \lambda (Tx|x) + (1-\lambda) (Ty|y)$$

I see in a paper that $W(S_1,S_2)$ is in general not convex and there are a cases in which $W(S_1,S_2)$ is convex. My question is the following:

In the cases in which $W(S_1,S_2)$ is convex. Do you think that $S=(S_1,S_2)$ has the convex property $(^*)$ or not?

I try as follows: Assume that $W(S_1,S_2)$ is convex. Let $\lambda_1,\lambda_2\in W(S)$, and $\lambda_3$ on the line segment joining $\lambda_1$ and $\lambda_2$. Thus there exist $z\in E$ such that $\|z\|=1$ and $(\langle S_1z\; ,\;z\rangle,\langle S_2z\; ,\;z\rangle)=\lambda_3$. Do you think that we can show that $z$ has the form $\alpha x+\beta y$ or not??

Thank you!!