Let $E$ be a complex Hilbert space. For $S=(S_1,S_2)\in \mathcal{L}(E)^2$, $W(S)$ is defined as $$W(S)=\{(\langle S_1 z\;,\;z\rangle,\langle S_2 z ,\;z\rangle)\,;\,z \in E,\;\;\|z\|=1\}.$$ The pair $S=(S_1,S_2)$ satisfy the property $(^*)$ if:

$\forall\,\lambda_1=(\langle S_1 x\; ,\;x\rangle,\langle S_2 x\; ,\;x\rangle),\;\lambda_2=(\langle S_1 y\; ,\;y\rangle,\langle S_2 y\; ,\;y\rangle)\in W(S)$, with $\|x\|=\|y\|=1$ and $\forall\,\lambda_3$ on segment joining $\lambda_1$ with $\lambda_2$, $\exists\,a,\,b\in \mathbb{C};\;$ $\|ax+b y\|=1$ and $$(\langle S_1(ax+by)\; ,\;a x+b y\rangle,\langle S_2(ax+by)\; ,\;ax+by\rangle)=\lambda_3.$$

For a single operator $A$, it is well known that $W(A)$ is convex and by the proof the Toeplitz-Hausdorff Theorem, the operator $A$ satisfy the property $(^*)$.

Assume that $W(S_1,S_2)$ is convex. Is $S=(S_1,S_2)$ has the property $(^*)$?

Thank you!!