Question related to the Toeplitz-Hausdorff Theorem 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 linear operator $A$, it is well known that $W(A)$ is convex and by the proof of the Toeplitz-Hausdorff Theorem, the operator $A$ satisfy the property $(^*)$.

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

Thank you!!
 A: Your question is basically if $W(S_1,S_2)$ is convex then is $W(PS_1P, PS_2P)$ convex over the space $PE$ where $P$ is a rank 2 projection. This is not true as demonstrated in the following example.
Fact: There exist operators $T_1, T_2$ on a separable Hilbert space such that $W(T_1, T_2) = \mathbb D^2$, the open unit bidisc.
For instance, it is known that if $T$ is the backward shift then $W(T) = \mathbb D$. If $\theta$ is an irrational rotation then define
$$
T_1 = \bigoplus_{n\geq 1} T \ \ \textrm{and} \ \ T_2 = \bigoplus_{n\geq 1} e^{in\theta}T.
$$
For each $n$ we have that $W(T_1,T_2) = \{(\lambda, e^{in\theta}\lambda): \lambda\in \mathbb D\}$ and so $W(T_1,T_2) = \mathbb D^2$.
Now to your problem. Define
$$
S_1 = \left[\begin{matrix}1/2 & 0 \\ 0 & 0\end{matrix}\right] \oplus T_1
\ \ \textrm{and} \ \ S_2 = \left[\begin{matrix}0 & 0 \\ 1/2 & 0\end{matrix}\right] \oplus T_2.
$$
Thus, $W(S_1,S_2) = \mathbb D^2$ is convex. However for $x = e_1 \oplus 0, y = e_2 \oplus 0$ we have
$$
(\langle S_1x,x\rangle, \langle S_2x,x\rangle) = (1/2,0) \ \ \textrm{and} \ \ (\langle S_1y,y\rangle, \langle S_2y,y\rangle) = (0,0)
$$
but for $z = \alpha x + \beta y$ with $\|z\|=1$, which implies $|\alpha|^2 + |\beta|^2 =1$, we have
$$
(\langle S_1z,z\rangle, \langle S_2z,z\rangle) = (1/2|\alpha|^2,1/2\alpha\bar\beta) \neq \frac{1}{2}(1/2,0) + \frac{1}{2}(0,0)
$$
for any choice of $\alpha$ and $\beta$ with $|\alpha|^2 + |\beta|^2 = 1$.
Therefore, $W(S_1,S_2)$ is convex but does not satisfy the convex property (*).
