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 $(^\*)$.



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> Assume that $W(S\_1,S\_2)$ is convex. Does $S=(S\_1,S\_2)$ satisfy the property $(^\*)$?
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Thank you!!



Question related to the Toeplitz-Hausdorff Theorem
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 (\*).


