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 For $S=(S_1,S_2)\in \mathcal{L}(E)^2$, the numerical range of $S$$W(S)$ is defined as
$$W(S)=\{(\langle S_1 x\;,\;x\rangle,\langle S_2 x ,\;x\rangle)\,;\,x \in E,\;\;\|x\|=1\}.$$$$W(S)=\{(\langle S_1 z\;,\;z\rangle,\langle S_2 z ,\;z\rangle)\,;\,z \in E,\;\;\|z\|=1\}.$$
We say thatThe pair $S=(S_1,S_2)$ satisfy the convex property $(^*)$ if for any $\lambda_1=(\langle S_1 x\; ,\;x\rangle,\langle S_2 x\; ,\;x\rangle),\;\lambda_2=(\langle S_1 y\; ,\;y\rangle,\langle S_2 y\; ,\;y\rangle)\in W(S)$:
$\forall\,\lambda_1=(\langle S_1 x\; ,\;x\rangle,\langle S_2 x\; ,\;x\rangle),\;\lambda_2=(\langle S_1 y\; ,\;y\rangle,\langle S_2 y\; ,\;y\rangle)\in W(S)$, with $\|x\|=\|y\|=1$ and for any point $\lambda_3$$\forall\,\lambda_3$ on the line segment joining $\lambda_1$ andwith $\lambda_2$, $\exists\,\alpha,\,\beta\in \mathbb{C}$ such that$\exists\,a,\,b\in \mathbb{C};\;$
$\|\alpha x+\beta y\|=1$$\|ax+b y\|=1$ and $(\langle S_1(\alpha x+\beta y)\; ,\;\alpha x+\beta y\rangle,\langle S_2(\alpha x+\beta y)\; ,\;\alpha x+\beta y\rangle)=\lambda_3$.$$(\langle S_1(ax+by)\; ,\;a x+b y\rangle,\langle S_2(ax+by)\; ,\;ax+by\rangle)=\lambda_3.$$
IfFor a single operator $T\in \mathcal{L}(E)$$A$, it is well known that $W(A)$ is convex and by the proof the Toeplitz-Hausdorff Theorem for linear operators that $W(T)$ is convex. One of the shortest proofs I have seen is one page by K Gustafson.
Moreover, the operator $T$$A$ 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
$$\langle Tz\;, \;z\rangle=\lambda\langle Tx\;, \;x\rangle+(1-\lambda)\langle Ty\;, \;y\rangle.$$
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 whichAssume that $W(S_1,S_2)$ is convex. Do you think thatIs $S=(S_1,S_2)$ has the convex property $(^*)$ or not?
Assume that $W(S_1,S_2)$ is convex and let $\lambda_1,\lambda_2\in W(S)$. Thus, there exists $x,y\in E$ with $\|x\|=\|y\|=1$ and $\lambda_1=(\langle S_1 x\; ,\;x\rangle,\langle S_2 x\; ,\;x\rangle),\;\lambda_2=(\langle S_1 y\; ,\;y\rangle,\langle S_2 y\; ,\;y\rangle)$. So, if $\lambda_3$ belongs to the line segment joining $\lambda_1$ and $\lambda_2$. Thus, there exists $z\in E$
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!!
