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!!