Convexity of $W(S_1,S_2)$ Let $M\in \mathcal{B}(F)^+$ ( i.e. $\langle Mx, x\rangle \geq$ for all $x\in F$), where $F$ is an infinite-dimensional complex Hilbert space.

An operator $A\in \mathcal{B}(F)$ is said to be $M$-self adjoint if $MA$ is self adjoint.

For $S_1,S_2\in \mathcal{B}(F)$  We consider
 $$W(S_1,S_2)=\{(\langle MS_1x, x\rangle,\langle MS_2x, x\rangle);\;\;x\in F,\;\langle Mx,x \rangle=1\}.$$

Assume that $S_1$ and $S_2$ are $M$-self adjoint such that $S_1S_2=S_2S_1$ and $MS_k=S_kM,\,k=1,2.$ It is possible to show that $W(S_1,S_2)$ is convex?

Since $S_1$ and $S_2$ are commuting and $M$-self adjoint such that $MS_k=S_kM,\,k=1,2$, then $\exists\,(X,\mu)$,  $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that
$$U(MS_k)U^*h=\varphi_kh,\;\forall h\in L^2(\mu),\,k=1,2.$$
Let $\lambda=(\lambda_1,\lambda_2)$, $\eta=(\eta_1,\eta_2)$ be any pair of point in  $W(S_1,S_2)$, then there exist $f,g \in L^2(X, \mu)$ such that for all $k= 1,2$ we have
$$\lambda_k=\displaystyle\int_X \varphi_k|f|^2d\mu\;\;\text{and}\;\;\eta_k=\displaystyle\int_X \varphi_k|g|^2d\mu.$$
Let $\xi=t\lambda+(1-t)\eta,\;t\in[0,1]$. So
$$\xi_k=t\displaystyle\int_X \varphi_k|f|^2d\mu+(1-t)\displaystyle\int_X \varphi_k|g|^2d\mu.$$

It is possible to show that $\xi\in W(S_1,S_2)$?

 A: You was on the right way to prove it. Since $M$, $S_1M$ and $S_2M$ are bounded self-adjoint operators which commute, you can find $(X, \mu)$, $\psi, \phi_1, \phi_2 \in L^{\infty} (\mu)$ and an unitary operator $U :\mathcal  F \to L^2(\mu)$ such that : $$UMU^*h = \psi h, US_kMU^*h = \phi_kh, \forall h\in L^2(\mu).$$ Now when you take the $\lambda = (\lambda_1, \lambda_2), \eta = (\eta_1, \eta_2) \in W(S_1,S_2)$, you have $f, g \in L^2(\mu)$ such that $$\lambda_k=\displaystyle\int_X \varphi_k|f|^2d\mu,\;\;\eta_k=\displaystyle\int_X \varphi_k|g|^2d\mu,\;\;\text{and}\;\; 1 = \displaystyle\int_X \psi|g|^2d\mu = \displaystyle\int_X \psi|f|^2d\mu.$$
Let $\xi=t\lambda+(1-t)\eta,\;t\in[0,1]$ and $h = \sqrt {t|f|^2 + (1-t)|g|^2}$. So
$$\xi_k=t\displaystyle\int_X \varphi_k|f|^2d\mu+(1-t)\displaystyle\int_X \varphi_k|g|^2d\mu = \displaystyle\int_X \varphi_k|h|^2d\mu$$ and $$ \displaystyle\int_X \psi|h|^2d\mu = t\displaystyle\int_X \psi|f|^2d\mu + (1-t) \displaystyle\int_X \psi|g|^2d\mu = t+1-t = 1.$$ This proves also that $\xi \in W(S_1,S_2)$
