Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$. Let $S_1,S_2\in \mathcal{B}(F)$.

I claim that if $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$, then there exists a suitable measure space $(X,\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that each $MS_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$U(MS_k)U^*f=\varphi_kf,\;\forall f\in F,\,k=1,2.$$ Do you think that this claim is true?

This claim is true if $MS_k=S_kM, k=1,2.$

Note that, if $F$ is separable, then $(X,\mu)$ will be $\sigma-$ finite measure space.

Thank you.

Let $M\in \mathcal{B}(F)^+$ and $S_1,S_2\in \mathcal{B}(F)$.

I claim that if $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$, then $\exists\,(X,\mu)$,   $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that $$U(MS_k)U^*h=\varphi_kh,\;\forall h\in F,\,k=1,2.$$ Do you think that this claim is true?

Thanks.

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$. Let $S_1,S_2\in \mathcal{B}(F)$.

I claim that if $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$, then there exists a suitable measure space $(X,\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that each $MS_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$U(MS_k)U^*f=\varphi_kf,\;\forall f\in F,\,k=1,2.$$ Do you think that this claim is true?

This claim is true if $MS_k=S_kM, k=1,2.$

Thank you.

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $M^*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$. Let $S_1,S_2\in \mathcal{B}(F)$.

I claim that if $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$, then there exists a suitable measure space $(X,\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that each $MS_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$U(MS_k)U^*f=\varphi_kf,\;\forall f\in F,\,k=1,2.$$ Do you think that this claim is true?

This claim is true if $MS_k=S_kM, k=1,2.$

Note that, if $F$ is separable, then $(X,\mu)$ will be $\sigma-$ finite measure space.

Thank you.