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.



Generalization to the spectral Theorem of self-adjoint operators
There is a 2x2 matrix counter-example.


$$ M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad
S\_1 = \begin{pmatrix} 1 & 2 \\ 1 & 4 \end{pmatrix}, 
S\_2 = \begin{pmatrix} a & 2c \\ c & d \end{pmatrix}. $$
Then
$$ MS\_1 = \begin{pmatrix} 1 & 2 \\ 2 & 8 \end{pmatrix}, \quad
MS\_2 = \begin{pmatrix} a & 2c \\ 2c & 2d \end{pmatrix} $$
so $MS\_1$ is self-adjoint, and $MS\_2$ is when $a,c, d\in\mathbb R$. Then
$$ S\_1S\_2 = \begin{pmatrix} a + 2c & 2c+2d \\ a+4c & 2c+4d \end{pmatrix},
\quad S\_2S\_1 = \begin{pmatrix} a+2c & 2a+8c \\ c+d & 2c+4d \end{pmatrix}. $$
So these commute when $d = a + 3c$. But then
$$ MS\_2 = \begin{pmatrix} a & 2c \\ 2c & 2a+6c \end{pmatrix} $$
Tediously multiplying out shows these commute only when $a=c$.


