Let be T,S two self adjoint linear Operator on a Hilbert Space $\mathcal{H}$ with pure point spectrum.
Then T,S commute if and only if they have a complete set of common eigenvectors.

This is the full statement i want to prove. In the case of bounded Operators this would be fine. My Problem now is, the Unbounded Case.I may tell you my thoughts for the first implication:

Let be $\phi\in D(T)$ an eigenvector from T with eigenvalue $\lambda$, then we know, since T and S commute: $$ TS\phi=ST\phi=S\lambda \phi=\lambda S\phi $$ Now, here is my problem: Can i assume (and why) that $\phi\in D(S)$ with just known that $T,S$ commute and $\phi$ is an eigenvector fron $T$? Maybe this is quite simple but i dont get it at the Moment...

Let $T$, $S$ be two self-adjoint linear operators on a Hilbert space $\mathcal{H}$ with pure point spectrum.
Then $T$ and $S$ commute if and only if they have a complete set of common eigenvectors.

This is the full statement I want to prove. In the case of bounded operators this would be fine. My problem now is the unbounded case. I may tell you my thoughts for the first implication:

Let be $\phi\in D(T)$ an eigenvector from $T$ with eigenvalue $\lambda$, then we know, since $T$ and $S$ commute: $$ TS\phi=ST\phi=S\lambda \phi=\lambda S\phi $$ Now, here is my problem: Can I assume (and why) that $\phi\in D(S)$ just given that $T,S$ commute and $\phi$ is an eigenvector of $T$? Maybe this is quite simple but I don't get it at the moment...

Let be T,S two self adjoint linear Operator on a Hilbert Space $\mathcal{H}$ with pure point spectrum.
Then T,S commute if and only if they have a complete set of common eigenvectors.

This is the full statement i want to prove. In the case of bounded Operators this would be fine. My Problem now is, the Unbounded Case.I may tell you my thoughts for the first implication:

Let be $\phi$ an eigenvector from T with eigenvalue $\lambda$, then we know, since T and S commute: $$ TS\phi=ST\phi=S\lambda \phi=\lambda S\phi $$ Now, here is my problem: Can i assume (and why) that $\phi\in D(S)$ with just known that $T,S$ commute and $\phi$ is an eigenvector fron $T$? Maybe this is quite simple but i dont get it at the Moment...

Let be T,S two self adjoint linear Operator on a Hilbert Space $\mathcal{H}$ with pure point spectrum.
Then T,S commute if and only if they have a complete set of common eigenvectors.

This is the full statement i want to prove. In the case of bounded Operators this would be fine. My Problem now is, the Unbounded Case.I may tell you my thoughts for the first implication:

Let be $\phi\in D(T)$ an eigenvector from T with eigenvalue $\lambda$, then we know, since T and S commute: $$ TS\phi=ST\phi=S\lambda \phi=\lambda S\phi $$ Now, here is my problem: Can i assume (and why) that $\phi\in D(S)$ with just known that $T,S$ commute and $\phi$ is an eigenvector fron $T$? Maybe this is quite simple but i dont get it at the Moment...