Common Eigenvectors for commuting unbounded Operators 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 $\phi\in D(S)$? Maybe this is quite simple but i dont get it at the Moment...

Your assumption is valid.
Since $TS = ST$, we have $D(TS) = D(ST)$.  Since $\phi \in D(TS)$, it must be the case that $\phi \in D(S)$ and $S\phi \in D(T)$. In particular, we see that $D(TS) \subset D(S)$ and $D(TS) = D(ST) \subset D(T)$.   That is, $D(ST) \subset D(S) \cap D(T)$
The equation you wrote only makes sense if we have $\phi \in D(TS)$.  If $\phi \in D(TS)$, it must be the case $\phi \in D(S)$.
