Let $T$ be a (possibly unbounded) self-adjoint operator on domain $D(T)$ of a Hilbert space $H$. We say that a closed subspace $M \subset H$ is reducing if $D(T) = (M\cap D(T))+(M^{\perp}\cap D(T))$ and both $M$ and $M^{\perp}$ are invariant under $T$ in the sense that $TM \subset M$.
I want to prove the following: $M$ is reducing iff $PT\subset TP$, where $P$ is the projection operator onto $M$.
My proof: First, suppose $M$ reducing and $x \in D(T)$. By hypothesis, $x = x_{1}+x_{2}$ where $x_{1} \in M\cap D(T)$ and $x_{2} \in M^{\perp}\cap D(T)$. Moreover: $$Tx = Tx_{1}+Tx_{2} \Rightarrow PTx = PTx_{1} = Tx_{1}.$$ On the other hand, $$Px = Px_{1}+Px_{2} = Px_{1} = x_{1} \Rightarrow TPx = Tx_{1}.$$ Because the domain $D(T)$ is a subset of $H$, which is the domain of $H$, we get $PT \subset TP$.
Let us prove the converse. Because $M$ is closed, write $H = M\oplus M^{\perp}$. If $x \in D(T)$, this implies $x=x_{1}+x_{2}$ where $x_{1}\in M\cap D(T)$ and $x_{2}\in M^{\perp}\cap D(T)$. In addition: $$PTx = TPx = Tx,$$ so it follows that $Tx \in M$. Analogously for $M^{\perp}$.
Is my proof correct?