Let $H_{i}$ be Hilbert spaces for $i=1,2,3$. Let $T_{21}:H_{1} \rightarrow H_{2}$ and $T_{32}: H_{2} \rightarrow H_{3}$ be closed, densely defined, unbounded operators. What are appropriate conditions to guarantee that $T_{32}T_{21}$ is closed or closeable?
I had some options in mind that seemed reasonable:
I expect that if $T_{21}\mathrm{dom}(T_{21}) = \mathrm{dom}(T_{32})$, then $T_{32}T_{21}$ is closed. This was discussed in this question, but not proved. Is there a reference for this statement?
Next, suppose that $H_{1}'$ is a core for $T_{21}$ and $H_{2}'$ is a core for $T_{32}$. Further suppose that $T_{21}H_{1}' = H_{2}'$. Is it then true that $T_{32}T_{21}$ is closeable?
An operator $T$ is closeable, if and only if for every sequence $x_{i} \in \mathrm{dom}(T)$ with $x_{i} \rightarrow 0$, the only accumulation point of $Tx_{i}$ is zero. Because we know that $T_{21}|_{H_{1}'}$ and $T_{32}|_{H_{2}'}$ are closeable, it suffices to prove that if $x_{i} \in H_{1}'$ with $x_{i} \rightarrow 0$, then the only accumulation point of $T_{32}T_{21}x_{i}$ is zero. However, I can't quite get this to work. We know that the only accumulation point of $T_{21}x_{i}$ is zero, but this seems insufficient to prove that the only accumulation point of $T_{32}T_{21}x_{i}$ is zero.
Claim: In the notation of the question, if $H_{1}'$ is a core for $T_{21}$ and $H_{2}'$ is a core for $T_{32}$, and $T_{21}H_{1}' = H_{2}'$, then $T_{32}T_{21}$ is closeable, and $H_{1}'$ is a core for its closure.
Proof: Let $\{ x_{i} \}_{i \in X}$ be a sequence in $H_{1}'$ which converges to $0$. Suppose that $w \in H_{3}$ is an accumulation point of $\{T_{32}T_{21}x_{i} \}_{i \in X}$. This means that there is a subsequence $\{x_{i} \}_{i \in X'}$ such that $\{ T_{32}T_{21} x_{i} \}_{i \in X'}$ converges to $w$. Because the sequence $\{x_{i} \}_{i \in X'}$ converges to $0$, we have that the only accumulation point of $\{T_{21} x_{i} \}_{i \in X'}$ is zero. Thus, there is a (sub)subsequence $\{x_{i} \}_{i \in X''}$ such that $\{T_{21}x_{i} \}_{i \in X''}$ converges to $0$. In turn, the only accumulation point of $\{T_{32}T_{21}x_{i} \}_{i \in X''}$ is zero. But $\{T_{32}T_{21}x_{i} \}_{i \in X''}$ converges to $w$, thus $w=0$.