Closure of tensor product /tensor product semigroup In this reference the following claim is made in Remark 2
Let $A,B$ be closable operators on Banach spaces $X,Y$, then  $A \otimes 1$ and $1 \otimes B$ are closable operators on the Banach space $X \otimes^{\pi} Y.$ Here $1$ denotes the identity operator.
Moreover, $A \otimes 1+1 \otimes B$ is closable and its closure is given by
$$ \overline{A \otimes 1+1 \otimes B} = \overline{A \otimes 1}+\overline{1 \otimes B}.$$
There is also a reference this one  mentioned for parts of this result, but I could not find this Theorem 1.1 in this reference.
I noticed that there are a few typos in the paper, not necessarily major once, but I wanted to ask whether anybody can verify confirm
$$ \overline{A \otimes 1+1 \otimes B} = \overline{A \otimes 1}+\overline{1 \otimes B}.$$
and ideally also provide a reference.
The result that suprises me is that the closure of the sum of the two operators is again closed, which is not true in general. Perhaps anybody knows how this follows here?
 A: So...  It seems to me that the 1st claim is in Lemma 6 of the main paper.  This actually references the following:
Ichinose, Takashi
Operators on tensor products of Banach spaces.
Trans. Amer. Math. Soc. 170 (1972), 197–219. MR0322553.  Available on the TAMS archive.
This paper does seem to have that if $A$ is closable then $A\otimes 1$ is closable, but under the additional assumption that $X \otimes^\pi Y \rightarrow X \otimes^\epsilon Y$ (the injective tensor product) is injective.  Morally speaking this means you need $X$ or $Y$ to have the approximation property (this is not quite an if and only if, but is close).
(I messed about trying to prove the result myself, and this condition is very natural, as you want to slice out by a member of $Y^*$, but to make use of this, you need to know that if $\tau\in X \otimes^\pi Y$ with $(\textrm{id}\otimes\mu)(\tau)=0$ for all $\mu\in Y^*$ then $\tau=0$).
I think then Theorem 3.1 of this paper shows that $\overline{A\otimes 1 + 1\otimes B}$ is the closure of $\overline{A\otimes 1} + \overline{1\otimes B}$.  This all under the assumption on $X,Y$ and under a spectral condition on $A,B$.  This is also not quite what you want.
However, this particular care is considered more in Section 4.2 (bottom, page 213).  I don't see any definitive conclusion here.  However, it suggests very strongly that you are correct to doubt this claim.
(I can't currently get access to the other reference.)
