If $B$ is bounded with $0< B\leqslant 1$ and $T$ is closed, is $BT$ closable? Let $H$ be a separable Hilbert space, let $B$ be a bounded operator on $H$ with $0< B \leqslant 1$ and let $T$ be a closed, densely defined operator in $H$. The notation $0<B\leqslant 1$ signifies that $0<\langle x,Bx\rangle \leqslant 1$ for all $x\in H$ with $\lVert x \rVert =1$, i.e. $B$ is self-adjoint, positive definite and satisfies $\lVert B \rVert \leqslant 1$.
Then $TB$ is closed (even without the assumption $0< B \leqslant 1$). Is it also true that $BT$ is closable?
If $B$ is boundedly invertible, then $BT$ is closed. I expect that $BT$ may not be closable if $0\in\sigma(B)$, but have not been able to come up with an example. We have $(BT)^*=T^*B$, so one may equivalently find a closed, densely defined operator $S$ such that $SB$ is not densely defined.

This answer is based on Nate's example. The operator $B$ is made injective by adding a compact multiplication operator. The coefficients had to be adjusted to make it self-adjoint, bounded, and positive definite. Moreover, the diagonal part is made to vanish faster than then rank-one part for $n\to\infty$.
Take an orthonormal base $(e_n)$. Define $B$ by
$$
Be_n = n^{-4} e_n + n^{-3} e_1\quad \text{ for } n>1
$$
and
$$
Be_1= a_1e_1 + \sum_{n=2}^\infty n^{-3} e_n.
$$
Then $B$ is bounded and self-adjoint. The element $a_1$ can be adjusted to make it positive definite:
Take $x\in X$ and set $x_n=\langle x,e_n\rangle$. Then
$$
\langle x,Bx\rangle = a_1x_1^2+ \sum_{n=2}^\infty n^{-4}x_n^2 + 2\sum_{n=2}^\infty n^{-3}x_1x_n\\
= a_1x_1^2 + \sum_{n=2}^\infty n^{-4}x_n^2 + 2\sum_{n=2}^\infty n^{-1}x_1 n^{-2}x_n\\
\ge x_1^2 (a_1 - \sum_{n=2}^\infty n^{-2}).
$$
Define
$$
Te_n =n^4e_n
$$
with obvious domain. Then
$$
BT(n^{-1}e_n) = n^{-1} e_n + e_1 \to e_1
$$
and $BT$ is not closable.
