Let $A$ be a $C^\*$-algebra and $E$ be a (right) Hilbert $C^\*$-module over $A$. Assume $F$ is a closed submodule of $E$ such that $F^\perp := \{x \in E: \langle x, F\rangle=0\}$ is orthogonally complemented, i.e. we have $F^\perp \oplus F^{\perp \perp} = E.$ Can we conclude that $F$ is orthogonally complemented, and if this is the case do we have $F= F^{\perp \perp}$?


I know that in general we don't need to have $F= F^{\perp \perp}$. However, when $F$ is orthogonally complemented, i.e. $F\oplus F^\perp = E$ this is obvious. For Hilbert spaces (i.e. $A = \mathbb{C}$), there are no counterexamples.



Complemented submodules of a Hilbert C*-module
If $F^\perp =\{0\}$ then $F^\perp \oplus F^{\perp\perp} = E$ is automatic, so all you need is an example of this where $F \neq E$. For instance, $C[0,1]$ as a Hilbert module over itself with $F =$ the functions which vanish at 0.


