Let $M$ be a closed linear subspace of an incomplete inner product space $X$ and let $M + M^\perp \neq X$ then is it true that $M \neq M^{\perp\perp}$. If true then how to prove it and if not then do we have a counterexample.
I know the converse of the above statement is true, that is if $M + M^\perp = X$ then $M = M^{\perp\perp}$. Any help is greatly appreciated.
Let V be the space of continuous functions on [-1,1] with the $L_2$ inner product, i.e $$ \langle f, g \rangle = \int_{-1}^1 f\overline{g} $$ V is an incomplete inner product space as it is a non-closed subspace of the Hilbert space $L_2([-1,1])$.
Let M be the subspace of V consisting of all functions that are 0 on [0,1]. Then M is closed subspace of V and $M^{\perp}$ is precisely the set of functions that are 0 on [-1,0]. We can see that $M + M^{\perp} \neq V$ because if $f \in M + M^{\perp}$ then $f(0)=0$. At the same time $M^{\perp \perp} = M$.
No matter what, it is $M\subset M^{\bot\bot}$. Since $M^{\bot\bot}$ is always a closed set, we also have $\overline{M}\subset M^{\bot\bot}$. If the space is not complete, or, more generally, if $M+M^\bot\neq X$, this is the best one could say.
As stated in the answer of Josh Messing, we can find examples where $M+M^\bot\neq X$ but $M^{\bot\bot}=\overline{M}$ (and further $=M$ if $M$ is closed). On the other hand there are numerous examples where $\overline{M}$ is a proper subset of $M^{\bot\bot}$. The easiest way to construct one is to find a proper subspace $M$ where $M^\bot=0$, so $M^{\bot\bot}=X$.
