$T(M^\perp)\subseteq M^\perp \Longleftrightarrow T^*(M)\subseteq M?$

Question

Let $E$ be an infinite-dimensional complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators from $E$ to $E$.

Let $M$ be a subspace of $E$ and $T\in \mathcal{L}(E)$. It is true that $$T(M^\perp)\subseteq M^\perp \Longleftrightarrow T^*(M)\subseteq M?$$

Answer 1

\begin{gather*}T(M^{\perp})\subset M^{\perp}\iff \left(x\in M^{\perp}\Rightarrow Tx\in M^{\perp}\right) \iff \\ \iff \left[x\in M^{\perp} \Rightarrow \left\langle Tx,y\right\rangle=0\;\forall y\in M\right]\iff \left[x\in M^{\perp} \Rightarrow \left\langle x,T^*y\right\rangle=0\;\forall y\in M\right]\iff \\ \iff T^*(M)\subset (M^{\perp})^{\perp} \end{gather*} Now, if $M$ is closed, $(M^{\perp})^{\perp}=M$, and we are done. Otherwise, $(M^{\perp})^{\perp}=\bar{M}\neq M$, so the equivalence does not hold in general.

Answer 2

The answer to your question is "No."

Let $H=\ell^2(\mathbb{N})$. Let $M$ consist of all finite sequences $(x_0,x_1,\cdots,x_n,0,0,0,0,\cdots)$ such that $\sum_{n=0}^{\infty} x_n=0$. This is a dense subspace of $H$. To see why, suppose $x \perp M$. Then $x \perp (1,-1,0,0,0)$, $x\perp (1,0,-1,\cdots)$, etc. leads to the conclusion that $x_1=x_2=x_3=\cdots$, which is impossible unless $x=0$. Hence, $M^{\perp}=\{0\}$.

Because $M^{\perp}=\{0\}$, the shift operator $T(x_0,x_1,x_2\cdots)=(0,x_0,x_1,x_2,\cdots)$ trivially satisfies $$ TM^{\perp} \subseteq M^{\perp}. $$ $T$ has adjoint $T^*(x_0,x_1,x_2,\cdots)=(x_1,x_2,\cdots)$. So $T^*M\not\subseteq M$ because the imagine sequence may not sum to $0$. Summarizing, $$ TM^{\perp}\subseteq M^{\perp},\;\; T^*M\not\subseteq M $$