Closed subspace bounded linear operator.

Question

Let $H_1$ and $H_2$ be Hilbert spaces and $T:H_1\rightarrow H_2$ a bounded linear operator and let $M_1\subset H_1$ and $M_2\subset H_2$ be the closed subspaces. If $T^*({M_2}^\perp)\subset{M_1}^\perp$, then $T(M_1)\subset M_2$. Here is my proof.

Suppose that $T^*({M_2}^\perp)\subset{M_1}^\perp$. Let $y\in T(M_1)$. Then there exists $x\in M_1$ such that $Tx=y$. Then for any $k\in {M_2}^\perp$, $$\langle y,k\rangle=\langle Tx,k\rangle=\langle x,T^*k\rangle=0$$since $x\in M_1$ and $T^*k\in T^*({M_2}^\perp)\subset {M_1}^\perp$ Hence, $y\perp {M_2}^\perp$ and thus $y\in M_2$. Therefore, $T(M_1)\subset M_2$ as required.

In my proof, I did not use the condition $M_1, M_2$ being closed. Is there any mistake in my proof?

Answer 1

You have used the fact that $(M_2^{\perp})^{\perp} =M_2$. This is true iff the subspace $M$ is closed.

For this result $M_1$ need not be closed.