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Here's Prob. 4, Sec. 3.9 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $H_1$ and $H_2$ be Hilbert spaces, and let $T \colon H_1 \to H_2$ be a bounded linear operator. If $M_1 \subset H_1$ and $M_2 \subset H_2$ and if $T(M_1) \subset M_2$, show that $T^*(M_2^\perp) \subset M_1^\perp$.

Here $T^*$ denotes the Hilbert adjoint operator of $T$.

Here's my proof:

Let $x \in M_2^\perp$. Then $$\langle x, v \rangle = 0 \ \mbox{ for all } \ v \in M_2.$$ Thus, in particular, we have $$\langle x, Tu \rangle = 0 \ \mbox{ for all } \ u \in M_1 $$ because $Tu \in M_2$ whenever $u \in M_1$.

But $\langle x, Tu \rangle = \langle T^*x, u \rangle$, by the definition of $T^*$. So we can conclude that $$\langle T^* x , u \rangle = 0 \ \mbox{ for all } \ u \in M_1,$$ and therefore $T^* x \in M_1^\perp$ whenever $x \in M_2^\perp$, as required.

Is the above proof correct?

Now can we derive the following result?

If $M_1 \subset H_1$, $M_2 \subset H_2$, and $T(M_1) \subset M_2$, then $T^*(M_2) \subset M_1$.

I know that this does indeed hold if $M_1$ and $M_2$ are closed subspaces of $H_1$ and $H_2$, respectively; in fact for closed subspaces, we can even make the if and only if statement.

But does this result hold otherwise?

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  • $\begingroup$ Just a friendly (I hope!) note: to me, what I'm looking for in a title is some indication of the mathematical content. Having the full author name, book title, and question number doesn't seem broadly useful to me. $\endgroup$ – Hoot Jun 1 '15 at 18:13
  • $\begingroup$ I would like to keep this title. So please don't edit it!! $\endgroup$ – Saaqib Mahmood Jun 2 '15 at 5:44
  • $\begingroup$ I have no intention of doing so. $\endgroup$ – Hoot Jun 2 '15 at 6:03
  • $\begingroup$ @Hoot, so nice of you!! $\endgroup$ – Saaqib Mahmood Jun 2 '15 at 6:15
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Your initial part of the proof is correct. But even for closed subspace $M_1$ and $M_2$, you can have a bounded operator $T$ on $H$ such that $T(M_1)\subset M_2$ but $T^*(M_2)\nsubseteq M_1$.

For example take $H = l^2(\mathbb{N})$ and choose $M_1= M_2 =\{(a_n)\in l^2(\mathbb{N})|a_1=0\}$. Now consider the forward shift operator as $T$ i.e \begin{align} T\big((a_1,a_2,a_3,\cdots) \big)= (0,a_1,a_2,a_3,...) \end{align} So we will have $T^*$ as backward shift operator i.e \begin{align} T\big((a_1,a_2,a_3,\cdots) \big)= (a_2,a_3,a_4,...) \end{align} Note that $T(M_1) = \{(a_n)\in l^2(\mathbb{N})|a_1=0,a_2=0\}$. But $M_2= M_1$ is not invariant under $T^*$. In fact it turns out that in this case, $T^*(M_2)= H$.

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  • $\begingroup$ how do we know that the operator $(a_1, a_2, a_3, \ldots ) \mapsto (a_2, a_3, a_4, \ldots)$ is indeed the Hilbert adjoint operator of $(a_1, a_2, a_3, \ldots) \mapsto (0, a_1, a_2, a_3, \ldots)$ on $\ell^2$? $\endgroup$ – Saaqib Mahmood Jun 2 '15 at 5:33
  • $\begingroup$ If you call the operator $(a_1,a_2,a_3...)\to (a_2,a_3,a_4,...)$ by $S$. Then observe that we have $\langle Ta,b \rangle = \langle a,Sb \rangle$, for all $a,b \in l^2(\mathbb{N})$. This gives us that $S$ is the adjoint of $T$. $\endgroup$ – Timon Jun 2 '15 at 19:42

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