Antilinear unbounded operator has closed graph

Question

Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation $$\langle T^*\eta, \xi\rangle = \langle T\xi,\eta\rangle$$ for all $\eta \in D(T^*)$ and all $\xi \in D(T)$, where $D(T^*)$ is the subspace of $K$ of all $\eta\in K$ such that $ D(T)\ni \xi \mapsto \langle \eta, T\xi\rangle$ is a bounded linear functional.

I want to prove that $T^*$ has closed graph in a direct way.

Let me quickly revise the idea of the proof for unbounded linear operators. We introduce the unitary $$V: H \oplus K \to K \oplus H: (\xi, \eta)\mapsto (\eta, -\xi).$$ Then, one shows the equality $$G(T^*) = V(G(T))^\perp$$ where $G(T)$ is the graph of $T$ and $G(T^*)$ is the graph of $T^*$. From this, it is clear that $G(T^*)$ is a closed subset of $K \oplus H$.

Is there a way to repair this argument for antilinear operators? If $T$ is antilinear and densely defined, I can prove that $T^*$ has closed graph using an argument with the 'adjoint' Hilbert space where the scalar multiplication is conjugated and reduce it to the linear case where I already know the result. However, I'm interested to see if we can save the above argument used to prove the linear case or do something similar and give a direct proof.

Thanks in advance for any ideas or suggestions.

Answer 1

I'm not sure what you can expect here. Notice that if $T$ is antilinear then defining the "graph" as $$ G(T) = \{ (T\xi, \xi) : \xi\in D(T) \} $$ does not give a subspace: it's not closed under (complex) scalar multiplication. The obvious way to fix this is to consider $T$ as a linear map $H\supseteq D(T)\rightarrow \overline K$, that is, use the conjugate Hilbert space construction.

However, the OP asks for other (not using the conjugate Hilbert space construction) options. If all you want to prove is that $T$ densely-defined implies $T^*$ is closed, this can be argued directly. First, notice that $T$ being densely-defined is required to show that $T^*$ is well-defined. By translating, we need only show that $T^*$ is closed at $0$. That is, if $(\eta_n)$ is a sequence in $D(T^*)$ with $\eta_n\rightarrow 0$ and $T^*(\eta_n)\rightarrow\alpha$, we wish to show that $\alpha=0$. However, then $$ \langle \alpha,\xi \rangle = \lim_n\langle T^*\eta_n,\xi\rangle=\lim_n \langle T\xi, \eta_n\rangle =0 \qquad (\xi\in D(T)). $$ As $D(T)$ is dense, this shows that $\alpha=0$.

---

In the comments, it's asked if we can show $D(T^*)$ is densely-defined when $T$ is closed. I don't know how to do this if you want to avoid the conjugate Hilbert space. The usual proof has at its heart the fact that if $V\subseteq H$ is a subspace of a Hilbert space then $V^{\perp\perp}$ is the closure of $H$. The analogous result for (say, reflexive) Banach spaces uses Hahn-Banach. You want to apply this to $H\oplus\overline K$, and I'm not sure how to avoid this. (Or to do something terribly artificial, essentially sneaking in the conjugate Hilbert space.)