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.



Antilinear unbounded operator has closed graph
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 $\alpha=0$.




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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.)


