For an assignment I have to show that exists a densely defined operator on a infinite dimensional separable Hilbert space, such that its adjoint is the zero operator on the zero subspace.
To show this there is a reference to exercise 13.3 in the book Functional Analysis of Rudin. The exercise is as follows
By theorem 13.8, $\mathscr{D}(T^*) =\{0\}$ for a densely defined operator $T$ in $H$ if and only if $\mathscr{G}(T)$ is dense in $H\times H$. Show that this can actually happen. Suggestion: Let $\{e_n:n=1,2,3, \dots\}$ be an orthonormal basis of $H$; let $\{x_n\}$ be a dense subset of $H$; define $Te_n=x_n$; and extend $T$ linearly to $\mathscr{D}(T)$, the set of all finite linear combinations of the basis vector $e_n$. Show that the graph of this $T$ is dense in $H \times H$.
The exercise has been proved here. So now we know that $\mathscr{G}(T)$ is dense in $H \times H$. Then $V\,\mathscr{G}(T)$ is also dense, where $V$ is given by $V\{a,b\}=\{-b,a\}$. According to theorem 13.8 we have
$ \mathscr{G}(T^*) = [V\,\mathscr{G}(T)]^{\perp}. $
Since $V\,\mathscr{G}(T)$ is dense we have that $\mathscr{G}(T^*)$ must be the zero vector. Since
$ \mathscr{G}(T^*) = \{\,\{x,T^*x\} : x \in \mathscr{D}(T^*)\}, $
we can conclude that $\mathscr{D}(T^*) = 0$ and $T^*x=0$, so that $T^*$ is the zero operator. I thought that I'm done here, however the original assignment has the remark:
Once you have the correct idea and also realize that $\lim_{n \rightarrow \infty}(x,e_n)=0$ for all $x\in H$, the solution is almost immediate.
If $y \in \mathscr{D}(T^*)$ then there exists a $T^*y \in H$ s.t.
$ (Tx,y) = (x,T^*y) \qquad [x \in \mathscr{D}(T)] $
and $x \rightarrow (Tx,y)$ must be continuous in $x$. Letting $x=e_n$ we get
$ (x_n,y) = (e_n,T^*y) $
then using the remark above we see that $(x_n,y) \rightarrow 0$. Using the continuity condition it follows that $\mathscr{D}(T^*)=\{0\}$.
If this is correct it is just another way of proving the assignment. Am I missing something here?
