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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?

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simon
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  • What exactly was the wording of the original assignment? Did it directly refer to exercise 13.3, or was it only "Show that there is an operator with $\mathscr{D}(T^\ast) = {0}$", and then the remark? Once you have shown that the graph of the $T$ constructed in Rudin's suggestion is dense, you are indeed done. – Daniel Fischer Dec 29 '13 at 17:14
  • It indeed referred directly to exercise 13.3. That's why I thought that the suggestion in exercise 13.3 doesn't show that $T^$ is also the zero operator only that $\mathscr{D}(T^)$ is zero. – simon Dec 29 '13 at 17:38
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    The adjoint is linear on its domain. So, if the adjoint has the trivial domain {0}, then the adjoint must be 0 on that domain. – Disintegrating By Parts Dec 30 '13 at 04:43

1 Answers1

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If $\{a,b\}$ is an element in the orthocomplement of $\mathscr{G}(T)$, then for $x\in\mathscr{D}(T)$ we have $(x,a)+(Tx,b)=0$. In particular, for $x=e_n$ you get $(e_n,a)+(x_n,b)=0$. Taking the limit $n\rightarrow\infty$, we see that the first term cancels because of the remark. What's left is $\lim_{n\rightarrow\infty}(x_n,b)=0$. Using the density of the set $\{x_n\}$ it's easy to show that this implies $b=0$. Indeed: just take a sequence $x_n$ which converges to $b$.

This just leaves $(x,a)=0$, from which $a=0$ follows since $\mathscr{D}(T)$ is dense. So the orthocomplement of $\mathscr{G}(T)$ is just the zero subspace, from which the density of $\mathscr{G}(T)$ follows.

ScroogeMcDuck
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