I am studying closed/closable operators and their adjoints and I am confused:
is it true to say that the domain of the adjoint of an unbounded linear functional is necessarily $\{0\}\subset \mathbb{R}$?
Let $H$ be a Hilbert space and let $T\colon D(T)\subset H\to \mathbb{R}$ be a densely defined linear operator.
- If $T$ is unbounded on $H$, then $T^*$ is not densely defined:
See e.g., Kato's book "Perturbation Theory for Linear Operators" Problem III.5.18
Every closable operator with finite rank is bounded. (Thus an unbounded linear form is never closable.)
On the other hand,
- For $T$ densely defined, the adjoint operator $T^*$ from $\mathbb{R}$ to $H$ is densely defined if and only if $T$ is closable (e.g., Pedersen, Analysis Now, Theorem 5.1.5.).
But since $$D(T^*)=\{ r \in \mathbb{R}\mid \forall h\in D(T),\exists g\in H\text{ such that } r(Th)=(g,h)_H \},$$ if $0\neq r_1\in D(T^*)$ then for $r_2$ we have $ r_2(Th)=(\frac{r_2}{r_1}g,h)_H$. Hence, if there is one non-zero element $r$ in $D(T^*)$, then we have $D(T^*)=\mathbb{R}$, isn't it?
But $D(T^*)=\mathbb{R}$ will imply that $T$ is closable, and because its range is $\mathbb{R}$, it's bounded.
Does it mean that if $T$ is an unbounded linear functional, the domain of its adjoint is necessarily $\{0\}\subset \mathbb{R}$? (Similar argument works for $\mathbb{R}^n$ for the range of $T$)
Am I missing something?
If $T:D(T)\to\Bbb K^n$ is unbounded then the domain of $T^*$ will be some subspace that is not dense.
If $n=1$ there are precisely two linear subspaces of $\Bbb K^n$, namely $\{0\}$ and $\Bbb K$. Since $\Bbb K$ is dense this leaves just $\{0\}$. On the other hand if $n\geq2$ there are many more subspaces, and all except $\Bbb K^n$ are not dense. In fact any proper subspace $V\subset \Bbb K^n$ can be realised as the domain of the adjoint of some unbounded operator $T:D(T)\to\Bbb K^n$.
Example. Let $f_1:D\to \Bbb K$ be an unbounded functional and $f_2:H\to \Bbb K$ a bounded functional. Then $(f_1,f_2):D\to \Bbb K^2$, $x\mapsto (f_1(x), f_2(x))$ has $$\{0\}\times\Bbb K\to H, \qquad (0,y)\mapsto f_2^*(y)$$ as an adjoint, since $$\langle(f_1,f_2)(x), (0,y)\rangle_{\Bbb K^2} = f_2(x)\cdot y = \langle x,f_2^*(y)\rangle_H$$ holds for all $(0,y)\in\{0\}\times \Bbb K$.