3
$\begingroup$

I've encountered the following:

Consider the usual Hilbert space $L^2([0,1],dx)$ and the dense subspace $\mathcal{D}=\mathcal{C}[0,1]$. Define $T$ on $\mathcal{D}$ by $T(f)=f(0)$. This is a densely defined operator, but it its adjoint is not densely defined.

I'm not so familiar with computing adjoints. Could someone give me a hint how one can find and see that the adjoint is not densely-defined?

I'm also interested if there are other 'simple' examples of non-closable operators

Thanks

$\endgroup$
2
$\begingroup$

Generally, given $T\colon \mathcal{D}(T) \to H_2$, where $H_1, H_2$ are Hilbert spaces, $\mathcal{D}(T)$ is a dense subspace of $H_1$, and $T$ is a linear operator, the adjoint of $T$ is defined on the subspace $\mathcal{D}(T^\ast) \subset H_2$ of elements $y$ such that there exists a $z\in H_1$ with $\langle Tx,y\rangle_2 = \langle x, z\rangle_1$ for all $x\in \mathcal{D}(T)$, then $T^\ast(y) = z$. In short,

$$\langle Tx,y\rangle_2 = \langle x, T^\ast y\rangle_1$$

for all $x\in \mathcal{D}(T),\, y \in \mathcal{D}(T^\ast)$. The denseness of $\mathcal{D}(T)$ ensures the well-definedness of $T^\ast$.

In the situation at hand, the codomain of $T$ is $\mathbb{K}$ (whether that's $\mathbb{R}$ or $\mathbb{C}$ doesn't matter), so there are two possibilities for $\mathcal{D}(T^\ast)$; it can be either $\{0\}$ or $\mathbb{K}$. If $\mathcal{D}(T^\ast) = \mathbb{K}$, then $T^\ast\colon \mathbb{K}\to L^2$ is continuous, and hence it has a continuous adjoint $T^{\ast\ast}\colon L^2 \to \mathbb{K}$. But we then have $T \subset T^{\ast\ast}$, so $T$ itself would be continuous. The given $T\colon f \mapsto f(0)$ is not continuous, hence $\mathcal{D}(T^\ast) = \{0\}$, i.e. $T^\ast$ is not densely defined.

The argument shows, with minor modifications, that a densely defined operator with finite-dimensional codomain has a densely defined adjoint if and only if it is continuous, since the only dense subspace of a finite-dimensional Hausdorff topological vector space is the entire space, and every linear operator $\mathbb{K}^n \to V$, where $V$ is a topological vector space, is continuous.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.