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



1 Answer 1


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.


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