# An example of a non-closable operator

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

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.