# Is this counterexample for $T$ closed, symmetric $\iff T$ self-adjoint valid?

Lemma Let $$T: \text{dom}(T) \to \mathscr{H}$$, where $$\mathscr{H}$$ is a Hilbert space, be densely defined. Then $$T$$ is closed and symmetric if and only if $$T = T^{**} \subset T^*$$.

To better get a feeling for this statement I wanted to construct an operator $$T$$ which was closed and symmetric but not self-adjoint. Is the following valid to see the necessity of "densely defined"

Let $$T: D\to \mathscr{H}$$, where $$D$$ is not closed (then by the closed graph theorem we get that $$T$$ is closed, right?) be the zero operator, i.e. $$T x = 0$$ for all $$x \in D$$. Then $$\langle T x, y \rangle = \langle 0, y \rangle = 0 = \langle x, 0 \rangle = \langle x, T y \rangle$$ holds for all $$x,y \in D$$, hence $$T$$ is symmetric.

Furthermore, by definition, $$\text{dom}(T^*) = \{ y \in \mathscr{H}: x \mapsto \langle T x, y \rangle \text{ is continuous on } D\}$$ Let $$\Phi: D \to \mathbb{C}$$, $$x \mapsto \langle T x, y \rangle$$. Then $$\Phi \equiv 0$$, and is trivially continuous, hence dom$$(T^*) = \mathscr{H} \supsetneq D$$. Therefore, we can not have $$T = T^*$$, i.e. $$T$$ is not self-adjoint.

Questions

1. Is this counterexample valid?
2. How could $$D$$ look like?
• The zero operator is closed only if its domain is closed. Mar 6 '20 at 22:14
• @s.harp So the counterexample is valid if I choose $D$ to be a closed strict subset of $\mathscr{H}$? Mar 6 '20 at 22:17
• And ... because the operator has to be densely-defined, then ... Mar 7 '20 at 0:06

The original study of symmetric operators started with differential operators. The differentiation operator $$\partial=\frac{1}{i}\frac{d}{dx}$$ is a good example, with different properties on $$L^2(-\infty,\infty),L^2[0,\infty), L^2[0,2\pi]$$.
Because of needing complete spaces, differentiability is a little tricky to define. For example, how do you define a differentiable function when functions are the same if they are changed on a set of measure $$0$$, which could change where or if the function is differentiability. The problem is that a function $$f\in L^2[0,\infty)$$ is really an equivalence class $$[f]$$ of functions that are equal a.e. to the given function $$f$$. The standard way of handling this is to say that $$[f]\in\mathcal{D}(D)$$ on $$[0,\infty)$$ (or the other intervals,) if there is an element $$g$$ of the equivalence class $$[f]$$ such that $$g$$ is absolutely continuous on $$[0,\infty)$$ with $$g'\in L^2[0,\infty)$$. If there are two such elements in the equivalence class, they're actually the same function. So everything is well-defined. And so is the value of $$g(0)$$. Therefore it makes sense to talk about $$f(0)$$, provided this sort of meaning is understood.
Then it makes sense to define $$\partial : \mathcal{D}(\partial)\subset L^2[0,\infty)\rightarrow L^2[0,\infty)$$ by $$\partial f=-if'$$. If you impose the condition $$f(0)$$ on $$\mathcal{D}(\partial)$$, then you have an operator that is symmetric, but not selfadjoint. It is not selfadjoint because the adjoint $$\partial^*$$ is the same operator, but without the requirement that the functions in the domain vanish at $$x=0$$. It is easy to verify through integration by parts that $$\langle \partial f,g\rangle = \langle f,\partial^*g\rangle,\;\;\;f\in\mathcal{D}(\partial),g\in\mathcal{D}(\partial^*).$$ The operator $$\partial$$ is closed, densely-defined and symmetric, but it is not selfadjoint because the adjoint $$\partial^*$$ has a strictly larger domain.