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.


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


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