Given a Hilbert space.

Symmetric operators can be described by $$\overline{\mathcal{D}(A)}=\mathcal{H}:\quad A\subseteq A^*\iff\langle A\varphi,\psi\rangle=\langle\varphi,A\psi\rangle\quad(\varphi,\psi\in\mathcal{D}(A))$$

So that essentially selfadjoint operators can be characterized by: $$A\subseteq A^*:\quad\overline{A}=A^*\iff A^*\subseteq A^{**}$$ (The existence of adjoints being implicitely required.)

Does it fail if symmetry assumption on the original operator is dropped?

More short, can it happen: $A^*\subsetneq A^{**}=\overline{A}$

  • $\begingroup$ Let $A$ be any closed densely-defined linear operator with $A \prec A^{\star}$ ($\prec$ denotes strict graph inclusion.) Then $B=A^{\star}$ satisfies $B^{\star} \prec B^{\star\star}=B=\overline{B}$. $\endgroup$ – DisintegratingByParts Feb 15 '15 at 22:41
  • $\begingroup$ @T.A.E.: Ah right so any not (essentially) selfadjoint operator would serve as an example. $\endgroup$ – C-Star-W-Star Feb 15 '15 at 22:54
  • $\begingroup$ @T.A.E.: Wanna give your comment as short answer? $\endgroup$ – C-Star-W-Star Feb 15 '15 at 23:37

Let $B : \mathcal{D}(B)\subset X \rightarrow X$ be any closed densely-defined linear operator on a Hilbert space $X$, with $B \prec B^{\star}$ (I am using $\prec$ to denote strict graph inclusion.) Then $A=B^{\star}$ satisfies $A^{\star} \prec A^{\star\star}$. Indeed, because $B$ is closed and densely-defined, then $B^{\star\star}=\overline{B}=B$, which gives the strict graph inclusion $$ A^{\star} =(B^{\star})^{\star}=B \prec B^{\star}= A=\overline{A}=A^{\star\star}, $$ and is as you stated: $$ A^{\star} \prec A^{\star\star} = \overline{A}. $$ An example is $X=L^{2}[0,\infty)$ and $B=-i\frac{d}{dx}$ defined on the domain $\mathcal{D}(B)$ consisting of all absolutely continuous functions $f \in X$ for which $f' \in X$ and $f(0)=0$. Then, $$ (Bf,g) = (f,Bg),\;\;\; f,g\in \mathcal{D}(B). $$ The operator $B$ is closed and densely-defined with adjoint $B^{\star}=-i\frac{d}{dx}$ on the domain $\mathcal{D}(B^{\star})$ consisting of all absolutely continuous $f \in X$ for which $f' \in X$ (without any endpoint condition at $0$.) The opertor $B$ is symmetric, but $A=B^{\star}$ is not.


Any essentially selfadjoint operator is symmetric. So, if you consider as definition for essential selfadjointness $\bar{A}=A^*$ then, automatically, $A\subseteq A^*$ ($A$ is symmetric) since $A$ is contained in its closure $\bar{A}$.

The converse is not true. Some counter-examples and theoretical results can be found here: http://www.math.umn.edu/~garrett/m/fun/adjointness_crit.pdf.

  • $\begingroup$ Yes but the converse might fail. $\endgroup$ – C-Star-W-Star Feb 15 '15 at 16:28
  • $\begingroup$ Of course, there are examples of symmetric operators that are not essentially selfadjoint. $\endgroup$ – aly Feb 15 '15 at 16:31
  • $\begingroup$ Yes but the symmetry is on the adjoint: $\overline{A}=A^*\iff A^*\subseteq A^{**}$ $\endgroup$ – C-Star-W-Star Feb 15 '15 at 16:32
  • $\begingroup$ Does it also address the case: $A^*\subsetneq\overline{A}$ $\endgroup$ – C-Star-W-Star Feb 15 '15 at 16:39
  • $\begingroup$ Yes, if you use $A^*\subseteq\bar{A}$ as a definition then you must also add that $A$ is symmetric. This is because, when you pass to adjoints in $A\subseteq A^*$ you will get the converse inclusion $\bar{A}\subseteq A^*$. $\endgroup$ – aly Feb 15 '15 at 16:49

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