I am not an expert in functional analysis but I was studying some, motivated from some mathematical physics considerations. I am not quite sure whether this is research-level, but let me state some context first:
An old result of functional analysis tells us that a symmetric (in the sense that $(Ax,y)=(x,Ay)$, for all $x,y \in H$), unbounded operator $A$, acting on a Hilbert space $H$, cannot be defined on the whole space but only in a dense subspace of it. This is a direct consequence of the Hellinger-Toeplitz theorem. (see also: Riesz-Nagy, "Functional Analysis", 1955, p.296 and also Reed-Simon, "Methods of Modern Mathematical Physics", 1975, p.84). Since the operators of interest in physics are self-adjoint (and thus symmetric) they fall into this.

On the other hand, it is well known that any linear map from a subspace of a Banach space $X$ to another Banach space $Y$ can be extended to a linear map $X\to Y$ defined on the whole of $X$ using Zorn's Lemma (see for example: Unbounded linear operator defined on $l^2$).

Now the question is: Since the extension of a linear, unbounded operator to the whole of the space through the AC, will produce a -still- unbounded, linear operator, does the previous remark imply that the extension of linear, self-adjoint, unbounded operators on the whole of the space, produces non-self-adjoint operators? What would be a concrete relevant example?

Related question: Invertible unbounded linear maps defined on a Hilbert space

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    $\begingroup$ Any extension or restriction of a self-adjoint operator produces a non-self-adjoint operator (whether to the whole space or not). $\endgroup$ – Christian Remling Jan 27 '17 at 16:00
  • $\begingroup$ you mean any self-adjoint operator with no further assumptions ? $\endgroup$ – Konstantinos Kanakoglou Jan 29 '17 at 0:54
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    $\begingroup$ @KonstantinosKanakoglou: Yes, just as in Robert Israel's answer. You can show, directly from the definition of adjoint, that if $B$ is a proper extension of $A$, so that $D(A) \subsetneq D(B)$ and $A=B$ on $D(A)$, then $D(B^*) \subseteq D(A^*)$. So if $A$ is self-adjoint, we have $D(B^*) \subseteq D(A^*) = D(A) \subsetneq D(B)$; i.e. $D(B^*) \ne D(B)$. A similar argument applies to restrictions. $\endgroup$ – Nate Eldredge Jan 29 '17 at 2:22

Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. (In fact you can produce many such operators; the extension is highly non-unique.) By Hellinger–Toeplitz, $A'$ cannot be symmetric. So it definitely isn't self-adjoint. Another way to see this that $A'$ cannot be self-adjoint is to note that, by the closed graph theorem, $A'$ cannot be closed.

Since you used Zorn's lemma in an essential way, you won't get a "concrete" description of such an $A'$. There's a strong sense in which this is true. A common working definition of "concrete" is "something whose existence you can prove using only the axiom of dependent choice (DC)". There's a famous theorem of Solovay (extended by Shelah) that it's consistent with DC that every set of reals has the property of Baire (BP); i.e. there are models of set theory in which DC and BP both hold (but full AC necessarily fails). But from BP you can prove that every everywhere defined operator on any Banach space is bounded. So in such models, $A$ won't have any extension to all of $H$. Put another way, you can't even prove $A'$ exists without using AC in an essential way, so you certainly can't construct it concretely.

You can read more about these ideas in Schechter's Handbook of Analysis and its Foundations.

  • $\begingroup$ Thank you for analyzing the word "concrete" in this sense. And for the reference as well. I was not aware of it and it seems quite interesting! $\endgroup$ – Konstantinos Kanakoglou Jan 27 '17 at 2:54
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    $\begingroup$ Regarding to the last sentence of your first paragraph, since non symmetric implies non self-adjoint, one doesn't really need the closed graph theorem.Hellinger–Toeplitz would be enough. Do i miss something? $\endgroup$ – Konstantinos Kanakoglou Jan 29 '17 at 0:50
  • $\begingroup$ Nope - I'm giving two separate reasons why your operator cannot be self-adjoint. $\endgroup$ – Nate Eldredge Jan 29 '17 at 1:40
  • $\begingroup$ ok! thanks for editing. it is clear now. $\endgroup$ – Konstantinos Kanakoglou Jan 29 '17 at 4:45

Yes, of course. By definition of the adjoint operator, $\{[-A^* y, y]: y \in \mathscr D(A^*)\}$ is the orthogonal complement in $H \oplus H$ of the graph $\{[x, Ax]: x \in \mathscr D(A)\}$ of $A$. Thus if $A$ is unbounded and self-adjoint, i.e. $A = A^*$, the adjoint of any extension of $A$ would have a smaller domain than that of $A$.

It's fairly easy to give concrete examples with an extension by finitely many dimensions. However, an extension to the whole space requires something like the Axiom of Choice, so you won't get a concrete example of that.


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