If $X$ is a Hilbert space and $A$ is an unbounded self-adjoint operator on $X$, is it necessarily that $A^2$ is self-adjoint as well?(admittedly, $A^2$ is densely defined)

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    $\begingroup$ How do you define "self-adjoint" for an operator on a Banach space? $\endgroup$ Apr 27, 2012 at 3:11
  • $\begingroup$ Dunno if this is standard, but you could try to extend the definition by analogy with the SVD. Say that $A:X \rightarrow X$ is "self adjoint" if there exists a bijective isometry $U:X \rightarrow X$ and a multiplication operator $\Sigma:X \rightarrow X$ such that $A=U \Sigma U^{-1}$. $\endgroup$
    – Nick Alger
    Apr 27, 2012 at 5:37
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    $\begingroup$ @NickAlger, it is indeed nonstandard (if only for the reason that there is no standard notion of "multiplication operator" on a Banach space). $\endgroup$ Apr 27, 2012 at 6:21
  • $\begingroup$ Isn't there a theorem that for every Banach space $X$, there exists a topological vector space $S$ such that $X$ is equivalent to a subspace of $B(S)$ (the space of bounded continuous functions on $S$ characterized by the sup-norm)? In that case one might try using multiplication operators on $B(S)$ where they can be unambiguously defined.. $\endgroup$
    – Nick Alger
    Apr 27, 2012 at 22:32

1 Answer 1


Yes. More generally whenever $A$ is closed and densely defined operator on Hilbert space, the operator $A^* A$ is also closed and densely defined, and in fact is self-adjoint. (Wikipedia calls this von Neumann's theorem but I do not think this is a standard name for it.)

As plausible as this sounds, it is highly nontrivial. (If you think a little bit about what is actually being asserted, it is far from obvious why the domain of $A^* A$ must include even one nonzero vector, let alone be dense.)

For the details of this and a great deal more on unbounded operators, consult your local library. Any textbook that treats unbounded operators on Hilbert space from a mathematically coherent point of view (and not simply to motivate applications to physics or some non-analysis-oriented field) should include a proof. It is key, for example, in constructing the polar decomposition of a closed operator.

There is also a functional calculus for unbounded self-adjoint operators of which this discussion is a very special case (if $f$ is any real-valued function on $\mathbb{R}$ one can, with a great deal of work, give meaning to a self-adjoint operator $f(A)$ in a way that respects the usual algebra of functions, and in particular, in a way that assigns to $f(x) = x^2$ the operator $A^2$).

A standard reference for this kind of thing is Volume 1 of Reed and Simon's "Methods of modern mathematical physics" but many other books cover it too. The Wikipedia entry for unbounded operator is also worth a look (although much of the mathematical internet's treatment of unbounded operators is nonrigorous, it is an OK start).


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