# The exponent of self-adjoint operator

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)

• How do you define "self-adjoint" for an operator on a Banach space? Apr 27, 2012 at 3:11
• 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 $A=U \Sigma U^{-1}$. Apr 27, 2012 at 5:37
• @NickAlger, it is indeed nonstandard (if only for the reason that there is no standard notion of "multiplication operator" on a Banach space). Apr 27, 2012 at 6:21
• 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.. Apr 27, 2012 at 22:32

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.)
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 $A^2$).