# Spectral theorem for unbounded self-adjoint operators on REAL Hilbert spaces

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In all books that I have checked the spectral theorem (every self-adjoint unbounded operator on a Hilbert space is unitary equivalent to a multiplication operator on some $L_2(\mu)$) is only stated for complex Hilbert spaces (and the use of the Cayley transformation for the reduction to the bounded case requires indeed complex scalars). However, since the spectrum of a a self-adjoint operator is real the theorem should (or could) be true in real Hilbert spaces. I could imagine an argument by complexification but there are a number of things to do, in particucar, I think that the unitary operator between the complexified Hilbert space and the complex $L_2(\mu)$ does not automatically map the original real Hilbert space to the real-valued functions). Hence the question:

Is the the spectral theorem in real Hilbert spaces true? (In this case I would also like to have a reference.)

• If you use the version of the spectral theorem where the operator is written as an integral of its spectral measure then there is no longer any problems with comparing the initial Hilbert space and the space of real valued functions. – Simon Henry Jan 17 '14 at 10:30
• Try the brilliant proof by Leinfelder in his article "A geometric proof of the spectral theorem for unbounded self-adjoint operators" in Math. Ann. vol. 242 which does not use the complexity of the Hilbert space. – 7891user Jan 17 '14 at 10:39
• Although the complexification is not the unique way, it is a general and natural tool, so I'd suggest to consider it fixing all details. – Pietro Majer Jan 17 '14 at 10:52
• I am not so sure whether things work out well: link.springer.com/article/10.1007%2FBF01364454#page-1. – Marc Palm Jan 17 '14 at 12:02
• Yes because, as you mentioned the spectrum is real and if $U$ is a subset of $\mathbb{R}$ then $E_{x,y}(U)$ is defined as $\langle x ,P y \rangle$ where $P$ is the spectral projection of $T$ corresponding to $U$, and as far as i know you don't need complex number to define them. – Simon Henry Jan 17 '14 at 13:16

Suggestion: As you point out, Leinfelder does in fact use Cayley. However, he shows that if you have a self-adjoint operator, then you can represent the space as a Hilbert sum of closed subspaces, each of which is invariant under $A$ (or any operator which commutes with it), and on which it is bounded. If you apply this to the complexification of $A$ as in your problem, then you get a corresponding splitting. Once you have that, then it is a simple matter to get the result you are looking for.