# Self adjoint operators on incomplete spaces

I am actually a physics student.

The definition of self adjoint operators which I have studied is

Definition. A densely defined $A: D_A \rightarrow \mathcal{H}$ is self adjoint if it coincides with its adjoint, where the adjoint is given by $$D_{A^*} = \{\psi \in \mathcal{H}|\exists \eta \;\forall \alpha \in D_A: \langle \psi, A\alpha\rangle = \langle \eta, \alpha\rangle \}$$ and $A^* \psi = \eta$

This suggests to me that the underlying space $\mathcal{H}$ needs to be a Hilbert space, necessarily, since

(i) the domain lies dense in $\mathcal{H}$ (thus we need complete space (?))

(ii) The definition uses inner product structure

Is this true, or can we still define them on incomplete spaces? The reason I ask this is, suppose I see a random differential equation (excuse me for the typical physics notation) $$D_x f(x) = g(x);\; x\in (a,b)$$ Given nothing more, the reference I use says $$\int_a^b dx |x\rangle\langle x| = 1$$ which is actually analogue of something like integration over projection valued measure (of some self adjoint operator) $$\int_\mathbb{R} dP = id_{\mathcal{H}}$$

According to my knowledge, if we can't define self adjoint operators (here, projection valued measure) without Hilbert space structure, this cannot be done. And if indeed this is to be true, $g \in L^2(a,b)$ and differential operator must be defined on domain of $L^2$.

Can you tell whether I am right or wrong( and why?)

You can get around using complete spaces for classical problems where separation of variables works. As an example, consider the following on $L^2[0,2\pi]$ consisting of continuous functions: $$A = \frac{1}{i}\frac{d}{dx}$$ The domain of $A$ may be defined as all continuously differentiable functions $f$ on $[0,2\pi]$ with $f(0)=f(2\pi)$. This $L^2[0,2\pi]$ space is not complete, but it is a valid inner product space, without having to group according to equivalence classes of functions. And $A$ has a classical resolvent operator $(A-\lambda I)^{-1}$ that can be explicitly obtained by solving a classical ODE: $$g = (A-\lambda I)^{-1}f \iff -ig'-\lambda g=f,\; g(0)=g(2\pi).$$ This has a unique solution for $\lambda\ne 0,\pm 1,\pm 2,\cdots$ given by $$(A-\lambda I)^{-1}f = \frac{e^{i\lambda x}}{e^{-2\pi i\lambda}-1}\int_{0}^{2\pi}ie^{-i\lambda t}f(t)dt+e^{i\lambda x}\int_{0}^{x}ie^{-i\lambda t}f(t)dt.$$ The resolvent has simple poles at the eigenvalues $\lambda=0,\pm 1,\pm 2,\cdots$ with residues $$R_n f = \left(\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-in t}dt\right)e^{inx}$$ You will no doubt recognize this as the projection of $f$ onto the eigenfunction $e^{inx}$. It is Stone's formula that connects the resolvent with the spectral object: $$\frac{1}{2}(E[a,b]f+E(a,b)f) \\=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}(A-(u-iv)I)^{-1}f-(A-(u+iv)I)^{-1}f du$$ And this formula allows you to construct the spectral measure on all finite intervals in a classical way. The above works for all sefaldjoint operators $A$ on a Hilbert space. The convergence of the limit is in $L^2$, but this stays classical in classical ODE problems, even when there is continuous spectrum. Some of the first general classical proofs of convergence of Fourier expansions relied on this method. For the continuous spectrum case on $(-\infty,\infty)$, the problem is formulated with continuous $f\in L^2$ as $$-ig-\lambda g = f,\;\; g\in L^2(\mathbb{R}).$$ This gives continuous spectrum equal to $\mathbb{R}$ and Stone's formula gives $$E[a,b]f = \mathcal{F}^{-1}(\chi_[a,b]\mathcal{F}f),$$ where $\mathcal{F}$ is the Fourier transform. The substitute for complete spaces is the existence of the resolvent operator on the full space of continuous functions, for example.
The Hilbert space constructions for a general selfadjoint operator force you into complete spaces, but still the derivation of the spectral measure on an interval follows Stone's formula, and the spectral integrals may be formed from limits of classical Riemann integrals of vector-valued holomorphic functions $R(\lambda)f=(A-\lambda I)^{-1}f$.
• @physicophilic : That's correct: the example is essentially selfadjoint in the context of the Hilbert space $L^2$. Selfadjoint is not completely meaningful when working in incomplete spaces, such as the one I suggested. Requiring symmetry and the existence of the resolvent $R(\lambda)$ for non-real $\lambda$ provides a way to circumvent Hilbert space and selfadjointness. By the way, the resolvent was the earliest way of dealing with this. Fredholm used the resolvent as a foundation; spectral theory, compactness and complex analysis of the resolvent all grew out his work on integral equations. Jan 26, 2018 at 7:09