A symmetric half-line momentum with no self-adjoint extension

Given: The half-line Hilbert space and operator in the Statement.

Countable Choice, the Axiom of Choice, and Dependent Choice are the hypotheses of the interval Sobolev and integration-by-parts suppliers. The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) The Axiom of Choice The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain

$H^1(0,\mathrm{\infty })$ consists of the $L^2$ classes with weak derivative in $L^2$. On each bounded subinterval the class has the unique absolutely continuous representative of One-dimensional Sobolev spaces on intervals and Absolutely continuous representative of an L2 function with L2 weak derivative. Uniqueness makes these representatives agree on overlaps, defining one locally absolutely continuous representative on $(0,\mathrm{\infty })$. The half-line integration-by-parts identity is clause (d) of Integration by parts and the endpoint-vanishing space on intervals.

The adjoint is characterized by $⟨S_{c}f,g⟩=⟨f,S_c^{\ast }g⟩$ on the dense domain $C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$; a densely defined symmetric operator is closable and the adjoint of its closure equals its adjoint (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators, Closability is equivalent to density of the adjoint domain). The weak derivative convention is the test-function identity of Weak derivative of a locally integrable function.

Absolute continuity on a compact real interval means that small total length of a finite family of pairwise interior-disjoint subintervals forces small total endpoint increment; complex-valued absolute continuity is componentwise (Absolute continuity on a compact interval, One-dimensional $W^{1,p}$ functions have unique absolutely continuous representatives). For real $x$, the complex exponential equals the real exponential, and the complex entire theorem gives its derivative as itself (The complex exponential is entire and its complex derivative is itself, $\exp (z+w)=\exp z\exp w$, and the complex exponential extends the real exponential). The real derivative theorem and real chain rule give $(e^x{)}^{\mathrm{\prime }}=e^x$ and $(e^{-x}{)}^{\mathrm{\prime }}=-e^{-x}$ (The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$). The $H^1$ representative's classical derivative equals its weak derivative a.e., and its increments are the integrals of that derivative, by the representative result cited above. On each compact interval, the real exponential is $C^1$ with bounded derivative, hence Lipschitz and absolutely continuous by $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation. The exponential times the representative is then absolutely continuous by The product of two absolutely continuous functions is absolutely continuous applied to real and imaginary parts. At every point where the representative is differentiable, the real product rule from Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$ gives the derivative of that product; hence the formula holds almost everywhere. For a real absolutely continuous function, zero derivative almost everywhere implies constancy by the fundamental theorem of calculus (Fundamental theorem of calculus for absolutely continuous functions); apply this to the real and imaginary parts for complex-valued products.

For a closed densely defined symmetric operator, a self-adjoint extension exists exactly when its two deficiency indices agree (Closed symmetric operators and deficiency data, Unitary deficiency parameterization at a reference scale).

$C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$ is dense in $L^2(0,\mathrm{\infty })$ by the smooth-test density theorem on open subsets of $\mathbb{R}$ (Smooth compactly supported functions of an open set are dense in $L^2$).