Unbounded Self Adjoint Operators and Stones Theorem

Throughout this page $H$ is a complex Hilbert space (Hilbert space) with inner product linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

A (possibly unbounded) linear operator on $H$ is a linear map $T:D(T)\to H$ (Linear map between vector spaces over the same field) whose domain $D(T)\subseteq H$ is a linear subspace (Linear subspace of a vector space). The domain is part of the data: $T=S$ means $D(T)=D(S)$ and $Tx=Sx$ for all $x\in D(T)$. One writes $T\subseteq S$, and calls $S$ an extension of $T$, when $D(T)\subseteq D(S)$ and $Sx=Tx$ for every $x\in D(T)$.

The graph of $T$ is $\mathrm{\Gamma }(T):=\{(x,Tx):x\in D(T)\}\subseteq H\oplus H\mathrm{.}$ The direct sum $H\oplus H$ is read as the complex vector space of pairs with coordinatewise operations and the inner product $⟨(x,y),(u,v)⟩=⟨x,u⟩+⟨y,v⟩$, whose induced norm is $\mathrm{\parallel }(x,y)\mathrm{\parallel }=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2};$ it is complete because $H$ is (Complete metric space: every Cauchy sequence converges in the space). The graph norm on $D(T)$ is $\mathrm{\parallel }x{\mathrm{\parallel }}_T:=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Tx{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2},$ so that $x\mapsto (x,Tx)$ is an isometric isomorphism of $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ onto $\mathrm{\Gamma }(T)$ with the norm restricted from $H\oplus H$.

$T$ is closed when $\mathrm{\Gamma }(T)$ is a closed subset of $H\oplus H$. A linear operator is determined by its graph, and the following elementary translations are used silently below: $\mathrm{\Gamma }(T)$ is a linear subspace of $H\oplus H$; $\mathrm{\Gamma }(T)\cap (\{0\}\oplus H)=\{(0,0)\}$; the image of $\mathrm{\Gamma }(T)$ under the first coordinate projection is $D(T)$; and $T\subseteq S$ if and only if $\mathrm{\Gamma }(T)\subseteq \mathrm{\Gamma }(S)$. In particular a closed operator is exactly one whose graph is a closed subspace of $H\oplus H$.

Notation. No boundedness of $T$ is assumed, and $T$ is frequently called unbounded to stress this; a bounded everywhere defined operator on $H$ is the special case $D(T)=H$ in which $\mathrm{\Gamma }(T)$ is a closed subspace by continuity. The letter $I$ denotes the identity operator on $H$ with domain $H$.

Completeness of $H\oplus H$. If $(x_n,y_n)$ is a Cauchy sequence in $H\oplus H$, then $\mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }\le \mathrm{\parallel }(x_n-x_m,y_n-y_m)\mathrm{\parallel }$ and $\mathrm{\parallel }{y}_n-y_m\mathrm{\parallel }\le \mathrm{\parallel }(x_n-x_m,y_n-y_m)\mathrm{\parallel }$ show that $(x_n)$ and $(y_n)$ are Cauchy in $H$; let $x,y$ be their limits. Then $\mathrm{\parallel }(x_n,y_n)-(x,y){\mathrm{\parallel }}^2=\mathrm{\parallel }{x}_n-x{\mathrm{\parallel }}^2+\mathrm{\parallel }{y}_n-y{\mathrm{\parallel }}^2\to 0$, so $H\oplus H$ is complete. Consequently a sequence in $H\oplus H$ converges exactly when its two coordinate sequences converge in $H$.

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a linear operator on $H$ with domain $D(T)$ and graph $\mathrm{\Gamma }(T)$ (Unbounded linear operators: domain, graph and extension).

$T$ is densely defined when $D(T)$ is a dense subset of $H$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets); closed when $\mathrm{\Gamma }(T)$ is a closed subset of $H\oplus H$; and closable when $T$ has a closed extension. Recall that the graph norm is $\mathrm{\parallel }x{\mathrm{\parallel }}_T=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Tx{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$.

The graph-norm dictionary. The following three statements are part of the definition's content and are proved in the remarks below rather than assumed:

1. $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$ is a norm on $D(T)$, and $x\mapsto (x,Tx)$ is an isometric isomorphism of $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ onto $\mathrm{\Gamma }(T)$;
2. $T$ is closed if and only if $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space);
3. if $T$ is closed, then $D(T)$ is a Hilbert space for the inner product $⟨x,y{⟩}_T=⟨x,y⟩+⟨Tx,Ty⟩$ whose induced norm is $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$.

A linear subspace $D_0\subseteq D(T)$ is a core for a closed operator $T$ when $D_0$ is dense in $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$, equivalently when $\overline{\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})}=\mathrm{\Gamma }(T)$, where the closure is taken in $H\oplus H$ and $T{\mathrm{\mid }}_{D_0}$ denotes the restriction of $T$ to $D_0$.

Claim 1. $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T=\mathrm{\parallel }\cdot \mathrm{\parallel }$ holds on $\ker T$, and $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x{\mathrm{\parallel }}_T$ always. The map $J:x\mapsto (x,Tx)$ is linear, and $\mathrm{\parallel }x{\mathrm{\parallel }}_T=\mathrm{\parallel }Jx{\mathrm{\parallel }}_{H\oplus H}$. Hence homogeneity and the triangle inequality for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$ are the corresponding norm properties in $H\oplus H$ pulled back along $J$. It is definite because $\mathrm{\parallel }x{\mathrm{\parallel }}_T=0$ forces $\mathrm{\parallel }x\mathrm{\parallel }=0$ and $x=0$. The map $J$ is therefore linear and isometric, and its image is $\mathrm{\Gamma }(T)$ with $\mathrm{\parallel }(x,Tx){\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_T^2$; a linear isometry is injective, so it is a bijection onto $\mathrm{\Gamma }(T)$.

Claim 2. The space $H\oplus H$ is complete with $\mathrm{\parallel }(x,y)\mathrm{\parallel }$ (Complete metric space: every Cauchy sequence converges in the space, Hilbert space), while $\mathrm{\Gamma }(T)$ carries the subspace metric. If $\mathrm{\Gamma }(T)$ is closed, every Cauchy sequence in it converges in $H\oplus H$ and its limit remains in $\mathrm{\Gamma }(T)$, so the graph is complete. Conversely suppose $\mathrm{\Gamma }(T)$ is complete and $q\in \overline{\mathrm{\Gamma }(T)}$. Countable Choice selects $q_n\in \mathrm{\Gamma }(T)$ with $\mathrm{\parallel }{q}_n-q\mathrm{\parallel }<1\mathrm{/}(n+1)$ for every $n\in \mathbb{N}$. Then $(q_n)$ is Cauchy, so it converges to a point of $\mathrm{\Gamma }(T)$; uniqueness of metric limits makes that point $q$. Thus $\mathrm{\Gamma }(T)$ is closed. Since the map of claim 1 is an isometry onto $\mathrm{\Gamma }(T)$, the space $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ is complete exactly when $\mathrm{\Gamma }(T)$ is closed, that is, exactly when $T$ is closed.

Claim 3. If $T$ is closed then $\mathrm{\Gamma }(T)$ is a closed subspace of the Hilbert space $H\oplus H$, hence a Hilbert space in its own right, and the inner product $⟨(x,Tx),(y,Ty)⟩=⟨x,y⟩+⟨Tx,Ty⟩$ transfers to $D(T)$ through the isometry of claim 1, making $D(T)$ a Hilbert space with inner product $⟨x,y{⟩}_T$ and induced norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$.

Core. The isometry $J:x\mapsto (x,Tx)$ of claim 1 is a homeomorphism from $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ onto $\mathrm{\Gamma }(T)$ and maps $D_0$ onto $\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})$. Hence it carries the closure of $D_0$ onto the closure of $\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})$ inside $\mathrm{\Gamma }(T)$. Therefore $D_0$ is graph-norm dense exactly when $\overline{\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})}=\mathrm{\Gamma }(T)$; this topological argument does not replace density by sequential density.

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a linear operator on $H$ with domain $D(T)$ and graph $\mathrm{\Gamma }(T)$ (Unbounded linear operators: domain, graph and extension). Then the following are equivalent:

1. $T$ is closable (Densely defined, closed and closable operators, and cores);
2. whenever $x_n\in D(T)$, $x_n\to 0$ and $T{x}_n\to y$, one has $y=0$.

If either condition holds, then the closure of $\mathrm{\Gamma }(T)$ in $H\oplus H$ is the graph of a linear operator $\bar{T}$, the closure of $T$; it is the least closed extension of $T$, and $T$ is closed if and only if $\bar{T}=T$. Necessity of the hypothesis is never claimed: both conditions hold automatically for a closed operator.

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a densely defined linear operator on $H$, with domain $D(T)$ dense in $H$ (Densely defined, closed and closable operators, and cores, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets). Since $D(T)$ is dense, and since $H$ is complete (Hilbert space), every bounded linear functional on the normed space $(D(T),\mathrm{\parallel }\cdot \mathrm{\parallel })$ (A bounded linear operator between normed spaces) extends uniquely to a bounded linear functional $\tilde{\phi }$ on $H$ of the same norm, by setting $\tilde{\phi }(z)={\lim }_n\phi (z_n)$ for any sequence $z_n\in D(T)$ with $z_n\to z$ in $H$; the limit exists because $\phi$ is bounded on a dense subspace and $H$ is complete, and it does not depend on the sequence because a bounded functional is uniformly continuous.

A vector $y\in H$ belongs to the adjoint domain $D(T^{\ast })$ when the linear functional ${\phi }_y:D(T)\to \mathbb{C}$, ${\phi }_y(x)=⟨Tx,y⟩$, is bounded on $(D(T),\mathrm{\parallel }\cdot \mathrm{\parallel })$. In that case Hilbert space Riesz representation (Riesz representation for Hilbert spaces) applied to the extension $\widetilde{{\phi }_y}$ produces a unique vector $T^{\ast }y\in H$ with $T^{\ast }y\text{ is the unique }w\in H\text{ such that }⟨Tx,y⟩=⟨x,w⟩\text{for all }x\in D(T),$ equivalently $\widetilde{{\phi }_y}(z)=⟨z,T^{\ast }y⟩$ for all $z\in H$. The map $T^{\ast }:D(T^{\ast })\to H$, $y\mapsto T^{\ast }y$, is the adjoint of $T$.

Well-definedness. The functional ${\phi }_y$ is linear in $x$ for each $y$, so its domain of boundedness $D(T^{\ast })$ is a linear subspace: if ${\phi }_y,{\phi }_{y^{\mathrm{\prime }}}$ are bounded so is ${\phi }_{ay+b{y}^{\mathrm{\prime }}}$ for scalars $a,b$, because the first-variable-linear convention gives ${\phi }_{ay+b{y}^{\mathrm{\prime }}}=\bar{a}{\phi }_y+\bar{b}{\phi }_{y^{\mathrm{\prime }}}$. On $D(T^{\ast })$ the map $T^{\ast }$ is linear: if $y,y^{\mathrm{\prime }}$ are represented by $w,w^{\mathrm{\prime }}$, then $ay+b{y}^{\mathrm{\prime }}$ is represented by $aw+b{w}^{\mathrm{\prime }}$, since $⟨x,aw+b{w}^{\mathrm{\prime }}⟩=\bar{a}⟨x,w⟩+\bar{b}⟨x,w^{\mathrm{\prime }}⟩$; the representing vector is unique. By Riesz representation for the Hilbert space $H$ a vector $w\in H$ is determined by the values $⟨z,w⟩$ with $z$ ranging over $H$, and those values are determined by the functional $\widetilde{{\phi }_y}$. Equivalently, if $w,w^{\mathrm{\prime }}$ both represent ${\phi }_y$, then $⟨x,w-w^{\mathrm{\prime }}⟩=0$ for every $x\in D(T)$; density of $D(T)$ and continuity of the inner product extend this equality to every $x\in H$, and taking $x=w-w^{\mathrm{\prime }}$ gives $w=w^{\mathrm{\prime }}$. Finally only ambient-norm boundedness of ${\phi }_y$ on $D(T)$ is required: no extension, no closure and no closedness of $T$ is presupposed.

Assume Countable Choice. For a densely defined linear operator $T$ on $H$ the adjoint $T^{\ast }$ is well defined and linear with linear domain $D(T^{\ast })$; $T^{\ast }$ is closed; for every $z\in \mathbb{C}$ $\mathrm{ran}(T-z{)}^{\perp }=\ker (T^{\ast }-\bar{z}),$ and if $S\subseteq T$ and both are densely defined, then $T^{\ast }\subseteq S^{\ast }$.

Assume Countable Choice. Let $T$ be a densely defined linear operator on $H$. Then $T$ is closable if and only if $D(T^{\ast })$ is dense in $H$. In that case $\bar{T}=T^{\ast \ast }$ and $T^{\ast }={\bar{T}}^{\ast }$; moreover if $T\subseteq T^{\ast }$, then $T$ is closable.

Assume Countable Choice. Let $T$ be a densely defined linear operator on $H$.

• $T$ is symmetric when $T\subseteq T^{\ast }$, that is, $D(T)\subseteq D(T^{\ast })$ and $Tx=T^{\ast }x$ for all $x\in D(T)$; equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in D(T)$.
• $T$ is self-adjoint when $T=T^{\ast }$, that is, the domains and the values agree.
• $T$ is essentially self-adjoint when its closure $\bar{T}$ is self-adjoint; by Closability is equivalent to density of the adjoint domain this presupposes that $T$ is closable.

Consequences, with proofs. These are part of the content of the definition.

1. A symmetric $T$ is closable. $T\subseteq T^{\ast }$ and $T^{\ast }$ is closed (The adjoint is well defined, closed, and reverses inclusions), so $T^{\ast }$ is a closed extension of $T$. In particular the closure $\bar{T}$ exists and $\bar{T}\subseteq T^{\ast }$.
2. The closure of a symmetric operator is symmetric. If $T\subseteq T^{\ast }$, the inclusion reversal of The adjoint is well defined, closed, and reverses inclusions applied to $T\subseteq T^{\ast }$ gives $(T^{\ast }{)}^{\ast }\subseteq T^{\ast }$, that is $T^{\ast \ast }\subseteq T^{\ast }$. Applying it once more to the inclusion $T^{\ast \ast }\subseteq T^{\ast }$ gives $(T^{\ast }{)}^{\ast }\subseteq (T^{\ast \ast }{)}^{\ast }$, that is $T^{\ast \ast }\subseteq T^{\ast \ast \ast }$. Since $\bar{T}=T^{\ast \ast }$ and $(\bar{T}{)}^{\ast }=T^{\ast \ast \ast }$ (Closability is equivalent to density of the adjoint domain), this reads $\bar{T}\subseteq (\bar{T}{)}^{\ast }$: the closure is symmetric. (Each application is legitimate because the adjoint is defined once its operator is densely defined, and $D(T^{\ast \ast })$ is dense because $T\subseteq T^{\ast \ast }=\bar{T}$, so that $D(T^{\ast \ast })$ contains the dense domain $D(T)$.)
3. A self-adjoint operator has no proper symmetric extension. If $T=T^{\ast }$ and $T\subseteq S\subseteq S^{\ast }$, then $S^{\ast }\subseteq T^{\ast }$ by inclusion reversal, so $T\subseteq S\subseteq S^{\ast }\subseteq T^{\ast }=T$ and all inclusions are equalities.
4. A self-adjoint operator is closed, being equal to the adjoint $T^{\ast }$ of the densely defined $T$, and an essentially self-adjoint operator has exactly one self-adjoint extension, namely $\bar{T}$: a self-adjoint extension $S$ of $T$ is closed, hence contains the least closed extension $\bar{T}$, and then $S\subseteq S^{\ast }\subseteq (\bar{T}{)}^{\ast }=\bar{T}$ by item 2 and symmetry, so $S=\bar{T}$.

Assume the Axioms of Countable Choice and Dependent Choice. On the Hilbert space $H=L^2((0,1);\mathbb{C})$ (The space $L^p(\mu )$ as the quotient by null functions) let $T:=-id\mathrm{/}dx$ with domain $D(T)=\{f\in AC[0,1]:f^{\mathrm{\prime }}\in L^2(0,1),f(0)=f(1)=0\},$ where complex-valued absolute continuity and the derivative are read on real and imaginary parts (Absolute continuity on a compact interval). Domain notation means the $L^2$ classes having the indicated absolutely continuous representative; endpoint values refer to that representative. Then:

1. $T$ is densely defined, closed and symmetric, so $T$ refutes the reading "closed and symmetric implies self-adjoint";
2. $D(T^{\ast })=\{g\in AC[0,1]:g^{\mathrm{\prime }}\in L^2(0,1)\}$ and $T^{\ast }g=-i{g}^{\mathrm{\prime }}$, so $T^{\ast }$ is a proper closed extension of $T$; $T^{\ast }$ itself is not symmetric, and $T\mathrm{\ne }{T}^{\ast }$;
3. the periodic domain $D_2=\{g\in AC[0,1]:g^{\mathrm{\prime }}\in L^2(0,1),g(0)=g(1)\}$ carries a closed symmetric extension of $T$, strictly between $T$ and $T^{\ast }$.