Hermitian adjoint

Let the inner product ${\displaystyle \langle \cdot ,\cdot \rangle }$ be linear in the first argument. A densely defined operator A from a complex Hilbert space H to itself is a linear operator whose domain D(A) is a dense linear subspace of H and whose values lie in H.[3] By definition, the domain D(A^∗) of its adjoint A^∗ is the set of all y ∈ H for which there is a z ∈ H satisfying

${\displaystyle \langle Ax,y\rangle =\langle x,z\rangle \quad {\mbox{for all }}x\in D(A).}$

Owing to the density of ${\displaystyle D(A)}$ and Riesz representation theorem, ${\displaystyle z}$ is uniquely defined, and, by definition, ${\displaystyle A^{*}y=z.}$[4]

Properties 1.–5. hold with appropriate clauses about domains and codomains. For instance, the last property now states that (AB)^∗ is an extension of B^∗A^∗ if A, B and AB are densely defined operators.[5]

For every ${\displaystyle y\in \ker A^{*},}$ the linear functional ${\displaystyle x\mapsto \langle Ax,y\rangle =\langle x,A^{*}y\rangle }$ is identically zero, and hence ${\displaystyle y\in (\operatorname {im} A)^{\perp }.}$

Conversely, the assumption that ${\displaystyle y\in (\operatorname {im} A)^{\perp }}$ causes the functional ${\displaystyle x\mapsto \langle Ax,y\rangle }$ to be identically zero. Since the functional is obviously bounded, the definition of ${\displaystyle A^{*}}$ assures that ${\displaystyle y\in D(A^{*}).}$ The fact that, for every ${\displaystyle x\in D(A),}$ ${\displaystyle \langle Ax,y\rangle =\langle x,A^{*}y\rangle =0}$ shows that ${\displaystyle A^{*}y\in D(A)^{\perp }={\overline {D(A)}}^{\perp }=\{0\},}$ given that ${\displaystyle D(A)}$ is dense.

This property shows that ${\displaystyle \operatorname {ker} A^{*}}$ is a topologically closed subspace even when ${\displaystyle D(A^{*})}$ is not.

If ${\displaystyle H_{1}}$ and ${\displaystyle H_{2}}$ are Hilbert spaces, then ${\displaystyle H_{1}\oplus H_{2}}$ is a Hilbert space with the inner product

${\displaystyle {\bigl \langle }(a,b),(c,d){\bigr \rangle }_{H_{1}\oplus H_{2}}{\stackrel {\text{def}}{=}}\langle a,c\rangle _{H_{1}}+\langle b,d\rangle _{H_{2}},}$

where ${\displaystyle a,c\in H_{1}}$ and ${\displaystyle b,d\in H_{2}.}$

Let ${\displaystyle J\colon H\oplus H\to H\oplus H}$ be the symplectic mapping, i.e. ${\displaystyle J(\xi ,\eta )=(-\eta ,\xi ).}$ Then the graph

${\displaystyle G(A^{*})=\{(x,y)\mid x\in D(A^{*}),\ y=A^{*}x\}\subseteq H\oplus H}$

of ${\displaystyle A^{*}}$ is the orthogonal complement of ${\displaystyle JG(A):}$

${\displaystyle G(A^{*})=(JG(A))^{\perp }=\{(x,y)\in H\oplus H:{\bigl \langle }(x,y),(-A\xi ,\xi ){\bigr \rangle }_{H\oplus H}=0\;\;\forall \xi \in D(A)\}.}$

The assertion follows from the equivalences

${\displaystyle {\bigl \langle }(x,y),(-A\xi ,\xi ){\bigr \rangle }=0\quad \Leftrightarrow \quad \langle A\xi ,x\rangle =\langle \xi ,y\rangle ,}$

and

${\displaystyle {\Bigl [}\forall \xi \in D(A)\ \ \langle A\xi ,x\rangle =\langle \xi ,y\rangle {\Bigr ]}\quad \Leftrightarrow \quad x\in D(A^{*})\ \&\ y=A^{*}x.}$

An operator ${\displaystyle A}$ is closed if the graph ${\displaystyle G(A)}$ is topologically closed in ${\displaystyle H\oplus H.}$ The graph ${\displaystyle G(A^{*})}$ of the adjoint operator ${\displaystyle A^{*}}$ is the orthogonal complement of a subspace, and therefore is closed.

An operator ${\displaystyle A}$ is closable if the topological closure ${\displaystyle G^{\text{cl}}(A)\subseteq H\oplus H}$ of the graph ${\displaystyle G(A)}$ is the graph of a function. Since ${\displaystyle G^{\text{cl}}(A)}$ is a (closed) linear subspace, the word "function" may be replaced with "linear operator". For the same reason, ${\displaystyle A}$ is closable if and only if ${\displaystyle (0,v)\notin G^{\text{cl}}(A)}$ unless ${\displaystyle v=0.}$

The adjoint ${\displaystyle A^{*}}$ is densely defined if and only if ${\displaystyle A}$ is closable. This follows from the fact that, for every ${\displaystyle v\in H,}$

${\displaystyle v\in D(A^{*})^{\perp }\ \Leftrightarrow \ (0,v)\in G^{\text{cl}}(A),}$

which, in turn, is proven through the following chain of equivalencies:

${\displaystyle {\begin{aligned}v\in D(A^{*})^{\perp }&\Longleftrightarrow (v,0)\in G(A^{*})^{\perp }\Longleftrightarrow (v,0)\in (JG(A))^{\text{cl}}=JG^{\text{cl}}(A)\\&\Longleftrightarrow (0,-v)=J^{-1}(v,0)\in G^{\text{cl}}(A)\\&\Longleftrightarrow (0,v)\in G^{\text{cl}}(A).\end{aligned}}}$

The closure ${\displaystyle A^{\text{cl}}}$ of an operator ${\displaystyle A}$ is the operator whose graph is ${\displaystyle G^{\text{cl}}(A)}$ if this graph represents a function. As above, the word "function" may be replaced with "operator". Furthermore, ${\displaystyle A^{**}=A^{\text{cl}},}$ meaning that ${\displaystyle G(A^{**})=G^{\text{cl}}(A).}$

To prove this, observe that ${\displaystyle J^{*}=-J,}$ i.e. ${\displaystyle \langle Jx,y\rangle _{H\oplus H}=-\langle x,Jy\rangle _{H\oplus H},}$ for every ${\displaystyle x,y\in H\oplus H.}$ Indeed,

${\displaystyle {\begin{aligned}\langle J(x_{1},x_{2}),(y_{1},y_{2})\rangle _{H\oplus H}&=\langle (-x_{2},x_{1}),(y_{1},y_{2})\rangle _{H\oplus H}=\langle -x_{2},y_{1}\rangle _{H}+\langle x_{1},y_{2}\rangle _{H}\\&=\langle x_{1},y_{2}\rangle _{H}+\langle x_{2},-y_{1}\rangle _{H}=\langle (x_{1},x_{2}),-J(y_{1},y_{2})\rangle _{H\oplus H}.\end{aligned}}}$

In particular, for every ${\displaystyle y\in H\oplus H}$ and every subspace ${\displaystyle V\subseteq H\oplus H,}$ ${\displaystyle y\in (JV)^{\perp }}$ if and only if ${\displaystyle Jy\in V^{\perp }.}$ Thus, ${\displaystyle J[(JV)^{\perp }]=V^{\perp }}$ and ${\displaystyle [J[(JV)^{\perp }]]^{\perp }=V^{\text{cl}}.}$ Substituting ${\displaystyle V=G(A),}$ obtain ${\displaystyle G^{\text{cl}}(A)=G(A^{**}).}$

For a closable operator ${\displaystyle A,}$ ${\displaystyle A^{*}=\left(A^{\text{cl}}\right)^{*},}$ meaning that ${\displaystyle G(A^{*})=G\left(\left(A^{\text{cl}}\right)^{*}\right).}$ Indeed,

${\displaystyle G\left(\left(A^{\text{cl}}\right)^{*}\right)=\left(JG^{\text{cl}}(A)\right)^{\perp }=\left(\left(JG(A)\right)^{\text{cl}}\right)^{\perp }=(JG(A))^{\perp }=G(A^{*}).}$

Let ${\displaystyle H=L^{2}(\mathbb {R} ,l),}$ where ${\displaystyle l}$ is the linear measure. Select a measurable, bounded, non-identically zero function ${\displaystyle f\notin L^{2},}$ and pick ${\displaystyle \varphi _{0}\in L^{2}\setminus \{0\}.}$ Define

${\displaystyle A\varphi =\langle f,\varphi \rangle \varphi _{0}.}$

It follows that ${\displaystyle D(A)=\{\varphi \in L^{2}\mid \langle f,\varphi \rangle \neq \infty \}.}$ The subspace ${\displaystyle D(A)}$ contains all the ${\displaystyle L^{2}}$ functions with compact support. Since ${\displaystyle \mathbf {1} _{[-n,n]}\cdot \varphi \ {\stackrel {L^{2}}{\to }}\ \varphi ,}$ ${\displaystyle A}$ is densely defined. For every ${\displaystyle \varphi \in D(A)}$ and ${\displaystyle \psi \in D(A^{*}),}$

${\displaystyle \langle \varphi ,A^{*}\psi \rangle =\langle A\varphi ,\psi \rangle =\langle \langle f,\varphi \rangle \varphi _{0},\psi \rangle =\langle f,\varphi \rangle \cdot \langle \varphi _{0},\psi \rangle =\langle \varphi ,\langle \varphi _{0},\psi \rangle f\rangle .}$

Thus, ${\displaystyle A^{*}\psi =\langle \varphi _{0},\psi \rangle f.}$ The definition of adjoint operator requires that ${\displaystyle \mathop {\text{Im}} A^{*}\subseteq H=L^{2}.}$ Since ${\displaystyle f\notin L^{2},}$ this is only possible if ${\displaystyle \langle \varphi _{0},\psi \rangle =0.}$ For this reason, ${\displaystyle D(A^{*})=\{\varphi _{0}\}^{\perp }.}$ Hence, ${\displaystyle A^{*}}$ is not densely defined and is identically zero on ${\displaystyle D(A^{*}).}$ As a result, ${\displaystyle A}$ is not closable and has no second adjoint ${\displaystyle A^{**}.}$