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Unbounded operator
In mathematics, more specifically functional analysis and operator theory, the notion of unbounded operator provides an abstract framework for dealing with differential operators, unbounded observables in quantum mechanics, and other cases.

The term "unbounded operator" can be misleading, since
 * "unbounded" should be understood as "not necessarily bounded";
 * "operator" should be understood as "linear operator" (as in the case of "bounded operator");
 * the domain of the operator is a linear subspace, not necessarily the whole space (in contrast to "bounded operator");
 * this linear subspace is not necessarily closed; often (but not always) it is assumed to be dense;
 * in the special case of a bounded operator, still, the domain is usually assumed to be the whole space.

In contrast to bounded operators, unbounded operators on a given space do not form an algebra, nor even a linear space, because each one is defined on its own domain.

The term "operator" often means "bounded linear operator", but in the context of this article it means "unbounded operator", with the reservations made above. The given space is assumed to be a Hilbert space. Some generalizations to Banach spaces and more general topological vector spaces are possible.

Linear operator

 * Let $$V$$ und $$W$$ vector spaces over the field $$\mathbb{K}=\{ \R, \mathbb{C}\}$$. A map $$T:V \to W$$ is called a linear operator, if for any $$x, y \in V$$ and any $$\lambda \in \mathbb{K}$$ the following properties hold:
 * $$T$$ is homogeneous : $$T (\lambda x) = \lambda T(x)$$
 * $$T$$ is additive: $$T (x + y) = T(x) + T(y)$$.


 * For two normed vector spaces $$V$$ and $$W$$ and a linearer Operator $$T:V\to W$$ the operator norm of $$T$$ is defined by:
 * $$ \|T\| = \inf \left\{C\ge 0 \mid \forall x\in V\colon \|Tx\| \le C\,\|x\|\right\}=\sup_{\|x\| = 1} \|Tx\| \in [0, \infty].$$


 * An operator is called bounded, if $$ \| T \| < \infty. $$ Otherwise it is said to be unbounded.


 * Let $$T:V \to W$$ be linearer operator between two normed vector spaces $$V$$ and $$W$$. Then the following conditions are equivalent:


 * $$T$$ is bounded.
 * $$T$$ is uniformly continuous.
 * $$T$$ is continuous.
 * $$T$$ is continuous in some point $$x$$ in $$ V.$$


 * The set of all bounded and linear operators from $$V$$ to $$W$$ is denoted by $$\mathfrak{L}(V,W)$$ and is a normed vector space. If $$W$$ is a banach space, then so is $${L}(V,W).$$
 * The set $$\mathfrak{L}(V,\mathbb{K})$$ is the space of all continuous functionals on $$V$$ and is called the continuous dual space of $$V.$$ It is denoted by $$V'.$$

Examples of linear operators

 * Let $$A$$ be a real $$n \times m$$-matrix. Then the linear map $$A:x \mapsto Ax$$ is a linear operator between $$\R^m$$ and $$\R^n$$.


 * Let $$\Omega \subset \R$$ be a open set. Then the differential operator $$f \mapsto D f = f'$$ is a linear operator between $$C^1(\Omega) $$ (the set of all continuously differentiable functions from $$\Omega$$ to $$\Complex$$) and $$C(\Omega)$$ (the set of all continuous functions on $$\Omega$$).


 * Let $$\Omega_1, \Omega_2 \subset \R^n$$ be open and $$K : \Omega_1 \times \Omega_2 \to \Complex$$ be a measurable function. Then the integral operator $$Tf(x) = \int_{\Omega_1} K(t,x) f(t) \mathrm{d} t$$ is a linear operator between two vector spaces with the kernel function $$K.$$

Remark For a distribution one define the Schwartz kernel as follow:
 * Schwartz kernel theorem For $$\Omega \subset \R^n$$ open let $$C_c^{\infty }(\Omega)$$ be the set of all smooth functions with compact support on $$\Omega$$ and $$C_c^{\infty'}(\Omega)$$ its dual space.
 * For each linear operator $$T: C_c^{\infty }(\Omega_2) \to C_c^{\infty' }(\Omega_1)$$ there exists a unique distribution $$K \in C_c^{\infty' }(\Omega_1 \times \Omega_2)$$ such that $$ (T \phi)(\psi) = K(\phi, \psi)$$ for all $$\phi \in C_c^{\infty }(\Omega_1), \psi \in C_c^{\infty }(\Omega_2).$$ This distribution $$K$$ is called  Schwartz kernel.

Short history
The theory of unbounded operators was stimulated by attempts in the late 1920s to put quantum mechanics on a rigorous mathematical foundation. The systematic development of the theory is due to John von Neumann and Marshall Stone. The technique of using the graph to analyze unbounded operators was introduced by von Neumann in "Über Adjungierte Funktionaloperatoren".

Definitions
Let $$X$$ and $$Y$$ be Banach spaces. An unbounded linear operator (or simply operator)
 * $$T: \mathcal{D}(T) \subseteq X \to Y \; $$

is a linear map $$T$$ from a linear subspace $$\mathcal{D}(T)$$ of $$X$$ — the domain of $$T$$ — to the space $$Y.$$ Contrary to the usual convention, $$T$$ may not be defined on the whole space $$X.$$

An operator $$T$$ is said to be densely defined if $$D(T)$$ is dense in $$X.$$ This also includes operators defined on the entire space $$X,$$ since the whole space is dense in itself. The denseness of the domain is necessary and sufficient for the existence of the adjoint and the transpose (see below).

An operator $$T$$ is called closed if its graph $$\Gamma (T) = \{ (x, Tx) \mid x \in D(T) \}$$ of $$T$$ is a closed set in the direct sum $$X\oplus Y,$$. This means that for every sequence $$(x_n)_{n \in \mathbb{N}}$$ in  $$\mathcal{D}(T)$$ converging to $$x\in X$$ such that $$Tx_n\to y\in Y$$ as $$n\to\infty$$ one has $$x\in\mathcal{D}(T)$$ and $$Tx = y.$$

An operator $$T$$ is called closeable if the closure $$\overline{\Gamma(T})$$ of $$\Gamma(T)$$ is the graph of some operator $$\overline{T}.$$ In this case $$\overline{T}$$ is unique and is called the closure of $$T.$$

$$S$$ is an extension of an operator $$T$$ if $$\Gamma (T) \subseteq \Gamma (S)$$, i.e. $$D(T) \subseteq D(S)$$ and $$Tx = Sx$$ for $$x \in D(T).$$ Denote by $$T \subset S.$$

Two operators are equal if $$T \subset S$$ and $$S \subset T;$$ or equivalent: $$D(T)=D(S)$$ and $$Tx = Sx$$ for $$ x \in D(T).$$

Operations
The operations of unbounded operators are more complicated than in the bounded case, since one has take care of the domains of the operators. Let $$X,Y$$ and $$Z$$ be Banach spaces over $$\mathbb{K}=\{\mathbb{R},\mathbb{C}\}.$$

For an operator $$T: \mathcal{D}(T) \subseteq X \to Y$$ and an scalar $$\lambda \in \mathbb{K}$$ the operator $$ \lambda T$$ is given by
 * $$\mathcal{D}(\lambda T)=\mathcal{D}(T)$$ and $$(\lambda T)x=\lambda (Tx)$$ for $$x \in \mathcal{D}(T).$$

For two operator $$S,T:\mathcal{D}(T), \mathcal{D}(S) \subseteq X \to Y$$ one define the operator $$S + T$$ by
 * $$\mathcal{D}(S+T) = \mathcal{D}(S)\cap \mathcal{D}(S)$$ and $$(S+T)x = Sx + Tx$$ for $$x \in \mathcal{D}(S+T).$$

For an operators $$T: \mathcal{D}(T) \subseteq X \to Y$$ and an operator $$S:\mathcal{D}(S) \subseteq Y \to Z$$ the operator $$ST=S \circ T$$ is defined by
 * $$\mathcal{D}(ST) = \{ x \in \mathcal{D}(T) \mid Tx \in \mathcal{D}(S) \} $$ and $$(ST)x = S(Tx)$$ for $$x \in \mathcal{D}(ST).$$

The inverse of $$T$$ exists if $$\operatorname{ker}(T) = \{0\},$$ i.e. $$T$$ is injective. Then the operator $$T^{-1}$$ is defined by
 * $$\mathcal{D}(T^{-1}) = \operatorname{ran}(T), \operatorname{ran}(T^{-1}) = \mathcal{D}(T)$$ and $$T^{-1}(Tx)=x$$ for $$x \in \mathcal{D}(T)$$, where $$\operatorname{ran}(T)$$ is the range and $$\operatorname{ker}(T)$$ is the kernel of $$T.$$

Examples

 * Consider the classical differential operator
 * $$ Tf = f' \; $$
 * on $$L^2[0,1],$$ that is the Hilbert space of all square-integrable functions on $$[0,1]$$ (more exactly, equivalence classes; the functions must be measurable, either real-valued or complex-valued) with the norm $$ \|f \|_{L^2} = \sqrt{ \int_{0}^1 |f(x)|^2 \, dx} $$ defined on the domain $$\mathcal{D}(T)=C^1[0,1],$$the set of all continuously differentiable functions $$f$$ on the closed interval $$[0,1].$$ The definition of $$T$$ is correct, since a continuous (the more so, continuously differentiable) function cannot vanish almost everywhere, unless it vanishes everywhere.


 * This is a linear operator, since a linear combination $$af+bg$$ of two continuously differentiable functions $$f,g$$ is also continuously differentiable, and $$(af+bg)'=af'+bg'.\;$$


 * The operator is not bounded. For example, the functions $$(f_n)_{n \in \mathbb{N}}$$ defined on $$[0,1]$$ by $$ f_n(t) = \sin 2\pi n t \;$$ satisfy $$ \|f_n\|_{L^2} = 1/\sqrt2 $$ but $$ \|Tf_n\|_{L^2} = 2\pi n/\sqrt2 \to \infty. $$


 * The operator is densely defined, and not closed.


 * The same operator can be treated as an operator $$X \to X$$ for many Banach spaces $$X$$ and is still not bounded. However, it is bounded as an operator $$X \to Y$$ for some pairs of Banach spaces $$X, Y$$, and also as operator $$X \to X $$ for some topological vector spaces $$X$$. As an example consider $$T : ( C^1 ( I ), \| \cdot \|_{C^1} ) \rightarrow ( C ( I), \| \cdot \|_{\infty} )$$, for some open interval $$I \subset \mathbb{R}$$ and the $$C^1$$ norm being $$\| f \|_{C^1} = \| f \|_{\infty} + \| f' \|_{\infty},$$ where $$\|f\|_{\infty} = \sup \{ |f(x)| \mid x \in I\}$$ is the Supremum norm.


 * The multiplication operator
 * $$T(x_n)_{n\in \mathbb{N}} := (nx_n)_{n \in \mathbb{N}}$$
 * on the sequence space $$\ell^2$$ of all square-summable sequences with the norm $$\| x\|_{\ell^2} = \sqrt{\sum_{n \in \mathbb{N}} |x_n|^2}$$ defined on $$\mathcal{D}(T)=\{(x_n)_{n\in \mathbb{N}} \in \ell^2 | \sum_{n \in \mathbb{N}} n^2|x_n|^2 < \infty\}$$ is a linear closed operator, which is not bounded.

Closed linear operators
Closed linear operators are a class of linear operators on Banach spaces. They are more general than bounded operators, and therefore not necessarily continuous, but they still retain nice enough properties that one can define the spectrum and (with certain assumptions) functional calculus for such operators. Many important linear operators which fail to be bounded turn out to be closed, such as the derivative and a large class of differential operators.

Definition
Let $$X,Y$$ be two Banach spaces. A linear operator  $$T\colon\mathcal{D}(T)\subseteq X\to Y$$ is said to be closed if one of the following equivalent properties hold:
 * The graph $$\Gamma(T)$$ is closed in $$X \oplus Y.$$
 * $$\mathcal{D}(T)$$ is a complete space with respect to the graph norm defined by $$ \| x \|_T = \sqrt{ \|x \|^2 + \|Tx \|^2} $$ for $$x \in \mathcal{D}(T)$$.
 * For every sequence $$(x_n)_{n \in \mathbb{N}}$$ in  $$\mathcal{D}(T)$$ converging to $$x\in X$$ such that $$Tx_n\to y\in Y$$ as $$n\to\infty$$ one has $$x\in\mathcal{D}(T)$$ and $$Tx = y.$$

Properties
For a closed operator $$T$$ one has
 * $$T-\lambda I$$ is closed where $$\lambda$$ is a scalar and $$I$$ is the identity function.
 * $$ \operatorname{ker}(T)$$ is a closed subspace of $$X.$$
 * If $$T$$ is injective, then its inverse $$T^{-1}$$ is also closed.
 * If $$T$$ is densely defined and bounded on its domain, then it is defined on $$X.$$

Example
Consider the derivative operator
 * $$ T f = f'\,$$

on the Banach space $$X = C[a,b]$$ of all continuous functions on an interval $$[a,b]$$ with the Supremum norm. If one takes its domain $$\mathcal{D}(T)$$ to be $$\mathcal{D}(T)=C^{1}[a, b],$$ then $$T$$ is a closed operator. (Note that one could also set $$\mathcal{D}(T)$$ to be the set of all differentiable functions including those with non-continuous derivative. That operator is not closed!)

The Operator is not bounded. For example, for the sequence $$f_n(x)=\sin(nx)$$ one has $$\|f_n\|_{\infty} = 1,$$ but for $$Tf_n(x)=n\cos(nx)$$ it is $$\|Tf_n\|_{\infty} \to \infty$$ for $$ n \to \infty.$$

If one takes $$\mathcal{D}(T)$$ to be instead the set of all infinitely differentiable functions, $$T$$ will no longer be closed, but it will be closable, with the closure being its extension defined on $$C^{1}[a, b].$$

Definitions
For two Banch spaces $$X, Y$$ an operator $$T: \mathcal{D}(T) \subseteq X \to Y$$ is called  closable if the following equivalent properties hold:
 * $$T$$ has a closed extension.
 * The closure $$\overline{\Gamma(T)}$$ of the graph of $$T$$ is the graph of some operator.
 * For every sequence $$(x_n)_{n \in \mathbb{N}} \subseteq \mathcal{D}(T)$$ such that $$x_n \to 0$$ and $$Tx_n \to y$$ holds $$y=0.$$
 * For every pair of sequences $$(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}} \subseteq\mathcal{D}(T)$$ both converging to $$x$$ such that both $$(Tx_n)_{n \in \mathbb{N}}$$ and $$(Ty_n)_{n \in \mathbb{N}}$$ converge, one has $$\lim_n Tx_n = \lim_n Ty_n.$$

The operator with the graph $$\overline{\Gamma(T)}$$ is said to be the closure of $$T$$ and is denoted by $$\overline{T}.$$ It follows that $$T$$ is the restriction of $$\overline{T}$$ to $$\mathcal{D}(T).$$ Note, that other, non-minimal closed extensions may exist.

A core of a closable operator is a subset $$\mathcal{C}$$ of  $$\mathcal{D}(T)$$ such that the closure of the restriction of $$T$$ to  $$\mathcal{C}$$ is  $$\overline{T}.$$

Remark Not all operators are closable as the following example shows:

Example
Consider the Operator $$T$$ on $$ L^2[0,1]$$ defined on $$\mathcal{D}(T) = C[0,1]$$ and $$Tf(x)=f(0)$$. For the sequence $$(f_n)_{n \in \mathbb{N}}$$ in $$\mathcal{D}(T)$$ given by $$f_n(x) = (1-x)^n,$$ one has
 * $$\| f_n \|_{L^2} = \sqrt{\int_{0}^1 (1-x)^{2n} \, dx} = \sqrt{\frac 1{2n+1}} \to 0$$ for $$n \to \infty;$$

but $$Tf_n = 1 \neq 0.$$ Thus, $$T$$ is not closable.

Resolvent and spectrum
Let $$T: \mathcal{D}(T) \subseteq X \rightarrow X$$ be a densely defined operator on a Banach space $$X$$ and $$\mathbb{K} = \Complex.$$ Then $$\lambda \in \mathbb{C}$$ is called to be in the resolvent set of $$T,$$ denoted by $$\varrho(T),$$ if the operator $$ T - \lambda I$$ is bijective and $$(T- \lambda I)^{-1}$$ is a bounded operator. It follows by the closed graph theorem that the resolvent is bounded for all $$\lambda \in \varrho(T)$$ if $$T$$ is a closed operator. For $$\lambda \in \varrho(T)$$ the resolvent of $$T$$ is defined by $$R(T,\lambda)= (\lambda I -T)^{-1}.$$ The set $$\mathbb{C}\setminus \varrho(T)$$ is called the spectrum of $$T,$$ denoted by $$\sigma(T).$$ 

The spectrum $$\sigma(T)$$ of an unbounded operator $$T$$ can be divided into three parts in exactly the same way as in the bounded case:
 * The point spectrum is the set of eigenvalues and is defined by $$\sigma_p(T) = \{\lambda \in \mathbb{C} \mid (T - \lambda I)$$ is not injective$$ \}.$$
 * The continuous spectrum is given by $$\sigma_c(T)= \{ \lambda \in \mathbb{C} \mid (T - \lambda I)$$ is injective and has dense range, but not surjective$$ \}.$$
 * The residual spectrum is the set $$ \sigma_r(T)=\{ \lambda \in \mathbb{C} \mid (T - \lambda I)$$ is injective, but its range is not dense$$ \}.$$

Remark The spectrum of an unbounded operator can be any closed set, including $$\mathbb{C}$$ and $$ \emptyset.$$ The domain plays an important role as the following example shows:

Example
Consider the banach space $$C[0,1]$$ and the operators $$T_{1,2}$$ defined by $$T_{1,2}f = f'$$ and $$ \mathcal{D}(T_1)=C^1[0,1]$$ and $$ \mathcal{D}(T_2)= \{ f \in C^1[0,1] \mid f(0)=0 \}. $$ If $$ f(x)= e^{\lambda x}$$, then $$ T_1f - \lambda f = 0. $$ Thus, $$ \sigma(T_1) = \mathbb{C}.$$ For the linear differential equation $$(T_2- \lambda I)f = g, f(0)=0$$ exists a unique solution $$f(x) = e^{\lambda x} \int_0^x e^{\lambda s} g(s) \, ds,$$ which defines an inverse for $$(T_2- \lambda I).$$ Therefore $$\sigma(T_2)=\emptyset.$$

Definition
Let $$T: \mathcal{D}(T) \subseteq X \to Y $$ be an densely defined operator between Banach spaces and $$X'$$ the continuous dual space of $$X.$$ Using the notation $$ \langle x,x'\rangle = x'(x)$$ the transpose (or dual) $$T ': \mathcal{D}(T') \subseteq {Y}^' \to {X}^'$$ of $$T$$ is an operator satisfying:
 * $$ \langle Tx, y' \rangle = \langle x, T'y' \rangle $$ for all $$x \in X$$ and $$y' \in Y'.$$

The operator $$T'$$ is defined by
 * $$\mathcal{D}(T')=\{y' \in Y' \mid \exists x' \in X': \langle Tx, y' \rangle = \langle x, T'y' \rangle$$ for all $$ x \in X \}$$ and $$T'y'=x'$$ for $$ y' \in \mathcal{D}(T'). $$.

Remark The necessary and sufficient condition for the transpose of $$T$$ to exist is that $$T$$ is densely defined (for essentially the same reason as to adjoints, see below.)

Theorems
Hahn-Banach Theorem

Let $$V$$ be a vector space over the field $$\mathbb{K}= \{\mathbb{R}$$, $$\mathbb{C}\}, \, U$$ a linear subspace. Let $$p: V \to \mathbb{R}$$ be a sublinear function and $$f: U \to \mathbb{K}$$ be a linear functional with $$\operatorname{Re}f(u)\leq p(u)$$ for all $$u\in U$$ (where $$\operatorname{Re}z$$ is the real part of a complex number $$z$$).

Then, there exists a linear functional $$F: V \to \mathbb{K}$$ with
 * $$F|_U=f\,\,$$ and
 * $$\operatorname{Re}F(v)\leq p(v)$$ for all $$v\in V.$$

Satz von Banach-Steinhaus (Uniform boundedness principle)

Let $$X$$ be a banach space and $$Y$$ be a normed vector space. Suppose that $$F$$ is a collection of bounded linear operators from $$X$$ to $$Y.$$ The uniform boundedness principle states that if for all $$x$$ in $$X$$ we have $$\sup_{T \in F} \|T (x)\|  < \infty $$, then $$\ \sup_{T \in F} \|T\|  < \infty. $$

Open mapping theorem

Let $$X,Y$$ be banach spaces and $$T \in \mathfrak{L}(X,Y)$$ surjective. Then $$T$$ is an open map.

In particular: '''Bounded inverse theorem If $$T$$ bijective und bounded, then its inverse $$T^{-1}$$ is also bounded.

Closed graph theorem

Let $$X,Y$$ be banach spaces. If $$T: X \to Y$$ is linear and closed, then $$T$$ is bounded.

Closed range theorem

For a densely defined closed operator $$T: \mathcal{D}(T) \subseteq X \to Y$$ the following properties are equivalent:
 * $$\operatorname{ran}(T)$$ is closed in $$Y.$$
 * $$\operatorname{ran}(T')$$ is closed in $$X'.$$
 * $$\operatorname{ran}(T)=\operatorname{ker}(T')^{\perp}=\{y\in Y | \langle y, y' \rangle = 0 $$ for all $$ y'\in \operatorname{ker}(T')\}.$$
 * $$\operatorname{ran}(T')=\operatorname{ker}(T)^{\perp}=\{x'\in X' | \langle x, x'\rangle = 0 $$ for all $$ x\in \operatorname{ker}(T)\}.$$

Classes of unbounded operators on Hilbert spaces
In this section let $$(H_1,\langle . \mid . \rangle_1 )$$, $$(H_2,\langle . \mid . \rangle_2 )$$ and $$(H,\langle . \mid . \rangle)$$ be Hilbert spaces.

Definiton
For an unbounded operator $$T$$ the definition of the adjoint is more complicated than in the bounded case, since it is necessary to take care of the domains of the operators.

The adjoint of an unbounded operator can be defined in two equivalent ways. First, it can be defined in a way analogous to how we define the adjoint of a bounded operator.

For a densely defined operator $$T: \mathcal{D}(T) \subseteq H_1 \to H_2$$ its adjoint $$T^*: H_2 \to H_1$$ is defined by
 * $$\mathcal{D}(T^*)=\{y \in H_2 \mid x \mapsto \langle Tx \mid y \rangle_2$$ is a continuous functional on $$\mathcal{D}(T)\}.$$

Since $$ \mathcal{D}(T) $$ is dense in $$H_1,$$ the functional extends to the whole space $$H_1$$ via the Hahn–Banach theorem. Thus, one can find a unique $$z \in H_1$$ such that
 * $$\langle Tx \mid y \rangle_2 = \langle x \mid z \rangle_1$$ for all $$x \in \mathcal{D}(T).$$

Finally, let $$T^*y=z,$$ completing the construction of $$T^*$$ and it is
 * $$\langle Tx \mid y \rangle_2 = \langle x \mid T^*y \rangle_1$$ for all $$x \in \mathcal{D}(T), y \in \mathcal{D}(T^*).$$

Remark $$T^*$$ exists if and only if $$T$$ is densely defined.

The other equivalent definition of the adjoint can be obtained by noticing a general fact: define a linear operator
 * $$U: H_1 \oplus H_2 \to H_2 \oplus H_1$$ by $$U(x, y) = (-y, x)$$. (Since $$U$$ is an isometric surjection, it is unitary.)

We then have: $$U(\Gamma (T))^\bot$$ is the graph of some operator $$S$$ if and only if $$T$$ is densely defined. A simple calculation shows that this "some" $$S$$ satisfies
 * $$\langle Tx \mid y \rangle_2 = \langle x \mid Sy \rangle_1$$ for every $$x \in \mathcal{D}(T).$$

Thus, $$S$$ is the adjoint of $$T.$$

The definition of the adjoint can be given in terms of a transpose as follow: For any Hilbert space $$H$$ and its continuous dual space $$H',$$ there is the anti-linear isomorphism
 * $$J: H' \to H$$

given by $$Jf = y$$ where $$f(x) = \langle x \mid y \rangle$$ for $$x \in H$$ and $$f \in H'.$$ Through this isomorphism, the transpose $$T'$$ relates to the adjoint $$T^*$$ in the following way:
 * $$T^* = J_1 T' J_2^{-1}$$,

where $$J_j: H_j' \to H_j$$. (For the finite-dimensional case, this corresponds to the fact that the adjoint of a matrix is its conjugate transpose.)

Properties
By definition, the domain of $$T^*$$ could be anything; it could be trivial (i.e., contains only zero) It may happen that the domain of $$T^*$$ is a closed hyperplane and $$T^*$$ vanishes everywhere on the domain. Thus, boundedness of $$T^*$$ on its domain does not imply boundedness of $$T$$. On the other hand, if $$T^*$$ is defined on the whole space then $$T$$ is bounded on its domain and therefore can be extended by continuity to a bounded operator on the whole space. If the domain of $$T^*$$ is dense, then it has its adjoint $$T^{**}.$$

For a densely defined operator $$T: \mathcal{D}(T) \subseteq H_1 \to H_2$$
 * $$T^*$$ is closed.
 * $$T$$ is closable if and only if $$T^*$$ is densely defined. In this case $$ \overline T = T^{**} $$ and $$ (\overline T)^* = T^*. $$
 * If $$T^*$$ densely defined, then $$T\subset T^{**}.$$
 * $$T$$ is bounded if and only if $$T^*$$ is bounded. In this case $$\|T\|=\|T^*\|.$$

If $$T, S$$ densely defined and $$S \subset T$$, then $$T^* \subset S^*$$. Further if $$T+S, ST$$ are densely defined, then $$ S^* +T^* \subset (S+T)^*$$ and $$T^*S^* \subset (ST)^*.$$ In contrast to the bounded case, it is not necessary that we have: $$ (TS)^* = S^*T^*,$$ since, for example, it is even possible that $$(TS)^*$$ doesn't exist. This is, however, the case if, for example, $$T$$ is bounded.

Some well-known properties for bounded operators generalize to closed densely defined operators.
 * $$T$$ is closed and densely defined if and only if $$ T = T^{**}.$$
 * von Neumann's theorem $$T$$ densely defined and closed, then $$T^*T, TT^*$$ are self-adjoint and $$(I+T^*T)$$ and $$(I+TT^*)$$ both admit bounded inverses.
 * Closed range theorem For a densely defined closed operator $$T: \mathcal{D}(T) \subseteq H_1 \to H_2$$ the following properties are equivalent:
 * $$\operatorname{ran}(T)$$ is closed in $$H_2.$$
 * $$\operatorname{ran}(T^*)$$ is closed in $$H_1.$$
 * $$\operatorname{ran}(T)=\operatorname{ker}(T^*)^{\perp} =\{y\in H_2 | \langle y^* \mid y\rangle_2 = 0 $$ for all $$ y^*\in \operatorname{ker}(T^*)\}.$$
 * $$\operatorname{ran}(T^*)=\operatorname{ker}(T)^{\perp} = \{x^*\in H_1| \langle x^* \mid x\rangle_1 = 0 $$ for all $$ x\in \operatorname{ker}(T)\}.$$

In particular, if $$T^*$$ has trivial kernel, $$T$$ has dense range (by the above identity.) Moreover, $$T$$ is surjective if and only if there is a $$K > 0$$ such that (This is essentially a variant of the closed range theorem.)
 * $$\|x\|_2 \le K\|T^*x\|_1$$ for every $$x \in \mathcal{D}(T^*)$$.

Definitions
A densely defined operator $$T:\mathcal{D}(T) \subseteq H \to H$$ is called symmetric if $$ \langle Tx \mid y \rangle = \lang x \mid Ty \rang $$ for all $$x, y \in \mathcal{D}(T).$$

A symmetric operator is called maximal symmetric if it has no symmetric extensions, except for itself.

A symmetric operator $$T: \mathcal{D}(T) \subseteq H \to H$$ is called bounded (from) below if there exists a constant $$m \in \mathbb{R}$$ with $$\langle Tx \mid x \rangle \geq m \|x\|^2$$. The operator is said to be positve if $$\langle Tx \mid x \rangle \geq 0$$.

Properties

 * Hellinger-Toeplitz theorem An everywhere defined symmetric operator is closed, therefore bounded.


 * Every symmetric operator $$T$$ is closable, since $$T$$ is densely defined and $$T \subset T^*$$, therefore $$T \subset \overline{T} \subset T^*.$$


 * If $$T$$ is symmetric then $$ T \subset T^{**} \subset T^* . $$
 * If $$T$$ is closed and symmetric then $$ T = T^{**} \subset T^*.$$

An operator $$T:\mathcal{D}(T) \subseteq H \to H$$ is symmetric if it satisfies one of the following equivalent properties:
 * Its quadratic form is real, that is, the number $$ \langle Tx \mid x \rangle \in \mathbb{R}$$ for all $$x \in \mathcal{D}(T).$$
 * The subspace $$\Gamma(T)$$ is orthogonal to its image $$U(\Gamma (T)) $$ $$ ($$ where $$U$$ is an unitary operator on $$H \oplus H$$ defined by $$U(x,y)=(-y,x)).$$
 * $$T \subset T^*.$$

Remark The last condition does not cover non-densely defined closed operators. Non-densely defined symmetric operators can be defined directly or via graphs, but not via adjoint operators.

Examples

 * A densely defined, positive operator is symmetric.


 * The differential operator
 * $$Tf =-{\rm i}f'$$
 * on $$L^2[0,1]$$ defined on the domain $$\mathcal{D}(T)=\{f\in L^2[0,1] | f$$ is absolutely continuous and $$f(0)=f(1)=0 \}$$ is closed and symmetric, but not self-adjoint.

Definition
A densely defined operator $$ T: \mathcal{D}(T) \subseteq H \to H$$ is said to be self-adjoint if $$T^* = T. $$

Properites
For a densely defined closed operator $$T$$ one has:
 * If $$T$$ is self-adjoint, then it is closed, because $$T^*$$ is necessarily closed.
 * The operator $$T^*T$$ is self-adjoint, positive and $$\mathcal{D}(T^*T)$$ is a core for $$T$$
 * If $$T$$ symmetric, then $$T$$ is self-adjoint if and only if $$T^*$$ is symmetric. It may happen that it is not.

Let $$T$$ be a symmetric operator. Then follwing conditions are equivalent:
 * $$T$$ is self-adjoint.
 * $$T$$ is closed and $$\operatorname{ker}(T^* \pm iI) = \{0 \}$$.
 * $$\operatorname{ran}(T \pm iI) = H$$.

An operator $$ T: \mathcal{D}(T) \subseteq H \to H$$ is self-adjoint if the following equivalent properties hold:
 * $$T$$ is symmetric and $$\mathcal{D}(T^*)=\mathcal{D}(T).$$
 * The two subspaces $$ \Gamma(T)$$ and $$ U(\Gamma(T))$$ are orthogonal and their sum is the whole space $$ H \oplus H .$$ $$ ($$ where $$U$$ is an unitary operator on $$H \oplus H$$ defined by $$U(x,y)=(-y,x))$$
 * $$T$$ closed, symmetric and satisfies the condition: both operators $$ T \pm i$$ are surjective, that is, map the domain of $$T$$ onto the whole space $$H.$$ In other words: for every $$x \in H$$ there exist $$y, z \in D(T)$$ such that $$Ty -iy = x$$ and $$Tz + iz = x.$$

Remarks
 * For a bounded operator the terms symmetric and self-adjoint are equivalent.
 * The distinction between closed symmetric operators and self-adjoint operators is important, since only for self-adjoint operators the spectral theorem holds.

Example

 * Let $$(\Omega,\Sigma, \mu)$$ be a measure space, $$f : \Omega \to \R$$ a measurable function. Then the  multiplication operator
 * $$T_f g = f \cdot g$$
 * on $$L^2(\Omega)$$ with $$D(T_f) = \{g \in L^2(\Omega)| f \cdot g \in L^2(\Omega)\}$$ is densely defined and self-adjoint.

Normal operators
A densely defined, closed operator $$ T: \mathcal{D}(T) \subseteq H \to H$$ is called normal if it satisfies the following equivalent properties :
 * $$T^*T = TT^*.$$
 * $$\mathcal{D}(T) = \mathcal{D}(T^*)$$ and $$ \| Tx \| = \| T^* x \| $$ for every $$x \in \mathcal{D}(T).$$
 * There exist self-adjoint operators $$A, B$$ such that $$T= A + iB, T^*= A - iB$$ and $$ \| Tx \|^2 = \| Ax \|^2 + \| Bx \|^2 $$ for every $$x \in \mathcal{D}(T).$$

Remarks
 * Every self-adjoint operator is normal.
 * The spectral theorem applies to self-adjoint operators and moreover, to normal operators, but not to densely defined, closed operators in general, since in this case the spectrum can be empty.  In particulary, the spectral-theorem does not hold for closed symmetric operators.

Self-adjoint extensions of symmetric operators
Let $$T$$ a symmetric operator on a Hilbert space $$H$$.

Problem When does $$T$$ have self-adjoint extensions?

The Cayley transform of a symmetric operator $$T$$ is defined by $$V_T = (T - iI)(T + iI)^{-1}$$. $$V_T$$ is an isometry between $$\operatorname{ran}(T + iI)$$ and $$\operatorname{ran}(T - iI)$$ and the range $$\operatorname{ran}(I - V_T)$$ is dense in $$H.$$

Theorem $$T$$ is self-adjoint if and only if $$V_T$$ is unitary.

In particular: $$T$$ has self-adjoint extensions if and only if $$V_T$$ has unitary extensions.

'''Friedrichs extension theorem Every symmetric operator which is bounded from below has at least one self-adjoint extension with the same lower bound.

These operators always have a canonically defined self-adjoint extension which is called Friedrichs extension.

Remark An everywhere defined extension exists for every operator, which is a purely algebraic fact explained at General existence theorem and based on the axiom of choice. If the given operator is not bounded then the extension is a discontinuous linear map. It is of little use since it cannot preserve important properties of the given operator, and usually is highly non-unique

Definition
A symmetric operator $$T$$ is called essentially self-adjoint if $$T$$ has one and only one self-adjoint extension. Or equivalent, if its closure $$\overline{T}$$ is self-adjoint. . Note, that an operator may have more than one self-adjoint extension, and even a continuum of them.

Remark The importance of essentially self-adjointness is that one is often given a non-closed symmetric operator $$T.$$ If this operator $$T$$ is essential self-adjoint, then there is uniquely associated to $$T$$ a self-adjoint operator $$\overline{T} = T^{**}.$$

Properties

 * If $$T$$ is essentially self-adjoint then $$T \subset T^{**} = T^*. $$

Let $$T$$ be a symmetric operator. Then follwing conditions are equivalent:
 * $$T$$ is essentially self-adjoint.
 * $$\operatorname{ker}(T^* \pm iI) = \{0 \} $$.
 * $$\operatorname{ran}(T \pm iI)$$ is dense.

Remark For a bounded operator the terms self-adjoint, symmetric and essentially self-adjoint are equivalent.

Example
Let $$M$$ be complete Riemannian manifold. The Laplace operator
 * $$\Delta f = \operatorname{div} (\operatorname{grad} f)$$ (where $$\operatorname{grad}$$ is the gradient and $$\operatorname{div}$$ is the divergence)

on $$L^2(M)$$ with the domain $$\mathcal{D}(\Delta) = C_c^{\infty}(M)$$ the space of all smooth, compactly supported function on $$M$$ is essentially self-adjoint.

The importance of self-adjoint operators
The class of self-adjoint operators is especially important in mathematical physics. Every self-adjoint operator is densely defined, closed and symmetric. The converse holds for bounded operators but fails in general. Self-adjointness is substantially more restricting than these three properties. The famous spectral theorem holds for self-adjoint operators. In combination with Stone's theorem on one-parameter unitary groups it shows that self-adjoint operators are precisely the infinitesimal generators of strongly continuous one-parameter unitary groups, see Self-adjoint operator. Such unitary groups are especially important for describing time evolution in classical and quantum mechanics.