Inmathematics, more specificallyfunctional analysis andoperator theory, the notion ofunbounded operator provides an abstract framework for dealing withdifferential operators, unboundedobservables inquantum mechanics, and other cases.

The term "unbounded operator" can be misleading, since

• "unbounded" should sometimes 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 alinear subspace, not necessarily the whole space;
• this linear subspace is not necessarilyclosed; often (but not always) it is assumed to bedense;
• in the special case of a bounded operator, still, the domain is usually assumed to be the whole space.

In contrast tobounded operators, unbounded operators on a given space do not form analgebra, 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 theory of unbounded operators developed in the late 1920s and early 1930s as part of developing a rigorous mathematical framework forquantum mechanics.^[1] The theory's development is due toJohn von Neumann^[2] andMarshall Stone.^[3] Von Neumann introduced usinggraphs to analyze unbounded operators in 1932.^[4]

LetX,Y beBanach spaces. Anunbounded operator (or simplyoperator)T :D(T) →Y is alinear mapT from a linear subspaceD(T) ⊆X—the domain ofT—to the spaceY.^[5] Contrary to the usual convention,T may not be defined on the whole spaceX.

An operatorT is said to beclosed if its graphΓ(T) is aclosed set.^[6] (Here, the graphΓ(T) is a linear subspace of thedirect sumX ⊕Y, defined as the set of all pairs(x,Tx), wherex runs over the domain ofT .) Explicitly, this means that for every sequence{x_n} of points from the domain ofT such thatx_n →x andTx_n →y, it holds thatx belongs to the domain ofT andTx =y.^[6] The closedness can also be formulated in terms of thegraph norm: an operatorT is closed if and only if its domainD(T) is acomplete space with respect to the norm:^[7]

${\displaystyle \Vert x{\Vert }_T=\sqrt{\Vert x{\Vert }^2+\Vert Tx{\Vert }^2}.}$

An operatorT is said to bedensely defined if its domain isdense inX.^[5] This also includes operators defined on the entire spaceX, since the whole space is dense in itself. The denseness of the domain is necessary and sufficient for the existence of the adjoint (ifX andY are Hilbert spaces) and the transpose; see the sections below.

IfT :D(T) →Y is closed, densely defined andcontinuous on its domain, then its domain is all ofX.^{[nb 1]}

A densely defined symmetric^{[clarification needed]} operatorT on aHilbert spaceH is calledbounded from below ifT +a is a positive operator for some real numbera. That is,⟨Tx|x⟩ ≥ −a ||x||² for allx in the domain ofT (or alternatively⟨Tx|x⟩ ≥a ||x||² sincea is arbitrary).^[8] If bothT and−T are bounded from below thenT is bounded.^[8]

LetC([0, 1]) denote the space of continuous functions on the unit interval, and letC¹([0, 1]) denote the space of continuously differentiable functions. We equip${\displaystyle C([0,1])}$ with the supremum norm,${\displaystyle \Vert \cdot {\Vert }_{\mathrm{\infty }}}$, making it a Banach space. Define the classical differentiation operator⁠d/dx⁠ :C¹([0, 1]) →C([0, 1]) by the usual formula:

${\displaystyle \left(\frac{d}{dx}f\right)(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h},\mathrm{\forall }x\in [0,1].}$

Every differentiable function is continuous, soC¹([0, 1]) ⊆C([0, 1]). We claim that⁠d/dx⁠ :C([0, 1]) →C([0, 1]) is a well-defined unbounded operator, with domainC¹([0, 1]). For this, we need to show that${\displaystyle \frac{d}{dx}}$ is linear and then, for example, exhibit some${\displaystyle \{f_n{\}}_n\subset C^1([0,1])}$ such that${\displaystyle \Vert f_n{\Vert }_{\mathrm{\infty }}=1}$ and${\displaystyle \underset{n}{sup}\Vert \frac{d}{dx}{f}_n{\Vert }_{\mathrm{\infty }}=+\mathrm{\infty }}$.

This is a linear operator, since a linear combinationa f  +bg of two continuously differentiable functions f ,g is also continuously differentiable, and

${\displaystyle \left(\frac{d}{dx}\right)(af+bg)=a\left(\frac{d}{dx}f\right)+b\left(\frac{d}{dx}g\right).}$

The operator is not bounded. For example,

${\displaystyle \{\begin{array}{l}{f}_n:[0,1]\to [-1,1]\\ f_n(x)=\sin (2\pi nx)\end{array}}$

satisfy

${\displaystyle {\Vert f_n\Vert }_{\mathrm{\infty }}=1,}$

but

${\displaystyle {\Vert \left(\frac{d}{dx}{f}_n\right)\Vert }_{\mathrm{\infty }}=2\pi n\to \mathrm{\infty }}$

as${\displaystyle n\to \mathrm{\infty }}$.

The operator is densely defined (which can be shown by the Weierstrass approximation theorem, since the set of polynomial functions on [0,1] is contained inC¹([0, 1]), while also being dense inC([0, 1])) and closed.

The same operator can be treated as an operatorZ →Z for many choices of Banach spaceZ and not be bounded between any of them. At the same time, it can be bounded as an operatorX →Y for other pairs of Banach spacesX,Y, and also as operatorZ →Z for some topological vector spacesZ.^{[clarification needed]} As an example letI ⊂R be an open interval and consider

${\displaystyle \frac{d}{dx}:\left(C^1(I),\Vert \cdot {\Vert }_{C^1}\right)\to \left(C(I),\Vert \cdot {\Vert }_{\mathrm{\infty }}\right),}$

where:

${\displaystyle \Vert f{\Vert }_{C^1}=\Vert f{\Vert }_{\mathrm{\infty }}+\Vert f^{\prime }{\Vert }_{\mathrm{\infty }}.}$

The adjoint of an unbounded operator can be defined in two equivalent ways. Let${\displaystyle T:D(T)\subseteq H_1\to H_2}$ be an unbounded operator between Hilbert spaces.

First, it can be defined in a way analogous to how one defines the adjoint of a bounded operator. Namely, the adjoint${\displaystyle T^{\ast }:D\left(T^{\ast }\right)\subseteq H_2\to H_1}$ ofT is defined as an operator with the property:$${\displaystyle ⟨Tx\mid y{⟩}_2={⟨x\mid T^{\ast }y⟩}_1,x\in D(T).}$$More precisely,${\displaystyle T^{\ast }y}$ is defined in the following way. If${\displaystyle y\in H_2}$ is such that${\displaystyle x\mapsto ⟨Tx\mid y⟩}$ is a continuous linear functional on the domain ofT, then${\displaystyle y}$ is declared to be an element of${\displaystyle D\left(T^{\ast }\right),}$ and after extending the linear functional to the whole space via theHahn–Banach theorem, it is possible to find some${\displaystyle z}$ in${\displaystyle H_1}$ such that$${\displaystyle ⟨Tx\mid y{⟩}_2=⟨x\mid z{⟩}_1,x\in D(T),}$$sinceRiesz representation theorem allows the continuous dual of the Hilbert space${\displaystyle H_1}$ to be identified with the set of linear functionals given by the inner product. This vector${\displaystyle z}$ is uniquely determined by${\displaystyle y}$ if and only if the linear functional${\displaystyle x\mapsto ⟨Tx\mid y⟩}$ is densely defined; or equivalently, ifT is densely defined. Finally, letting${\displaystyle T^{\ast }y=z}$ completes the construction of${\displaystyle T^{\ast },}$ which is necessarily a linear map. The adjoint${\displaystyle T^{\ast }y}$ exists if and only ifT is densely defined.

By definition, the domain of${\displaystyle T^{\ast }}$ consists of elements${\displaystyle y}$ in${\displaystyle H_2}$ such that${\displaystyle x\mapsto ⟨Tx\mid y⟩}$ is continuous on the domain ofT. Consequently, the domain of${\displaystyle T^{\ast }}$ could be anything; it could be trivial (that is, contains only zero).^[9] It may happen that the domain of${\displaystyle T^{\ast }}$ is a closedhyperplane and${\displaystyle T^{\ast }}$ vanishes everywhere on the domain.^[10][11] Thus, boundedness of${\displaystyle T^{\ast }}$ on its domain does not imply boundedness ofT. On the other hand, if${\displaystyle T^{\ast }}$ is defined on the whole space thenT is bounded on its domain and therefore can be extended by continuity to a bounded operator on the whole space.^{[nb 2]} If the domain of${\displaystyle T^{\ast }}$ is dense, then it has its adjoint${\displaystyle T^{\ast \ast }.}$^[12] A closed densely defined operatorT is bounded if and only if${\displaystyle T^{\ast }}$ is bounded.^{[nb 3]}

The other equivalent definition of the adjoint can be obtained by noticing a general fact. Define a linear operator${\displaystyle J}$ as follows:^[12]$${\displaystyle \{\begin{array}{l}J:H_1\oplus H_2\to H_2\oplus H_1\\ J(x\oplus y)=-y\oplus x\end{array}}$$Since${\displaystyle J}$ is an isometric surjection, it is unitary. Hence:${\displaystyle J(\mathrm{\Gamma }(T){)}^{\mathrm{\perp }}}$ is the graph of some operator${\displaystyle S}$ if and only ifT is densely defined.^[13] A simple calculation shows that this "some"${\displaystyle S}$ satisfies:$${\displaystyle ⟨Tx\mid y{⟩}_2=⟨x\mid Sy{⟩}_1,}$$for everyx in the domain ofT. Thus${\displaystyle S}$ is the adjoint ofT.

It follows immediately from the above definition that the adjoint${\displaystyle T^{\ast }}$ is closed.^[12] In particular, a self-adjoint operator (meaning${\displaystyle T=T^{\ast }}$) is closed. An operatorT is closed and densely defined if and only if${\displaystyle T^{\ast \ast }=T.}$^{[nb 4]}

Some well-known properties for bounded operators generalize to closed densely defined operators. The kernel of a closed operator is closed. Moreover, the kernel of a closed densely defined operator${\displaystyle T:H_1\to H_2}$ coincides with the orthogonal complement of the range of the adjoint. That is,^[14]$${\displaystyle \ker (T)=\mathrm{ran}(T^{\ast }{)}^{\mathrm{\perp }}.}$$von Neumann's theorem states that${\displaystyle T^{\ast }T}$ and${\displaystyle T{T}^{\ast }}$ are self-adjoint, and that${\displaystyle I+T^{\ast }T}$ and${\displaystyle I+T{T}^{\ast }}$ both have bounded inverses.^[15] If${\displaystyle T^{\ast }}$ has trivial kernel,T has dense range (by the above identity.) Moreover:

T is surjective if and only if there is a${\displaystyle K>0}$ such that${\displaystyle \Vert f{\Vert }_2\le K{\Vert T^{\ast }f\Vert }_1}$ for all${\displaystyle f}$ in${\displaystyle D\left(T^{\ast }\right).}$^{[nb 5]} (This is essentially a variant of the so-calledclosed range theorem.) In particular,T has closed range if and only if${\displaystyle T^{\ast }}$ has closed range.

In contrast to the bounded case, it is not necessary that${\displaystyle (TS{)}^{\ast }=S^{\ast }{T}^{\ast },}$ since, for example, it is even possible that${\displaystyle (TS{)}^{\ast }}$ does not exist.^{[citation needed]} This is, however, the case if, for example,T is bounded.^[16]

A densely defined, closed operatorT is callednormal if it satisfies the following equivalent conditions:^[17]

• ${\displaystyle T^{\ast }T=T{T}^{\ast }}$;
• the domain ofT is equal to the domain of${\displaystyle T^{\ast },}$ and${\displaystyle \Vert Tx\Vert =\Vert T^{\ast }x\Vert }$ for everyx in this domain;
• there exist self-adjoint operators${\displaystyle A,B}$ such that${\displaystyle T=A+iB,}$${\displaystyle T^{\ast }=A-iB,}$ and${\displaystyle \Vert Tx{\Vert }^2=\Vert Ax{\Vert }^2+\Vert Bx{\Vert }^2}$ for everyx in the domain ofT.

Every self-adjoint operator is normal.

Let${\displaystyle T:B_1\to B_2}$ be an operator between Banach spaces. Then thetranspose (ordual)${\displaystyle {}^{t}T:{B_2}^{\ast }\to {B_1}^{\ast }}$ of${\displaystyle T}$ is the linear operator satisfying:$${\displaystyle ⟨Tx,y^{\prime }⟩=⟨x,\left({}^{t}T\right)y^{\prime }⟩}$$for all${\displaystyle x\in B_1}$ and${\displaystyle y\in B_2^{\ast }.}$ Here, we used the notation:${\displaystyle ⟨x,x^{\prime }⟩=x^{\prime }(x).}$^[18]

The necessary and sufficient condition for the transpose of${\displaystyle T}$ to exist is that${\displaystyle T}$ is densely defined (for essentially the same reason as to adjoints, as discussed above.)

For any Hilbert space${\displaystyle H,}$ there is the anti-linear isomorphism:$${\displaystyle J:H^{\ast }\to H}$$given by${\displaystyle Jf=y}$ where${\displaystyle f(x)=⟨x\mid y{⟩}_H,(x\in H).}$ Through this isomorphism, the transpose${\displaystyle {}^{t}T}$ relates to the adjoint${\displaystyle T^{\ast }}$ in the following way:^[19]$${\displaystyle T^{\ast }=J_1\left({}^{t}T\right)J_2^{-1},}$$where${\displaystyle J_j:H_j^{\ast }\to H_j}$. (For the finite-dimensional case, this corresponds to the fact that the adjoint of a matrix is its conjugate transpose.) Note that this gives the definition of adjoint in terms of a transpose.

Closed linear operators are a class oflinear operators onBanach spaces. They are more general thanbounded operators, and therefore not necessarilycontinuous, but they still retain nice enough properties that one can define thespectrum 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 thederivative and a large class ofdifferential operators.

LetX,Y be twoBanach spaces. Alinear operatorA :D(A) ⊆X →Y isclosed if for everysequence{x_n} inD(A)converging tox inX such thatAx_n →y ∈Y asn → ∞ one hasx ∈D(A) andAx =y. Equivalently,A is closed if itsgraph isclosed in thedirect sumX ⊕Y.

Given a linear operatorA, not necessarily closed, if the closure of its graph inX ⊕Y happens to be the graph of some operator, that operator is called theclosure ofA, and we say thatA isclosable. Denote the closure ofA byA. It follows thatA is therestriction ofA toD(A).

Acore (oressential domain) of a closable operator is asubsetC ofD(A) such that the closure of the restriction ofA toC isA.

Consider thederivative operatorA =⁠d/dx⁠ whereX =Y =C([a,b]) is the Banach space of allcontinuous functions on aninterval[a,b]. If one takes its domainD(A) to beC¹([a,b]), thenA is a closed operator which is not bounded.^[20] On the other hand ifD(A) =C^∞([a,b]), thenA will no longer be closed, but it will be closable, with the closure being its extension defined onC¹([a,b]).

An operatorT on a Hilbert space issymmetric if and only if for eachx andy in the domain ofT we have${\displaystyle ⟨Tx\mid y⟩=⟨x\mid Ty⟩}$. A densely defined operatorT is symmetric if and only if it agrees with its adjointT^∗ restricted to the domain ofT, in other words whenT^∗ is an extension ofT.^[21]

In general, ifT is densely defined and symmetric, the domain of the adjointT^∗ need not equal the domain ofT. IfT is symmetric and the domain ofT and the domain of the adjoint coincide, then we say thatT isself-adjoint.^[22] Note that, whenT is self-adjoint, the existence of the adjoint implies thatT is densely defined and sinceT^∗ is necessarily closed,T is closed.

A densely defined operatorT issymmetric, if the subspaceΓ(T) (defined in a previous section) is orthogonal to its imageJ(Γ(T)) underJ (whereJ(x,y):=(y,-x)).^{[nb 6]}

Equivalently, an operatorT isself-adjoint if it is densely defined, closed, symmetric, and satisfies the fourth condition: both operatorsT –i,T +i are surjective, that is, map the domain ofT onto the whole spaceH. In other words: for everyx inH there existy andz in the domain ofT such thatTy –iy =x andTz +iz =x.^[23]

An operatorT isself-adjoint, if the two subspacesΓ(T),J(Γ(T)) are orthogonal and their sum is the whole space${\displaystyle H\oplus H.}$^[12]

This approach 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.

A symmetric operator is often studied via itsCayley transform.

An operatorT on a complex Hilbert space is symmetric if and only if the number${\displaystyle ⟨Tx\mid x⟩}$ is real for allx in the domain ofT.^[21]

A densely defined closed symmetric operatorT is self-adjoint if and only ifT^∗ is symmetric.^[24] It may happen that it is not.^[25][26]

A densely defined operatorT is calledpositive^[8] (ornonnegative^[27]) if its quadratic form is nonnegative, that is,${\displaystyle ⟨Tx\mid x⟩\ge 0}$ for allx in the domain ofT. Such operator is necessarily symmetric.

The operatorT^∗T is self-adjoint^[28] and positive^[8] for every densely defined, closedT.

Thespectral theorem applies to self-adjoint operators^[29] and moreover, to normal operators,^[30][31] but not to densely defined, closed operators in general, since in this case the spectrum can be empty.^[32][33]

A symmetric operator defined everywhere is closed, therefore bounded,^[6] which is theHellinger–Toeplitz theorem.^[34]

By definition, an operatorT is anextension of an operatorS ifΓ(S) ⊆ Γ(T).^[35] An equivalent direct definition: for everyx in the domain ofS,x belongs to the domain ofT andSx =Tx.^[5][35]

Note that an everywhere defined extension exists for every operator, which is a purely algebraic fact explained atDiscontinuous linear map § General existence theorem and based on theaxiom of choice. If the given operator is not bounded then the extension is adiscontinuous linear map. It is of little use since it cannot preserve important properties of the given operator (see below), and usually is highly non-unique.

An operatorT is calledclosable if it satisfies the following equivalent conditions:^[6][35][36]

• T has a closed extension;
• the closure of the graph ofT is the graph of some operator;
• for every sequence (x_n) of points from the domain ofT such thatx_n → 0 and alsoTx_n →y it holds thaty = 0.

Not all operators are closable.^[37]

A closable operatorT has the least closed extension${\displaystyle \overline{T}}$ called theclosure ofT. The closure of the graph ofT is equal to the graph of${\displaystyle \overline{T}.}$^[6][35] Other, non-minimal closed extensions may exist.^[25][26]

A densely defined operatorT is closable if and only ifT^∗ is densely defined. In this case${\displaystyle \overline{T}=T^{\ast \ast }}$ and${\displaystyle (\overline{T}{)}^{\ast }=T^{\ast }.}$^[12][38]

IfS is densely defined andT is an extension ofS thenS^∗ is an extension ofT^∗.^[39]

Every symmetric operator is closable.^[40]

A symmetric operator is calledmaximal symmetric if it has no symmetric extensions, except for itself.^[21] Every self-adjoint operator is maximal symmetric.^[21] The converse is wrong.^[41]

An operator is calledessentially self-adjoint if its closure is self-adjoint.^[40] An operator is essentially self-adjoint if and only if it has one and only one self-adjoint extension.^[24]

A symmetric operator may have more than one self-adjoint extension, and even a continuum of them.^[26]

A densely defined, symmetric operatorT is essentially self-adjoint if and only if both operatorsT –i,T +i have dense range.^[42]

LetT be a densely defined operator. Denoting the relation "T is an extension ofS" byS ⊂T (a conventional abbreviation for Γ(S) ⊆ Γ(T)) one has the following.^[43]

• IfT is symmetric thenT ⊂T^∗∗ ⊂T^∗.
• IfT is closed and symmetric thenT =T^∗∗ ⊂T^∗.
• IfT is self-adjoint thenT =T^∗∗ =T^∗.
• IfT is essentially self-adjoint thenT ⊂T^∗∗ =T^∗.

The class ofself-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 famousspectral theorem holds for self-adjoint operators. In combination withStone'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, seeSelf-adjoint operator § Self-adjoint extensions in quantum mechanics. Such unitary groups are especially important for describingtime evolution in classical and quantum mechanics.

• Hilbert space § Unbounded operators
• Stone–von Neumann theorem
• Bounded operator

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