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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.

1 Short history
2 Definitions and basic properties
3 Adjoint
4 Transpose
5 Symmetric operators and self-adjoint operators
6 Extension-related
7 Importance of self-adjoint operators
8 See also
9 Notes
10 References

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 von Neumann^[1] and M. Stone^[2]. The technique of using the graph to analyze unbounded operators was introduced by von Neumann in ^[3]. (Reed & Simon 1980, Notes to Chapter VIII, page 305)

Definitions and basic properties

Let B₁ and B₂ be Banach spaces. An operator T : B₁ → B₂, here refers to a linear map T from a linear subspace D(T) of B₁ — the domain of T — to B₂.^[4] Contrary to the usual convention, T may not be defined on the whole space B₁. Two operators are equal if they have the common domain and they coincide on that common domain.^[4]

An operator $T$ is said to be closed if its graph is closed.^[5] (Here, the graph Γ ( T ) of an operator T, defined as the set of all pairs ( x, Tx ) where x runs over the domain of T, is a linear subspace of the direct sum $H_1 \oplus H_2$ .^[5]) Explicitly, this means that: for every sequence (x_n) of points from the domain of T such that x_n converge to some x and also Tx_n converge to some y it holds that x belongs to the domain of T and Tx = y.^[5] The closedness can also be formulated in terms of the graph norm: an operator T is closed if and only if its domain is a complete space (i.e., a Banach space) with respect to the norm:^[6]

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

If T : B₁ → B₂ is closed and densely defined and continuous on its domain, then it is defined on B₁.^[7]

An operator T is said to be densely defined if its domain is dense.^[4] The denseness of the domain is necessary and sufficient for the existence of the adjoint; see the next section.

A densely defined operator T is called bounded from below if T + a is a positive operator for some real number a. That is, $\langle Tx \mid x \rangle \ge -a \| x \|^2$ for all x in the domain of T.^[8] If both T and (–T) are bounded from below then T is bounded.^[8]

Adjoint

The adjoint of an unbounded operator can be defined in two equivalent ways. First, it can be define in a way analogous to how we define the adjoint of a bounded operator. Namely, the adjoint T^∗ : H₂ → H₁ of T is defined as an operator with the property:

$\langle Tx \mid y \rangle_2 = \langle x \mid T^*y \rangle_1, \quad (x \in D(T)).$

More precisely, T^∗ is defined in the following way. If y is such that $x \mapsto \langle Tx, y \rangle$ is a continuous linear functional on the domain of T, then, after extending it to the whole space via the Hahn-Banach theorem, we can find a z such that

$\langle Tx \mid y \rangle_2 = \langle x \mid z \rangle_1, \quad (x \in D(T))$

since the dual of a Hilbert space can be identified with the set of linear functionals given by the inner product. For each y, z is unique determined if and only if the linear functional is densely defined; i.e., T is densely defined. Finally, we let T^∗y = z, completing the construction of T^∗.^[9] Note that T^∗ exists if and only if T is densely defined.

By definition, the domain of T^∗ consists of elements $y \in H_2$ such that $x \mapsto \langle Tx, y \rangle$ is continuous on the domain of T. Consequently, the domain of T^∗ could be anything; it could be trivial (i.e., contains only zero)^[10] It may happen that the domain of T^∗ is a closed hyperplane and T^∗ vanishes everywhere on the domain.^[11]^[12] 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.^[13] If the domain of T^∗ is dense, then it has its adjoint T^∗∗.^[14] A closed densely defined operator T is bounded if and only if T^∗ is bounded.^[15]

The other equivalent definition of the adjoint can be obtained by noticing a general fact: define a linear operator $J: H_1 \oplus H_2 \to H_2 \oplus H_1$ by $J(x \oplus y) = -y \oplus x$ .^[14] (Since $J$ is an isometric surjection, it is unitary.) We then have: $J(\Gamma (T))^\bot$ is the graph of some operator S if and only if $T$ is densely defined.^[16] 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 the domain of T. Thus, S is the adjoint of T.

It follows immediately from the above definition that the adjoint T^∗ is closed.^[14] In particular, a self-adjoint operator (i.e., T = T^∗) is closed. An operator T is closed and densely defined if and only if T^∗∗ = T. ^[17]

Some well-know 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 T : H₁ → H₂ coincides with the orthogonal complement of the range of the adjoint. That is, ^[18]

$\operatorname{ker}(T) = \operatorname{ran}(T^*)^\bot.$

von Neumann's theorem states that T^∗T and TT^∗ are self-adjoint, and that I + T^∗T and I + TT^∗ both have bounded inverses.^[19] 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

$\|f\|_2 \le K\|T^*f\|_1$ for every $f \in D(T^*)$ .^[20]

(This is essentially a variant of the so-called closed range theorem.) In particular, T has closed range if and only if T^∗ has closed range.

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.^{[citation needed]} This is, however, the case if, for example, T is bounded.^[21]

A densely defined, closed operator T is called normal if it satisfies the following equivalent conditions:^[22]

T^∗T = T T^∗;
the domain of T is equal to the domain of T^∗, and $\| Tx \| = \| T^* x \|$ for every x in this domain;
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 the domain of T.

Every self-adjoint operator is normal.

Transpose

Let T : B₁ → B₂ be an operator between Banach spaces. Then the transpose (or dual) $T ': {B_2}^* \to {B_1}^*$ of T is an operator satisfying:

$\langle T x, y' \rangle = \langle x, T' y' \rangle$

for all x in B₁ and y in B₂^*. Here, we used the notation: $\langle x, x' \rangle = x'(x)$ .^[23]

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, as discussed above.)

For any Hilbert space H, there is the anti-linear isomorphism:

$J: H^* \to H$

given by $J f = y$ where $f(x) = \langle x \mid y \rangle_H, (x \in H)$ . Through this isomorphism, the transpose T^' relates to the adjoint T^∗ in the following way:

$T^* = J_1 T' J_2^{-1}$ ,^[24]

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.) Note that this gives the definition of adjoint in terms of a transpose.

Symmetric operators and self-adjoint operators

A densely defined operator T is symmetric if $\langle Tx \mid y \rangle = \lang x \mid Ty \rang$ for all elements x and y in the domain of T.^[25]

An operator T is said to be self-adjoint if T^∗ = T.^[25] Note that, when T is self-adjoint, the existence of the adjoint implies that T is dense and since $T *$ is necessarily closed, $T$ is closed.

A densely defined operator T is symmetric, if the subspace Γ ( T ) is orthogonal to its image J ( Γ ( T ) ) under J.^[26]

Equivalently, an operator T is self-adjoint if it is densely defined, closed, symmetric, and satisfies the fourth condition: both operators T – i, T + 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 and z in the domain of T such that Ty – iy = x and Tz + iz = x.^[27]

An operator T is self-adjoint, if the two subspaces Γ ( T ), J ( Γ ( T ) ) are orthogonal and their sum is the whole space $H \oplus H .$ ^[14]

A densely defined operator T is symmetric if T^∗ is an extension of T (see below).^[25]

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 its Cayley transform.

An operator T is symmetric if and only if its quadratic form is real, that is, the number $\langle Tx \mid x \rangle$ is real for all x in the domain of T.^[25]

A densely defined closed symmetric operator T is self-adjoint if and only if T^∗ is symmetric.^[28] It may happen that it is not.^[29]^[30]

A densely defined operator T is called positive^[8] (or nonnegative^[31]) if its quadratic form is nonnegative, that is, $\langle Tx \mid x \rangle \ge 0$ for all x in the domain of T. Such operator is necessarily symmetric.

The operator T^∗T is self-adjoint^[32] and positive^[8] for every densely defined, closed T.

The spectral theorem applies to self-adjoint operators ^[33] and moreover, to normal operators,^[34]^[35] but not to densely defined, closed operators in general, since in this case the spectrum can be empty.^[36]^[37]

A symmetric operator defined everywhere is closed, therefore bounded,^[5] which is the Hellinger–Toeplitz theorem.^[38]

Extension-related

By definition, an operator T is an extension of an operator S if Γ (S) ⊆ Γ (T).^[39] An equivalent direct definition: for every x in the domain of S, x belongs to the domain of T and Sx = Tx.^[4]^[39]

Note that an everywhere defined extension exists for every operator, which is a purely algebraic fact explained at Discontinuous linear map#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 (see below), and usually is highly non-unique.

An operator T is called closable if it satisfies the following equivalent conditions:^[5]^[39]^[40]

T has a closed extension;
the closure of the graph of T is the graph of some operator;
for every sequence (x_n) of points from the domain of T such that x_n converge to 0 and also Tx_n converge to some y it holds that y = 0.

Not all operators are closable.^[41]

A closable operator T has the least closed extension $\overline T$ called the closure of T. The closure of the graph of T is equal to the graph of $\overline T.$ ^[5]^[39]

Other, non-minimal closed extensions may exist.^[29]^[30]

A densely defined operator T is closable if and only if T^∗ is densely defined. In this case $\overline T = T^{**}$ and $(\overline T)^* = T^*.$ ^[14]^[42]

If S is densely defined and T is an extension of S then S^∗ is an extension of T^∗.^[43]

Every symmetric operator is closable.^[44]

A symmetric operator is called maximal symmetric if it has no symmetric extensions, except for itself.^[25]

Every self-adjoint operator is maximal symmetric.^[25] The converse is wrong.^[45]

An operator is called essentially self-adjoint if its closure is self-adjoint.^[44]

An operator is essentially self-adjoint if and only if it has one and only one self-adjoint extension.^[28]

An operator may have more than one self-adjoint extension, and even a continuum of them.^[30]

A densely defined, symmetric operator T is essentially self-adjoint if and only if both operators T – i, T + i have dense range.^[46]

Let T be a densely defined operator. Denoting the relation "T is an extension of S" by S ⊂ T (a conventional abbreviation for Γ(S) ⊆ Γ(T)) one has the following.^[47]

If T is symmetric then T ⊂ T^∗∗ ⊂ T^∗.
If T is closed and symmetric then T = T^∗∗ ⊂ T^∗.
If T is self-adjoint then T = T^∗∗ = T^∗.
If T is essentially self-adjoint then T ⊂ T^∗∗ = T^∗.

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#Self adjoint extensions in quantum mechanics. Such unitary groups are especially important for describing time evolution in classical and quantum mechanics.

Notes

^ von Neumann (1929-1930), "Allgemeine Eigenwerttheorie Hermitescher Functionaloperatoren", Math. Ann. 102: 49–131, doi:10.1007/BF01782338
^ Stone, M. (1932), "Linear transformations in Hilbert spaces and their applications to analysis", Amer. Math. Soc. Colloq. Publ. (New York) 15
^ von Neumann (1936), "Über Adjungierte Funktionaloperatoren", Ann. Math. (2) 33: 294–310, doi:10.2307/1968331
^ ^a ^b ^c ^d Pedersen 1989, 5.1.1
^ ^a ^b ^c ^d ^e ^f Pedersen 1989, 5.1.4
^ Berezansky, Sheftel & Us 1996, page 5
^ Suppose $f j$ is a sequence in the domain of T that converges to $g \in B_2$ . Since T is uniformly continuous on its domain, $T f j$ is Cauchy in $B 2$ . Thus, $(f j, T f j)$ is Cauchy and so converges to some $(f, T f)$ since the graph of T is closed. Hence, $f = g$ , and the domain of T is closed.
^ ^a ^b ^c ^d Pedersen 1989, 5.1.12
^ Verifying that T^∗ is linear trivial.
^ Berezansky, Sheftel & Us 1996, Example 3.2 on page 16
^ Reed & Simon 1980, page 252
^ Berezansky, Sheftel & Us 1996, Example 3.1 on page 15
^ Proof: being closed, the everywhere defined T^∗ is bounded, which implies boundedness of T^∗∗, the latter being the closure of T. See also (Pedersen 1989, 2.3.11) for the case of everywhere defined T.
^ ^a ^b ^c ^d ^e Pedersen 1989, 5.1.5
^ Proof: We have: $T * * = T$ . So, if $T *$ is bounded, then its adjoint $T$ is bounded.
^ Berezansky, Sheftel & Us 1996, page 12
^ Proof: If T is closed densely defined, then T^∗ exists and is densely defined. Thus, T^∗∗ exists. The graph of T is dense in the graph of T^∗∗; hence, T = T^∗∗. Conversely, since the existence of T^∗∗ implies that that of T^∗, which in turn implies T is densely defined. Since T^∗∗ is closed, T is densly defined and closed.
^ Brezis, pp. 28.
^ Yoshida, pp. 200.
^ If T is surjective, then $T: (\operatorname{ker}T)^\bot \to H_2$ has bounded inverse, which we denote by S. The estimate then follows since

$\|f\|_2^2 = |\langle TSf \mid f \rangle_2| \le \|S\| \|f\|_2 \|T^*f\|_1$

Conversely, suppose the estimate holds. Since $T *$ has closed range then, we have: $\operatorname{ran}(T) = \operatorname{ran}(TT^*)$ . Since $\operatorname{ran}(T)$ is dense, it suffices to show that $T T *$ has closed range. If $T T * f j$ is convergent, then $f j$ is convergent by the estimate since

$\|T^*f_j\|_1^2 = | \langle T^*f_j \mid T^*f_j \rangle_1| \le \|TT^*f_j\|_2 \|f_j\|_2.$

Say, $f_j \to g$ . Since $T T *$ is self-adjoint; thus, closed, (von Neumann's theorem), $TT^* f_j \to TT^* g$ . $\square$
^ Yoshida, pp. 195.
^ Pedersen 1989, 5.1.11
^ Yoshida, pp. 193.
^ Yoshida, pp. 196.
^ ^a ^b ^c ^d ^e ^f Pedersen 1989, 5.1.3
^ Follows from (Pedersen 1989, 5.1.5) and the definition via adjoint operators.
^ Pedersen 1989, 5.2.5
^ ^a ^b Reed & Simon 1980, page 256
^ ^a ^b Pedersen 1989, 5.1.16
^ ^a ^b ^c Reed & Simon 1980, Example on pages 257-259
^ Berezansky, Sheftel & Us 1996, page 25
^ Pedersen 1989, 5.1.9
^ Pedersen 1989, 5.3.8
^ Berezansky, Sheftel & Us 1996, page 89
^ Pedersen 1989, 5.3.19
^ Reed & Simon 1980, Example 5 on page 254
^ Pedersen 1989, 5.2.12
^ Reed & Simon 1980, page 84
^ ^a ^b ^c ^d Reed & Simon 1980, page 250
^ Berezansky, Sheftel & Us 1996, pages 6,7
^ Berezansky, Sheftel & Us 1996, page 7
^ Reed & Simon 1980, page 253
^ Pedersen 1989, 5.1.2
^ ^a ^b Pedersen 1989, 5.1.6
^ Pedersen 1989, 5.2.6
^ Reed & Simon 1980, page 257
^ Reed & Simon 1980, pages 255, 256

References

Pedersen, Gert K. (1989), Analysis now, Springer (see Chapter 5 "Unbounded operators").
Reed, Michael; Simon, Barry (1980), Methods of Modern Mathematical Physics, 1: Functional Analysis (revised and enlarged ed.), Academic Press (see Chapter 8 "Unbounded operators").
Berezansky, Y.M.; Sheftel, Z.G.; Us, G.F. (1996), Functional analysis, II, Birkhäuser (see Chapter 12 "General theory of unbounded operators in Hilbert spaces").
Yoshida, Kôsaku (1980), Functional Analysis (sixth ed.), Springer
Brezis, Haïm (1983) (in French), Analyse fonctionnelle - Théorie et applications, Paris: Mason

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Unbounded operator

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