unbounded operator



This page is about unbounded linear operators on Hilbert spaces. For operators on Hilbert spaces, “bounded” and “continuous” are synonymous, so the first question to be answered is: Why consider unbounded, i.e., discontinuous operators in a category that is a subcategory of Top? The reason is simple: It is forced upon us both by applications, such as quantum mechanics, and by the fact that simple and useful operators like differentiation are not bounded. Happily, in most applications the operators considered retain some sort of “limit property”, namely the property of being “closed”. Although that seems to be negligible compared to continuity, it allows the development of a rich and useful theory, and as a consequence there is a tremendous amount of literature devoted to this subject.

One way of dealing with unbounded operators is via affiliated operators, see there.

Example: differentiation is unbounded

Let \mathcal{H} be the Hilbert space L 2()L^2(\mathbb{R}), and let TT be the differentiation operator defined on the dense subspace of Schwartz functions ff by Tf(x)f(x)Tf(x) \coloneqq f'(x). One might hope TT has a continuous extension to all of L 2L^2, but consider the sequence f k(x):=exp(k|x|)f_k(x) := \exp(-k|x|) for kk \in \mathbb{N}. Then we have Tf kf k=k\frac{\|Tf_k\|}{\|f_k\|} = k, so TT is unbounded.

Note that the domain of definition of an unbounded operator will generally be given only on a dense subspace, as in this example. Indeed, the existence of unbounded operators defined everywhere (on a Hilbert space) is non-constructive, relying on the Hahn–Banach theorem and refutable in dream mathematics.


After the definition we will look at some concepts that can be transferred from the bounded context here.

We will talk about the von Neumann algebra that contains all spectral projections? here.

Finally we give some counterexamples, i.e., phenomena that contradict the intuition built from bounded operators here.


An unbounded operator TT on a Hilbert space \mathcal{H} is a linear operator defined on a subspace DD of \mathcal{H}. DD is necessarily a linear submanifold. Usually one assumes that DD is dense in \mathcal{H}, which we will do, too, unless we indicate otherwise.

In particular every bounded operator A:A: \mathcal{H} \to \mathcal{H} is an unbounded operator (red herring principle).

Domains: a first look

Unbounded operators are not defined on the whole Hilbert space, so it is essential that, when talking about a specific unbounded operator, we are actually talking about the pair (T,D T)(T, D_T) of an operator TT together with its domain D TD_T. In particular two unbounded operators T,ST, S are equal iff their domains are equal, D T=D SD_T = D_S, and for all xD S=D Tx \in D_S=D_T we have Tx=SxTx = Sx.

If the domain is not specified, the default definition of the domain of a given operator TT is simply D T:={x|Tx}D_T := \{x \in \mathcal{H} \vert Tx \in \mathcal{H} \}.

Warning: if one composes two unbounded operators TT and SS, it may happen that D TD S={0}D_T \cap D_S = \{0\}. If we insist that all our unbounded operators are densely defined, we need as an additional assumption that D TD SD_T \cap D_S is dense to make sense of the composite TST S.

The Hellinger-Toeplitz theorem

  • Theorem: Let AA be an everywhere defined linear operator on a Hilbert space \mathcal{H} that is symmetric. Then A is bounded.

For the definition of symmetric see below.

The Hellinger-Toeplitz theorem is a no-go theorem for quantum mechanics. Since it is known that operators essential for quantum mechanics are both symmetric and unbounded, we are led to conclude that they cannot be everywhere defined. This means that the problems that accompany only densely defined operators cannot be avoided.


This is a corollary to the closed graph theorem III.2 in the book

  • Reed, M.; Simon, B.: Methods of modern mathematical physics. Volume 1, Functional Analysis

Closedness, selfadjointness, resolvent


Recall that the graph of an operator TT (or any function, in general) is the subset 𝒢 T:={(x,y)×|Tx=y}\mathcal{G}_T := \{(x, y) \in \mathcal{H} \times \mathcal{H} \vert Tx = y \}. The graph of a given operator need not be closed (in the product topology of ×\mathcal{H} \times \mathcal{H}). The notion that will be a surrogate for continuity is “closable”, defined as follows:

  • Definition: Given an operator T with domain D TD_T, any operator TT' with larger domain that is equal to TT on D TD_T is called an extension of T, we write TTT \subset T'.

  • Definition: An operator is closed if its graph is closed.

  • Definition: An operator is closable if it has a closed extension. The smallest such extension is called the closure of TT and is denoted by T¯\overline{T}.

  • Proposition (closure of graph is graph of closure): If an operator TT is closable, then the closure of its graph 𝒢¯\overline{\mathcal{G}} is the graph of an operator, and this operator is its closure.

The last part deserves some elaboration: Given an operator TT, we can always form the closure of its graph 𝒢\mathcal{G}. How can the closure not be the graph of an operator? Given a sequence (x n)(x_n) in 𝒢\mathcal{G}, such that both limits xlim nx nx \coloneqq \lim_{n \to \infinity} x_n and ylim nTx ny \coloneqq \lim_{n \to \infinity} T x_n exist, we have that (x,y)(x, y) is in the closure of 𝒢\mathcal{G}. Now it may happen that there is another point (x,y)(x, y') in the closure with yyy \neq y', which implies that the closure cannot be the graph of a single valued function.

We may assume without loss of generality that lim nx n=0\lim_{n \to \infinity} x_n = 0, so that we get as a characterisation of closability: if ylim nTx ny \coloneqq \lim_{n \to \infinity} T x_n exists, then y=0y=0. It TT were continuous, we would not have to assume that (Tx n)(Tx_n) is convergent, so this additional assumption tells us in what respect closability is weaker than continuity.

  • Definition: For a closed operator TT a subset DD of D TD_T is called a core of TT if T| D¯=T\overline{T \vert_{D}} = T. In other words: if we restrict TT to DD and take the closure, we obtain again TT.

Example of an operator that is not closable


We let T *T^* be the adjoint of an operator TT. Note that for an only densely defined TT, the domain of the adjoint may be strictly larger.

  • Definition (selfadjoint et al.): An operator is symmetric (or Hermitian) if T=T *| D TT = T^* \vert_{D_T} (the adjoint is restricted to the domain of TT). It is selfadjoint if it is symmetric and D T=D T *D_T = D_{T^*}. A symmetric operator is essentially selfadjoint if its closure is selfadjoint.

The difference of being symmetric and being selfadjoint is crucial, although there is a famous anecdote that seems to indicate otherwise:

  • Anecdote of selfadjointness: Once upon a time John von Neumann thanked Werner Heisenberg for the invention of quantum mechanics, because this had led to the development of so much beautiful mathematics, adding that mathematics paid back a part of the debt by clarifying for example the difference of a selfadjoint operator and one that is only symmetric. Heisenberg replied: “What is the difference?”

Nevertheless theorems that assume an operator to be selfadjoint will be not applicable to an operator that is only symmetric. One example is the spectral theorem.

Example of a symmetric, but not selfadjoint, operator


The definition of the resolvent does not pose any problems compared to the bounded case:

  • Definition: let TT be a closed operator on a Hilbert space \mathcal{H}. A complex number λ\lambda is in the resolvent set ρ(T)\rho(T) if λ𝟙T\lambda \mathbb{1} - T is a bijection of D TD_T and \mathcal{H} with bounded inverse. The inverse operator is called the resolvent R λ(T)R_{\lambda}(T) of TT at λ\lambda.

  • Theorem: The resolvent set is an open subset of \mathbb{C} on which the resolvent is an analytic operator valued function. Resolvents at different points commute and we have

    R λ(T)R μ(T)=(μλ)R μ(T)R λ(T) R_{\lambda}(T) - R_{\mu}(T) = (\mu - \lambda) R_{\mu}(T) R_{\lambda}(T)

The proof can be done as in the bounded case.

Commuting operators

The concept of commuting operators, which is of no problem in the bounded case, presents a conceptual difficulty in the unbounded one: for given operators AA, BB we would like to be able to say whether they commute, although their composite may not have dense domain. For selfadjoint operators there is a solution to this problem: We know that in the bounded case, two selfadjoint operators commute iff their spectral projections commute. This suggests the

  • Definition: two (possibly unbounded) selfadjoint operators commute iff all their spectral projections commute.

The spectral theorem shows that all bounded Borel functions of two commuting operators will also commute.

The following theorem states the reverse for two of the most important functions and shows that the definition of commutation above is reasonable:

  • Theorem: let AA and BB be selfadjoint operators on a Hilbert space \mathcal{H}. Then the following three statements are equivalent:

(a) AA and BB commute

(b) if Imλ\operatorname{Im} \lambda and Imμ\operatorname{Im} \mu are 0\neq 0, then R λ(A)R_{\lambda}(A) and R μ(B)R_{\mu}(B) commute, as it is defined for bounded operators: R λ(A)R μ(B)=R μ(B)R λ(A)R_{\lambda}(A) R_{\mu}(B) = R_{\mu}(B) R_{\lambda}(A)

(c) for all t,st, s \in \mathbb{R} we have exp(itA)exp(isB)=exp(isB)exp(itA)\exp(itA) \exp(isB) = \exp(isB) \exp(itA)

Strongly continuous one-parameter semigroups

A strongly continuous one-parameter semigroup is an unitary representation of \mathbb{R} on \mathcal{H} where \mathbb{R} is seen as a topological group with respect to multiplication, see topological group. An explicit definition recalling these concepts is this:

  • Definition: a function U:()U : \mathbb{R} \to \mathcal{B}(\mathcal{H}) is a one-parameter semigroup if the semigroup condition U(t+s)=U(t)U(s)U(t+s) = U(t)U(s) holds. If for every xx \in \mathcal{H} and tt 0t \to t_0 we have U(t)xU(t 0)xU(t)x \to U(t_0)x then it is strongly continuous. If every UU is an unitary operator, it is an unitary semigroup.

In the following a semigroup will be understood to be a one-parameter unitary strongly continuous semigroup.

In physics, one-parameter semigroups of this kind often represent the time evolution of a physical system described by an evolution equation.

  • Theorem: a selfadjoint operator AA generates a semigroup via U(t):=exp(itA)U(t) := \exp(itA)

  • Theorem (Stone’s theorem): let UU be a semigroup, then there is a selfadjoint operator AA such that U(t)=exp(itA)U(t) = \exp(i t A). This operator is often called the infinitesimal generator of UU.

These two theorems are essential for the Schrödinger picture of quantum mechanics, which describes a system by the Schrödinger equation, we have now a one-to-one correspondence of selfadjoint operators which can be seen as Hamilton operators (only special operators will be seen as describing actual physical systems, of course), and semigroups which describe the time evolution generated by the Hamilton operator.

As a trivial observation we add that Stone’s theorem is a (huge) generalization of the Taylor series, let f:f: \mathbb{R} \to \mathbb{R} that is (real) analytic in a neighborhood of 00, then we get for xx small enough:

f(h)= k=0 f n(0)n!h n=exp(ih(iddx)) f(h) = \sum_{k=0}^{\infty} \frac{f^{n}(0)}{n!} h^n = \exp(i h(-i \frac{d}{dx} ))

This shows that the operator iddx-i \frac{d}{dx} generates the semigroup of translations on the real line. Now we could, for example, use Stone’s theorem to prove that iddx-i \frac{d}{dx} is selfadjoint by proving that the translation group is strongly continuous.

Subtleties resulting from domain issues

(This rather generic title will have to be revised.)

Nelson’s example of noncommuting exponentials


Nelson’s example shows that the rather involved definition of commutativity of two unbounded operators is well motivated, because a more naive one will have unwanted consequences. It is a counterexample to the following conjecture:

  • Conjecture (false!): Let AA and BB be selfadjoint operators on a Hilbert space \mathcal{H} and DD a dense subset such that both AA and BB restricted to DD are essentially selfadjoint. Suppose that for all xDx \in D we have ABxBAx=0 A B x - B A x = 0. Then AA and BB commute.

We have already seen that on 2\mathbb{R}^2 we can define two essentially selfadjoint operators i x-i \partial_x and i y-i \partial_y that generate translations along the x-axis and the y-axis respectively. Both the generated translations and the operators commute (the latter if applied to differentiable functions, of course).

The central idea of Nelson’s counterexample is to replace 2\mathbb{R}^2 by a Riemann surface with two sheets, such that walking east, then walking north takes you to one sheet, while walking north then walking east takes you to the other sheet.


We use the Riemann surface MM of f(z)=(z)f(z) = \sqrt(z). We give a brief exposition of its construction: take two copies of \mathbb{C}, two “sheets”, and call them I and II. Cut both sheets along (0,)(0, \infty), label the edge of the first quadrant along the cut as ++ and the edge of the fourth quadrant as -. Then attach the ++ edge of I with the - edge of II and the - of I with the ++ edge of II.

Let =L 2(M)\mathcal{H} = L^2(M) with respect to Lebesgue measure. As indicated above we define A = i x-i \partial_x and B = i y-i \partial_y, this is with respect to the canonical chart on each sheet that projects it onto \mathbb{C} and then identifies \mathbb{C} with 2\mathbb{R}^2.

Let DD be the set of all smooth functions with compact support not containing 00. Then:

(a) AA and BB are essentially selfadjoint on DD

(b) AA and BB map DD onto DD

(c) for all xDx \in D we have ABx=BAxA B x = B A x

(d) exp(itA)\exp(i t A) and exp(isB)\exp(i s B) do not commute.

Only part (a) needs further explanation…


  • Konrad Schmuedgen, Unbounded self-adjoint operators on Hilbert space, Springer GTM 265, 2012

Chapter VIII of the following classic volume is devoted to unbounded operators:

  • Reed, M.; Simon, B.: Methods of modern mathematical physics. Volume 1, Functional Analysis

Nelson’s example is taken from the above reference, the original reference is this:

  • Edward Nelson: Analytic Vectors Ann.Math. 70, p.572-615, 1959

category: analysis

Revised on March 1, 2015 22:31:36 by Gabriele Tornetta? (