Decomposition of spectrum (functional analysis)

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In functional analysis, the spectrum of an operator generalizes the notion of eigenvalues. Given an operator, it is sometimes useful to break up the spectrum into various parts. This article discusses a few examples of such decompositions.

1 Operators on Banach space
- 1.1 Examples
  - 1.1.1 Multiplication operator
  - 1.1.2 Shifts
- 1.2 Unbounded operators
2 Self adjoint operators on Hilbert space
3 See also
4 References

[edit] Operators on Banach space

Let X be a Banach space, L(X) the family of bounded operators on X.

A complex number λ is in the spectrum of T, σ(T), if T - λ has no bounded inverse. In general, T - λ ∈ L(X) is invertible if and only if it is bounded below and has dense range. Immediately from the definition, if λ is in σ(T), one of the following must be true:

T - λ is not injective.
T - λ is injective, and has dense range. But the densely defined linear map (T - λ) x → x is not bounded, therefore can not be extended to all of X.
T - λ is injective and does not have dense range. The map (T - λ) x → x may be bounded on Ran(T - λ), but can not be extended uniquely to all of X. Or, it may be that the map is just unbounded.

In cases 1 and 2, T - λ is not bounded below, while case 3 covers the possibility T - λ has no dense range.

Correspondingly, for T ∈ L(X), its spectrum σ(T) can be classified as follows:

The point spectrum of T consists of eigenvalues of T and will be denoted by σ_p(T). λ ∈ σ(T) is in the point spectrum if and only if T - λ is not injective.
If λ ∈ σ(T) is not an eigenvalue and the range of T - λ, Ran(T - λ), is dense in X, λ is said to be in the continuous spectrum, σ_c(T), of T.
If T - λ is injective and Ran(T - λ) is not dense, λ is in the residual spectrum of T, σ_r(T).

So σ(T) is the disjoint union

$\sigma(T) = \sigma_p (T) \cup \sigma_c (T) \cup \sigma_r (T).$

In particular, as a consequence of open mapping theorem, T - λ is has a bounded inverse if T - λ: X → X is bijective. The bijectivity condition is stronger than invertibility. If T - λ is not onto but has dense range, and (T - λ)^-1 is bounded on Ran(T -λ). For an element y not in Ran(T - λ), let (T - λ) x_n → y. Then (T - λ)(T - λ)^-1 y = lim (T - λ) x_n = y.

If X* is the dual space of X, and T* : X* → X* is the adjoint operator of T, then σ(T) = σ(T*).

Theorem For a bounded operator T, σ_r(T) ⊂ σ_p(T*) ⊂ σ_r(T) ∪ σ_p(T).

Proof The notation <·, φ> will denote an element of X*, i.e. x → <x, φ> is the action of a bounded linear functional φ. Let λ ∈ σ_r(T). So Ran(T - λ) is not dense in X. By Hahn-Banach theorem, there exists a non-zero φ ∈ X* that vanishes on Ran(T - λ). For all x ∈ X,

$\langle (T - \lambda)x, \phi \rangle = \langle x, (T^* - \lambda) \phi \rangle = 0.$

Therefore (T* - λ)φ = 0 ∈ X* and λ is an eigenvector of T*. The shows the first part of the inclusion. Next suppose that (T* - λ)φ = 0 where φ ≠ 0, i.e.

$\forall x \in X,\; \langle x, (T^* - \lambda) \phi \rangle = \langle (T - \lambda) x, \phi \rangle = 0.$

If Ran(T - λ) is dense, then φ must be the zero functional, a contradiction. The claim is proved.

In particular, when X is a reflexive Banach space, σ_r(T*) ⊂ σ_p(T**) = σ_p(T).

[edit] Examples

[edit] Multiplication operator

Given a σ-finite measure space (S, Σ, μ), consider the Banach space L^p(μ). A function h: S → C is called essentially bounded if h is bounded μ-almost everywhere. An essentially bounded h induces a bounded multiplication operator T_h on L^p(μ):

$(T_h f)(s) = h(s) \cdot f(s).$

The operator norm of T is the essential supremum of h. The essential range of h is defined in the following way: a complex number λ is in the essential range of T if for all ε > 0, the preimage of the open ball B_ε(λ) under h has strictly positive measure. We will show first that σ(T_h) coincides with the essential range of h and then examine its various parts.

If λ is not in the essential range of h, take ε > 0 such that h^-1(B_ε(λ)) has zero measure. The function g(s) = 1/(h(s) - λ) is bounded almost everywhere by 1/ε. The multiplication operator T_g satisfies T_g · T_{h - λ} = T_{h - λ} · T_g = I. So λ does not lie in spectrum of T_h. On the other hand, if λ lies in the essential range of h, consider the sequence of sets {S_n = h^-1(B_1/n(λ))}. Each S_n has positive measure. Let f_n be the characteristic function of S_n. We can compute directly

$\| (T_h - \lambda) f_n \|_p ^p = \| (h - \lambda) f_n \|_p ^p = \int_{S_n} | h - \lambda \; |^p d \mu \leq \frac{1}{n^p} \; \mu(S_n) = \frac{1}{n^p} \| f_n \|_p ^p.$

This shows T_h - λ is not bounded below, therefore not invertible.

If λ is an isolated point of the essential range of h, i.e. μ( h^-1({λ})) > 0, then λ lies in the point spectrum of T_h: Pick an open ball B_ε(λ) that contains only λ from the essential range. Let f be the characteristic function of h^-1(B_ε(λ)), then

$\forall s \in S, \; (T_h f)(s) = \lambda f(s).$

Any λ in the essential range of h that's not an isolated point is in the continuous spectrum of T_h. To show this is to show that T_h - λ has dense range for all such λ. Given f ∈ L^p(μ), again we consider the sequence of sets {S_n = h^-1(B_1/n(λ))}. Let g_n be the characteristic function of S - S_n. Define

$f_n(s) = \frac{1}{ h(s) - \lambda} \cdot g_n(s) \cdot f(s).$

Direct calculation shows that f_n ∈ L^p(μ), and, by the dominated convergence theorem,

$T_h f_n \rightarrow f$

in the L^p(μ) norm.

Therefore multiplication operators have no residual spectrum. In particular, by the spectral theorem, normal operators on a Hilbert space have no residual spectrum.

[edit] Shifts

In the special case when S is the set of natural numbers and μ is the counting measure. The corresponding L^p(μ) is denoted by l ^p. This space consists of complex valued sequences {x_n} such that

$\sum_{n \geq 0} | x_n |^p < \infty.$

For 1 < p < ∞, l ^p is reflexive. Define the left shift T : l ^p → l ^p by

$T(x_1, x_2, x_3, \cdots) = (x_2, x_3, x_4, \cdots).$

T is a partial isometry with operator norm 1. So σ(T) lies in the close unit disk of the complex plane.

T* is the right shift (or unilateral shift), which is an isometry, on l ^q where 1/p + 1/q = 1:

$T^*(x_1, x_2, x_3, \cdots) = (0, x_1, x_2, \cdots).$

For λ ∈ C with |λ| < 1,

$x = (1, \lambda, \lambda ^2, \cdots) \in l^p$

and T x = λ x. Consequently the point spectrum of T contains open unit disk. Invoking reflexivity and the theorem given above, we can deduce that the open unit disk lies in the residual spectrum of T*.

The spectrum of a bounded operator is closed, which implies the unit circle, {|λ| = 1} ⊂ C, is in σ(T). Also, T* has no eigenvalues, i.e. σ_p(T*) is empty. Again by reflexivity of l ^p and the theorem given above, we have that σ_r(T) is also empty. Therefore, for a complex number λ with unit norm, one must have λ ∈ σ_p(T) or λ ∈ σ_c(T). Now if |λ| = 1 and

$T x = \lambda x, \; i.e. \; (x_2, x_3, x_4, \cdots) = \lambda (x_1, x_2, x_3, \cdots),$

then

$x = x_1 (1, \lambda, \lambda^2, \cdots),$

which can not be in l ^p, a contradiction. This means the unit circle must be the continuous spectrum of T.

For the right shift T*, σ_r(T*) is the open unit disk and σ_c(T*) is the unit circle.

For p = 1, one can perform a similar analysis. The results will not be exactly the same, since reflexivity no longer holds.

[edit] Unbounded operators

The spectrum of an unbounded operator can be divided into three parts in exactly the same way as in the bounded case.

[edit] Self adjoint operators on Hilbert space

Hilbert spaces are Banach spaces, so the above discussion applies to bounded operators on Hilbert spaces as well, although possible differences may arise from the adjoint operation on operators. For example, let H be a Hilbert space and T ∈ L(H), σ(T*) is not σ(T) but rather its image under complex conjugation.

For a self adjoint T ∈ L(H), the Borel functional calculus gives additional ways to break up the spectrum naturally.

[edit] Borel functional calculus

Further information: Borel functional calculus

This subsection briefly sketches the developement of this calculus. The idea is to first establish the continuous functional calculus then pass to measurable functions via the Riesz-Markov representation theorem. For the continuous functional calculus, the key ingredients are the following:

1. If T is self adjoint, then for any polynomial P, the operator norm

$\| P(T) \| = \sup_{\lambda \in \sigma(T)} |P(\lambda)|.$

2. The Stone-Weierstrass theorem, which gives that the family of polynomials (with complex coefficients), is dense in C(σ(T)), the continuous functions on σ(T).

The family C(σ(T)) is a Banach algebra when endowed with the uniform norm. So the mapping

$P \rightarrow P(T)$

is an isometric homomorphism from a dense subset of C(σ(T)) to L(H). Extending the mapping by continuity gives f(T) for f ∈ C(σ(T)): let P_n be polynomials such that P_n → f uniformly and define f(T) = lim P_n(T). This is the continuous functional calculus.

For a fixed h ∈ H, we notice that

$f \rightarrow \langle h, f(T) h \rangle$

is a positive linear functional on C(σ(T)). According to the Riesz-Markov representation theorem that there exists a unique measure μ_h on σ(T) such that

$\int_{\sigma(T)} f d \mu_h = \langle h, f(T) h \rangle.$

This measure is sometimes called the spectral measure associated to h. The spectral measures can be used to extend the continuous functional calculus to bounded Borel functions. For a bounded function g that is Borel measurable, define, for a proposed g(T)

$\int_{\sigma(T)} g d \mu_h = \langle h, g(T) h \rangle.$

Via the polarization identity, one can recover (since H is assumed to be complex)

$\langle k, g(T) h \rangle.$

and therefore g(T) h for arbitrary h.

In the present context, the spectral measures, combined with a result from measure theory, give a decomposition of σ(T).

[edit] Decomposing the spectrum

Let h ∈ H and μ_h be its corresponding spectral measure on σ(T) ⊂ R. According to a refinement of Lebesgue's decomposition theorem, μ_h can be decomposed into three mutually singular parts:

$\, \mu = \mu_{pp} + \mu_{cont} + \mu_{sing}$

where μ_pp is a pure point measure, μ_cont is absolutely continuous with respect to the Lebesgue measure, and μ_sing is singular with respect to the Lebesgue measure.

All three types of measures are invariant under linear operations. Let H_cont be the subspace consisting of vectors whose spectral measures are absolutely continuous with respect to the Lebesgue measure. Define H_pp and H_sing in analogous fashion. These subspaces are invariant under T. For example, if h ∈ H_cont and k = T h. Let χ be the characteristic function of some Borel set in σ(T), then

$\langle k, \chi(T) k \rangle = \int_{\sigma(T)} \chi(\lambda) \cdot \lambda^2 d \mu_{h}(\lambda) = \int_{\sigma(T)} \chi(\lambda) \; d \mu_k(\lambda).$

$\lambda^2 d \mu_{h} = d \mu_{k}\,$

and k ∈ H_cont . Furthermore, applying the spectral theorem gives

$H = H_{cont} \oplus H_{pp} \oplus H_{sing}.$

This leads to the following definitions:

The spectrum of T restricted to H_cont is called the absolutely continuous spectrum of T, σ_cont(T).
The spectrum of T restricted to H_sing is called its singular spectrum, σ_sing(T).
The set of eigenvalues of T are called the pure point spectrum of T, σ_pp(T).

The closure of the eigenvalues is the spectrum of T restricted to H_pp. So

$\sigma(T) = {\bar \sigma_{pp}(T)} \cup \sigma_{cont}(T) \cup \sigma_{sing}(T).$

[edit] Comparison

As stated above, a bounded self adjoint operator on Hilbert space can be is a bounded operator on a Banach space.

Unlike the Banach space formulation, the union

$\sigma(T) = {\bar \sigma_{pp}(T)} \cup \sigma_{cont}(T) \cup \sigma_{sing}(T).$

need not be disjoint. It is disjoint when the operator T is of uniform multiplicity, say m, i.e. if T is unitarily equivalent to multiplication by λ on the direct sum

$\oplus _{i = 1} ^m L^2(\mathbb{R}, \mu)$

for some Borel measure μ. When more than one measure appears in the above expression, we see that it is possible for the union of the three types of spectra to not be disjoint. If λ ∈ σ_cont(T) ∩ σ_pp(T), λ is sometimes called an eigenvalue embedded in the absolutely continuous spectrum.

When T is unitarily equivalent to multiplication by λ on

$L^2(\mathbb{R}, \mu),$

the decomposition of σ(T) from Borel functional calculus is a refinement of the Banach space case.

[edit] Physics

The preceding comments can be extended to the unbounded self-adjoint operators since Riesz-Markov holds for locally compact Hausdorff spaces.

In quantum mechanics, observables are, not necessarily bounded, self adjoint operators and their spectra are the possible outcomes of measurements. Absolutely continuous spectrum of a physical observable corresponds to free states of a system, while the pure point spectrum corresponds to bound states. The singular spectrum correspond to physically impossible outcomes. An example of a quantum mechanical observable which has purely continuous spectrum is the position operator of a free particle moving on a line. Its spectrum is the entire real line. Also, since the momentum operator is unitarily equivalent to the position operator, via the Fourier transform, they have the same spectrum.

The discreteness of the spectrum is intimately related to the corresponding states being "localized". Consider a quantum mechanical Hamiltonian H on L²(R). If f is an eigenvector of H, f is square integrable by definition. This means that f must decay sufficiently fast outside some bounded region (Lebesgue-almost everywhere). Sometimes, when performing physical quantum mechanical calculations, one encounters "eigenvectors" that does not lie in L²(R), i.e. wave functions that are not localized. These are the free states of the system. As stated above, in the mathematical formulation, the free states correspond to the absolutely continuous spectrum. Alternatively, if it is insisted that the notion of eigenvectors and eigenvalues survive the passage to the rigorous, one can consider operators on rigged Hilbert spaces.

[edit] See also

Essential spectrum, spectrum of an operator modulo compact perturbations.

[edit] References

N. Dunford and J.T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.

M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1972.

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Category: Spectral theory