Decomposition of spectrum (functional analysis)

The spectrum of a linear operator $T$ that operates on a Banach space $X$ (a fundamental concept of functional analysis) consists of all scalars $\lambda$ such that the operator $T-\lambda$ does not have a bounded inverse on $X$. The spectrum has a standard decomposition into three parts:

• a point spectrum, consisting of the eigenvalues of $T$
• a continuous spectrum, consisting of the scalars that are not eigenvalues but make the range of $T-\lambda$ a proper dense subset of the space;
• a residual spectrum, consisting of all other scalars in the spectrum

This decomposition is relevant to the study of differential equations, and has applications to many branches of science and engineering. A well-known example from quantum mechanics is the explanation for the discrete spectral lines and the continuous band in the light emitted by excited atoms of hydrogen.

Definitions

For bounded Banach space operators

Let X be a Banach space, L(X) the family of bounded operators on X, and T ∈ L(X). By definition, a complex number λ is in the spectrum of T, denoted σ(T), if T − λ does not have an inverse in L(X).

If T − λ is one-to-one and onto, then its inverse is bounded; this follows directly from the open mapping theorem of functional analysis. So, λ is in the spectrum of T if and only if T − λ is either not one-to-one or not onto. One distinguishes three separate cases:

1. T − λ is not injective. That is, there exist two distinct elements x,y in X such that (T − λ)(x) = (T − λ)(y). Then z = x − y is a non-zero vector such that T(z) = λz. In other words, λ is an eigenvalue of T in the sense of linear algebra. In this case, λ is said to be in the point spectrum of T, denoted σ_p(T).
2. T − λ is injective, and its range is a dense subset R of X; but is not the whole of X. In other words, there exists some element x in X such that (T − λ)(y) can be as close to x as desired, with y in X; but is never equal to x. It can be proved that, in this case, T − λ is not bounded below (i.e. it sends far apart elements of X too close together). Equivalently, the inverse linear operator (T − λ)⁻¹, which is defined on the dense subset R, is not a bounded operator, and therefore cannot be extended to the whole of X. Then λ is said to be in the continuous spectrum, σ_c(T), of T.
3. T − λ is injective but does not have dense range. That is, there is some element x in X and a neighborhood N of x such that (T − λ)(y) is never in N. In this case, the map (T − λ)⁻¹ x → x may be bounded or unbounded, but in any case does not admit a unique extension to a bounded linear map on all of X. Then λ is said to be in the residual spectrum of T, σ_r(T).

So σ(T) is the disjoint union of these three sets,

$\sigma (T)=\sigma _{p}(T)\cup \sigma _{c}(T)\cup \sigma _{r}(T).$

Spectrum of dual operator

If X* is the dual space of X, and T* : X* → X* is the transpose 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 the Hahn–Banach theorem, there exists a non-zero φ ∈ X* that vanishes on Ran(T - λ). For all x ∈ X,

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

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

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

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

When X is a reflexive Banach space, we also get σ_r(T*) ⊂ σ_p(T) ⊂ σ_r(T*) ∪ σ_p(T*).

For unbounded operators

The spectrum of an unbounded operator can be divided into three parts in the same way as in the bounded case, but because the operator is not defined everywhere, the definitions of domain, inverse, etc. are more involved.

Examples

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 h 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 such that μ( h⁻¹({λ})) > 0, then λ lies in the point spectrum of T_h as follows. Let f be the characteristic function of the measurable set h⁻¹(λ), then by considering two cases, we find

$\forall s\in S,\;(T_{h}f)(s)=\lambda f(s),$

so λ is an eigenvalue of T_h.

Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of T_h. To show this, we must show that T_h − λ has dense range. 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), with $\|f_{n}\|_{p}\leq n\|f\|_{p}$. Then by the dominated convergence theorem,

$(T_{h}-\lambda )f_{n}\rightarrow f$

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.

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

$T(x_{1},x_{2},x_{3},\dots )=(x_{2},x_{3},x_{4},\dots ).$

T is a partial isometry with operator norm 1. So σ(T) lies in the closed 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:

$T^{*}(x_{1},x_{2},x_{3},\dots )=(0,x_{1},x_{2},\dots ).$

For λ ∈ C with |λ| < 1,

$x=(1,\lambda ,\lambda ^{2},\dots )\in l^{p}$

and T x = λ x. Consequently, the point spectrum of T contains the open unit disk. Now, T* has no eigenvalues, i.e. σ_p(T*) is empty. Thus, invoking reflexivity and the theorem given above (that σ_p(T) ⊂ σ_r(T*) ∪ σ_p(T*)), 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). Again by reflexivity of l ^p and the theorem given above (this time, that σ_r(T) ⊂ σ_p(T*)), 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

$Tx=\lambda x,\qquad i.e.\;(x_{2},x_{3},x_{4},\dots )=\lambda (x_{1},x_{2},x_{3},\dots ),$

then

$x=x_{1}(1,\lambda ,\lambda ^{2},\dots ),$

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

So for the left shift T, σ_p(T) is the open unit disk and σ_c(T) is the unit circle, whereas 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.

Self adjoint operators on Hilbert space

Hilbert spaces are Banach spaces, so the above discussion applies to bounded operators on Hilbert spaces as well. A subtle point concerns the spectrum of T*. For a Banach space, T* denotes the transpose and σ(T*) = σ(T). For a Hilbert space, T* normally denotes the adjoint of an operator T ∈ L(H), not the transpose, and 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.

Borel functional calculus

This subsection briefly sketches the development 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).

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:

$\,\mu =\mu _{{{\mathrm {ac}}}}+\mu _{{{\mathrm {sc}}}}+\mu _{{{\mathrm {pp}}}}$

where μ_ac is absolutely continuous with respect to the Lebesgue measure, and μ_sc is singular with respect to the Lebesgue measure and atomless, and μ_pp is a pure point measure.^[1]

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

$\langle k,\chi (T)k\rangle =\int _{{\sigma (T)}}\chi (\lambda )\cdot \lambda ^{2}d\mu _{{h}}(\lambda )=\int _{{\sigma (T)}}\chi (\lambda )\;d\mu _{k}(\lambda ).$

So

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

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

$H=H_{{{\mathrm {ac}}}}\oplus H_{{{\mathrm {sc}}}}\oplus H_{{{\mathrm {pp}}}}.$

This leads to the following definitions:

1. The spectrum of T restricted to H_ac is called the absolutely continuous spectrum of T, σ_ac(T).
2. The spectrum of T restricted to H_sc is called its singular spectrum, σ_sc(T).
3. 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

$\sigma (T)=\sigma _{{{\mathrm {ac}}}}(T)\cup \sigma _{{{\mathrm {sc}}}}(T)\cup {{\bar \sigma }_{{{\mathrm {pp}}}}(T)}.$

Comparison

A bounded self adjoint operator on Hilbert space is, a fortiori, a bounded operator on a Banach space. Therefore, one can also apply to T the decomposition of the spectrum that was achieved above for bounded operators on a Banach space. Unlike the Banach space formulation, the union

$\sigma (T)={{\bar \sigma }_{{{\mathrm {pp}}}}(T)}\cup \sigma _{{{\mathrm {ac}}}}(T)\cup \sigma _{{{\mathrm {sc}}}}(T).$

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

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

for some Borel measures $\mu _{i}$. 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_ac(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.

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.

Intuition may induce one to say that the discreteness of the spectrum is intimately related to the corresponding states being "localized". However, a careful mathematical analysis shows that this is not true. The following $f$ is an element of $L^2(\mathbb{R})$ and increasing as $x\to \infty$.

$f(x)={\begin{cases}n&{\text{if }}x\in \left[n,n+{\frac {1}{n^{4}}}\right],\\0&{\text{else.}}\end{cases}}$

However, the phenomena of Anderson localization and dynamical localization describe, when the eigenfunctions are localized in a physical sense. Anderson Localization means that eigenfunctions decay exponentially as $x\to \infty$. Dynamical localization is more subtle to define.

Sometimes, when performing physical quantum mechanical calculations, one encounters "eigenvectors" that do 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.

It was believed for some time that singular spectrum is something artificial. However, examples as the almost Mathieu operator and random Schrödinger operators have shown, that all types of spectra arise naturally in physics.

See also

• Essential spectrum, spectrum of an operator modulo compact perturbations.

References

1. Bogachev, Vladimir (2007). Measure Theory volume 1. Springer. p. 344.

• 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.