Holomorphic functional calculus

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In mathematics, holomorphic functional calculus is functional calculus with holomorphic functions. That is to say, given a holomorphic function $f$ of a complex argument z and an operator $T$ , the aim is to construct an operator

$f(T)\,$

which in a sense extends the function $f$ from complex argument to operator argument.

This article will discuss the case where $T$ is a bounded linear operator on some Banach space. In particular, $T$ can be a square matrix with complex entries, a case which will be used to illustrate functional calculus and provide some heuristic insights for the assumptions involved in the general construction.

[edit] Motivation

[edit] Need for a general functional calculus

In this section $T$ will be assumed to be a n × n matrix with complex entries.

If a given function f is of certain special type, there are natural ways of defining f(T). For instance, if

$p(z)= \sum_{i \geq 0} ^m a_i z^i$

is a complex polynomial, one can simply substitute T for z and define

$p(T) = \sum_{i \geq 0} ^m a_i T^i$

where T⁰ = I, the identity matrix. This is the polynomial functional calculus. It is a homomorphism from the ring of polynomials to the ring of n × n matrices.

Extending slightly from the polynomials, if

$f:{\mathbb C} \rightarrow {\mathbb C}$

is holomorphic everywhere, i.e. an entire function, with MacLaurin series

$f(z)= \sum_{i \geq 0} ^{\infty} a_i z^i,$

mimicking the polynomial case suggests we define

$f(T)= \sum_{i \geq 0} ^{\infty} a_i T^i.$

Since the MacLaurin series converges everywhere, the above series will converge, in a chosen operator norm. An example of this is the exponential of a matrix. Replacing z by T in the MacLaurin series of f(z) = e^z gives

$f(T) = e^T = I+T+\frac{T^2}{2!}+\frac{T^3}{3!}+\cdots$

The requirement that the MacLaurin series of f converges everywhere can be relaxed somewhat. From above it is evident that all that is really needed is the radius of convergence of the MacLaurin series be greater than

$\| T \|,$

the operator norm of T. This enlarges somewhat the family of f for which f(T) can be defined using the above approach. However it is not quite satisfactory. For instance, it is a fact from matrix theory that every non-singular T has a logarithm S in the sense that

$e^S = T.\,$

It is desirable to have a functional calculus that allows one to define, for a non-singular T,

$\ln (T)\,$

such that it coincides with S. This can not be done via power series. For example, the logarithmic series

$\ln(z+1) = z - \frac{z^2}{2} + \frac{z^3}{3} - \cdots$

converges only on the open unit disk. Substituting T for z in the series fails to give a well-defined ln (T + I) for any invertible T + I with

$\| T \| \geq 1.$

Thus a more general functional calculus is needed.

[edit] Functional calculus and the spectrum

It is expected that a necessary condition for f(T) to make sense is f be defined on the spectrum of T. For example, the spectral theorem for normal matrices states every normal matrix is unitarily diagonalizable. There leads to a definition of f(T) when T is normal. One encounters difficulties if f(λ) is not defined for some eigenvalue λ of T.

Other indications also reinforce the idea that f(T) can be defined only if f is defined on the spectrum of T. If T is not invertible, then 0 is an eigenvalue. Since the natural logarithm is undefined at 0, one would expect that ln (T) can not be defined naturally. This is indeed the case. As another example, for

$f(z)=\frac{1}{(z-2)(z-5)}$

the reasonable way of calculating $f (T)$ would seem to be

$f(T)=(T-2I)^{-1}(T-5I)^{-1}.\,$

However, this expression is not be defined if the inverses on the right-hand side do not exist, that is, if either 2 or 5 are eigenvalues of $T$ .

For a given matrix T, the eigenvalues of T dictates to what extent f(T) can be defined; f(λ) must be defined for all eigenvalue λ of T. For a general bounded operator this condition translates to "f must be defined on the spectrum of T". This assumption turns out to be an enabling condition such that the functional calculus map

$f \rightarrow f(T)$

has certain desirable properties.

[edit] Functional calculus for a bounded operator

The spectrum σ(T) in light blue and the path γ in red.

The case when the spectrum has multiple connected components and the corresponding path γ.

The case when the spectrum is not simply connected.

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

Recall the Cauchy integral formula from classical function theory. Let

$f \colon \mathbb{C} \rightarrow \mathbb{C}$

be holomorphic on some open subset D in the complex plane, and $Γ$ be a rectifiable Jordan curve in D, that is, a closed curve of finite length without self-intersections. Cauchy's integral formula states

$f(z)=\frac{1}{2\pi i}\int\limits_{\Gamma}^{}\! \frac{f(\zeta)}{\zeta-z}\,d\zeta$

for any z lying in the inside of Γ, i.e. the winding number of Γ about z is 1.

The idea is to extend this formula to functions taking values in the Banach space L(X). Cauchy's integral formula suggests the definition, purely formal at the moment,

$f(T)=\frac{1}{2\pi i}\int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta.$

where (ζ - T)^-1 is the resolvent of T at ζ.

Assuming this Banach space-valued integral is appropriately defined, this proposed functional calculus implies the following necessary conditions:

As the scalar version of Cauchy's integral formula applies to holomorphic f, we anticipate that is also the case for the Banach space case, where there should be a suitable notions of holomorphy for functions taking value in the Banach space L(X).
As the resolvent mapping ζ → (ζ - T)^-1 is undefined on the spectrum of T, σ(T), the Jordan curve Γ should not intersect σ(T). Furthermore, the resolvent mapping is holomorphic on the compliment of σ(T). So, to obtain a non-trivial functional calculus, Γ must enclose, at least part of, σ(T).
The functional calculus should be well-defined in the sense that f(T) is independent of Γ.

The full definition of the functional calculus is as follows: For T ∈ L(X), define

$f(T)=\frac{1}{2\pi i}\int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta.$

where f is a holomorphic function defined on an open set D in the complex plane which contains σ(T), and

$\Gamma = \{ \gamma_i \}_{i =1}^m$

is a collection of Jordan curves in D such that σ(T) lies in the inside of Γ, and each γ_i is oriented in the positive sense.

The open set D may vary with f and need not be connected, as shown by the figures on the right.

The following subsections make precise the notions invoked in the definition and show f(T) is indeed well defined under given assumptions.

[edit] Banach space-valued integral

Cf. Bochner integral

For a continuous function g defined in an open neighborhood of Γ and taking values in L(X), the contour integral ∫_Γg is defined in the same way as the scalar case. One can parametrize each γ_i ∈ Γ by a real interval [a, b], and the integral is the limit of the Riemann sums obtained from ever-finer partitions of [a, b]. The Riemann sums converge in the uniform operator topology. We define

$\int_{\Gamma} g = \sum_i \int_{\gamma_i} g.$

In the definition of the functional calculus, f is assumed to be holomorphic in an open neighborhood of Γ. It will be shown below that the resolvent mapping is holomorphic on the resolvent set. Therefore the integral

$\frac{1}{2\pi i}\int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta$

makes sense.

[edit] The resolvent mapping

The mapping

$\zeta \to (\zeta-T)^{-1}$

is called the resolvent mapping of T. It is defined on the compliment of σ(T), called the resolvent set of T and will be denoted by ρ(T).

Much of classical function theory depends on the properties of the integral

$\frac{1}{2\pi i}\int\limits_{\Gamma}^{}\! (\zeta-z)^{-1}\,d\zeta.$

The holomorphic functional calculus is similar in that the resolvent mapping plays a crucial role in obtaining properties one require from a nice functional calculus. This subsection outlines properties of the resolvent map that are essential in this context.

[edit] A resolvent formula

Direct calculation shows, for z₁, z₂ ∈ ρ(T),

$(z_1 - T)^{-1} - (z_2 - T)^{-1} = (z_1 - T)^{-1} (z_2 - z_1) (z_2 - T)^{-1}.\,$

Therefore

$(z_1 - T)^{-1} (z_2 - T)^{-1} = \frac{(z_1 - T)^{-1} - (z_2 - T)^{-1} }{(z_2 - z_1)}.$

This equation is called the first resolvent formula. The formula shows (z₁ - T)^-1 and (z₂ - T)^-1 commute, which hints at the fact that the image of the functional calculus will be a commutative algebra. Letting z₂ → z₁ shows the resolvent map is (complex-) differentiable at each z₁ ∈ ρ(T); so the integral in the expression of functional calculus converges in L(X).

[edit] Holomorphy

Stronger statement than differentiability can be made regarding the resolvent map. The resolvent set ρ(T) is actually an open set on which the resolvent map is holomorphic. This property will be used in subsequent arguments for the functional calculus. To verify this claim, let z₁ ∈ ρ(T) and notice the formal expression

$\frac{1}{z_2 - T} = \frac{1}{z_1 - T} \cdot \frac{1}{1 - \frac{z_1 - z_2}{z_1 -T}}$

suggests we consider

$(z_1 - T)^{-1} \sum _{n \geq 0} ((z_1 - z_2) (z_1 - T)^{-1})^n$

for (z₂ - T)^-1. The above series converges in L(X), which implies the existence of (z₂ - T)^-1, if

$|z_1 - z_2| < \frac{1}{ \|(z_1 - T)^{-1}\| } .$

Therefore the resolvent set ρ(T) is open and the power series expression on an open disk centered at z₁ ∈ ρ(T) shows the resolvent map is holomorphic on ρ(T).

[edit] Neumann series

Another expression for (z - T)^-1 will also be useful. The formal expression

$\frac{1}{z - T} = \frac{1}{z} \cdot \frac{1}{1 - \frac{T}{z}}$

leads one to consider

$\frac{1}{z} \sum _{n \geq 0} (\frac{T}{z})^n.$

This series, the Neumann series, converges to (z - T)^-1 if

$\| \frac{T}{z} \| < 1, \; i.e. \; |z| > \|T\|.$

[edit] Compactness of σ(T)

From the last two properties of the resolvent we can deduce that the spectrum σ(T) of a bounded operator T is a compact subset of the complex plane. Therefore for any open set D such that σ(T) ⊂ D, there exists a positively-oriented and smooth system of Jordan curves

$\Gamma = \{ \gamma_1, \cdots, \gamma_m\}$

such that σ(T) is in the inside of Γ and the compliment of D is contained in the outside of Γ. Hence, for the definition of the functional calculus, indeed a suitable family of Jordan curves can be found for each f that is holomorphic on some D.

[edit] Well-definedness

The previous discussion has shown that the integral makes sense, i.e. a suitable collection Γ of Jordan curves does exist for each f and the integral does converge in the appropriate sense. What has not been shown is that the definition of the functional calculus is unambiguous, i.e. does not depend on the choice of Γ. This issue we now try to resolve.

[edit] A preliminary fact

For a collection of Jordan curves Γ = {γ₁... γ_m} and a point a in the complex plane C, the winding number of Γ with respect to a is the sum of the winding numbers of its elements. In symbols we put n(Γ, a) = ∑_i n(γ_i, a). The following theorem is by Cauchy:

Theorem Let G ⊂ C be an open set and Γ ⊂ G. If g: C → C be holomorphic on G, and for all a in the complement of G, n(Γ, a) = 0, then the contour integral ∫_Γ g = 0.

We will need the vector-valued analog of this result when g takes values in L(X). To this end, let g: G → L(X) be holomorphic, with the same assumptions on Γ. The idea is use the dual space L(X)* of L(X), and pass to Cauchy's theorem for the scalar case.

Consider the integral ∫_Γ g ∈ L(X). If we can show, for all φ ∈ L(X)*, φ (∫_Γ g) = 0, then the claim, that ∫_Γ g = 0, follows. By boundedness of φ and the fact that the integral converges in norm,

$\phi (\int_{\Gamma} g) = \int_{\Gamma} \phi(g).\,$

But g is holomorphic implies the composition φ(g): G ⊂ C → C is holomorphic. So by Cauchy's theorem

$\phi (\int_{\Gamma} g) = \int_{\Gamma} \phi(g) = 0.\,$

[edit] Main argument

The well-definedness of functional calculus now follows as an easy consequence. Let D be an open set containing σ(T). Suppose Γ = {γ_i} and Ω = {ω_j} be two (finite) collections of Jordan curves satisfying the assumption given for the functional calculus. We wish to show

$\int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta = \int\limits_{\Omega}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta .$

Let Ω' be obtained from Ω by reversing the orientation of each ω_j, then

$\int\limits_{\Omega}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta = - \int\limits_{\Omega'}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta.$

Consider the union of the two collections Γ ∪ Ω' . Both Γ ∪ Ω' and σ(T) are compact. So there is some open set U containing Γ ∪ Ω' such that σ(T) lies in the compliment of U. Any a in the compliment of U has winding number n(Γ ∪ Ω', a) = 0 and the function

$\zeta \rightarrow f(\zeta) (\zeta-T)^{-1}$

is holomorphic on U. So the vector-valued version of Cauchy's theorem gives

$\int\limits_{\Gamma \cup \; \Omega' }^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta = 0$

i.e.

$\int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta + \int\limits_{\Omega' }^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta = \int\limits_{\Gamma}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta - \int\limits_{\Omega}^{}\! f(\zeta) (\zeta-T)^{-1}\,d\zeta = 0.$

Hence the functional calculus is well-defined.

Consequently, if f₁ and f₂ are two holomorphic functions defined on corresponding neighborhoods D₁ and D₂ of σ(T) and they are equal on an open set containing σ(T), then f₁(T) = f₂(T). Moreover, even though the D₁ may not be D₂, the operator (f₁ + f₂) (T) is well-defined. Same holds for the definition of (f₁·f₂)(T).

[edit] On the assumption that f be holomorphic over an open neighborhood of σ(T)

It should perhaps be noted that so far the full strength of this assumption has not been utilized. For convergence of the integral, only continuity was used. For well-definedness, we only needed f be holomorphic on some open set U containing the contours Γ ∪ Ω' but not σ(T). The assumption will be applied in its entirety for showing the homomorphism property of the functional calculus.

[edit] Properties

[edit] Polynomial case

The linearity of the map

$f \rightarrow f(T)$

follows from the convergence of the integral and that linear operations on a Banach space are continuous.

We recover the polynomial functional calculus when f(z) = ∑_{0 ≤ i ≤ m} a_i zⁱ is a polynomial. To prove this, it is sufficient to show, for k ≥ 0 and f(z) = z^k, it is true that f(T) = T^k, i.e.

$\frac{1}{2 \pi i} \int_{\Gamma} \zeta ^k ( \zeta - T)^{-1} d \zeta = T^k$

for any suitable Γ enclosing σ(T). Choose Γ to be a circle of radius greater than the operator norm of T. As stated above, on such Γ, the resolvent map admits a power series representation

$( z - T)^{-1} = \frac{1}{z} \sum_{n \geq 0} ( \frac{T}{z} )^n.$

Substituting gives

$f (T) = \frac{1}{2 \pi i} \int_{\Gamma} \sum_{n \geq 0} ( \frac{T^n}{\zeta^{n+1-k}} ) d \zeta$

, which is

$\sum_{n \geq 0} T^n \cdot \frac{1}{2 \pi i} \int_{\Gamma} ( \frac{1}{\zeta^{n+1-k}} ) d \zeta = \sum_{n \geq 0} T^n \cdot \delta_{nk} = T^k.$

The δ is the Kronecker delta symbol.

[edit] The homomorphism property

For any f₁ and f₂ satisfying the appropriate assumptions, the homomorphism property states

$f_1 (T) f_2(T) = (f_1 \cdot f_2)(T).$

We sketch an argument which invokes the first resolvent formula and the assumptions placed on f. Start by calculating directly

$f_1 (T) f_2(T) = \frac{1}{(2 \pi i)^2} \int _{\Gamma_1} f_1 (\zeta) (\zeta - T)^{-1} d \zeta \cdot \int _{\Gamma_2} f_2 (\omega) (\omega - T)^{-1} d \omega$ ,

where the Jordan curves are chosen such that Γ₁ lies in the inside of Γ₂. The reason for this will be clearly shortly. The above expression is equal to

$\frac{1}{(2 \pi i)^2} \int _{\Gamma_1} \int _{\Gamma_2} f_1 (\zeta) f_2 (\omega) \cdot (\zeta - T)^{-1} (\omega - T)^{-1} \; d \omega d \zeta.$

After applying the first resolvent formula, this becomes

$\frac{1}{(2 \pi i)^2} \int _{\Gamma_1} \int _{\Gamma_2} f_1 (\zeta) f_2 (\omega) \cdot \frac{(\zeta - T)^{-1} - (\omega - T)^{-1}}{\omega - \zeta} \; d \omega d \zeta$

$= \frac{1}{(2 \pi i)^2} \int _{\Gamma_1} f_1 (\zeta) (\zeta - T)^{-1} \; [\int_{\Gamma_2}\frac{f_2(\omega)}{\omega - \zeta} d\omega] \; d \zeta - \frac{1}{ (2 \pi i) ^2} \int _{\Gamma_2} f_2 (\omega) (\omega - T) ^{-1} \; [\int_{\Gamma_1}\frac{f_1(\zeta)}{\omega - \zeta} d\zeta] \; d \omega.$

But the second term from above vanishes because ω ∈ Γ₂ is in the outside of Γ₁ and because f₁ is holomorphic on some open neighborhood of σ(T).

Thus,

$f_1 (T) f_2 (T)= \frac{1}{2 \pi i} \int _{\Gamma_1} f_1 (\zeta) (\zeta - T)^{-1} \; [\frac{1}{2 \pi i}\int_{\Gamma_2}\frac{f_2(\omega)}{\omega - \zeta} d\omega] \; d \zeta .$

For the same reason as above,

$\frac{1}{2 \pi i}\int_{\Gamma_2}\frac{f_2(\omega)}{\omega - \zeta} d\omega = f_2(\zeta).$

Therefore

$f_1 (T) f_2 (T)= \frac{1}{2 \pi i} \int _{\Gamma_1} f_1 (\zeta) f_2 (\zeta)(\zeta - T)^{-1} d \zeta = (f_1 \cdot f_2)(T).$

[edit] Continuity with respect to compact convergence

Let G ⊂ C be open with σ(T) ⊂ G. Suppose a sequence {f_k} of holomorphic functions on G converges uniformly on compact subsets of G (this is sometimes called compact convergence). Then {f_k(T)} is convergent in L(X):

Assume for simplicity that Γ consists of only one Jordan curve. We estimate

$\| f_k(T) - f_l(T) \| = \frac{1}{2 \pi} \|\int_{\Gamma} (f_k - f_l)(\zeta) (\zeta - T)^{-1} d \zeta \|$

$\leq \frac{1}{2 \pi} \int_{\Gamma} |(f_k - f_l)(\zeta)| \cdot \| (\zeta - T)^{-1} \| d \zeta.$

By combining the uniform convergence assumption and various continuity considerations, we see that the above tends to 0 as k, l → ∞. So {f_k(T)} is Cauchy, therefore convergent.

[edit] Uniqueness

To summarize, we have shown the holomorphic functional calculus

$f \rightarrow f(T)$

has the following properties:

It extends the polynomial functional calculus.
It is an algebra homomorphism from the algebra of holomorphic functions defined on a neighborhood of σ(T) to L(X)
It preserves uniform convergence on compact sets.

It can be proved that a calculus satisfying the above properties is unique.

We note that, everything discussed so far holds verbatim if the family of bounded operators L(X) is replaced by a Banach algebra A. The functional calculus can be defined in exactly the same way for an element in A.

[edit] Spectral considerations

The above demonstrates the intimate relationship between the holomorphic functional calculus of a given T ∈ L(X) and σ(T). This is true in general. Under more restrictive assumptions, the spectral theorem for bounded normal operators (see below) can be reformulated in terms of a functional calculus. This section sketches some results in this direction.

[edit] Spectral mapping theorem

It is known that the spectral mapping theorem holds for the polynomial functional calculus: for any polynomial p, σ(p(T)) = p(σ(T)). This can be extended to the holomorphic calculus. To show f(σ(T)) ⊂ σ(f(T)), let μ be any complex number. By a result from complex analysis, there exists a function g holomorphic on a neighborhood of σ(T) such that

$f(z) - f(\mu) = (z - \mu)g(z). \,$

According to the homomorphism property, f(T) - f(μ) = (T - μ)g(T). Therefore μ ∈ σ(T) implies f(μ) ∈ σ(f(T)).

For the other inclusion, if μ is not in f(σ(T)), then the functional calculus is applicable to

$g(z) = \frac{1}{f(z) - \mu}.$

So g(T)(f(T) - μ) = I. Therefore μ does not lie in σ(f(T)).

[edit] Invariant subspace decomposition

If the spectrum σ(T) is not connected, X can be decomposed into invariant subspaces of T using the functional calculus. Let σ(T) be a disjoint union

$\sigma(T) = \cup_{i=1}^m F_i.$

Define e_i to be 1 on some neighborhood that contains only the component F_i and 0 elsewhere. By the homomorphism property, e_i(T) is a projection for all i. The relation e_i(T) T = T e_i(T) means the range of each e_i(T), denoted by X_i, is an invariant subspace of T. Since

$\sum_i e_i(T) = I,\,$

X can be expressed in terms of these complimentary subspaces:

$X = \sum_i X_i.\,$

Similarly, if T_i is T restricted to X_i, then

$T = \sum_i T_i.\,$

Consider the direct sum

$X' = \oplus_i X_i.$

With the norm

$\| \oplus x_i \| = \sum_i \|x_i\|,$

X' is a Banach space. The mapping R: X' → X defined by

$R(\oplus x_i) = \sum x_i$

is a Banach space isomorphism, and we see that

$R T R^{-1} = \oplus T_i.$

This can be viewed as a block diagonalization of T.

When X is finite dimensional, σ(T) = {λ_i} is a finite set of points in the complex plane. Choose e_i to be 1 on an open disc containing only λ_i from the spectrum. The corresponding block-diagonal matrix

$\oplus T_i$

is the Jordan canonical form of T.

[edit] Related results

With stronger assumptions, when T is a normal operator acting on a Hilbert space, the domain of the functional calculus can be broadened. When comparing the two results, a rough analogy can be made with the relationship between the spectral theorem for normal matrices and the Jordan canonical form. When T is a normal operator, a continuous functional calculus can be obtained, that is, one can evaluate f(T) with f being a continuous function defined on σ(T). Using the machinery of measure theory, this can be extended to functions which are only measurable (see Borel functional calculus). In that context, if E ⊂ σ(T) is a Borel set and E(x) is the characteristic function of E, the projection operator E(T) is a refinement of e_i(T) discussed above.

The Borel functional calculus extends to unbounded self-adjoint operators on a Hilbert space.

[edit] Unbounded operators

A holomorphic functional calculus can be defined in a similar fashion for unbounded closed operators with non-empty resolvent set.

[edit] See also

Resolvent formalism

Jordan canonical form, where the finite dimensional case is discussed in some detail.

[edit] References

N. Dunford and J.T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.
Steven G Krantz. Dictionary of Algebra, Arithmetic, and Trigonometry. CRC Press, 2000. ISBN 1-58488-052-X.

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Categories: Functional calculus | Complex analysis

Holomorphic functional calculus

From Wikipedia, the free encyclopedia

Contents

[edit] Motivation

[edit] Need for a general functional calculus

[edit] Functional calculus and the spectrum

[edit] Functional calculus for a bounded operator

[edit] Banach space-valued integral

[edit] The resolvent mapping

[edit] A resolvent formula

[edit] Holomorphy

[edit] Neumann series

[edit] Compactness of σ(T)

[edit] Well-definedness

[edit] A preliminary fact

[edit] Main argument

[edit] On the assumption that f be holomorphic over an open neighborhood of σ(T)

[edit] Properties

[edit] Polynomial case

[edit] The homomorphism property

[edit] Continuity with respect to compact convergence

[edit] Uniqueness

[edit] Spectral considerations

[edit] Spectral mapping theorem

[edit] Invariant subspace decomposition

[edit] Related results

[edit] Unbounded operators

[edit] See also

[edit] References

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