Đang tải... (xem toàn văn)
Dual Spaces and Locally Convex Topologies 39c Consider the linear operator A defined by DA ={φ ∈ 2Z :supp φ is compact} and [Aφ]n = nφn.. Dual Spaces and Locally Convex TopologiesWe concl
Trang 1(a) Prove H¨older’s inequality: If 1p + 1q= 1, φ∈ p(Z), and ψ ∈ q(Z),
then φψ∈ 1(Z) and
Hints: If either φ or ψ vanishes identically, the claim is trivial.
Otherwise, reduce to the case of unit vectors and apply Young’s inequality.
(b) Prove that · pis a norm Hint : First show thatφp= supφ, ψwhere ψ ranges over all unit vectors in qandφ, ψ =φ(n)ψ(n).(c) Prove that (p(Z), · p) is a Banach space.
Exercise 1.2.10 Given Hausdorff topological spaces X and Y , let C(X, Y )
denote the space of continuous functions from X to Y , C(X) = C(X,C) the
space of continuous complex-valued functions on X, and Cc(X) the set ofall f∈ C(X) such that supp(f) := {x : f(x) = 0} is compact.
(a) If X is compact, show that
defines a norm on C(X) that makes C(X) into a Banach space.(b) If X is not compact, show that Cc(X) is not complete Identify
the completion of Cc(X) as an explicit subspace of C(X) Hint :Consider the one-point compactification X := X∪ {∞}.
Exercise 1.2.11 Show that every Hilbert space has a maximal orthonormal
subset and that a Hilbert space is separable if and only if every orthonormal subset of H is countable Hint: Zorn.
defines an inner product with respect to whichH is a Hilbert space.
(b) Give necessary and sufficient conditions on Γ and (Hγ)γ∈Γ for H
Trang 21.3 Bounded Linear Operators 15
in the sense that there is a linear bijection between them that pre-serves inner products.
γ∈ΓHγthe (external) direct sum (or orthog-onal sum) If Γ ={1, 2, , n} for some n ∈ N, we write the direct sum asH1⊕ · · · ⊕ Hn.
Notes and Remarks Banach spaces received their name in the
mono-graph of Fr´echet [96], in honor of foundational work of Banach on normedlinear spaces [19] Hilbert spaces are named for fundamental contributionsof Hilbert to spectral theory, such as the proof of the spectral theorem [122].
Schoenflies [221] used the term “Hilbertscher Raum” to refer to the unit ball
of 2(N).
The name of Theorem 1.2.4 comes from contributions of Cauchy [53],who proved the finite-dimensional version, and Schwarz [224], who proved it
for functions in C([0, 1]) It turns out that Bouniakowsky had the inequality
on C([0, 1]) before Schwarz [43] Bessel’s inequality (formula (1.2.24) from
Theorem 1.2.14) gets its name from Bessel’s work [34] on trigonometric
The proof of Proposition 1.2.9 is an argument due to F Riesz [205] andRellich [200] Their arguments discuss complementation of closed subspaces
in a Hilbert space, but the changes needed to discuss convex sets are merely cosmetic.
Completeness of Lp is proved in most graduate-level real analysis books,
so we omit the proof here; see, e.g., Royden [210], Rudin [211], Simon [244],
or Stein and Shakarchi [250] Even though completeness of 2 is a corollary of this statement (by considering a space equipped with the counting mea-sure), we elected to present the proof, since it is central to the text, whereas
completeness of general Lp spaces does not figure as prominently here.
1.3 Bounded Linear Operators
We now introduce linear operators on normed spaces The primary goal of this section is to set notation and review background results to set the stage for the spectral theorem.
from V to W is a pair (D(A), A), where D(A) is a subspace of V, calledthe domain of A, and A is a linear map from D(A) toW, that is, we haveA(aφ + bψ) = aA(φ) + bA(ψ) for φ, ψ∈ D(A) and a, b ∈ C For linearoperators, we will generally drop the parentheses and write Aφ for A(φ).We will often just write A for the linear operator (D(A), A), especially if
the domain is clear from context.
Trang 3The range of A is
(1.3.1) Ran(A) ={Aφ : φ ∈ D(A)},and the kernel of A is
(1.3.2) Ker(A) ={φ ∈ D(A) : Aφ = 0}.A linear operator is called bounded if D(A) =V and
(1.3.3) A := sup{Aφ : φ ∈ V, φ = 1} < ∞.
The quantity in (1.3.3) is called the operator norm of A The space of
all bounded operators from V to W is denoted by B(V, W), in the event
that V = W, we say that A is a bounded operator in V and we writeB(V) := B(V, V).
The space B(V, W) inherits natural vector space operations To wit:given A, B∈ B(V, W) and a ∈ C, one can define A + B, aA ∈ B(V, W) via
(1.3.4) (A + B)v = Av + Bv,(aA)v = a(Av).
In addition to vector space operations, composition of functions endows spaces of bounded operators with a multiplicative structure Concretely, if V, W, and X are normed spaces, A ∈ B(V, W), and B ∈ B(W, X ),then BA∈ B(V, X ) (where (BA)v = B(Av)); indeed, since BAv ≤BAv ≤ BAv for all v ∈ V, one has
We denote by IVthe identity operator IVψ = ψ (and drop the subscriptwhen the underlying space is clear) An operator A∈ B(V, W) is calledinvertible if there is an element B∈ B(W, V) such that AB = IW and
BA = IV; we write B = A−1and call B the inverse of A.
collection of m× n matrices with complex entries.
It is often of interest to extend a linear operator from a subspace to the whole space If the operator is bounded on the subspace and the subspace is dense, there is always a unique bounded extension.
(D(A), A) is a linear operator inV such that D(A) is dense in V and
(1.3.6) AD(A):= sup{Ax : x ∈ D(A) and x = 1} < ∞.
There is a unique A∈ B(V) such that Ax = Ax for all x∈ D(A) Moreover, A = AD(A).
Trang 41.3 Bounded Linear Operators 17
D(A) such that ϕn→ ψ Since
Aϕn− Aϕm = A(ϕn− ϕm) ≤ AD(A)ϕn− ϕm,η = lim Aϕnexists by completeness Moreover, if (ϕ
n=1 is another
se-quence converging to ψ, the calculationAϕn− Aϕn = A(ϕn− ϕ
n) ≤ AD(A)ϕn− ϕnimplies that Aϕ
n→ η as well, so the limit is independent of the choice ofthe sequence converging to ψ Thus, we define Aψ = η One can check that
A∈ B(V, W) in norm if An− A → 0 as n → ∞ We say that An→ Astrongly, denoted
A = s-limn→∞An,
if Anψ→ Aψ for every ψ, that is (A − An)ψ → 0 as n → ∞ for every ψ.
The proof of the following proposition is left to the reader.
(a) B(V, W) is a vector space and (1.3.3) defines a norm on B(V, W)satisfying (1.3.5).
(b) IfW is a Banach space, so too is B(V, W).
(c) IfW is a Banach space and A1, A2, is a sequence of elements ofB(V, W) that is absolutely summable in the sense that
Trang 5Proof Exercise 1.3.2. Throughout the remainder of this section,V denotes a Banach space Abounded operator is continuous, that is, if φn→ φ, then Aφn→ Aφ As a
sample application of Proposition 1.3.5, we have the following.
ofB(V) If A ∈ G (V) and B < A−1−1, then A + B is invertible InparticularG (V) is an open subset of B(V).
I + A−1B By assumptionA−1B ≤ A−1B < 1, so one can check that
converges to the inverse of I + A−1B with the help of Proposition 1.3.5.
complex numbers z for which A− zI is invertible The spectrum of A is
(1.3.11) σ(A) =C \ ρ(A) = {z ∈ C : A − zI is not invertible}.
A scalar z∈ C is called an eigenvalue of A if Av = zv for some nonzerov∈ V; equivalently, z is an eigenvalue of A if and only if Ker(A − zI) isnontrivial Let us note that every eigenvalue of A necessarily belongs toσ(A), but the converse need not hold in infinite-dimensional spaces, as we
will see later.
The discrete spectrum of A consists of all isolated eigenvalues of A havingfinite multiplicity That is, z∈ σdisc(A) if and only if
(1.3.12) 0 < dim Ker(A− zI) < ∞
and B(z, δ)∩ σ(A) = {z} for some δ > 0 (recall that B(z, δ) denotes theopen ball of radius δ centered at z) The complement of σdisc(A) in σ(A) iscalled the essential spectrum of A and is denoted σess(A) = σ(A)\ σdisc(A).The spectral radius of A is given by
(1.3.13) spr(A) = sup{|z| : z ∈ σ(A)}.
Later on, we will see that σ(A) is compact, so sup can be replaced by max
in the definition of the spectral radius.
The resolvent of A is the following map
(1.3.14) R(A,·) : ρ(A) → B(V), z → (A − z)−1.
Trang 61.3 Bounded Linear Operators 19
Here and throughout the text, we adopt a standard convention from the mathematical physics literature and suppress occurrences of the
iden-tity operator, e.g., writing (A− z)−1instead of (A− zI)−1 The following
elementary identities will be very useful.
Proposition 1.3.8 (Resolvent identities).
(a) If A∈ B(V) and z, z∈ ρ(A), then
R(A, z)− R(A, z) = (z− z)R(A, z)R(A, z) = (z− z)R(A, z)R(A, z).
(b) If A, B∈ B(V) and z ∈ ρ(A) ∩ ρ(B), then
R(A, z)− R(B, z) = R(A, z)(B − A)R(B, z)= R(B, z)(B− A)R(A, z).
Let us note the following pleasant consequence of Proposition 1.3.8 The
function z→ R(A, z) defines an analytic map from the open set ρ(A) ⊂ C
toB(H) Concretely, (1.3.15) implies thatd
dzR(A, z) := limw→z
R(A, w)− R(A, z)
w− z= R(A, z)2
for any z∈ ρ(A) Many facts about analytic functions carry over from the
scalar case to the case of Banach-space-valued functions; we make use of this in the following proofs.
and σ(A) is a nonempty compact subset ofC.
One can check that R∈ B(V) and R(A − z) = (A − z)R = I, whencez∈ ρ(A) Thus, σ(A) is bounded and (1.3.17) holds By Theorem 1.3.6,ρ(A) is open Thus, σ(A) is compact.
It remains to see that σ(A) is nonempty To that end, suppose to thecontrary that σ(A) =∅ and define g(z) = (A − z)−1for each z∈ ρ(A) = C.
By (1.3.18), one has g(z) = O(1/|z|) as |z| → ∞ Thus, by Liouville’stheorem (Exercise 1.3.15), g(z) = 0 for all z∈ C, a contradiction.
Trang 7If A∈ B(V) and p(z) = cnzn+· · ·+c1z +c0 is a polynomial, it is natural to define
(1.3.19) p(A) = cnAn+· · · + c1A + c0I.
It is helpful to note that p(A) and q(A) commute with one another for allpolynomials p and q and that the spectrum of A transforms in a natural
fashion under this operation.
p is a polynomial, then
(1.3.20) σ(p(A)) = p(σ(A)) :={p(z) : z ∈ σ(A)}.
σ(p(A)) = σ(c0I) ={c0} = p(σ(A)).
Henceforth, assume deg(p)≥ 1 First, suppose w ∈ p(σ(A)), so w = p(E)with E∈ σ(A) One can factor
p(z)− w = (z − E)q(z)
for a nonzero polynomial q Since A− E is not invertible, neither isp(A)− p(E) = (A − E)q(A) = q(A)(A − E),
proving w = p(E)∈ σ(p(A)) and thus p(σ(A)) ⊆ σ(p(A)).Conversely, let w∈ C \ p(σ(A)), and write
for each n, proving
n→∞An1/n.
Trang 81.3 Bounded Linear Operators 21
Then, it remains to show
n→∞An1/n≤ spr(A).
To that end, consider the function f (z) = (I− zA)−1, which is analytic on D ={z ∈ C : z spr(A) < 1}, and which enjoys the series representation
thereupon Denoting the radius of convergence of this series by R, Cauchy’sformula for R (Exercise 1.3.15) yields
lim sup
n→∞An1/n = 1
operations are continuous in B = B(V, W) That is, if an→ a ∈ C,An→ A ∈ B, and Bn→ B ∈ B, then An+ Bn→ A + B, AnBn→ AB,and anAn→ aA.
Exercise 1.3.2 Prove Proposition 1.3.5.Exercise 1.3.3 Prove Proposition 1.3.8.
n=1is a sequence of bounded linear operators in a Banach space V If supnAn < ∞, show that (An)n≥1 converges strongly if there is a dense subset V0⊆ V such that Anψ is convergent inVfor all ψ∈ V0.
n=1 be a bounded sequence of complex
num-bers, and define a linear operator D : 2(N) → 2(N) by [Dψ](n) = dnψ(n).Characterize σ(D) in terms of the sequence d.
Trang 9Exercise 1.3.7 Let (Ω,B, μ) be a σ-finite measure space, and let g ∈L∞(μ) Define a linear operator M = M
g: L2(μ)→ L2(μ) by [M f ](ω) =g(ω)f (ω) Show that M is a bounded operator Characterize σ(M ) in termsof g.
I− AB is invertible if and only if I − BA is invertible Hint: Looking at
the case when AB < 1 might be helpful (though it is not necessary).
(a) Show that Ker(A) and Ran(A) are linear subspaces ofV and W,
(b) True or false: If V and W are Banach spaces and A is bounded,then Ker(A) and Ran(A) are closed subspaces (Give a proof or
counterexample in each case.)
such that the following hold.
(a) A has an isolated eigenvalue of infinite multiplicity.
(b) A has eigenvalues of finite multiplicity that are dense in [0, 1].(c) A has eigenvalues of infinite multiplicity that are dense in [0, 1].
Show that
dist(z, σ(A)).
Recall that the distance from a point to a set is given by dist(z, σ(A)) :=
inf{|z − s| : s ∈ σ(A)} Hint: Theorem 1.3.6.
γ∈ Γ, let H =γ∈ΓHγ denote their direct sum (cf Definition 1.2.15), and
suppose we are given bounded operators Aγ∈ B(Hγ) for each γ∈ Γ suchCharacterize σ(A) in terms of σ(Aγ).
Definition 1.3.12 We denote the operator constructed in Exercise 1.3.12
by
γ∈ΓAγand call it the direct sum of the Aγ.
Trang 101.3 Bounded Linear Operators 23
real vector space V, and suppose p : V → R≥0 is positive homogeneous and
subadditive Recall that p is positive homogeneous if and only ifp(av) = ap(v) for all a≥ 0 and v ∈ V
and is subadditive if and only if
p(v + w)≤ p(v) + p(w) for all v, w ∈ V.
Suppose 0 : U → R is a linear functional such that 0(u)≤ p(u) for allu∈ U Show that 0 enjoys an extension to all ofV that is also bounded byp in two steps:
(a) If w∈ V \ U, let W denote the subspace generated by w and U.Show that there exists :W → R linear such that |U= 0 and
(x)≤ p(x) for all x ∈ W Hints: The extension is uniquely definedby (w) Show that one can choose (w) in such a manner that
(u1+ w)≤ p(u1+ w) for all u1∈ U,
(u2− w) ≤ p(u2− w) for all u2∈ U,
and that this defines the desired .
(b) Show that there exists an extension of 0 to all of V that is alsobounded by p Hint : Zorn.
of a complex vector space V, and suppose p : V → R≥0 is subadditive and
satisfies p(av) =|a|p(v) for all a ∈ C and v ∈ V Suppose 0 :U → C is a
linear functional such that |0(u)| ≤ p(u) for all u ∈ U Show that 0 enjoys an extension to all ofV that is also bounded by p Hint: Write 0 = 1+ i2 for real linear functionals j, and show that 2 uniquely determines 1 via (1.3.34) 2(v) =−1(iv),v∈ V.
set A function f : D→ V is said to be continuously differentiable on D if
f (w)− f(z)w− z
exists for all z∈ D and defines a continuous map D → V.
(a) Formulate and prove a Banach-space-valued version of Liouville’s
theorem Hint : If f :C → V is bounded and continuously differen-tiable, the same is true of ◦ f for any ∈ B(V, C).
(b) Formulate and prove a Banach-space-valued version of the Cauchy
integral formula Hint : Reduce toC via linear functionals as in the previous part.
Trang 11(c) Formulate and prove a Banach-space-valued version of Cauchy’s
formula for the radius of convergence of a power series Hint :
Start-ing from the Cauchy integral formula, this should look fairly similar to the standard proof from complex analysis.
Notes and Remarks Theorem 1.3.11 is due to Gelfand [100] The
ex-tension result in Exercise 1.3.13 is a version of the Hahn–Banach theorem,
named for work of Hahn in 1927 [113] and Banach in 1929 [20, 21] The
extension to complex functionals in Exercise 1.3.14 is generally credited to
Bohnenblust and Sobczyk [39], Murray [179], Sukhomlinov [255]; the key
observation (1.3.34) appeared earlier in work of L¨owig [170].1.4 Self-Adjoint Operators and the Spectral Theorem
We now arrive at the fundamental result of the first portion of the chapter: the spectral theorem for bounded self-adjoint operators In fact, we will see several theorems, each of which might be considered a facet of the spectral theorem Each version of the spectral theorem will be used in an essential manner throughout the text The multiplicity of results and perspectives will, in sum, comprise our understanding of “the spectral theorem”.
We begin by defining fundamental objects, namely the dual space and adjoint of a bounded operator.
dual space ofV; elements of V∗are called bounded linear functionals onV In view of Proposition 1.3.5, V∗ is a Banach space with respect to the
operator norm.
Throughout the remainder of this section,H denotes a separable Hilbertspace For each φ∈ H,
defines an element of H∗ In fact, every element ofH∗ is of this form.
= φ The map T : φ→ φis an isometric anti-linear1 bijection fromHontoH∗.
subspace of H Choose a unit vector χ ∈ [Ker ]⊥and take φ = (χ)χ Wehave = φand φ is the unique element ofH with this property
(Exer-cise 1.4.1) Isometry and anti-linearity follow from the properties defining an inner product, which we leave to the reader
1By “anti-linear”, we mean T (φ + ψ) = T φ + T ψ and T (aφ) = aT φ for all φ, ψ∈ H and alla∈ C.
Trang 121.4 Self-Adjoint Operators and the Spectral Theorem 25
there is a unique element ψ∈ H such that
for all φ∈ H The map A∗: ψ→ ψ is called the adjoint of A.
B(Cn) = Cn×n In this case, one can check that the adjoint operation coincides with the conjugate transpose That is, (A∗)
jk= Akjfor A∈ Cn×n.
Let us collect a few of the main properties of the adjoint operation The following proposition is left to the reader.
unitary if AA∗= A∗A = I, normal if AA∗= A∗A, and idempotent ifA2= A.
self-adjoint, then σ(A) is a compact subset ofR Moreover, for any normalA∈ B(H),
Proof Suppose A is self-adjoint In view of Proposition 1.3.9, we need only
show that σ(A)⊆ R to prove the first statement Given z = x + iy withy = 0, notice that, for any ψ, one has
(1.4.7) (A − z)ψ2 =(A − x)ψ2+|y|2ψ2≥ |y|2ψ2.
In particular, this implies that (A− z) is one-to-one Since Im ¯z = 0also, this implies Ker(A− ¯z) = {0} as well.
Additionally, (1.4.7) implies that Ran(A− z) is closed For, if φn∈Ran(A− z) and φn→ φ, writing φn= (A− z)ψn and making use of (1.4.7) gives us
ψn− ψm2≤ |y|−2(A − z)(ψn− ψm)2=|y|−2φn− φm2,
Trang 13showing that ψn→ ψ for some ψ ∈ H, and hence φ = (A − z)ψ belongs toRan(A− z).
Since Ran(A− z) is closed, we may employ Theorem 1.2.11 to arrive atRan(A− z) = Ran(A − z) =Ran(A− z)⊥⊥
Ker(A∗− ¯z)⊥={0}⊥=H.
Thus, A− z is bijective Finally, note that (1.4.7) implies (A − z)−1 is
bounded: R(A, z) ≤ |y|−1 Thus, z∈ ρ(A), as desired.2
Now, if A is normal, using (1.4.4) gives us
A2 = (A∗)2A21/2=(A∗A)21/2 =A∗A = A2.
Inductively, A2n
= A2n
for all n∈ Z+ Thus, A = spr(A) follows
Orthogonal projections (see Definition 1.2.10) play an important role in the spectral theorem Let us collect a few important properties here.
orthogonal projection ontoV.
where we explicitly allow the case in which both sides are∞ Inparticular, the sum does not depend on the choice of basis.
is precisely the orthgonal projection onto the closed subspace V⊥) For any ψ, since Qψ∈ V⊥, we have
ψ2 =P ψ2+Qψ2 ≥ P ψ2,
showingP ≤ 1, and hence P ∈ B(H), since P is readily seen to be linear,
which proves (a).
For any φ∈ V, P φ = φ by definition In particular, given ψ ∈ H,P ψ∈ V, so the previous observation implies P2ψ = P ψ, showing P2 = P Next, using P ψ⊥ Qφ and P φ ⊥ Qψ, one notes
φ, P ψ = P φ + Qφ, P ψ = P φ, P ψ = P φ, P ψ + Qψ = P φ, ψ,showing P = P∗ and concluding the proof of (b).
2By the closed graph theorem, one does not actually need to check boundedness of the inversehere We elected to directly check it since we have not stated or proved the closed graph theoremin the text.
Trang 141.4 Self-Adjoint Operators and the Spectral Theorem 27
To prove (1.4.8), first note that φ, P φ = P φ2 ≥ 0 for any φ, so the
sum on the right-hand-side of (1.4.8) consists solely of nonnegative terms,
and hence exists as an element of [0,∞] Let N denote the dimension of V
(which may be finite or infinite), let {ηk}N
k=1 be an orthonormal basis ofV,
and let {φn} be an orthonormal basis for H Using Theorem 1.2.14 in the
first line and Parseval’s formula in the penultimate line, one has
Definition 1.4.9 The quantity on the right-hand-side of (1.4.8) is called
the trace of P and is denoted Tr P
With the basic facts in hand, we now move on to discuss the spectral the-orem for self-adjoint operators on a separable, infinite-dimensional Hilbert space The situation is more subtle than the finite-dimensional setting In-deed, we will discuss three results that one can think of as versions of the spectral theorem: the functional calculus, the existence of spectral mea-sures, and the equivalence between self-adjoint operators and multiplication operators.
To motivate what follows and elucidate the subtleties, let us briefly recall the spectral theorem for self-adjoint operators on Cn (i.e., Hermitian
ma-trices) If A∈ Cn×nis self-adjoint, then there is a unitary matrix U and adiagonal matrix D such that U∗AU = D The columns of U are normalizedeigenvectors of A, and the diagonal entries of D are the corresponding
eigen-values Now, if H is an infinite-dimensional Hilbert space and A ∈ B(H) isself-adjoint, A may have no eigenvalues whatsoever! The reader is invitedto verify that the discrete Laplacian Δ : 2(Z) → 2(Z), given by
(1.4.9) [Δψ](n) = ψ(n + 1) + ψ(n− 1), ψ ∈ 2(Z), n ∈ Z,
is an example of such an operator (Exercise 1.4.7) That is, Δ∈ B(2(Z)), Δ∗ = Δ, and Δ has no eigenvalues.
Thus, we need to reformulate the finite-dimensional spectral theorem in a fashion that is amenable to the present situation To that end, given a
Trang 15self-adjoint matrix A∈ Cn×n, write P
λ for the orthogonal projection onto
Vλ:= Ker(A− λ) Then, the spectral theorem asserts that
λ∈σ(A)λPλ
and that PλPμ= PμPλ= 0 whenever λ = μ This allows one to see that,for a polynomial function f , f (A) =
f (λ)Pλ, that is, f (A) depends onlyon the values assumed by f on the spectrum of A.
This is the first point of view that we develop Suppose that A∈ B(H)is self-adjoint We wish to apply a function f to A in order to form a newoperator f (A) Recall that if f (z) = cnzn+· · · + c1z + c0 is a polynomial,
we defined f (A) as follows:
(1.4.11) f (A) = cnAn+· · · + c1A + c0.
In order to pass from polynomials to more general continuous functions, one would like to understand the behavior of this construction with respect to uniform approximation on compact sets The following proposition supplies the key input.
(1.4.12) p(A) = sup{|p(z)| : z ∈ σ(A)}.
Proof This follows from Propositions 1.3.10 and 1.4.7, and the observation
Thus, if A is self-adjoint, it is then possible to define f (A) for functions fthat are continuous on σ(A) Concretely, given f∈ C(σ(A)), let p1, p2, bea sequence of polynomials that converges uniformly to f on σ(A) Naturally,
one would like to define
The first version of the spectral theorem states that this is well-defined and has the kinds of properties one might desire from such an apparatus.
Theorem 1.4.11 (Spectral theorem: continuous functional calculus) Let
A∈ B(H) be self-adjoint, and let A ⊆ B(H) denote the algebra generatedby A (that is, the smallest closed subalgebra ofB(H) containing A).
(a) The association from (1.4.13) is well-defined, i.e., the limit existsand does not depend on the choice of approximating polynomials.
Trang 161.4 Self-Adjoint Operators and the Spectral Theorem 29
(b) The map C(σ(A)) f → f(A) ∈ A is a ∗-homomorphism Thatis to say, for all continuous f and g and all a∈ C, one has
(f + g)(A) = f (A) + g(A),
where denotes the function that is identically equal to 1 on σ(A).(c) For all f∈ C(σ(A)), f(A) = f∞.
n=1is a sequence of polynomials with pn→ f uniformly onσ(A), then (pn(A))∞
n=1 is a Cauchy sequence inB(H) by Proposition 1.4.10,
and hence the limit in (1.4.13) exists by completeness ofB(H) If (qn)∞n=1is
another sequence of polynomials that converges uniformly to f on σ(A), thenqn(A) and pn(A) converge to the same limit (again by Proposition 1.4.10).
The proofs of (b) and (c) are left to the reader Having constructed the continuous functional calculus, we are now able
to describe the spectral measures of A Once again, the analogy with the
finite-dimensional case is a useful guide Let A be a self-adjoint n× ncomplex matrix with spectral decomposition as in (1.4.10) For v∈ Cn and
f a continuous function, to calculate the corresponding matrix element off (A), we have
f (λ)Pλv2.
Thus, we can think of the matrix element v, f(A)v as being obtained byintegrating f against the Dirac measure supported on σ(A) with the weightat λ∈ σ(A) given by Pλv2 In the infinite-dimensional setting, one can no longer guarantee the existence of a basis of eigenvectors, but one can still speak of the measures defined by the left-hand side of (1.4.19).
Theorem 1.4.12 (Spectral theorem: existence of spectral measures) Let
A∈ B(H) be self-adjoint For all φ, ψ ∈ H, there is a complex Borelmeasure μ = μAφ,ψsuch that
Trang 17for all f∈ C(σ(A)) In particular, if ψ = φ, then μ = μA
φ:= μAφ,φis ameasure.3
φ, f(A)ψ defines a bounded linear functional on C(σ(A)), and hence the
existence of a complex Borel measure satisfying (1.4.20) follows from the Riesz–Markov–Kakutani representation theorem.
Moreover, if φ = ψ and f∈ C(σ(A)) is positive, there is a uniquecontinuous g such that g2= f and g≥ 0 One then has
(f ) =φ, f(A)φ = φ, g(A)2φ = g(A)φ2≥ 0,
and hence is also a positive functional Thus, in this case μ is also a
The measures constructed in Theorem 1.4.12 are called the spectral mea-sures of A Whenever A is clear from context, we suppress it from thenotation, writing, e.g., μφ,ψand μφfor μAφ,ψand μAφ.
At this point, let us note a helpful consequence of Theorem 1.4.12 re-garding the norm of the resolvent operator.
dist(z, σ(A)).
dist(z, σ(A)), and let ψ be a unit vector By Theorem 1.4.12, we have
proving the desired upper bound on R(A, z) The opposite inequality
holds more generally and follows from Theorem 1.3.6 (see Exercise 1.3.11) The construction of spectral measures in turn allows one to extend the
map f→ f(A) and the formula (1.4.20) to all bounded Borel measurablefunctions on σ(A) Concretely, given
f∈ B(σ(A)) := {g : σ(A) → C : g is bounded and Borel measurable} ,we define f (A) by specifying its matrix elements,
(1.4.22) φ, f(A)ψ =
f (E) dμAφ,ψ(E).
3We adopt the convention in this text that “measure” always means “nonnegative regularBorel measure”, using modifiers (e.g., “signed” or “complex”) to distinguish variants.
Trang 181.4 Self-Adjoint Operators and the Spectral Theorem 31
B(H) be self-adjoint.
(a) For each f∈ B(σ(A)), (1.4.22) defines an element f(A) ∈ B(H).(b) This mapping is a∗-homomorphism from B(σ(A)) to B(H) (cf.
Theorem 1.4.11(b)).
(c) Additionally, if fn, n≥ 1 is a sequence in B(σ(A)) that is uni-formly bounded in the supremum norm and fn→ f pointwise onσ(A), then fn(A)→ f(A) strongly (i.e., fn(A)ψ→ f(A)ψ for ev-ery ψ∈ H).
By Theorem 1.4.14(b), let us note that if S⊆ R is a Borel set, then χS =
χ2S= χS implies that P(S) := χS(A) is an orthogonal projection, knownas the spectral projection of A on S Here, χSdenotes the characteristicfunction of S, given by χS(x) = 1 for x∈ S and 0 otherwise Using the
Borel functional calculus and the Cauchy–Schwarz inequality, one can check that
|μφ,ψ(S)| = |φ, χS(A)ψ| ≤ μφ(S)1/2μψ(S)1/2,
for all Borel S⊆ R, so μφ,ψ is absolutely continuous with respect to both
μφand μψ The family of these projections comprises a projection-valued-measure (or a resolution of the identity) in thatP(∅) = 0, P(R) = I,P(S ∩ T ) = P(S)P(T ), and, if {Sn: n∈ N} is a countable collectionof pairwise disjoint Borel sets and S =
Let us discuss one final manifestation of the spectral theorem To
mo-tivate this, consider again a self-adjoint A∈ Cn×n For simplicity, assume that every eigenvalue of A is simple (this is not necessary, but it makes this
discussion more transparent) Consider the space
H = 2(σ(A)) ={f : σ(A) → C}
and map v∈ Cnto f∈ H via f(λ) = eλ, v, where eλ is a normalized
eigenvector corresponding to the eigenvalue λ Then, the spectral theoremfor finite matrices establishes that the map U taking v to f as above isunitary and [U AU∗g](λ) = λg(λ) Our final instance of the spectral theorem
generalizes this perspective.
Trang 19Theorem 1.4.15 (Spectral theorem: equivalence to multiplication
opera-tors) Suppose A∈ B(H) is self-adjoint Then, there exist at most countablymany φn∈ H such that with μn= μφn, there is a unitary map
Here we write g(E) ={gn(E)}nfor an element g∈nL2(R, μn).
The central notion here is the decomposition ofH into cyclic subspaces.
generated by A and φ is the smallest closed subspace ofH containing{Anφ : n = 0, 1, 2, } An element φ of H is called cyclic for the oper-ator A if the corresponding cyclic subspace is all ofH.
and φ is a cyclic vector for A We wish to show that (1.4.24) can be definedwith just the measure μ = μφ To define a map U :H → L2(R, μφ) with the
desired property, one needs to invert the map f→ f(A) discussed above.Namely, if f∈ C(σ(A)), set U(f(A)φ) = f; the reader should check thatthis is well-defined (i.e., if f (A)φ = g(A)φ for f, g∈ C(σ(A)), then f = g).This defines U on a dense set One can then verify that U preserves normsand that (U AU−1f )(E) = Ef (E) for such f ’s and then extend the definition
and the relation to the whole space.
This gives the statement of the spectral theorem if there is a cyclic vector If not, a Zornification shows that one can decompose H = Hn
where each Hnis a cyclic subspace generated by A and some φn∈ H.Denote μn= μAφ
nthe spectral measure and Un:H → L2(R, μn) the unitary
equivalence f (A)φn→ f Then U =Un is the desired unitary map The multiplication operators encountered in Theorem 1.4.15 can be thought of as generalizations of direct sums Let us make this explicit For this discussion, we only state theorems and leave proofs to the exercises.
a fixed separable Hilbert space A function A : Ω→ B(H) will be calledweakly measurable if for all φ, ψ∈ H, the function gφ,ψ : Ω→ C,
(1.4.25) gφ,ψ: ω→ φ, Aωψ
is measurable; following a convention for operator-valued functions that will
be standard later in the text, we write Aωrather than A(ω) for the value
Trang 201.4 Self-Adjoint Operators and the Spectral Theorem 33
of A at ω Similarly, a function ϕ : Ω→ H is said to be measurable4 if
ψ, ϕ(ω) is measurable for all ψ ∈ H.
The (constant-fiber) direct integral , denoted by L2(Ω, μ;H) or sometimes
consists of measurable maps f : Ω→ H (as always, modulo the subspace of
measurable functions that vanish almost everywhere) for which
L2(Ω, μ;H) is a Hilbert space An operator A ∈ B(L2(Ω, μ;H)) is calleddecomposable if there is a weakly measurable family{Aω} such that
A·∞:= esssup
ω∈ΩAω < ∞
(recall that esssup denotes the μ-essential supremum, defined in (1.2.11))
(1.4.29) [Af ](ω) = Aω(f (ω)),μ-a.e ω∈ Ω.
In analogy to the notation from (1.4.26), we will sometimes write this as
⊕
Let us make a few observations about this construction First, if μ is
the counting measure on Ω, then we see that
the direct sum of copies of H (cf Definition 1.2.15), and the decomposable
operators are precisely those that can be expressed as direct sums of opera-tors inH (cf Definition 1.3.12) Thus, the direct integral can be viewed as
a generalization of the direct sum.
4For consistency, we should also call this notion “weak measurability” in order to distinguishit from the ostensibly stronger notion of measurability given by equippingH with the σ-algebra
generated by the metric topology thereupon However, these two notions are equivalent (Exer-cise 1.12.1).
Trang 21Given a measurable family {Aω}ω∈Ω with A·∞<∞, one can check(Exercise 1.4.13) that there is a unique decomposable operator A for whichone has (1.4.29) and the operator norm of A satisfies
Let us conclude with a versatile criterion for detecting the spectrum of
a self-adjoint operator On the one hand, every eigenvalue of an operator Abelongs to the spectrum of A On the other hand, as we noted earlier, there
are self-adjoint operators (such as Δ, the discrete Laplacian) that have no eigenvalues at all (Exercise 1.4.7) However, for self-adjoint operators, one has the following extremely helpful replacement: one can characterize the
spectra of self-adjoint operators as precisely those z for which the operatorenjoys a sequence of approximate eigenvectors Let us make this precise.
a sequence (ψn)∞
n=1where ψn∈ H for all n, ψn = 1 for all n, and
n→∞(A − z)ψn = 0.
z∈ σ(A) If A is self-adjoint, the converse holds as well.
Trang 221.4 Self-Adjoint Operators and the Spectral Theorem 35
1 =ψn = (A − z)−1(A− z)ψn ≤ (A − z)−1(A − z)ψn.
Thus, (A − z)ψn ≥ R(A, z)−1and hence (ψn)∞
n=1 cannot be a Weyl
sequence for A at z.
Conversely, if E∈ σ(A) and A = A∗, then E∈ R and E + iδ ∈ ρ(A)for all δ > 0 Choose a sequence δn↓ 0 By Theorem 1.4.13, we can finda sequence (φn)n≥1 of unit vectors such that (A − E − iδn)−1φn → ∞ as
Let us note the following helpful reformulation.
(1.4.36) dist(z, σ(A)) = inf
v =1(A − z)v.
right-hand side vanishes by Theorem 1.4.20 If z∈ ρ(A), the claim follows by
not-ing that the left-hand side is equivalent to R(A, z)−1 by Theorem 1.4.13
and the right-hand side is equivalent to the same quantity by a direct
Additionally, we can use the Weyl sequence formalism to bound spectra of strong limits from above.
is a unit vector ψ for which(A − E)ψ < ε By strong convergence, we get(An− E)ψ < ε
for all large n For such n, this means that dist(E, σ(An)) < ε by Corol-lary 1.4.21 Consequently, E belongs to the right-hand side of (1.4.37). We caution the reader that the previous bound is one-sided In
gen-eral, σ(A) can be much smaller than the right-hand side of (1.4.37)
(Exer-cise 1.4.12).
Trang 23Let us note that there is a characterization of the essential spectrum
via Weyl sequences comprised of orthonormal vectors similar to that of
Theorem 1.4.20.
A at z is an orthonormal sequence of vectors (ψn)∞
n=1, where ψn∈ H for alln,ψn = 1 for all n, and
n→∞(A − z)ψn = 0.
is a singular Weyl sequence for A at E.
multiplic-ity, then one can simply take (ψn)∞
n=1 to be an orthonormal basis of the corresponding eigenspace Otherwise, with
one has Tr (Pn)≥ 1 for infinitely many n ≥ 1 Thus, the desired orthonormalsequence can be extracted by choosing a normalized vector from Ran(Pn)for each n with Tr (Pn)≥ 1.
Conversely, assume that (ψn)∞
n=1is a singular Weyl sequence for A at E.By Theorem 1.4.20, E∈ σ(A), so it suffices to show that E /∈ σdisc(A) As-sume to the contrary that E∈ σdisc(A), which implies Pε:= χ(E−ε,E+ε)(A)is a finite-rank projection for some ε > 0 By a direct calculation usingorthonormality, Pεψn→ 0 as n → ∞ On the other hand, writing μψn for
the spectral measure associated to the pair (A, ψn), one has
which is incompatible with ψn = 1 and Pεψn→ 0 Thus, we obtain a
This has the following consequence: the essential spectrum is invariant under finite-rank perturbations.5
Corollary 1.4.25 If A and R are self-adjoint and R is finite-rank, then
(1.4.39) σess(A + R) = σess(A).
5In fact, the essential spectrum is invariant under compact self-adjoint perturbations, butwe do not state that result since we have not discussed compact operators in the text Moreover,there are further generalizations beyond the self-adjoint case.
Trang 241.4 Self-Adjoint Operators and the Spectral Theorem 37
σess(A + R) when R is rank-one, say R = λφ, ·φ for some λ and someφ∈ H Given E ∈ σess(A), Theorem 1.4.24 implies the existence of a singu-lar Weyl sequence (ψn)∞
n=1for A at E By a direct calculation, we have(A + R− E)ψn= (A− E)ψn+ λφ, ψnφ → 0.
The first term goes to zero because the ψncomprise a Weyl sequence for Aat E The second term goes to zero by orthonormality Thus, (ψn)∞
n=1 is
also a singular Weyl sequence for A + R at E and hence E∈ σess(A + R),
let φ = (χ)χ with χ a unit vector in Ker()⊥ Show that = φ Conclude the proof of Proposition 1.4.2 by showing that ψ→ ψ is an anti-linear isometry that maps H onto H∗ Hint : For any ψ, (ψ)φ− (φ)ψ ∈ Ker().
Exercise 1.4.2 Prove Proposition 1.4.5.
e−itAis unitary for every t∈ R.
σ(S) = ∂D.
Exercise 1.4.7 Let Δ be the operator defined in (1.4.9).
(a) Show that Δ is a bounded, self-adjoint operator with σ(Δ) =
[−2, 2].
(b) Show that Δ has no eigenvalues.
(c) Let μ = μΔδ0 denote the spectral measure associated with the pair Hint : With S as in Exercise 1.4.6, Δ = S + S−1.
Trang 25Exercise 1.4.8 If H is separable and A ∈ B(H) is self-adjoint, show thatA may have at most countably many eigenvalues.
and [Bψ]n= ψn−1(with the convention ψ0 = 0).
(a) Show A = B∗, AB = I, and BA = I.
(b) Show that every z∈ D := {w ∈ C : |w| < 1} is an eigenvalue of A.(c) Show that σ(A) = σ(B) =D.
(d) Show that there is no Weyl sequence for B at 0.
(e) Construct X, Y∈ B(H) such that XY = I and Y X is an
orthog-onal projection onto a subspace of infinite codimension.
Exercise 1.4.10 Prove Theorem 1.4.14 Hint : For part (c), use dominated
Exercise 1.4.11 Following the notation in Exercise 1.3.6, when is D
self-adjoint? Unitary? Normal? An orthogonal projection?
K, show that there exist self-adjoint operators Anand A such that An→A strongly, σ(An) = K for all n, and σ(A) ={a} Hint: Try diagonaloperators on 2(N).
show that there is a unique decomposable operator A for which one has
(1.4.29) and (1.4.32).
Exercise 1.4.14 Prove Theorem 1.4.18.
Exercise 1.4.15 (Unbounded operators) Let (D(A), A) be a linear
opera-tor in a Hilbert space H for which D(A) is dense in H (in this case, we sayA is densely defined ) Notice that we do not assume that A is bounded.
(a) Define
D(A∗) ={φ ∈ H : ψ → φ, Aψ
extends to a bounded linear functional onH}.Prove that, for φ∈ D(A∗), there is a unique element A∗φ∈ H with
the property that
A∗φ, ψ = φ, Aψ
and that this determines a linear operator (D(A∗), A∗) in H.(b) Show by example that D(A∗) ={0} is possible Hint: Heuristically,
A∗ functions like the conjugate transpose of a matrix Use this as
a starting point for a precise construction.
Trang 261.5 Dual Spaces and Locally Convex Topologies 39
(c) Consider the linear operator A defined by D(A) ={φ ∈ 2(Z) :
supp φ is compact} and [Aφ](n) = nφ(n) Calculate A∗.
(d) One says that A is self-adjoint if (D(A), A) = (D(A∗), A∗) (typi-cally abbreviated simply as A = A∗) Find a self-adjoint extension of the operator A from part (c), that is, a self-adjoint operator(D(B), B) such that D(B)⊃ D(A) and B|D(A)= A.
Notes and Remarks Proposition 1.4.2 is due to F Riesz [205] and is
sometimes called the Riesz representation theorem The spectral theorem
for bounded operators goes back to Hilbert [122], and the extension tounbounded operators is due to von Neumann [262].
The exposition in this section draws inspiration from Reed and Simon
[196], Simon [243], and Teschl [257].
1.5 Dual Spaces and Locally Convex Topologies
We conclude the discussion of functional analysis in general spaces with a brief excursion into dual spaces and alternative topologies that one can place on vector spaces For ease of exposition and since it suffices for our purposes,
we restrict our attention to real vector spaces and real linear functionals in
the present section.
We say that V is a topological vector space if the topology is Hausdorff and
the natural vector space operations are continuous That is,
We callV a locally convex topological vector space if in addition there is
a neighborhood basis for the topology at 0 comprised of open convex sets If V is a normed linear space, the collection N = {B(0, r) : r > 0} is a
neighborhood basis for the norm topology at the origin In particular, since balls are convex, every normed vector space is locally convex in the norm topology.
Let us briefly recall some terminology and notions from topology If
T1 and T2 are topologies on a set X, we say thatT1 is weaker thanT2
(equivalently, T2 is stronger thanT1) if every set open with respect to
T1 is open with respect to T2 Since the intersection of any collection of
Trang 27topologies is itself a topology, we can in certain situations precisely speak of
the weakest topology enjoying specified properties For instance, if X is a
set and {fα: α∈ A } is a collection of functions from X into a topologicalspace Y , one can construct on X the weakest topology for which every fα
is continuous The locally convex topologies of interest are those that are generated in precisely this fashion by families of linear functionals.
set of linear functions from V to R We define T (V, L) to be the weakest
topology on V in which every ∈ L is continuous; equivalently, T (V, L) is
the intersection of all topologies for which every element of L is continuous.
To construct the topology T = T (V, L), we note that a set U ⊆ V is open
with respect to T if and only if for every x ∈ U, there are finitely many1, , n∈ L and intervals (aj, bj) such that
We say that L is a separating family of linear functionals if for anyu = v in V, there is some ∈ L such that (u) = (v) Naturally, if L is
a separating family of linear functionals on V, then T (V, L) is a Hausdorff
topology on V.
In the particular case in which V is a Banach space and L = V∗, we
callT (V, V∗) the weak topology onV In the case V = U∗ for some normed
space U, each u ∈ U induces a linear functional on V via Eu(v) = v(u).
Identifying U with the collection of evaluation functionals {Eu: u∈ U}, we
callT (U∗,U) the weak-∗ topology on V.
weak-∗ topology on V∗ are locally convex.
functionals, then the topologyT (V, L) makes V a locally convex topologicalvector space.
n∈ N} be an orthonormal basis of H Sinceψ2=
|φn, ψ|2<∞,
we have φn, ψ → 0 In particular, φn→ 0 weakly However,
orthonor-mality implies that φn− φm =√2 whenever n = m, so (φn)∞
n=1 is not convergent in the norm topology.
Trang 281.5 Dual Spaces and Locally Convex Topologies 41
xn→ 0, equipped with the supremum norm One can identify U∗ with V = 1(N) and V∗with ∞(N) (Exercise 1.5.2) One can check that δnconverges to 0 as n→ ∞ in the weak-∗ topology on 1(N), but not in the
weak topology on 1(N) In fact, one can show that a sequence (yn)∞n=1 of
elements of 1(N) converges in the weak topology if and only if it converges in the norm topology (Exercise 1.5.3).
C(X,R) be the space of continuous functions equipped with the supremum norm By the Riesz–Markov–Kakutani representation theorem, V∗ may be
isometrically identified withM(X), the space of signed measures on X with
the total variation norm In view of these identifications, the weak-∗
topol-ogy on M(X) is then precisely the weakest topology for which every linear
functional of the form
f dμ
is continuous.
One advantage afforded by the weak-∗ topology is that it makes the
closed unit ball of the dual space a compact space that is metrizable when-ever V is separable First, we prove compactness.
is compact in the weak-∗ topology.
∈ B to an element x = (xv)v∈V∈ S = !v∈V[−v, v], namely bychoosing the element x∈ S whose vth coordinate is (v) Equipping S withthe product topology, S is compact by Tychonoff’s theorem The mappingf taking ∈ B to x ∈ S is a homeomorphism from B onto its image, by
the definitions of weak-∗ topology and product topology One can confirmthat the image of B under f is closed Thus, B is homeomorphic to a closed
subset of a compact space and hence is itself compact
Next, we discuss metrizability of the unit ball, which holds whenever the initial Banach space is separable.
ofV∗is metrizable in the weak-∗ topology.
Trang 29Proof Let S ={vj: j∈ N} be a dense subset of V The reader can verifythat the function d given on the closed unit ball ofV∗ by
is a metric that generates the weak-∗ topology thereupon.
is a compact metrizable space in the weak topology.
Proof This follows by combining Theorems 1.5.8 and 1.5.9 with
Proposi-tion 1.4.2 (which implies that the weak and weak-∗ topologies on H
The next fundamental result we want to prove is the Krein–Milman theorem, which shows that a compact convex set is precisely the closed convex hull of its extreme points First, we need a definition.
convex A nonempty subset E ⊆ K is called an extreme subset of K if it isconvex and whenever ax + (1−a)y ∈ E with 0 < a < 1 and x, y ∈ K, one hasx, y∈ E A point x ∈ K is an extreme point of K if {x} is an extreme subset
of K The convex hull of a set S ⊆ V is the smallest convex set containingS, that is, the intersection of all convex sets containing S Similarly, theclosed convex hull ofS is the smallest closed convex set containing S.
locally convex spaceV.
(a) K has at least one extreme point.
(b) K is the closed convex hull of its extreme points.
To prove this, it will be helpful to develop a bit of theory regarding continuous linear functionals on a locally convex space.
If K ⊆ V contains 0 and is nonempty, open, and convex, the gauge of K is
defined by
(1.5.5) pK(v) = inf{c ∈ (0, ∞) : v/c ∈ K}.
One can view this as a generalization of the notion of a norm, for, ifV is
a normed space andK is the open unit ball in V, then one has pK(x) =xfor all x∈ V.
Trang 301.5 Dual Spaces and Locally Convex Topologies 43
convex, and let p = pKdenote its gauge Then p has the following properties.
(a) 0≤ p(v) < ∞ for all v ∈ V.(b) p(x) < 1 for all x∈ K.
(c) p(z)≥ 1 for all z /∈ K.
(d) p(av) = ap(v) for all a≥ 0 and v ∈ V.(e) p(u + v)≤ p(u) + p(v) for all u, v ∈ V.
(f) If :V → R is a linear functional, then is continuous if and onlyif ≤ pKfor some nonempty, open, convexK containing 0.
any open, convexK ⊆ V and any y ∈ V \ K, there is a continuous linearfunctional onV such that
Since y /∈ K, we have p(y) ≥ 1 In particular, if we define a linearfunctional on the span of y by (ay) = a, we have
By Lemma 1.5.14, p is positive homogeneous and subadditive, so we may
apply the Hahn–Banach theorem (cf Exercise 1.3.13) to construct a linear
functional for which (y) = 1 and (v)≤ p(v) for all v ∈ V In particular,(x)≤ p(x) < 1 = (y) for every x ∈ K By Lemma 1.5.14, is continuous,
so the first statement is proved.
For the second statement, suppose x = y By local convexity, there isan open convex neighborhood of x that does not meet y Apply the firstpart of the lemma to this set and y yields a continuous linear functional
We can push just a bit further and also get a stronger separation
state-ment for closed convex sets.
Trang 31Corollary 1.5.16 SupposeV is a locally convex topological vector space.For any closed, convexK ⊆ V and any y ∈ V \ K, there is a continuouslinear functional onV and a scalar α such that
(1.5.8) (x) < α < (y)for all x∈ K.
topology of V is locally convex, there is an open convex set G containing
0 that is disjoint from K In particular, the set K1 = K − G is open and
convex (cf Exercise 1.5.4) and does not contain 0, so Lemma 1.5.15 implies
that there is a linear functional with (x) < (y) = (0) = 0 for all x∈ K1.
Since does not vanish identically andG is a neighborhood of 0, there existsz∈ G with (z) = α < 0 Since (x) < 0 for all x ∈ K1, we get for all w∈ K
(1.5.9) (w) = (z) + (w− z) < α + 0 < 0 = (y),
extreme subsets of K By compactness, the intersection of any totally
or-dered subcollection ofE is nonempty Thus, by a Zornification, there exists
a minimal elementS of E We claim that S consists of a single point Sup-pose on the contrary that x1 = x2 belong to S By Lemma 1.5.15, thereexists a continuous linear functional defined onV such that (x1) = (x2).
By compactness, attains a maximal value M onS One can check thatthe set of x∈ S for which (x) = M is an extreme subset of K that must be
strictly smaller than S, since is nonconstant on S; a contradiction Thus,S consists of a single point and therefore K has at least one extreme point.
(b) Let E ⊆ K denote the collection of all extreme points of K, and
let K denote the closed convex hull of E Naturally, K ⊆ K, so it remainsto show the opposite inclusion To that end, assume that u∈ V \ K ByCorollary 1.5.16, there is a continuous linear functional such that (v) <(u) for all v∈ K By compactness, attains a maximum value M on K.The set of x∈ K for which (x) = M is a closed extreme subset of K andcontains an element y ofE by a repetition of the Zornification in part (a).In particular, for all z∈ K, one has (z) ≤ M = (y) < (u), showing u /∈ K,
Exercise 1.5.1 Prove Theorem 1.5.4.
Trang 321.5 Dual Spaces and Locally Convex Topologies 45
n=1: xn∈ R, xn→ 0 asn→ ∞} consist of all real-valued sequences converging to zero, equippedwith the ∞ metric.
(a) Show that every y∈ 1(N) defines a bounded linear functional on
(b) Show that this gives an isometric bijection of c0(N)∗with 1(N).
(c) In a similar fashion, show that ∞(N) may be isometrically
identi-fied with 1(N)∗.
(d) Show that δn→ 0 in the weak-∗ topology on 1(N) but not in the weak topology.
n=1 denote a sequence of elements of V Prove that yn→ y in the weak topology on V if and only if yn→ y
in the norm topology on V Hints: For the hard direction, assume to thecontrary that yn→ 0 weakly in 1 and yn 1 → 0 Pass to a subsequencezk= ynksuch that “most” of the norm of zkis concentrated on a set Sk in
such a way that the Sk’s are disjoint.
(a) If K1,K2 ⊆ V are convex, show that K1− K2={x1− x2 : xj∈ Kj}
is convex Moreover, if V is a topological vector space, show thatK1− K2 is open wheneverK1 is open.
(b) If T :V → W is linear and K ⊆ V is convex, show that T (K) is
(c) If T :V → W is linear and K ⊆ W is convex, show that T−1(K) is
Exercise 1.5.5 Prove Lemma 1.5.14.
set of linear functions fromV to R If U ⊆ V is open with respect to T (V, L)
and 0∈ U, show that there are linearly independent 1, , n∈ L and δ > 0
set of linear functions fromV to R Show that a linear functional : V → R
is continuous with respect to the topology T (V, L) if and only if is a
finite linear combination of elements of L Hint: Start by showing that if
Trang 33{1, , n} is a linearly independent subset of L, then the range of the mapv→ (1(v), , n(v)) is all of Rn.
Notes and Remarks The exposition in this section is inspired by Lax
[166] and Simon [244].
Theorem 1.5.8 is usually named the Banach–Alaoglu theorem after
con-tributions of those authors [5,21] The result was discovered in various formsindependently by many authors Hilbert [122], Riesz [203], and Helly [120]had special cases before the general result, while Bourbaki [44], Shmulyan[228], and Kakutani [137] proved general results independently around the
same time as Alaoglu Discussing the number of authors with theorems related to this result, Pietsch remarks:
To be historically complete, one should speak of the Ascoli–Hilbert–Fr´echet–Riesz–Helly–Banach–Tychonoff– Alaoglu–Cartan–Bourbaki–Shmulyan–Kakutani theorem.
Theorem 1.5.12 is due to Krein and Milman [157].
1.6 Spectral Decompositions and Quantum Dynamics
Quantum mechanics suggests the study of the Schr¨odinger equation, which is a time-dependent problem A classical and very important approach to this problem is based on a detailed spectral analysis of the Schr¨odinger operator appearing in the Schr¨odinger equation, which is in fact a time-independent problem This has led to the development of the study of suitable spectral decompositions, which may be investigated for any self-adjoint operator These decompositions can then be connected to corresponding behaviors in the time-dependent setting In this section we develop the traditional aspects of this approach and present the decomposition of spectral measures and spectra into their absolutely continuous, singular continuous, and pure point parts, and we furthermore discuss how each piece of the decomposition gives information about the behavior of the solution of the time-dependent Schr¨odinger equation via the so-called RAGE theorem After this section, the next two sections address refinements in the pure point and singular continuous subspaces.
1.6.1 Decompositions of Measures and Operators Recall that any
measure μ on R has a unique Lebesgue decomposition (1.6.1) μ = μpp+ μsc+ μac,
where μppis a pure point measure, μscis a singular continuous measure, and
μac is an absolutely continuous measure This means that μpp(R \ C) = 0
Trang 341.6 Spectral Decompositions and Quantum Dynamics 47
for some countable set C, μsc({E}) = 0 for every E ∈ R, μsc(R \ S) = 0for some set S of zero Lebesgue measure, and μac(N ) = 0 for every set Nof zero Lebesgue measure We also write μc = μsc+ μac for the continuouspart and μs = μpp+ μsc for the singular part of μ.
In general, we write μ ν to mean μ is absolutely continuous withrespect to ν, that is, μ(S) = 0 whenever ν(S) = 0 If μ and ν are mutuallyabsolutely continuous, that is, μ ν and ν μ, we say that μ and ν areequivalent We write μ⊥ ν to mean μ and ν are mutually singular, that is,there are disjoint sets A and B such that μ(R \ A) = ν(R \ B) = 0.
is self-adjoint Define the following spaces:
Whenever we can make statements about multiple pieces of the spectral decomposition simultaneously, we may employ• as a stand-in for ac, sc, pp,
c, and s Concretely, we could rephrase the definitions above by declaring (1.6.7) H• ={φ : μφ= μφ,•}, • ∈ {ac, sc, pp, c, s}.
(a) The setsH•,• ∈ {ac, sc, pp, c, s}, are A-invariant closed subspaces
{ac, sc, pp, c, s} is closed and A-invariant, so, for each •, we can define alinear operator A•:= A|H•∈ B(H•) by restricting A toH• Define
(1.6.8) σ•(A) = σ(A•),• ∈ {ac, sc, pp, c, s}.
Respectively, the sets σac(A), σsc(A), σpp(A), σc(A), and σs(A) are called
the absolutely continuous, singular continuous, pure point, continuous, and
singular spectra of A Finally, letP•=P•(A) be the orthogonal projection
onto H• for• ∈ {ac, sc, pp, c, s}.
Trang 35We say that A has purely absolutely continuous spectrum ifHac = H,purely singular continuous spectrum ifHsc = H, pure point spectrum6 if
Hpp =H, purely continuous spectrum if Hc=H, and purely singular spec-trum ifHs =H.
or-thogonal, σac(A), σsc(A), and σpp(A) need not be pairwise disjoint
(Exer-cise 1.6.3).
A nonzero vector φ∈ H is called an eigenvector of an operator A ifthere is a so-called eigenvalue E∈ C with the property Aφ = Eφ We writeσp(A) for the set of eigenvalues of A Eigenvalues are intimately connected
to point spectrum:
(a) Every eigenvector of A belongs toHpp.(b) Every eigenvalue of A belongs to σpp(A).
(c) Aψ = Eψ if and only if ψ∈ Ran χ{E}(A).
(d) Hppis the smallest closed linear subspace ofH containing all eigen-vectors of A.
(e) The pure point spectrum of A is given by
E By induction one has f (A)ψ = f (E)ψ for any polynomial f and henceany continuous f by approximation Thus, for any continuous f ,
f dμψ =ψ, f(A)ψ = ψ, f(E)ψ = f(E)ψ2,
showing that μψ is ψ2 times a Dirac δ measure at E Thus, μψ is a point
measure, implying ψ∈ Hpp, and in particular that (App− E)ψ = 0, whichproves (a) and (b) Moreover, since μψ =ψ2δE,
(1.6.11) χ{E}(A)ψ2=ψ, χ{E}(A)ψ =
proving that ψ∈ Ran χ{E}(A) in this case and one direction of (c) For theother direction, assume ψ∈ Ran χ{E}(A) and note that for any φ, one has
(1.6.12) φ, ψ = φ, χ{E}(A)ψ =
χ{E}dμφ,ψ= μφ,ψ({E}),
6Strict parallelism would insist that we refer to this as “purely pure point spectrum”, butthe usage here is standard.
Trang 361.6 Spectral Decompositions and Quantum Dynamics 49
which yields in turn
φ, Aψ = φ, Aχ{E}(A)ψ =
xχ{E}(x) dμφ,ψ(x)= E· μφ,ψ({E})
=φ, Eψ.
Since φ, Aψ = φ, Eψ for all φ, Aψ = Eψ, concluding the proof of (c).
By (a), Hpp contains every eigenvector, hence it contains their closed
linear span Conversely, given ψ∈ Hpp, μψ is a point measure, so we can
which, in view of (c), shows that ψ belongs to the closed linear span of the
eigenvectors, concluding the proof of (d) Then (e) follows from (b) and (d)
since by the spectral theorem, σpp(A) is the smallest closed set that supports
If the operator A has a cyclic vector φ, then the spectral theorem assertsthat A is unitarily equivalent to multiplication by the independent variableon L2(R, μφ) via the canonical unitary map U : g(A)φ→ g By choosingdisjoint supports of μφ,pp, μφ,sc, and μφ,ac, we obtain the orthogonal sum
(1.6.14) L2(μφ) = L2(μφ,pp)⊕ L2(μφ,sc)⊕ L2(μφ,ac).
Moreover, this decomposition corresponds to the orthogonal splitting H =Hac⊕ Hsc⊕ Hpp in the sense that the unitary equivalence U mapsH• onto
L2(R, μφ,•) for • ∈ {ac, sc, pp} In particular, we can recover the point,singular continuous, and absolutely continuous spectra of A from supportsof the decomposition of μφ In general, a support of a measure μ is anyset whose complement has zero μ-measure We also define the topologicalsupport of μ, supp μ, to be the intersection of all closed supports of μ, orequivalently, the complement of the union of all open sets of zero μ-measure.
(1.6.15) σ(A) = supp(μφ),σ•(A) = supp(μφ,•),• ∈ {ac, sc, pp}.
Proof We write the proof of the first statement and leave the modifications
for the other three to the reader By the spectral theorem, A is unitarilyequivalent to the operator M : L2(μφ)→ L2(μφ) given by [M g](x) = xg(x)
Trang 37and hence it suffices to show σ(M ) = supp(μφ) If E /∈ supp(μφ), one can
check that M− E is invertible with inverse
sogn−1gnis a Weyl sequence for M at E, showing E∈ σ(M). Let us note that Theorem 1.6.6 suggests how one can go about construct-ing operators with prescribed spectral properties For instance, to find an operator with singular continuous spectrum equal to a given compact set
K⊆ R, it suffices to construct a finite Borel measure μ on R such thatsupp(μsc) = K and to consider the operator [Af ](x) = xf (x) in L2(μ).
1.6.2 The RAGE Theorem The Lebesgue decomposition of a spectral
measure μφ is also of importance in the study of the long-time asymptotics of solutions to the Schr¨odinger equation
(1.6.16) i∂tφ(t) = Aφ(t), φ(0) = φ.
Equation (1.6.16) has a unique solution given by
which implies that the evolution is unitary (compare Exercise 1.4.3) The following theorem establishes precise relations between spectral measures
and solutions of (1.6.16) for operators in 2(Z).
φ∈ 2(Z), and φ(t) is a solution of the Schr¨odinger equation (1.6.16).(a) We have μφ= μφ,ppif and only if for every ε > 0, there is N∈ Z+
Trang 381.6 Spectral Decompositions and Quantum Dynamics 51
The RAGE theorem has its roots in the following celebrated theorem of
Wiener regarding Fourier transforms of measures The Fourier transformof a finite Borel measure μ on R is defined by
so (1.6.22) follows by the dominated convergence theorem We now prove a technical result that will be useful in the proof of the RAGE theorem and also later on in the text In fact, we will prove the “only
if” direction of (b) in the course of proving this lemma For N∈ Z+, let
PN denote projection onto coordinates in [−N, N] That is, PNδn= δn for
|n| ≤ N and PNδn= 0 otherwise.
denoteP• =P•(A) Then, we have
proof of (1.6.24) Let φ, ψ be given, and abbreviate φ• =P•φ and ψ•=P•ψ
for• ∈ {c, pp}.
Trang 39If φ = φpp and ψ = ψpp, then, by Proposition 1.6.5, we may write φ andψ as convergent sums of eigenvectors of A Having done so, we deduce
from a calculation that we leave to the reader (Exercise 1.6.6).
Now, suppose φ = φc, that is, φ∈ 2(Z)c Denote by Πk projection onto
the coordinate k∈ Z so that PN = Since μδk,φis absolutely continuous with respect to μφ= μφ,c, Wiener’stheorem implies that the final line tends to zero as T→ ∞ The sameargument holds true if ψ∈ 2(Z)c (and φ is arbitrary) Consequently, for
whenever at least one of φ or ψ is in 2(Z)c (Notice that this already proves
the “only if” direction of Theorem 1.6.7(b) by taking ψ = φ.)At last, given general φ, ψ, we have
by (1.6.27) and (1.6.26), which completes the proof of (1.6.24)
Proof of Theorem 1.6.7 (a) For the “only if” direction, suppose we are
given φ∈ 2(Z)pp, and δ > 0 We can choose a finite sum of eigenfunctionsη = η1+· · · + ηk so that
Aηj= λjηj, andφ − η ≤ δ.
(I − PN)φ(t) = (I − PN)e−itAφ
≤ (I − PN)e−itAη + (I − PN)e−itA(φ− η)
Trang 401.6 Spectral Decompositions and Quantum Dynamics 53
For all N sufficiently large, the first term of the bottom line is no largerthan δ for all t Consequently,
For the converse, if for every ε > 0 there exists an N so that (1.6.18)holds, then setting ψ = φ in (1.6.25) yieldsPcφ = 0, whence μφ= μφ,pp.
(b) Taking φ = ψ in (1.6.27) proves the “only if” implication, so itremains to show the “if” direction If (1.6.19) holds for all N , then (1.6.24)
implies Pppφ = 0, which implies μφ= μφ,c.
(c) This follows from the same argument as in the forward direction of (b), but with the Riemann–Lebesgue lemma replacing Wiener’s theorem
In the mathematical formulation of quantum mechanics, A is given by
a so-called Schr¨odinger operator and for a solution φ(·) of the associated
Schr¨odinger equation, |δn, φ(t)|2 corresponds to the probability of finding
the system at site n at time t Thus, in this situation, we can interpret the
results of the RAGE theorem as follows The spectral measure of the initial state is pure point if and only if with arbitrarily high probability, the system can be found in some fixed compact region in space at any given time; the spectral measure of the initial state is purely continuous if and only if the system leaves any fixed compact region in space in a time-averaged sense as time goes to infinity; and if the spectral measure of the initial state is purely absolutely continuous, then the system leaves any fixed compact region in space as time goes to infinity.
1.6.3 Dynamical Averages via Resolvent Averages We conclude
this section with a useful lemma that connects the time-averaged quantum evolution to an integral involving the resolvent function at energies with
constant nonzero imaginary part For the proof, let us define the Fourier