Spectral TheorySummer 2019

Peer Christian KunstmannKarlsruher Institut fur Technologie (KIT)

Institut fur AnalysisEnglerstr. 2, 76131 Karlsruhe

e-mail: peer.kunstmann@kit.edu

These lecture notes contain a summary of the course. They are intended to be used parallelto the lecture. They are not meant to be used as a text book or for self studies. Attendingthe lectures cannot be substituted by reading these lecture notes.

The lecture itself is a natural continuation of the Functional Anylsis lecture in Winter2018/19. Sections are numbered consecutively.

1

8 The spectrum of closed operators

Notation: i) When nothing else is said, X, Y , and Z are complex Banach spaces.

ii) L(X, Y ) := T : X → Y : T is linear and bounded. For T ∈ L(X, Y ) we have theoperator norm

‖T‖ := sup‖x‖X≤1

‖Tx‖Y ,

and (L(X, Y ), ‖ · ‖) is a Banach space. For the following we also refer to FA1 2.14.

8.1. Definition: A linear operator A from X to Y is a linear operator A : D(A) → Ywhere D(A), the domain of A, is a linear subspace of X. We also write A : X ⊇ D(A)→ Y .Further we use

R(A) := Ax : x ∈ D(A) range of A,

N(A) := x ∈ D(A) : Ax = 0 kernel (null space) of A.

A linear operator A from X to Y is called closed if its graph

gr(A) := (x,Ax) : x ∈ D(A) ⊆ X × Y

is a closed subspace of X × Y .

Recall: Since A is linear, its graph gr(A) is a linear subspace of X ×Y . The space X ×Y ,equipped with the norm given by ‖(x, y)‖ := ‖x‖X + ‖y‖Y is a Banach space.

8.2. Lemma: Let A : X ⊇ D(A)→ Y be a linear operator. The following are equivalent:

(i) A is closed,

(ii) D(A) is a Banach space for the graph norm given by

‖x‖A := ‖x‖X + ‖Ax‖Y .

(iii) For every sequence (xn)n∈N in D(A) and x ∈ X, y ∈ Y such that xn → x in X andAxn → y in Y one has x ∈ D(A) and Ax = y.

Notation: We write [D(A)] := (D(A), ‖ · ‖A). Then obviously A ∈ L([D(A)], Y ).

Proof. By definition, closedness of A is equivalent to closedness of gr(A), and since X × Yis a Banach space, this is equivalent to completeness of gr(A). Since the map

J : [D(A)]→ gr(A), x 7→ (x,Ax),

is linear, bijective, and isometric, the latter is equivalent to [D(A)] being a Banach space.On the other hand, closedness of gr(A) means

1FA means the Functional Analysis lecture in winter 2018/19.

2

for any sequence (xn, Axn) in gr(A) and (x, y) ∈ X × Y such that (xn, Axn)→(x, y) one has (x, y) ∈ gr(A),

which clearly is equivalent to (iii).

Examples: 1) Let I ⊆ R an open interval, p ∈ [1,∞], X = Y = Lp(I), A := ddx

withD(A) = W 1,p(I) = f ∈ Lp(I) : f ′ ∈ Lp(I) (weak derivative). Then A is closed. Recallfirst (→ FA 5.1):

f ∈ L1loc(I) has a weak derivative in I if there exists g ∈ L1

loc(I) such that, forany ϕ ∈ C∞c (I), we have

−∫I

fϕ′ dx =

∫I

gϕ dx.

We show that A is closed: Let (fn) be a sequence in W 1,p(I) with fn → f and f ′n → g withrespect to ‖ · ‖p. Then for ϕ ∈ C∞c (I) ⊆ Lp

′(I) (where 1

p+ 1

p′= 1) we have, by Holder,

fnϕ′ → fϕ′ and gnϕ→ gϕ in ‖ · ‖1. This implies

−∫I

fϕ′ dx = limn−∫I

fnϕ′ dx = lim

n

∫I

gnϕdx =

∫I

gϕ dx.

Hence f has weak derivative g ∈ Lp(I), so f ∈ W 1,p(I) and f ′ = g.

2) Let X = Y = L1(0, 1), A := ddx

with D(A) = C1[0, 1]. Then A is not closed: Takeg ∈ L1(0, 1) \ C[0, 1], say g = 1[1/2,1] − 1[0,1/2), and approximate g in ‖ · ‖1 by a sequence(gn) in C[0, 1]. Then let

f(t) :=∣∣∣t− 1

2

∣∣∣ and fn(t) :=1

2+

∫ t

0

gn(s) ds.

We obtain fn ∈ C1[0, 1] and f ′n = gn → g in ‖ · ‖1, fn → f in ‖ · ‖∞, hence in particularfn → f in ‖ · ‖1. But f 6∈ D(A). [As a concrete gn one can take

gn(t) :=

−1 , t ∈ [0, 1

2− 1

n)

n(t− 12

, |t− 12| ≤ 1

n

1 , t ∈ (12

+ 1n, 1]

, n ≥ 3.]

3) Let X = L1(0, 1), Y = C, Af := f(0) with D(A) = C[0, 1]. Then A is not closed: Takefn(t) := (1 − nt)1[0,1/n](t). Then fn → 0 =: f in ‖ · ‖1 and Afn = fn(0) = 1 → 1. Heref ∈ D(A), but Af = f(0) = 0 6= 1.

Comment: L1[0, 1] is a Banach space whose elements are equivalence classes of functionsthat coincide almost everywhere (a.e.). Strictly speaking, C[0, 1] should be considered here

3

as the set of equivalence classes in L1[0, 1] that contain a continuous function. This contin-uous function is then unique within its class. A evaluates the unique continuous functionin the class at 0. In fact, a similar interpretation has to be given to C1[0, 1] as a subspaceof L1[0, 1] in Example 2) above.

4) Let p ∈ [1,∞], X = Y = Lp(Ω, µ) where (Ω, µ) is a σ-finite measure space. Letm : Ω→ C be measureable and

Af := mf with D(A) := f ∈ Lp(Ω, µ) : mf ∈ Lp(Ω, µ).

Then A is closed: If fn → f and mfn → g with respect to ‖ · ‖p, then we find subsequencessuch that fk(n) → f and mfk(l(n)) → g almost everywhere. Hence mf = g a.e., and byg ∈ Lp we have f ∈ D(A), Af = g.

We also recall the following (cp. FA 2.15).

8.3. Closed Graph Theorem: Let A : X → Y be a linear operator. Then

A is closed ⇐⇒ A is bounded.

Notation: Let A and B be linear operators form X to Y and C be a linear operator fromY to Z. We define

A+B by (A+B)x := Ax+Bx for x ∈ D(A+B) := D(A) ∩D(B),

CA by (CA)x := C(Ax) for x ∈ D(CA) := x ∈ D(A) : Ax ∈ D(C).

A+B is a linear operator from X to Y , and CA is a linear operator from X to Z.

Sum and product of linear operators are associative, but not distributive in general.

In general, sums or products of closed operators are not closed.

Example: Take X = l1 and, for x = (xn) define A by (Ax)n = nxn−1 if n is even and(Ax)n = 0 if n is odd, and let D(A) := x ∈ l1 : Ax ∈ l1.Then A is closed: If x(n) is a sequence in D(A) such that x(n) → x and Ax(n) → y, then

xk = limn x(n)k for each k, and yk = limn kx

(n)k−1 for even k, yk = 0 for odd k. We conclude

that yk = kxk−1 for even k. Hence we have Ax = y, and x ∈ D(A) by y ∈ l1.

However, B := A + (−A) = 0 with D(B) = D(A) and C := AA = 0 with D(C) = D(A)are not closed (otherwise D(A) would be a closed subspace of l1, but D(A) 6= l1 is densein l1).

8.4. Lemma (Properties of closed operators): Let A be a closed linear operator fromX to Y , T ∈ L(X, Y ), and S ∈ L(Z,X). Then:

(a) B := A+ T is closed (here D(B) = D(A)).

4

(b) C := AS is closed (here D(C) = z ∈ Z : Sz ∈ D(A)).(c) If A is injective then A−1 is closed (here D(A−1) = R(A)).

(d) If R is injective and closed from Y to Z such that R−1 ∈ L(Z, Y ) then D := RA isclosed (here D(D) = x ∈ D(A) : Ax ∈ D(R)).

Proof. (a) Take (xn) in D(A) such that xn → x, Bxn → y. Since T is bounded we haveTxn → Tx. Now (A+T )xn → y implies that Axn → y−Tx. By closedness of A we obtainx ∈ D(A) and Ax = y − Tx, i.e. Bx = y. end Tue

23.04.19(b) Let (zn) be a sequence in D(C) such that zn → z, Czn → y. Since S is bounded we haveSzn → Sz, A(Szn)→ y. But (Szn) is a sequence in D(A), thus Sz ∈ D(A), A(Sz) = y byclosedness of A. We have shown z ∈ D(C), Cz = y.

(c) follows from gr(A−1) = (y, x) : (x, y) ∈ gr(A).(d) Let (xn) be a sequence in D(D) such that xn → x and Dxn → z. By R−1 ∈ L(Z, Y ) wehave Axn = R−1Dxn → R−1z. Since A is closed and D(D) ⊂ D(A), we obtain x ∈ D(A)and Ax = R−1z ∈ R(R−1) = D(R). Hence x ∈ D(D) and Dx = z.

8.5. Definition (spectrum and resolvent): Let A : X ⊇ D(A) → X be a linearoperator. Then

ρ(A) := λ ∈ C : λI − A : D(A)→ X is bijective and (λI − A)−1 ∈ L(X)

is called the resolvent set of A and

σ(A) := C \ ρ(A)

is the spectrum of A. The map

ρ(A)→ L(X), λ 7→ (λI − A)−1,

is called the resolvent of A, and for λ ∈ ρ(A), the operator

R(λ,A) := (λI − A)−1

is called the resolvent operator (at λ).

Remark: It is common to write λ − A, (λ − A)−1 instead of λI − A, (λI − A)−1. Forλ ∈ ρ(A) we have

(λ− A)R(λ,A) = IX , R(λ,A)(λ− A) = ID(A).

Further Remarks: (a) If ρ(A) 6= ∅ then A is closed.

(b) If A is closed then ρ(A) = λ ∈ C : λI − A : D(A)→ X is bijective.

5

Proof. (a) We find λ0 ∈ ρ(A). Then −R(λ0, A) = −(λ0−A)−1 is closed. By 8.4(c), A−λ0Iis closed, and by 8.4(a), A = (A− λ0I) + λ0I is closed.

(b) “⊆” is clear. For “⊇” assume that λ − A : D(A) → X is bijective. By 8.4(a), λ − Ais closed, and by 8.4(c), (λ − A)−1 : X → X is closed. By the Closed Graph Theorem weobtain (λ− A)−1 ∈ L(X), i.e. λ ∈ ρ(A).

Examples: 1) X = C[0, 1], A = ddx

, D(A) = C1[0, 1]. Let λ ∈ C. Then f := eλ(·) ∈ D(A)and Af = f ′ = λeλ(·) = λf , hence λ − A : D(A) → X is not injective. We have shownσ(A) = C.

2) X = C[0, 1], A0 = ddx

, D(A0) = f ∈ C1[0, 1] : f(0) = 0. Let λ ∈ C, g ∈ C[0, 1]. ByODE-Theory there is a unique solution f ∈ C1[0, 1] to the initial value problem

λf − f ′ = g in [0, 1]; f(0) = 0,

given by f(t) := −∫ t

0eλ(t−s)g(s) ds, t ∈ [0, 1]. Thus λ−A0 : D(A0)→ X is bijective. Since

A0 is closed (!), we have shown σ(A0) = ∅, ρ(A) = C. Moreover, we have

|f(t)| ≤∫ t

0

eReλ(t−s)|g(s)| ds ≤∫ 1

0

eReλs ds ‖g‖∞ =

eReλ−1Reλ

‖g‖∞ , Re λ 6= 0‖g‖∞ , Re λ = 0

,

which means

‖R(λ,A0)‖ ≤

eReλ−1Reλ

, Re λ 6= 01 , Re λ = 0

.

Notice that this implies σ(A0) = ∅ without using closedness of A0.

8.6. Proposition (Properties of the resolvent): Let A be a closed linear operator inX. Then:

(a) For all λ, µ ∈ ρ(A):

R(λ,A)−R(µ,A) = (µ− λ)R(λ,A)R(µ,A) (resolvent equation).

In particular, R(λ,A)R(µ,A) = R(µ,A)R(λ,A) for all λ, µ ∈ ρ(A).

(b) If λ ∈ ρ(A) then B(λ, 1/‖R(λ,A)‖) ⊆ ρ(A) and

R(µ,A) =∞∑k=0

(−1)kR(λ,A)k+1(µ− λ)k, µ ∈ B(λ, 1/‖R(λ,A)‖),

where the series converges in operator norm. In particular, ρ(A) is open and σ(A) is closed,and, for any λ ∈ ρ(A),

‖R(λ,A)‖ ≥ 1

d(λ, σ(A)).

6

Proof. (a) Write

R(λ,A)−R(µ,A) = R(λ,A) (µ− A)R(µ,A)︸ ︷︷ ︸=IX

−R(λ,A)(λ− A)︸ ︷︷ ︸=ID(A)

R(µ,A) = (µ−λ)R(λ,A)R(µ,A).

(b) For µ ∈ C we have

µ− A = µ− λ+ λ− A = ((µ− λ)R(λ,A) + I)(λ− A).

For |µ− λ| < 1/‖R(λ,A)‖ the operator

I + (µ− λ)R(λ,A) ∈ L(X)

is invertible in L(X) by a Neumann series and

(I + (µ− λ)R(λ,A))−1 =∞∑k=0

(−1)kR(λ,A)k(µ− λ)k.

We obtain for these µ that µ− A : D(A)→ X is bijective and

(µ− A)−1 = R(λ,A)(I + (µ− λ)R(λ,A))−1 =∞∑k=0

(−1)kR(λ,A)k+1(µ− λ)k

as claimed.

— — Detour: Analyticity — —

Definition: Let Ω ⊆ C be open and X a complex Banach space. A function f : Ω→ X iscalled analytic (or holomorphic) in Ω if, for every z0 ∈ Ω the limit

f ′(z0) := limz→z0

f(z)− f(z0)

z − z0

exists. In this case, the function f ′ : Ω→ X is called the (complex) derivative of f .

Remark: A holomorphic function is continuous.

Power series: A (formal) power series in X has the form

∞∑k=0

ak(z − z0)k

where (ak)k∈N0 is a sequence in X and z0 ∈ C.

Radius of convergence: Define R ∈ [0,∞] by 1R

:= lim supk→∞ ‖ak‖1/kX . Then the power

series above converges absolutely and uniformly on compact subsets of B(z0, R) and definesa holomorphic function f in B(z0, R) which satisfies

f (k)(z0) = k!ak, k ∈ N0.

Moreover, for R > R, f cannot be extended to a holomorphic function on B(z0, R).

7

Proof. Let ε ∈ (0, R/2) and choose k0 ∈ N such that

‖ak‖1/k ≤ 1

R− εfor all k ≥ k0.

For |z − z0| ≤ R− 2ε we then have

∞∑k=k0

‖ak‖|z − z0|k ≤∞∑

k=k0

(R− 2ε

R− ε

)k<∞.

Differentiability follows as for scalar-valued power series.

Now let R ≥ R such that f has an analytic extension g : B(z0, R) → X. For arbitraryϕ ∈ X ′, the function ϕ g : B(z0, R)→ C is analytic and

∞∑k=0

ϕ(ak)(z − z0)k

converges for all |z − z0| < R. Hence for any δ ∈ (0, R),

supk|ϕ(ak)|(R− δ)k <∞.

By the Uniform Boundedness Principle we conclude that

Cδ := supk‖ak‖(R− δ)k <∞ for any δ ∈ (0, R),

which implies

‖ak‖1/k ≤ C1/kδ

R− δfor all δ ∈ (0, R), k ∈ N,

and1

R= lim sup

k‖ak‖1/k ≤ 1

R− δfor all δ ∈ (0, R).

Letting δ → 0, we obtain R ≤ R.end Thu25.04.19Remark: The proof indicates the two basic ways of approaching holomorphic functions

with values in a Banach space. The first is to proceed as in the complex valued case, andthe second is to apply linear functionals and resort to results on complex valued functions.

The following will be an exercise:

Proposition: A function f : Ω → X is analytic if and only if it is weakly analytic, i.e.ϕ f : Ω→ C is analytic for every ϕ ∈ X ′.

— — End of the Detour — —

For the applications we need another lemma.

8

8.7. Lemma: Let (cn) be a real sequence such that 0 ≤ cn+m ≤ cncm for all n,m ∈ N.Then n

√cn → c := infk k

√ck as n→∞.

Proof. Let ε > 0 and choose m such that m√cm < c + ε. Let b := maxc1, . . . , cm and

write n > m as n = km+ r with k ∈ N and r ∈ 1, . . . ,m. Then

c1/nn = (ckm+r)

1/n ≤ (ckmcr)1/n ≤ (c+ ε)km/nb1/n = (c+ ε)(c+ ε)−r/nb1/n ≤ c+ 2ε

for large n, since (c + ε)−r/nb1/n → 1 as n → ∞ (here we assumed b > 0 without loss ofgenerality).

8.8. Corollary: If A is closed linear operator in X, then its resolvent ρ(A) → L(X),λ 7→ R(λ,A), is an analytic function and

dk

dλkR(λ,A) = (−1)kk!R(λ,A)k+1, k ∈ N0.

For λ ∈ ρ(A) we have

d(λ, σ(A)) =1

infk ‖R(λ,A)k‖1/k. (+)

Proof. The first part is clear from 8.6(b) and analyticity of power series. Applying 8.7 tock := ‖R(λ,A)k‖ we see that the right hand side of (+) is the radius of convergence R forthe power series in 8.6(b). By the above, ∂B(λ,R) ∩ σ(A) 6= ∅ (otherwise the resolventwould be holomorphic on a strictly larger ball which is impossible).

We give an application to bounded operators.

8.9. Proposition and Definition: Let T ∈ L(X) and X 6= 0. Then σ(T ) 6= ∅, and thespectral radius of T ,

r(T ) := sup|λ| : λ ∈ σ(T )satisfies

r(T ) = limk→∞‖T k‖1/k = inf

k‖T k‖1/k.

In particular, σ(T ) is a non-empty, compact subset of B(0, ‖T‖).

Proof. We begin with the observation that, for |λ| > ‖T‖,

λI − T = λ(I − T

λ

),

which is invertible by a Neumann series

(λI − T )−1 = λ−1

∞∑k=0

T k

λk=∞∑k=0

T kλ−(k+1).

9

The latter is a power series in λ−1 with radius of convergence R := 1/ infk ‖T k‖1/k. Henceit converges for |λ| > 1/R, and σ(T ) ⊂ B(0, 1/R), i.e. r(T ) ≤ 1/R. The above on powerseries also shows that r(T ) = 1/R if 1/R > 0.

We now show that σ(T ) 6= ∅ (then σ(T ) = 0 in case 1/R = 0). From the series repre-sentation above we also obtain that R(λ, T ) → 0 as |λ| → ∞. If σ(T ) = ∅ then, for anyϕ ∈ L(X)′, λ→ ϕ R(λ, T ) is an entire function, which is bounded (since it tends to 0 as|λ| → ∞). By Liouville’s Theorem, this function is constant. We conclude ϕ R(λ, T ) = 0for any λ and any ϕ. But then R(λ, T ) = 0 which implies X = 0, and this case wasexcluded.

Bemerkung: There exist operators T with ‖T‖ = 1, r(T ) = 0, and hence σ(T ) = 0,

e.g. T =

(0 10 0

)in X = C2 (equipped with Euclidean norm).

8.10. Definition (Fine structure of the spectrum): Let A be a closed linear operatorin X. We define

σp(A) := λ ∈ C : λI−A is not injective = λ ∈ C : N(λI−A) 6= 0 (point spectrum).

Any λ ∈ σp(A) is called an eigenvalue of A and any x ∈ N(λI − A) \ 0 is called aneigenvector for the eigenvalue λ. We further define

σr(A) := λ ∈ C : R(λI − A) is not dense in X (residual spectrum)

σc(A) := λ ∈ C : R(λI − A) is dense but not closed in X (continuous spectrum)

σap(A) := λ ∈ C : there exists a sequence (xn) in D(A) with ‖xn‖ = 1 for all n

and (λI − A)xn → 0 (approximate point spectrum).

A sequence (xn) as in the definition of the approximate point spectrum is sometimes calledan approximate eigenvector.

Remark: σp(A) ⊆ σap(A) (take xn = x eigenvector).

Examples: 1) If dimX < ∞ and A ∈ L(X) then σ(A) = σp(A) = σr(A) (A is a finite-dimensional matrix for which injectivity is equivalent to surjectivity).

2) Let X = C[0, 1], A = ddx

with D(A) = C1[0, 1]. We saw above that σ(A) = σp(A) = C.On the other hand, for any λ ∈ C the ordinary differential equation λf−f ′ = g has a (non-unique) solution f ∈ C1[0, 1] for each right hand side g ∈ C[0, 1]. Hence R(λI − A) = Xfor each λ ∈ C and σr(A) = σc(A) = ∅.

8.11. Proposition: Let A be a closed operator in X. Then:

(a) σap(A) = σp(A) ∪ λ ∈ C : λI − A is injective and (λI − A)−1 is not bounded .(b) σ(A) = σap(A) ∪ σr(A).

(c) ∂σ(A) ⊆ σap(A).

10

Proof. (a) Let λ ∈ C and let λ− A be injective, Y := R(λ− A). Then:

(λ− A)−1 : Y → X is not bounded

⇔ there exists (yn) in Y such that ‖yn‖ = 1 and ‖(λ− A)−1yn‖ → ∞⇔ there exists (zn) in Y such that zn → 0 and ‖(λ− A)−1zn‖ = 1

⇔ there exists (xn) in D(A) s.t. ‖xn‖ = 1 and (λ− A)xn → 0

To see this put zn = yn/‖(λ−A)−1yn‖, yn = zn/‖zn‖, and xn = (λ−A)−1zn, zn = (λ−A)xn,respectively. Thus we have shown

σap(A) \ σp(A) = λ ∈ C : λI − A is injective and (λI − A)−1 is not bounded .

(b) “⊇” is clear. For the proof of “⊆” let λ ∈ σ(A) \ σap(A). By (a), λ−A is injective and

(λ−A)−1 is bounded. In particular, R(λ−A) is closed in X. Hence R(λ− A) = R(λ−A) 6=X (otherwise λ 6∈ σ(A)!). We have shown λ ∈ σr(A).

(c) Let λ ∈ ∂σ(A). We find a sequence (λn) in ρ(A) with λn → λ. By 8.6(b) we have‖R(λn, A)‖ → ∞. By the Uniform Boundedness Principle we find y ∈ X such that αn :=‖R(λn, A)y‖ → ∞ (observe that we have αn > 0 for any n). Let xn := R(λn, A)y/αn,n ∈ N. Then (xn) is a sequence in D(A) with ‖xn‖ = 1 for all n. Moreover,

(λ− A)xn = (λ− λn)xn + (λn − A)xn = (λ− λn)xn + y/αn → 0 (n→∞).

Example: Let X = l2 = (xn)n∈N : ‖(xn)‖l2 :=(∑

n |xn|2)1/2

< ∞ and let L ∈ L(l2)

be the left shift given by L(xn) := (xn+1), i.e. L(xn) = (x2, x3, x4, . . .). Clearly, we have‖L‖ = 1, ‖Lk‖ = 1 for every k ∈ N, and hence r(L) = 1. For λ ∈ C with |λ| ≤ 1 we have

(λ− L)(xn) = 0⇔ ∀n : λxn − xn+1 = 0⇔ ∀n : xn+1 = λnx1.

For |λ| < 1 we have (1, λ, λ2, . . .) ∈ l2, hence λ ∈ σp(L). For |λ| = 1 we have (1, λ, λ2, . . .) 6∈l2, and λ 6∈ σp(L).

We conclude σp(L) = |λ| < 1, σ(L) = σap(L) = |λ| ≤ 1, and σap(L) \ σp(L) = |λ| =1 = ∂σ(L). end Tue

30.04.19

8.12. Dual operators: Recall that, for T ∈ L(X, Y ), its dual (or adjoint)2 operatorT ′ ∈ L(X ′, Y ′) is given by

T ′φ := φ T, φ ∈ Y ′.2In this course we will reserve the term “adjoint” for the Hilbert space adjoint and thus speak of “dual

operators” here.

11

Using duality brackets

〈y, φ〉Y×Y ′ := φ(y), for all y ∈ Y , φ ∈ Y ′,

this can be written as

〈x, T ′φ〉X×X′ = 〈Tx, φ〉Y×Y ′ for all x ∈ X, φ ∈ Y ′,

Example: Let X = Y = l2 and L be the left shift. We calculate L′. For X = l2 we haveX ′ = l2 where the duality bracket is given by

〈(xn), (yn)〉l2×l2 =∑n

xnyn.

For (xn), (yn) ∈ l2 we then have

〈L(xn), (yn)〉 =∞∑n=1

xn+1yn =∞∑n=2

xnyn−1 = 〈(xn), (yn−1)〉

if we let y0 := 0. Hence L′ = R where R is the right shift given by

R(yn) := (0, y1, y2, y3, . . .).

Clearly, we have ‖R(yn)‖l2 = ‖(yn)‖l2 , ‖R‖ = 1, r(T ) = 1, and σ(R) ⊆ |λ| ≤ 1. For|λ| ≤ 1 we have

(λ−R)(yn) = 0⇔ λy1 = 0,∀n ≥ 2 : λyn = yn−1 ⇔ (yn) = 0.

Hence σp(R) = ∅.

Rules for dual operators: The following rules are easily verified:

(IX)′ = IX′ , (S + T )′ = S ′ + T ′, (αT )′ = αT ′, (ST )′ = T ′S ′

where T ∈ L(X, Y ), α ∈ C, S ∈ L(X, Y ) for the sum and S ∈ L(Y, Z) for the product,respectively.

Remark: For T ∈ L(X, Y ) we have

T : X → Y is an isomorphism⇐⇒ T ′ : Y ′ → X ′ is an isomorphism.

Moreover, we have (T ′)−1 = (T−1)′ in this case.

Proof. “⇒”: We have

T ′(T−1)′ = (T−1T )′ = (IX)′ = IX′ , (T−1)′T ′ = (TT−1)′ = (IY )′ = IY ′ ,

12

which proves invertibility of T ′ and (T ′)−1 = (T−1)′.

“⇐”: Suppose that T ′ has an inverse S : X ′ → Y ′. Then T ′′ := (T ′)′ is an isomorphismX ′′ → Y ′′ by the first part of the proof. Since X is a closed subspace of X ′′, T ′′(X) is aclosed subspace of Y ′′. But T ′′(X) = T (X) ⊆ Y , so T (X) is a closed subspace of Y .

If φ ∈ X ′ satisfies φ|T (X) = 0 then

〈x, T ′φ〉 = 〈Tx, φ〉 = 0, x ∈ X,

i.e. T ′φ = 0. Since T ′ is injective we conclude φ = 0. By Hahn-Banach we thus have shownthat T (X) is dense in Y . Since T (X) is also closed in Y we obtain T (X) = Y , i.e. T issurjective.

Since T ′′ is injective, T = T ′′|X is injective. We have shown that T : X → Y is bijective.By the Open Mapping Theorem, T : X → Y is an isomorphism.

Recall for the previous proof: The bidual X ′′ = (X ′)′ = L(X ′,C) of X is a Banachspace, and the map

J : X → X ′′, x 7→ δx where δx : X ′ → C, φ 7→ δx(φ) := φ(x)

is (by Hahn-Banach) an isometric injection. Written with duality brackets we have

〈φ, δx〉X′×X′′ = 〈x, φ〉X×X′ , x ∈ X,φ ∈ X ′.

It is common to identify X with the closed subspace J(X) of X ′′. The space X is calledreflexive if J(X) = X ′′.

If T ∈ L(X, Y ) then T ′′ := (T ′)′ ∈ L(X ′′, Y ′′) and T = T ′′|X , since for x ∈ X, φ ∈ Y ′ wehave

〈φ, T ′′δx〉 = 〈T ′φ, δx〉 = 〈x, T ′φ〉 = 〈Tx, φ〉 = 〈φ, δTx〉,

which means T ′′δx = δTx for all x ∈ X, i.e. T ′′x = Tx, x ∈ X, if we identify X and J(X).

8.13. Definition: Let A be a linear operator from X to Y with D(A) dense in X. Wedefine the linear operator A′ : Y ′ ⊇ D(A′)→ X ′ by letting for φ ∈ Y ′, ψ ∈ X ′:

φ ∈ D(A′), A′φ = ψ :⇐⇒ ∀x ∈ D(A) : 〈Ax, φ〉Y×Y ′ = 〈x, ψ〉X×X′

Then D(A′) is the set of all φ ∈ Y ′ such that the map

D(A)→ C, x 7→ 〈Ax, φ〉,

has a continuous extension ψ ∈ X ′. By denseness of D(A) in X, this extension is uniquelydetermined (if it exists). Then A′φ is this unique extension ψ.

13

Remark: A′ is a always closed: Let (φn) be a sequence in D(A′) such that φn → φ in Y ′

and A′φn → ψ in X ′. Then we have, for any x ∈ D(A):

〈Ax, y〉 = limn〈Ax, φn〉 = lim

n〈x,A′φn〉 = 〈x, ψ〉,

which means φ ∈ D(A′) and A′φ = ψ.

Example: X = L2(0, 1), A = ddx

, D(A) = C10 [0, 1] := ϕ ∈ C1[0, 1] : ϕ(0) = ϕ(1) = 0.

Then

D(A′) = f ∈ L2(0, 1) : ∃g ∈ L2(0, 1)∀ϕ ∈ C10 [0, 1] :

∫ 1

0

fϕ′ dx =

∫ 1

0

gϕ dx.

In particular we see that any f ∈ D(A′) has a weak derivative in L2(0, 1) and that A′f =−f ′, D(A′) ⊆ W 1,2(0, 1). On the other hand, if f ∈ W 1,2(0, 1) then we have∫ 1

0

fϕ′ dx = −∫ 1

0

f ′ϕdx

for all ϕ ∈ C∞c (0, 1). Any ϕ ∈ C10 [0, 1] can be approximated by a sequence (ϕn) in C∞c (0, 1)

in W 1,2-norm. This yields D(A′) = W 1,2(0, 1), A′ = − ddx

(weak derivative).

Rules: (λ−A)′ = λ−A′ for λ ∈ C and X = Y (more general one has (A+T )′ = A′+T ′ ifT ∈ L(X, Y )). If B is an extension of A then A′ is an extension of B′: A ⊆ B ⇒ B′ ⊆ A′.

8.14. Proposition: Let A be a closed linear operator in X which is densely defined. Thenσ(A′) = σ(A) and σp(A

′) = σr(A).

Proof. We start with the second assertion. For λ ∈ C we have

λ ∈ σr(A) ⇔ (λ− A)X 6= y ⇔ ∃φ ∈ Y ′ \ 0∀x ∈ D(A) : 〈(λ− A)x, φ〉 = 0

⇔ ∃φ ∈ D(A′) \ 0 : (λ− A′)φ = 0 ⇔ λ ∈ σp(A′).

σ(A′) ⊆ σ(A): Let λ ∈ ρ(A). By what we have just proved, λ 6∈ σr(A) = σp(A′) and λ−A′

is injective. Now let ψ ∈ X ′ and set φ := R(λ,A)′ψ. For x ∈ D(A) we then have

〈(λ− A)x, φ〉 = 〈R(λ,A)(λ− A)x, ψ〉 = 〈x, ψ〉.

This means φ ∈ D((λ− A)′) = D(A′) and (λ− A′)φ = ψ. Hence λ− A′ is also surjective,and λ ∈ ρ(A).

σ(A) ⊆ σ(A′): Let λ ∈ ρ(A′). Then λ 6∈ σp(A′) = σr(A), and thus Y := (λ − A)(D(A)) isdense in X.

14

We let S := R(λ,A′) ∈ L(X ′). Then S ′ ∈ L(X ′′) and we define R := S ′|X . We claim thatR ∈ L(X) is the inverse of λ− A. For x ∈ D(A) and φ ∈ X ′ we have

〈φ, S ′δ(λ−A)x〉 = 〈Sφ, δ(λ−A)x〉 = 〈(λ− A)x, Sφ〉.

Since Sφ ∈ D(A′) we can continue

〈(λ− A)x, Sφ〉 = 〈x, (λ− A′)Sφ〉 = 〈x, φ〉 = 〈φ, δx〉.

We thus have shown S ′δ(λ−A)x = δx for all x ∈ D(A). In particular R maps the dense rangeof λ − A into X, hence R ∈ L(X). Moreover, R(λ − A)x = x for all x ∈ D(A). Henceλ − A is injective and Ry = (λ − A)−1y for all y ∈ Y . Thus we see that (λ − A)−1 is abounded operator. Since it is also closed, its domain Y has to be a closed subspace of X.By denseness of Y we thus obtain Y = X, λ ∈ ρ(A) and R = R(λ,A).

Example: Let X = l2, L be the left shift, and R be the right shift. Since L′ = R, weobtain σr(L) = σp(R) = ∅.

The spectrum of an operator is related to the spectrum of its resolvents. This is calledspectral mapping.

8.15. Proposition: Let A be a closed linear operator in X and λ0 ∈ ρ(A). Then

σ(R(λ0, A)) \ 0 = 1

λ0 − z: z ∈ σ(A)

and 0 ∈ σ(R(λ0, A)) if and only if A 6∈ L(X).

Proof. We have

0 6∈ σ(R(λ0, A)) ⇔ R(λ0, A) : X → X surjective ⇔ D(A) = X ⇔ A ∈ L(X),

where we used the Closed Graph Theorem for the last equivalence. Now let z 6= λ0. Then

z−A = z−λ0+λ0−A = ((z−λ0)R(λ0, A)+I)(λ0−A) =(R(λ0, A)− 1

λ0 − z

)(z−λ0)(λ0−A),

where (z − λ0)(λ0 − A) : D(A) → X is bijective. Hence z − A : D(A) → X is bijectiveif and only if R(λ0, A) − (λ0 − z)−1 : X → X is bijective. We obtain that z ∈ σ(A) isequivalent to (λ0 − z)−1 ∈ σ(R(λ0, A)).

Remark: From the proof we see that, for λ0, z ∈ ρ(A) with z 6= λ0:( 1

λ0 − z−R(λ0, A)

)−1

= (λ0 − z)(λ0 − A)R(z, A).

What can be done if a linear operator A from X to Y is not closed? Of course, one canconsider the closure gr(A) in X × Y . But this need not be the graph of an operator.

15

8.16. Definition: A linear operator A from X to Y is called closable if it has a closedlinear extension B : X ⊇ D(B)→ Y . In this case, A has a smallest closed linear extensionA which is called the closure of A.

The following properties are equivalent (this will be an exercise):

• A is closable,

• gr(A) is the graph of a function,

• if (xn) is a sequence in D(A) such that xn → 0 in X and Axn → y in Y then y = 0.

In this case, one has gr(A) = gr(A). end Thu02.05.19

8.17. Proposition: Let A be a densely defined linear operator in X. If D(A′) is dense inX ′ then A is closable. The converse holds if X is reflexive.

Proof. If D(A′) is dense in X ′ then A′′ := (A′)′ is a closed linear operator in X ′′. We showthat A′′ is an extension of A: For x ∈ D(A), φ ∈ D(A′) we have

〈Ax, φ〉 = 〈x,A′φ〉 = 〈A′φ, δx〉,

which shows that δx ∈ D(A′′) and A′′δx = δAx for all x ∈ D(A).

Now let X be reflexive and assume that D(A′) is not dense in X ′. By Hahn-Banach wefind y ∈ X ′′ = X, y 6= 0, such that

〈y, φ〉 = 0 = 〈0, A′φ〉 for all φ ∈ D(A′).

We claim that (0, y) ∈ gr(A). If (ψ, φ) ∈ X ′×X ′ such that (ψ, φ)|gr(A) = 0. Then φ ∈ D(A′)and A′φ = −ψ. From the property of y above we get (ψ, φ)(0, y) = 0. By Hahn-Banach weconclude that (0, y) ∈ gr(A) as claimed. By y 6= 0, A is not closable.

16

9 Spectral projections and holomorphic functional

calculus

When nothing else is said, X denotes a complex Banach space and A is a closed linearoperator in X.

9.1. Spectral projections: Suppose that σ(A) = σ0∪σ1 where σ0∩σ1 = ∅, σ0 is compact,and σ1 is closed. Then we find finitely many closed piecewise C1-curves Γ in C \σ(A) suchthat

n(z,Γ) =

1 , z ∈ σ0

0 , z ∈ σ1,

where n(z,Γ) := 12πi

∫Γ

dζζ−z denotes the winding number. Let

P :=1

2πi

∫Γ

R(λ,A) dλ =1

2πi

∫ 1

0

R(γ(t), A)γ(t) dt

where γ : [0, 1]→ C is a parametrization of Γ and the integral is a Riemann integral (or aBochner integral).

The operator P has the following properties:

(a) P ∈ L(X) and P 2 = P , i.e. P is a projection.

(b) P commutes with A, i.e. for x ∈ D(A) we have Px ∈ D(A) and APx = PAx.Moreover, X0 := R(P ) ⊆ D(A).

(c) If we let A0 := A|X0 then A0 ∈ L(X0) and σ(A0) = σ0.

(d) Let X1 := N(P ) and define A1 as the restriction of A to D(A1) := D(A)∩X1. ThenA1 is a closed linear operator in X1 and σ(A1) = σ1.

This means that P induces a decomposition of the space X into the subspaces X0 and X1

which are complements of each other and invariant under A.

For the proof we shall need the following.

9.2. Holomorphic functions with values in a Banach space: Cauchy’s theorem andthe Cauchy integral formula hold for X-valued holomorphic functions f : Ω→ X:∫

Γ

f(z) dz = 0 and1

2πi

∫Γ

f(z)

z − z0

dz = f(z0)

if Γ is a finite family of closed piecewise C1-curves such that n(z,Γ) = 0 for all z 6∈ Ω andn(z0,Γ) = 1.

17

Proof of 9.1. (a) P ∈ L(X) is well defined since t 7→ R(γ(t), A)γ(t) is piecewise continuouswith values in L(X). By Cauchy’s theorem, Γ does not depend on the special choice of Γ,so we choose another family Γ′ such that n(λ,Γ′) = 1 for all λ ∈ Γ (Γ′ encircles all pointsof Γ once) and n(µ,Γ) = 0 for all µ ∈ Γ′. This means that Γ is “inside” Γ′. Then, by theresolvent equation and 9.2,

P 2 =( 1

2πi

)2∫

Γ

∫Γ′R(λ,A)R(µ,A) dµ dλ

=( 1

2πi

)2∫

Γ

∫Γ′

R(λ,A)−R(µ,A)

µ− λdµ dλ

=1

2πi

∫Γ

1

2πi

∫Γ′

1

µ− λdµ︸ ︷︷ ︸

=1

R(λ,A) dλ+( 1

2πi

)2∫

Γ′

∫Γ

R(µ,A)

µ− λdλ︸ ︷︷ ︸

=0

dµ

=1

2πi

∫Γ

R(λ,A) dλ = P.

(b) Actually, t 7→ R(γ(t), A)γ(t) is piecewise continuous with values in L(X, [D(A)]), soP ∈ L(X, [D(A)]), in particular R(P ) ⊆ D(A). We thus have

AP =1

2πi

∫Γ

AR(λ,A) dλ,

where t 7→ AR(γ(t), A)γ(t) is pieceweise continuous with values in L(X). For x ∈ D(A)we have AR(λ,A)x = R(λ,A)Ax, which implies APx = PAx.

(c) and (d): Since P ∈ L(X) is a projection,X0 = R(P ) is a closed subspace ofX. By (b) wehave A0 : X0 → X0. By closedness of A, this operator is closed, hence A0 ∈ L(X0). Clearly,P commutes with resolvents of A, which implies that X0 is invariant under resolvents ofA. By an exerxcise we thus have σ(A0) ⊆ σ(A), and R(λ,A0) = R(λ,A)|X0 for λ ∈ ρ(A).

On the other hand, for µ ∈ σ1 let

Rµ :=1

2πi

∫Γ

R(λ,A)

µ− λdλ.

Then Rµ|X0 ∈ L(X0), and by AR(λ,A) = λR(λ,A)− IX we have

(µ− A0)Rµ|X0 =1

2πi

∫Γ

(µ− A)R(λ,A)|X0

µ− λdλ = P |X0 −

1

2πi

∫Γ

IX0

µ− λdλ = IX0 .

Clearly, Rµ|X0 commutes with µ−A0, so also Rµ|X0(µ−A0) = IX0 . Hence µ ∈ ρ(A0) andR(µ,A0) = Rµ|X0 . We thus have shown σ(A0) ⊆ σ0.

Similarly, X1 is invariant under resolvents of A, and thus σ(A0) ⊆ σ(A). For µ ∈ σ0 let

Rµ :=1

2πi

∫Γ

R(λ,A)

µ− λdλ.

18

Then Rµ ∈ L(X, [D(A)]) commutes with resolvents of A, hence with P , and Rµ|X1 ∈L(X1). For x ∈ X1 we have

(µ− A)Rµx =1

2πi

∫Γ

R(λ,A)x dλ︸ ︷︷ ︸=Px=0

+1

2πi

∫Γ

x

µ− λdλ = −x.

Since Rµ commutes with A we obtain µ ∈ ρ(A1) and R(µ,A1) = −Rµ|X1 . This means thatwe have shown σ(A1) ⊆ σ1.

The decomposition X = X0⊕X1 leads to the decomposition of A as(A0

00A1

)with domain

X0 ⊕D(A1). It is thus clear that σ(A) = σ(A0) ∪ σ(A1) (this union is closed since σ(A0)is compact). Since σ(Aj) ⊆ σj (j = 0, 1) and σ0 ∩ σ1 = ∅ this is only possible if σ(Aj) = σjfor j = 0, 1.

Example: Let X = Cn and A ∈ Cn×n, and let λ0 ∈ σp(A0), σ0 = λ0, σ1 = σ(A) \ λ0.Then X0 equals the generalized eigenspace

⋃nk=1N

((λ0 − A)k

)and

X1 =⊕λ∈σ1

n⋃k=1

N((λ− A)k

).

end Tue07.05.19

9.3. The Dunford functional calculus: Let A be bounded in X. Then σ(A) is compact.

(1) Let U ⊆ C be an open neighborhood of σ(A). Then we find a finite family of closedpiecewise C1-curves Γ such that n(z,Γ) = 1 if z ∈ σ(A) and n(z,Γ) = 0 for z 6∈ U . Letf : U → C be a holomorphic function. Define

f(A) :=1

2πi

∫Γ

f(λ)R(λ,A) dλ.

Then f(A) ∈ L(X) and the definition does not depend on the special choice of Γ.

(2) The map Ψ : f 7→ f(A) is an algebra homomorphism, and

pn(A) = An for all n ∈ N0, where pn(λ) := λn,

i.e. Ψ is a functional calculus for the operator A for functions that are holomorphic in aneighborhood of σ(A).

Proof. (1) is clear. (2): Linearity is clear. The proof of multiplicativity is similar to whatwe have done in the proof of 9.1(a). This implies

pn(A) =1

2πi

∫Γ

λnR(λ,A), dλ = p1(A)n, n ∈ N0.

Moreover, Ap0(A) = p1(A)p0(A). Thus it rests to show p0(A) = IX . Apply 9.1 with σ0 =σ(A) and σ1 = ∅. Then A1 ∈ L(X1) (since A ∈ L(X)), and σ(A1) = σ1 = ∅. HenceX1 = 0, X0 = X, and p0(A) = P = IX .

19

9.4. Remark: Let A ∈ L(X) and U be an open neighborhood of B(0, r(A)) ⊇ σ(A). Inthis case we can choose Γ = ∂B(0, r) for a suitable r > r(A). By 8.9 and its proof we know

R(λ,A) =∞∑k=0

Ak

λk+1, |λ| > r(A).

If f : U → C is holomorphic then we obtain

f(A) =1

2πi

∫Γ

f(λ)R(λ,A) dλ =1

2πi

∫Γ

∞∑k=0

Ak

λk+1dλ =

∞∑k=0

fk(0)

k!Ak,

where we interchanged summation and integration by uniform convergence and usedCauchy’s integral formula. This means that we get f(A) for such f by inserting A forz in the power series expansion of f around 0. Of course, then pn(A) = An for n ∈ N0 isclear.

20

10 Fourier transform on Rd

10.1. Definition: For f ∈ L1(Rd) and ξ ∈ Rd we define

f(ξ) :=

∫Rde−2πixξf(x) dx,

where xξ = x1ξ1 + x2ξ2 + . . .+ xdξd is the scalar product in Rd. The function f : Rd → Cis called the Fourier transform of f .

10.2. Lemma: (a) For f ∈ L1(Rd) the function f is bounded and uniformly continuousand

‖f‖∞ ≤ ‖f‖1.

The map f → f is linear and continuous L1(Rd)→ L∞(Rd).

(b) Let f ∈ L1(Rd) and y ∈ Rd. For τyf := f(· − y) we have τyf(ξ) = e−2πiyξf(ξ), ξ ∈ Rd.

(c) For f, g ∈ L1(Rd) and λ > 0 we have∫f(ξ)g(λξ) dξ =

∫f(λx)g(x) dx.

Proof. (a) The estimate is clear. For ξ, η ∈ Rd we have

|f(ξ)− f(ξ + η)| ≤∫|e−2πixξ − e−2πix(ξ+η)| |f(x)| dx

=

∫|1− e−2πixη| |f(x)| dx indep. of ξ.

By 1 − e−2πixη → 0 (η → 0) for fixed x ∈ Rd and |1 − e−2πixη| ≤ 2 the integral convergesto 0 as η → 0 by dominated convergence. Hence f is uniformly continuous.

(b) We substitute x = u+ y and obtain∫e−2πixξf(x− y) dx = e−2πiyξ

∫e−2πiuξf(u) du = e−2πiyξf(ξ).

(c) We substitute x = λy and λξ = η and obtain by Fubini∫ ∫e−2πixξf(x)g(λξ) dx dξ =

∫f(λy)

∫e−2πiyηg(η) dη︸ ︷︷ ︸

=g(y)

dy.

10.3. Example: Let d = 1 and h = 1[a,b]. Then h(0) = b− a and for ξ 6= 0 we have

h(ξ) =

∫ b

a

e−2πixξ dx =e−2πibξ − e−2πiaξ

−2πiξ.

21

Hence h(ξ)→ 0 (|ξ| → ∞), but h 6∈ L1(R)!

For d > 1 and h(x) = 1∏nj=1[aj ,bj ](x) =

∏nj=1 1[aj ,bj ](xj) we obtain

h(ξ) =n∏j=1

1[aj ,bj ](ξj), ξ = (ξ1, . . . , ξn) ∈ Rd.

Hence also here we have h(ξ)→ 0 fur |ξ| → ∞.

10.4. Riemann-Lebesgue-Lemma: For each f ∈ L1(Rd) we have f(ξ)→ 0 fur |ξ| → ∞.

Proof. The space

C0(Rd) := f : Rd → C stetig : f(x)→ 0 (|x| → ∞)

is a closed linear subspace w.r.t. the norm ‖ · ‖∞ of the Banach space

BUC(Rd) := g : Rd → C : g is bounded and uniformly continuous .

By Example 10.3 we have h ∈ C0(Rd) for functions h of the form h =∑

k ck1Qk , where theQk are cartesian products of compact intervals. The set of these simple functions is densein L1(Rd), and we employ 10.2 (a).

10.5. Rules: Let f ∈ L1(Rd). Then the following hold:

(a) If a ∈ R \ 0 then F (F (a·))(ξ) = 1|a|n f( ξ

a), ξ ∈ Rd.

(b) If b ∈ Rd then F (e2πib(·)f)(ξ) = f(ξ − b), ξ ∈ Rd.

(c) If in addition x 7→ xjf(x) ∈ L1(Rd) then f has a continuous partial derivative w.r.t ξjand

∂

∂ξjf(ξ) = F (x 7→ (−2πixj)f(x))(ξ), ξ ∈ Rd.

Proof. The proof of (a) is similar to the one for 10.2(c). For the proof of (b) we write∫e−2πiξxe2πibx︸ ︷︷ ︸

=e−2πi(ξ−b)x

f(x) dx = f(ξ − b).

ad (c): For ξ ∈ Rd, j ∈ 1, . . . , d and h 6= 0 we have

f(ξ + hej)− f(ξ)

h−F (x 7→ (−2πixj)f(x))(ξ)

=

∫ (e−2πi(ξ+hej)x − e−2πiξx

h+ 2πixje

−2πiξx)f(x) dx.

22

For fixed x ∈ Rd the term in (. . .) tends to 0 as h→ 0. Moreover we have

e−2πiξx(e−2πihxj − 1

h+ 2πixj

)= e−2πiξx

(1

h

∫ h

0

(−2πixj)e−2πitxj dt+ 2πixj

)= e−2πiξx2πixj

(1− 1

h

∫ h

0

e−2πitxj dt).

Here the last term in (. . .) is ≤ 2 in modulus. The assertion follows by dominated conver-gence.

end Thu09.05.19

10.6. Proposition (weak derivatives and Fourier transform): Let f ∈ L1(Rd) andlet g ∈ L1(Rd) such that ∫

f(x)∂jϕ(x) dx = −∫g(x)ϕ(x) dx

for all ϕ ∈ C∞c (Rd), i.e. f has weak partial derivative g = ∂jf . Then

g(ξ) = 2πiξj f(ξ), ξ ∈ Rd.

Remark: The assumption is in particular satisfied if f is continuously differentiable withg := ∂jf ∈ L1(Rd).

Proof. Let ξ ∈ Rd. The idea is to take ϕ(x) = e−2πixξ, but ϕ 6∈ C∞c . Thus we approximatethis function. we choose ψ ∈ C∞c with 0 ≤ ψ ≤ 1 and ψ(x) = 1 for |x| ≤ 1 and setψk(x) := ψ(x/k) and ϕk(x) := e−2πixξψk(x). Dominated convergence yields∫

gϕk dx =

∫g(x)e−2πixξψk(x) dx→ g(ξ)

since ψk(x)→ 1 for each x ∈ Rd and |ψk| ≤ 1. Moreover we have, by ϕk ∈ C∞c ,∫gϕk dx = −

∫f∂jϕk dx = −

∫f(−2πiξj)ϕk dx︸ ︷︷ ︸→2πiξj f(ξ)

−∫f(x)e−2πixξ∂jψk(x) dx︸ ︷︷ ︸

=:A(k)

.

Here ∂jψk(x) = 1k(∂jψ)(x/k) and thus

|A(k)| ≤∫|x|≥k|f(x)| dx · 1

k· ‖∂jψ‖∞ → 0 (k →∞),

since ∂jψ is bounded.

23

10.7. Example: For φ(x) = e−π|x|2, x ∈ Rd, we have

φ(ξ) = e−π|ξ|2

, ξ ∈ Rd.

By φ(x) =∏n

j=1 e−π|xj |2 it suffices to study the case d = 1. Then we have for each x ∈ R:

φ′(x) = −2πxφ(x).

By 10.5(c) and 10.6 we obtain for each ξ ∈ R:

2πiξφ(ξ) = φ′(ξ) = −2πxφ(x)(ξ) = −i ddξφ(ξ),

hence(φ)′(ξ) = −2πξφ(ξ).

This means that φ and φ solve the same homogeneous linear differential equation. Byuniqueness of solutions of the initial value problem we thus have

φ(ξ) = φ(0)φ(ξ) = e−π|ξ|2

, ξ ∈ R,

since φ(0) =∫R e−π|x|2 dx = 1.

10.8. Fourier inversion formula: Let f ∈ L1(Rd)∩Cb(Rd) such that f ∈ L1(Rd). Thenwe have for any x ∈ Rd:

f(x) =

∫Rde2πixξf(ξ) dξ.

Remark: Actually, the formula holds in an a.e.-sense for all f ∈ L1(Rd) with f ∈ L1(Rd).(→ below).

Proof. By 10.2(b) it suffices to study x = 0. Let h(x) := e−π|x|2, x ∈ Rd. By 10.2(c) we

have for any a > 0: ∫Rdf(ξ)h(aξ) dξ =

∫Rdf(ax)h(x) dx.

As a → 0+, the left hand side tends to∫f(ξ) dξ (use dominated convergence and the

assumption f ∈ L1). Since f is continuous at 0, the integrand in the right hand sideconverges pointwise to f(0)h(x). Since f is bounded, dominated convergence yields thatthe right hand side tends to f(0)

∫h(x) dx = f(0).

24

10.9. Example: f(x) = e−a|x| for x ∈ R where a > 0. We have

f(ξ) =

∫ ∞−∞

e−2πiξxe−a|x| dx

and∫ ∞0

e−2πixξe−ax dx =1

a+ 2πiξ,

∫ 0

−∞e−2πixξeax dx =

∫ ∞0

e2πiyξe−ay dx =1

a− 2πiξ.

Hence

f(ξ) =1

a+ 2πiξ+

1

a− 2πiξ=

2a

a2 + 4π2ξ2, ξ ∈ R.

The functions f is bounded and continuous and f ∈ L1(R). Hence we have by 10.8:

F(x 7→ 2a

a2 + 4π2x2

)(ξ) = e−a|ξ|, ξ ∈ R.

10.10. Definition and Proposition (Convolution): Let f, g ∈ L1(Rd). For a.e. x ∈ Rd

we have y 7→ f(y)g(x− y) ∈ L1(Rd), and for the function h, given by

h(x) :=

∫Rdf(y)g(x− y) dy for a.e. x ∈ Rd,

we have h ∈ L1(Rd) and‖h‖1 ≤ ‖f‖1‖g‖1.

The function h is called convolution of f and g and is denoted by f ∗ g.

Remark: In FA 5.2 we defined the convolution of L1loc-functions with C∞c -functions. It is

clear that this coincides with the convolution defined here if the L1loc-function is actually

in L1.

Proof. The function F : (x, y) 7→ f(y)g(x− y) is measurable and∫ ∫|f(y)g(x− y)| dx dy =

∫|f(y)|

∫|g(x− y)| dx︸ ︷︷ ︸

=‖g‖1

dy = ‖f‖1‖g‖1.

By Fubini-Tonelli we thus have F ∈ L1(Rd × Rd), F (x, ·) ∈ L1(Rd) for a.e. x ∈ Rd, andx 7→

∫F (x, y) dy ∈ L1(Rd).

10.11. Lemma: The convolution ∗ : L1(Rd)×L1(Rd)→ L1(Rd) is bilinear and continuous,commutative, and associative. For f, g ∈ L1(Rd) and ξ ∈ Rd we have

f ∗ g(ξ) = f(ξ)g(ξ).

25

Proof. Bilinearity and continuity are clear by 10.10. Commutativity and associativenessfollow by Fubini. We show the formula. By Fubini we have∫

e−2πixξf ∗ g(x) dx =

∫ ∫e−2πixξf(y)g(x− y) dx dy

=

∫e−2πiyξf(y)

∫e−2πi(x−y)ξg(x− y) dx︸ ︷︷ ︸

=g(ξ)

dy = f(ξ)g(ξ).

10.12. Lemma: For f, f ∈ L1(Rd) we have the Fourier inversion formula 10.8. In partic-ular, we then have f, f ∈ C0(Rd).

Proof. is an exercise.

10.13. Theorem (Plancherel): For f ∈ L1(Rd) with f ∈ L1(Rd) we have f, f ∈ L2(Rd)and

‖f‖22 = ‖f‖2

2.

Proof. By ‖g‖22 ≤ ‖g‖∞‖g‖1 we get from 10.12: f, f ∈ L2(Rd). By 10.2 (dancing hat) we

have ∫Rn|ψ(ξ)|2 dξ =

∫Rnψ(ξ)ψ(ξ) dξ =

∫Rnψ(x)(ψ)∧(x) dx,

and by 10.8 we get

(ψ)∧(x) =

∫Rne−2πiξxψ(ξ) dξ =

∫Rne2πiξxψ(ξ) dξ = ψ(x).

end Tue14.05.19

10.14. Theorem: Let S := f ∈ L1(Rd) : f ∈ L1(Rd). The Fourier transform S → S,f 7→ f , has a unique continuous extension F : L2(Rn) → L2(Rn). This extension F islinear and bijective, and we have

‖Ff‖22 = ‖f‖2

2, f ∈ L2(Rd),

andF−1f = σFf, f ∈ L2(Rd), F 4 = IdL2 .

Moreover, we have for all f, g ∈ L2(Rd):

(f |g)L2 = (f |g)L2 , i.e.

∫Rdf(x)g(x) dx =

∫Rdf(ξ)g(ξ) dξ.

26

Proof. We only have to show that S is dense in L2(Rd). To see this, it suffices to approxi-mate f ∈ L2(Rn) with compact support by functions in S, which can be done with molli-fiers: Let φ(x) = e−π|x|

2, φk(x) = kdφ(k·), and fk := f ∗ φk. Then fk ∈ L1, fk = f φk ∈ L1

(since f ∈ BUC) and fk → f in ‖ · ‖L2 (this is shown as in FA).

10.15. Definition: The Schwartz space S (Rd) is given by

S (Rd) := ψ : Rd → C : ψ is C∞ and ∀α, β ∈ Nd0 : x 7→ xβ∂αψ(x) bounded ,

where we use multi-index notation: For α = (α1, α2, . . . , αd), β = (β1, β2, . . . , βd) ∈ Nd0 we

have

∂α :=∂α1

∂xα11

∂α2

∂xα22

. . .∂αd

∂xαdd,

|α| := α1 + α2 + . . .+ αd,

α! := α1!α2! · . . . · αd!,β ≤ α :⇔ β1 ≤ α1, β2 ≤ α2, . . . , βd ≤ αd,(α

β

):=

α!

β!(α− β)!=

(α1

β1

)(α2

β2

)· . . . ·

(αdβd

), β ≤ α,

xα := xα11 x

α22 · . . . · x

αdd , x = (x1, x2, . . . , xd) ∈ Cd.

Remark: We recall the Leibniz rule for products

∂α(ϕ · ψ) =∑γ≤α

(α

γ

)∂γϕ∂α−γψ,

which can be hown by induction on |α|.

Remark: For ψ ∈ S (Rd) and any α ∈ Nd0 we have that ∂αψ and x 7→ xαψ(x) belong to

S (Rd).

10.16. Proposition: For ϕ, ψ ∈ S (Rd) we have ϕ · ψ ∈ S (Rd).

Proof. Use the Leibniz rule

xβ∂α(ϕ · ψ) =∑γ≤α

(α

γ

)xβ∂γϕ︸ ︷︷ ︸

bounded

∂α−γψ︸ ︷︷ ︸bounded

.

10.17. Theorem: The Fourier transform ψ 7→ ψ is linear and bijective S (Rd)→ S (Rd).

Proof. is an exercise.

27

10.18. Proposition: For ϕ, ψ ∈ S (Rd) we have ϕ ∗ ψ ∈ S (Rd).

Proof. By ϕ ∗ ψ = ϕ · ψ this follows from 10.17 and 10.16.

10.19. Sobolev spaces via Fourier transform: First we show that, for any f ∈ L2(Rd)and j ∈ 1, . . . , d, we have

f has a partial weak derivative ∂jf ∈ L2(Rd) ⇐⇒ ξ 7→ ξj f(ξ) ∈ L2(Rd),

and in this case ∂jf(ξ) = 2πiξj f(ξ) for a.e. ξ ∈ Rd. This is an L2-version of 10.5(c) and10.6.

Proof. Let f have partial weak derivative ∂jf ∈ L2(Rd). For ϕ ∈ C∞c (Rd) we then have by10.14 and 10.5(c) ∫

(∂jf)ϕdx = −∫f ∂jϕdx = 2πi

∫f ξjϕ dξ.

This implies ∣∣∣ ∫ f ξjϕ dξ∣∣∣ ≤ 1

2π‖∂jf‖L2‖ϕ‖L2 ,

which in turn yields ξj f ∈ L2(Rd) (we again use Plancherel here). If, on the other hand,

ξj f ∈ L2(Rd) then we have g := F−1(2πiξj f(ξ)) ∈ L2(Rd) and obtain, for any ϕ ∈ C∞c ,∫gϕ dx =

∫gϕ dξ = 2πi

∫f ξjϕ dξ = −

∫f ∂jϕdξ = −

∫f(∂jϕ) dx.

But this means g = ∂jf as a weak derivative. Observe g(ξ) = 2πiξj f(ξ).

For s ≥ 0 we define

Hs(Rd) := f ∈ L2(Rd) : ξ 7→ (1 + 4π2|ξ|2)s/2f(ξ) ∈ L2(Rd)

and

‖f‖Hs := ‖ξ 7→ (1 + 4π2|ξ|2)s/2f(ξ)‖L2 =(∫

Rd(1 + 4π2|ξ|2)s|f(ξ)|2 dξ

)1/2

.

Then Hs(Rd) is a Hilbert space, where the scalar product is given by

(f |g)Hs =

∫Rd

(1 + 4π2|ξ|2)sf(ξ) g(ξ) dξ.

We claim that Hm(Rd) = Wm,2(Rd) for any m ∈ N with equivalent norms. This is provedvia the following observation which follows from Plancherel:

28

Let m : Rd → C be measurable. Then Tmf := F−1(mf) defines a boundedlinear operator on L2(Rd) if and only if m ∈ L∞(Ω).

Proof. Let m ∈ N. Recall

Wm,2(Rd) := f ∈ L2(Rd) : ∀α ∈ Nd0 : ∂αf ∈ L2(Rd)

= f ∈ Wm−1,2(Rd) : ∀j ∈ 1, . . . , d : ∂jf ∈ Wm−1,2(Rd)

It thus suffices to show H1 = W 1,2 (then one can use a simple induction). If f ∈ H1 thenalso

ξ 7→ ξj f(ξ) =ξj

(1 + 4π2|ξ|2)1/2︸ ︷︷ ︸bounded

(1 + 4π2|ξ|2)1/2f(ξ) ∈ L2(Rd).

If ∂jf ∈ L2 for all j then ξj f ∈ L2 for all j (as we have seen above). By

(1 + |ξ|2)1/2 ≤ 1 + |ξ| ≤ 1 +∑j

|ξj|

and the observation we conclude that f ∈ H1. The estimates also show equivalence ofnorms.

10.20. The Laplace operator in L2(Rd): We define the Laplace operator ∆ on L2(Rd)by ∆f :=

∑dj=1 ∂

2j f for f ∈ D(∆) := W 2,2(Rd) = H2(Rd).

Then −∆ = F−1MmF where Mm is the multiplication operator given by Mmf := mfwith m(ξ) = 4π2|ξ|2, which has domain D(Mm) = f ∈ L2(Rd) : mf ∈ L2(Rd) (here weuse 10.19). Since the Fourier transform is a unitary operator on L2(Rd) this means that ∆is unitarily equivalent to a multiplication operator.

Since Mm is closed, we conclude by 8.4 that −∆ is closed. But we also obtain that σ(−∆) =[0,∞) and that

R(λ,−∆)f = (λ+ ∆)−1f = F−1(ξ 7→ (λ− 4π2|ξ|2)−1︸ ︷︷ ︸

∈L∞

f(ξ)), f ∈ L2(Rd),

for any λ ∈ C \ [0,∞).

Moreover, we can easily define a functional calculus for −∆ by

L∞(0,∞)→ L(L2(Rd)), F 7→ F (−∆) := TFm = F−1MFmF .

i.e. by lettingF (−∆)f = F−1

(ξ 7→ F (4π2|ξ|2)f(ξ)

).

29

This is an algebra homomorphism, i.e. F 7→ F (−∆) is linear and (FG(−∆) =F (−∆)G(−∆), and we have

‖F (−∆)‖L(L2(Rd)) = ‖ξ 7→ F (4π2|ξ|2)‖L∞(Rd) = ‖F‖L∞(0,∞).

We observe that F (−∆) = I for F (z) = 1 and R(λ,−∆) = Fλ(−∆) where Fλ(z) =(λ− z)−1. We also observe that F (−∆) is self-adjoint if F is real-valued:

(F (−∆)f |g) = (F m · f |g) = (f |F m · g) = (f |F (−∆)g).

The functional calculus for −∆ is but a special case of the general spectral theorem forunbounded self-adjoint operators that we shall study lateron.

end Thu16.05.19

10.21. Theorem (Sobolev embedding): For s > d/2 and γ ∈ (0, s − d/2) ∩ (0, 1) thespace Hs(Rd) is continuously embedded into the space of Holder-continuous functions

Cγ(Rd) = f ∈ C(Rd) : ‖f‖∞ <∞, supx 6=y

|f(x)− f(y)||x− y|γ

<∞.

Remark: The space Cδ(Rd) is a Banach space w.r.t. the norm

‖f‖Cδ := ‖f‖∞ + supx6=y

|f(x)− f(y)||x− y|δ

,

where δ ∈ (0, 1].

Proof. For f ∈ Hs(Rd) we write

Ff(ξ) = (1 + 4π2|ξ|2)−s/2︸ ︷︷ ︸∈L2

(1 + 4π2|ξ|2)s/2Ff(ξ)︸ ︷︷ ︸∈L2

∈ L1,

from which we get by Fourier inversion that f ∈ C0(Rd) and that

‖f‖∞ ≤ C1‖f‖Hs(Rd).

Moreover, we have |ξ|γFf(ξ) ∈ L1 and

‖ξ 7→ |ξ|γFf(ξ)‖L1(Rd) ≤ C2‖f‖Hs(Rd).

Now we write

f(x)− f(y) =

∫Rd

(e2πix·ξ − e2πiy·ξ)F (ξ) dξ

and obtain by

|e2πix·ξ − e2πiy·ξ| ≤ min2, c|x− y||ξ| ≤ C|x− y|γ|ξ|γ,

that|f(x)− f(y)| ≤ C ′|x− y|γ ‖| · |γFf‖L1︸ ︷︷ ︸

≤C2‖f‖Hs

.

We thus have proved the continuous embedding Hs(Rd) ⊆ Cγ(Rd) for γ ∈ (0,min1, s −d/2).

30

10.22. Theorem (Hausdorff-Young): Let p ∈ [1, 2] and let p′ be the dual exponent,given by 1

p+ 1p′

= 1. Then the Fourier transform F is a bounded operator Lp(Rd)→ Lp′(Rd).

Proof. By 10.2(a) the assertion holds for p = 1. By 10.14 it holds for p = 2. We obtain itfor p ∈ (1, 2) by Riesz-Thorin interpolation (→ FA).

31

11 Fredholm operators and the spectral theory of

compact operators

Again, X and Y are Banach spaces if nothing else is said.

Recall from Linear Algebra: If dimX = dim Y <∞ and T ∈ L(X, Y ) then

dimN(T ) = codimR(T )︸ ︷︷ ︸=:dimY/R(T )

<∞.

11.1. Definition: An operator T ∈ L(X, Y ) is called a Fredholm operator if

α(T ) := dimN(T ) <∞ and R(T ) is closed with β(T ) := codimR(T ) <∞.

In this case, the index of T is defined by

indT := α(T )− β(T ).

The set of all Fredholm operators from X to Y is denoted by Φ(X, Y ).

Remark: If dimX = dimY < ∞, any T ∈ L(X, Y ) is a Fredholm operator of index0. If X and Y are arbitrary and T ∈ L(X, Y ) is an isomorphism from X to Y thenα(T ) = β(T ) = 0 and T is a Fredholm operator of index 0.

The following proposition shows that closedness of R(T ) can be omitted from the definition.

11.2. Proposition: Let T ∈ L(X, Y ). If codimR(T ) <∞ then R(T ) is closed.

Proof. We find a basis [y1], . . . , [yn] of Y/R(T ) and let H := span y1, . . . , yn. Then Y =R(T ) + H and R(T ) ∩H = 0. Since H is complete, X ×H is a Banach space (e.g. forthe sum-norm) and the map

T : X ×H → Y, (x, y) 7→ Tx+ y

is continuous and surjective. By the Open Mapping Theorem we have

γ := inf ‖T (x, y)‖Yd((x, y), N(T ))

: (x, y) 6∈ N(T )> 0.

Since

N(T ) = (x, y) ∈ X ×H : Tx = −y = (x, 0) : Tx = 0 = N(T )× 0

we have for x 6∈ N(T ):

‖Tx‖Y = ‖T (x, 0)‖Y ≥ γd((x, 0), N(T )) = γd(x,N(T )).

From this we obtain closedness of R(T ).

32

Remark: We recall that, if Y is a Banach space and Z is a closed linear subspace, thenthe quotient space Y/Z is a Banach space defined by

Y/Z := y + Z : y ∈ Y , ‖y + Z‖Y/Z := inf‖y − z‖Y : z ∈ Z = d(y, Z).

If Z is clear from the context, it is common to write [y] instead of y + Z and q for thequotient map Y → Y/Z, y 7→ [y]. We also recall that an operator T ∈ L(X, Y ) has aunique factorization T = S q where S ∈ L(X/N(T ), Y ) is injective and q : X → X/N(T )is the quotient map. S is given by S[x] = Tx.

Remark: (a) Suppose that W and Z are subspaces of a Banach space X such that W∩Z =0 and X = W + Z. We write X = W ⊕ Z if both W and Z are closed subspaces of X.Notice that X = W ⊕ Z if and only if W × Z → X, (w, z) 7→ w + z, is an isomorphism ifand only if there exists a projection P ∈ L(X) such that P (X) = Z and N(P ) = W .

A closed subspace Z is called complemented in X if there is a (closed) subspace W of Xsuch that X = W ⊕ Z.

(b) Any finite-dimensional subspace Z of X is complemented in X: Choose a basis z1, . . . , znof Z and linear functionals ψ1, . . . , ψn ∈ Z ′ such that ψj(xk) = δjk. Extend the ψk by Hahn-Banach and obtain φ1, . . . , φn ∈ X ′. Then W :=

⋂nk=1 N(φk) is a complement of Z in X.

11.3. Theorem: Φ(X, Y ) is open in L(X, Y ) and ind : Φ(X, Y )→ Z is continuous.

Proof. Let T ∈ Φ(X, Y ). We shall show the existence of ε > 0 such that, for S ∈ L(X, Y ),we have

‖S − T‖ < ε =⇒ S ∈ Φ(X, Y ) and indS = indT.

First we find closed subspaces G of X and H of Y such that

X = N(T )⊕G, Y = R(T )⊕H, dimH <∞.

Notice that R(T ) = T (X) = T (G). For any S ∈ L(X, Y ) we define

S : G×H → Y, (g, h) 7→ Sg + h.

Then we have, for S1, S2 ∈ L(X, Y ):

‖S1 − S2‖ = sup‖(g,h)‖=1

‖S1g + h− (S2g + h)‖ ≤ ‖S1 − S2‖.

We claim that T : G×H → Y is an isomorphism. Indeed, T is surjective by R(T ) = T (G)

and the choice of H, and T is injective, since T (g, h) = 0 implies Tg = −h ∈ R(T ) ∩H =

0, and T |G is injective. Hence we find ε > 0 such that, for S ∈ L(G×H,Y ), ‖S− T‖ < ε end Tue21.05.19implies that S : G×H → Y is an isomorphism.

Now let S ∈ L(X, Y ) with ‖S − T‖ < ε. Then ‖S − T‖ < ε, and S : G × H → Y is anisomorphism. We now show

33

(i) α(S) ≤ α(T ) <∞,

(ii) β(S) ≤ β(T ) <∞,

(iii) ind (S) = ind (T ).

(i): If g ∈ N(S) ∩G then S(g, 0) = 0, and g = 0. Hence N(S) ∩G = 0 and

α(S) = dimN(S) ≤ codimG = dimN(T ) = α(T ) <∞.

(ii): G×0 is closed in G×H, so S(G) = S(G×0) is closed in Y . Since S is surjective

we have Y = S(G) + H. If y = S(g) ∈ S(G) ∩ H then S(g,−y) = S(g) − S(g) = 0 and

g = 0, y = 0, since S is injective. Hence S(G)⊕H = Y = T (G)⊕H and

β(S) = codimR(S) ≤ codimS(G) = codimT (G) = codimR(T ) = β(T ) <∞.

(iii): By what we have already shown we know that N(S) ⊕ G is a closed subspace ofX of finite codimension. We thus find a subspace W of X with dimW < ∞ such thatX = W ⊕N(S)⊕G. We then have

α(T ) = dimN(T ) = dimN(S)⊕ dimW = α(S) + dimW.

On the other hand, we have

R(S) = S(X) = S(G⊕W ) = S(G)⊕ S(W )

since both subspaces on the right hand side are closed and their intersection is trivial (byinjectivity of S|G⊕W ). Hence

β(T ) = codimT (X)(ii)= codimS(G) = codimS(X) + dimS(W ) = β(S) + dimW,

and we conclude that

indT = α(T )− β(T ) = α(S) + dimW − (β(S) + dimW ) = indS,

which ends the proof.

11.4. Compact Operators: Recall (cf. FA) that a linear operator T : X → Y is calledcompact if T (BX) is compact in Y where BX := x ∈ X : ‖x‖ ≤ 1 denotes the closed unitball in X. The set of all compact linear operators from X to Y is denoted by K(X, Y ). Werecall the following properties (→ FA):

• Any compact linear operator T : X → Y is bounded, i.e. K(X, Y ) ⊆ L(X, Y ).

• dimT (X) <∞ =⇒ T is compact.

34

• IX is compact ⇐⇒ dimX <∞.

• K(X, Y ) is a closed linear subspace of L(X, Y ).

• If W , Z are Banach spaces and S ∈ L(W,X), R ∈ L(X, Y ) and T ∈ K(X, Y ) thenRT ∈ K(X,Z) and TS ∈ K(W,Y ) (ideal property).

• Schauder’s Theorem: For any T ∈ L(X, Y ): T ∈ K(X;Y ) ⇐⇒ T ′ ∈ K(Y ′, X ′).

11.5. Proposition: Let K ∈ K(X). Then I −K ∈ Φ(X) and ind (I −K) = 0.

Proof. On N := N(I − K) we have IN = K|N ∈ K(N) and thus dimN < ∞. We letT := I − K and show that R(T ) is closed. To this end we factorize T = S q whereq : X → X/N is the quotient map and S : X/N → X. Then R(T ) = R(S) and S isinjective. We claim that there exists η > 0 such that ‖Sw‖ ≥ η‖w‖ for all w ∈ X/N . Fromthis, closedness of R(S) = R(T ) follows.

Assume that such an η does not exist. Then we find a sequence (wn) such that ‖wn‖ = 1 andSwn → 0, and a sequence (xn) in X such that q(xn) = wn, ‖xn‖ ≤ 2. Then Txn = Swn → 0and d(xn, N) = ‖q(xn)‖ = 1.

Since K is compact we have Kxk(n) → y ∈ X for a subsequence, which implies xk(n) =Txk(n) + Kxk(n) → y and Swk(n) = Txk(n) → Ty. So Ty = 0 and y ∈ N . However,d(y,N) = 1, a contradiction.

By Schauder’s Theorem (cf. above) K ′ ∈ K(X ′), hence dimN((I − K)′) < ∞. SinceR(I −K) is closed we obtain

codimR(I −K) = dimX/R(I −K) = dim (X/R(I −K))′

= dim φ ∈ X ′ : φ|R(I−K) = 0 = dim (N(I −K)′) <∞.

It remains to show ind (I − K) = 0. We consider γ : [0, 1] → Z, t 7→ ind (I − tK) = 0(observe that tK ∈ K(X) and hence I−tK ∈ Φ(X) for any t ∈ R). By 11.3, γ is continuous,hence constant and ind (I −K) = γ(1) = γ(0) = ind I = 0.

11.6. Corollary: Let K ∈ K(X) and dimX =∞. Then 0 ∈ σ(K) and for λ ∈ σ(K)\0one has that λ−K is a Fredholm operator of index 0. For λ ∈ C\0 one has in particular

λ−K is injective ⇐⇒ λ−K is surjective.

and, more precisely, the following Fredholm alternative:

Either (λ −K)x = 0 has just the trivial solution; in this case (λ −K)x = y has a uniquesolution for any y ∈ X,

or (λ − K)x = 0 has exactly n ∈ N linearly independent solutions and also the dualequation (λ − K ′)φ = 0 has exactly n linearly independent solutions; in this case(λ−K)x = y has solutions if and only if φ(y) = 0 for any φ ∈ N(λ−K ′).

35

11.7. Corollary: Let T ∈ L(X, Y ) be an isomorphism from X to Y and K ∈ K(X, Y ).Then T −K ∈ Φ(X, Y ) and ind (T −K) = 0.

Proof. We have T−1K ∈ K(X), hence by Proposition 11.5, T−1(T −K) = IX − T−1K ∈Φ(X) and indT−1(T −K) = 0. Since T−1 is an isomorphism from Y to X we obtain theassertion.

The assertion in 11.7 is the keystone for the following powerful perturbation property ofFredholm operators.

11.8. Theorem: Let T ∈ Φ(X, Y ) and K ∈ K(X, Y ). Then T + K ∈ Φ(X, Y ) andind (T +K) = indT .

end Thu23.05.19

Proof. We take up the construction and notation of the proof of 11.3 and write

X = N(T )⊕G, Y = R(T )⊕H, dimH <∞.

where G is a closed subspace of X. We let S := T −K and consider the operators T , S :G×H → Y . Again, T is an isomorphism, and we have S = T − K where K(g, h) := Kg.

The operator K is compact, thus by 11.7 we have S ∈ Φ(G × H,Y ) and ind S = 0, i.e.

α(S) = β(S). Clearly,

(N(S) ∩G)× 0 = (g, 0) ∈ G×H : Sg = 0 ⊆ N(S),

which implies

α(S) = dimN(S) ≤ codimG+ α(S) = α(T ) + α(S) <∞.

Moreover, we have S(G × H) = S(G) + H, and S(G) has finite codimension and is thusclosed by 11.2. We obtain

β(S) = codimS(X) ≤ codimS(G) ≤ codim S(G×H) + dimH = β(S) + β(T ) <∞,

and S(X) is closed by 11.2.

For the calculation of the index we have to look closer. We let G0 := N(S) ∩ G, find acomplement G1 of G0 in G, a complement W0 of G0 in N(S), and a complement W1 ofN(S) +G = W0 ⊕G0 ⊕G1 in X. Then we have

α(T ) = dimW1 + dimW0,

β(T ) = dimH,

α(S) = dimW0 + dimG0,

β(S) = codimS(X) = codimS(G1)− dimW1,

36

and

β(S) = codim (S(G) +H) = codimS(G)− dimH + dim (S(G) ∩H)

= codimS(G)− dimH + dim (S(G1) ∩H).

Now

dim (S(G1) ∩H) = dim g ∈ G1 : Sg ∈ H = dim (g, h) ∈ G1 ×H : Sg = −h︸ ︷︷ ︸=:V

,

and V is a complement of G0 × 0 in N(S). We thus obtain

α(S) = dimN(S) = dimG0 + dimV = dimG0 + dim (S(G1) ∩H).

By α(S) = β(S) we obtain

dimG0 = codimS(G1)− dimH.

Putting everything together, we have

indS = α(S)− β(S)

= dimW0 + dimG0 − codimS(G1) + dimW1

= α(T )− codimS(G1) + dimG0

= α(T )− dimH = α(T )− β(T ) = indT,

which ends the proof.

11.9. Theorem: Let K ∈ K(X) and λ ∈ σ(K) \ 0, and denote, for k ∈ N, Nk :=N((λ−K)k) and Rk := R((λ−K)k). Then one has

(a) For each k ∈ N, Nk and Rk are closed and dimNk = codimRk <∞.

(b) There is a minimal p ∈ N such that Np = Np+1.

(c) Np+k = Np and Rp+k = Rp for each k ∈ N.

(d) X = Np ⊕ Rp, (λ − K)Np ⊆ Np, ((λ − K)|Np)p = 0 and λ − K : Rp → Rp is anisomorphism.

In particular, σ(K) \ 0 ⊆ σp(K). Moreover, σ(K) \ 0 is finite or σ(K) \ 0 = λn :n ∈ N where λn → 0.

Assertion (d) means that X = Np⊕Rp where Np and Rp are invariant under the operator

K. Thus K corresponds to a matrix operator

(K|Np 0

0 K|Rp

): Np×Rp → Np×Rp where

σ(K|Np) = λ and σ(K|Rp) = σ(K) \ λ. The space Np is finite-dimensional and K|Npcorresponds to a matrix with single eigenvalue λ.

37

Proof. (a) Writing (λ − K)k = λk(I − KMk) where Mk ∈ L(X) we see by 11.5 that(λ−K)k ∈ Φ(X) and ind (λ−K)k = 0.

(b) If not, we find a sequence (xn) in X such that, for each n, xn ∈ Nn+1 \ Nn and1 = d(xn, N) ≤ ‖xn‖ ≤ 2. For n > m we then have

‖Kxn −Kxm‖ = ‖λxn − ((λ−K)xn +Kxm)︸ ︷︷ ︸=:y

‖

and(λ−K)ny = (λ−K)n+1xn +K(λ−K)nxm = 0,

i.e. y ∈ Nn. Hence

‖Kxn −Kxm‖ = ‖λxn − y‖ ≥ |λ|d(xn, Nn) = |λ| > 0,

in contradiction with K ∈ K(X).

(c) If (λ−K)p+2x = 0 then (λ−K)x ∈ Np+1 = Np and (λ−K)p+1x = 0, i.e. x ∈ Np+1 = Np.We have shown Np+2 = Np. Now we iterate and use (a).

(d) For each k, λ − K : Rk → Rk+1 is surjective. Hence λ − K : Rp → Rp is surjective.But by 11.5, (λ − K)|Rp ∈ Φ(Rp) has index 0, so λ ∈ ρ(K|Rp). On the other hand,λ − K : Nk+1 → Nk for each k, hence λ − K : Np → Np and ((λ − K)|Np)p = 0. SincedimNp <∞ we obtain from linear algebra that σ(K|Np) = λ.In particular, there exists ε > 0 such that B(λ, ε) \ λ ⊂ ρ(A). This implies that the onlypossible accumulation point of σ(K) \ 0 is 0.

Example: Let X = C[0, 1] and define V ∈ L(X) by V f(t) :=∫ t

0f(s) ds. Then V : X → X

is compact, but σp(V ) = ∅, since λf = V f implies f(0) = 0 and λf ′ = f , hence f = 0. By11.6, σ(V ) = 0.

Via (V nf)(t) =∫ t

0(t−s)n−1

(n−1)!f(s) ds it is also possible to show ‖V n‖ ≤ 1

n!which implies

r(V ) = 0 for the spectral radius of V .

We want to have a corresponding result for unbounded operators, and this can be achievedvia resolvents.

11.10. Lemma: Let A be a closed operator in X with non-empty resolvent set. Thefollowing are equivalent:

(i) I : [D(A)]→ X is compact,

(ii) there exists λ0 ∈ ρ(A) such that R(λ0, A) ∈ K(X),

(iii) R(λ,A) ∈ K(X) for all λ ∈ ρ(A).

In this case we say that A has compact resolvents.

38

Proof. (i) ⇒ (ii): We choose λ0 ∈ ρ(A) and factorize R(λ0, A) = R(λ0, A) I whereR(λ0, A) ∈ L(X, [D(A)]) and I ∈ K([D(A)], X).

(ii) ⇒ (iii): Use the resolvent equation.

(iii) ⇒ (i): Choose λ0 ∈ ρ(A) and factorize I = R(λ0, A)(λ0 − A) where λ0 − A ∈L([D(A)], X) and R(λ0, A) ∈ K(X).

end Tue28.05.19

11.11. Lemma: Let A be a closed operator in X, λ0 ∈ ρ(A) and z ∈ C \ λ0. For allk ∈ N we then have

N((z−A)k

)= N

(((λ0−z)−1−R(λ0, A)

)k), R

((z−A)k

)= R

(((λ0−z)−1−R(λ0, A)

)k),

In particular, z − A ∈ Φ([D(A)], X) if and only if (λ0 − z)−1 − R(λ0, A) ∈ Φ(X). In thiscase, one has ind (z − A) = ind ((λ0 − z)−1 −R(λ0, A)).

Proof. is an exercise.

Combination of 11.9, 11.10, and 11.11 yields the following version of Theorem 11.9 forunbounded operators.

11.12. Theorem: Let A be a closed linear operator in X with ρ(A) 6= ∅ and compactresolvents. Let z ∈ σ(A) and denote, for k ∈ N, Nk := N((z−A)k) and Rk := R((z−A)k).Then one has

(a) For each k ∈ N, Nk and Rk are closed and dimNk = codimRk <∞.

(b) There is a minimal p ∈ N such that Np = Np+1.

(c) Np+k = Np and Rp+k = Rp for each k ∈ N.

(d) X = Np⊕Rp, (z−A)Np ⊂ Np, ((z−A)|Np)p = 0 and A|Rp with D(A|Rp) = D(A)∩Rp

is a closed linear operator in Rp with z ∈ ρ(A|Rp).

In particular, σ(A) = σp(A). Moreover, σ(A) is finite or σ(A) = zn : n ∈ N where|zn| → ∞.

Observe that, for a fixed λ0 ∈ ρ(A), the operator (λ0 − z)−1 − R(λ0, A) : Rp → Rp is by11.11 and 11.9 an isomorphism and that R(λ0, A)(Rp) = Rp ∩D(A) = D(A|Rp).

Example: Let X = f ∈ C[0, 1] : f(0) = f(1) and A = ddx

with D(A) = f ∈ X ∩C1[0, 1] : f ′ ∈ X. Cleraly, A is closed and, by Arzela-Ascoli, the embedding I : [D(A)]→X is compact. We show λ ∈ ρ(A) for λ 6∈ 2πiZ: The equation (λ − A)f = g meansλf − f ′ = g, and the general solution is given by

f(x) = ceλx − eλx∫ x

0

e−λyg(y) dy, x ∈ [0, 1],

39

where c ∈ C. The condition f(0) = f(1) leads to

c = c(g) =eλ

eλ − 1

∫ 1

0

e−λyg(y) dy,

observe that eλ 6= 1 by λ 6∈ 2πiZ. Hence λ−A is bijective and λ ∈ ρ(A). For λ ∈ 2πiZ wesee that f(x) = eλx, x ∈ [0, 1], satisfies f ∈ D(A) and (λ− A)f = 0 and spans N(λ− A).The operator A has compact resolvents and σ(A) = σp(A) = 2πiZ. For z ∈ σ(A) we haveα(z − A) = 1 = β(z − A).

Regarding Fredholm operators as generalizations of isomorphisms leads to the followingnotion.

11.13. Definition: Let A be a closed linear operator inX. We define the essential spectrumof A by

σess(A) := λ ∈ C : λ− A 6∈ Φ([D(A)], X).

Clearly, σess(A) ⊆ σ(A). Any λ ∈ C \ σess(A) is called a Fredholm point for A.

Example: If dimX <∞ then σess(A) = ∅ for any A ∈ L(X). If K ∈ K(X) and dimX =∞ then σess(K) = 0 (R(K) is either finite dimensional or not closed). If A is a closedoperator in X with non-empty resolvent set and compact resolvents then σess(A) = ∅.

Remarks: (a) Observe that σc(A) ⊆ σess(A).

(b) By 11.5, σess(A) is closed, C \ σess(A) is open, and the map C \ σess(A) → Z, λ 7→ind (λ− A) is continuous.

(c) By 11.11 we have: If λ0 ∈ ρ(A) then

σess(R(λ0, A)) \ 0 = 1

λ0 − z: z ∈ σess(A)

.

Example: Let X = f ∈ C[0,∞) : limx→∞ f(x) = 0 with sup-norm ‖ · ‖∞, and A = ddx

,D(A) = f ∈ X ∩ C1[0,∞) : f ′ ∈ X. For λ ∈ C and f ∈ D(A) we have (λ− A)f = 0 ifand only if f ′ = λf , i.e. f = ceλ(·). We thus obtain σp(A) = Reλ < 0.For Reλ > 0 and g ∈ X the unique solution f ∈ D(A) of (λ− A)f = g is given by

f(x) = ceλx − eλx∫ x

0

e−λtg(t) dt,

where c = c(g) has to be chosen such that limx→∞ f(x) = 0, i.e.

c(g) =

∫ ∞0

e−λtg(t) dt,

40

so that ((λ− A)−1g

)(x) =

∫ ∞x

eλ(x−t)g(t) dt =

∫ ∞0

e−λsg(x+ s) ds,

observe that, by dominated convergence, the integral on the right hand side tends to 0 forx → ∞. We obtain Reλ > 0 ⊆ ρ(A). For Reλ < 0 and g ∈ X the unique solution of(λ− A)f = g, f(0) = 0 is given by

f(x) =

∫ x

0

eλ(x−t)g(t) dt.

Observe that f ∈ D(A): extending g by 0 to a function on R we have

|f(x)| ≤∫ ∞

0

e(Reλ)y|g(x− y)| dy

which tends to 0 as x → ∞ by dominated convergence. This means that R(λ − A) = X,β(λ−A) = 0, λ−A ∈ Φ(X) and ind (λ−A) = 1 for any Reλ < 0. Continuity of the indexthen yields that σess(A) = iR. end Tue

04.06.1911.14. Lemma: Let T ∈ Φ(X, Y ). Then T ′ ∈ Φ(Y ′, X ′) and

α(T ′) = β(T ), β(T ′) = α(T ), indT ′ = −indT.

Proof. is an exercise.

From this lemma we obtain via spectral mapping:

11.15. Corollary: Let A be a closed and densely defined linear operator in X with ρ(A) 6=∅. Then σess(A

′) = σess(A) and, for z ∈ C \ σess(A),

α(z − A′) = β(z − A), β(z − A′) = α(z − A), ind (z − A′) = −ind (z − A).

11.16. Proposition: Let A be a closed operator in X and K ∈ K([D(A)], X). Then theBanach spaces [D(A)] and [D(A+K)] are isomorphic and one has

σess(A+K) = σess(A).

Proof. Clearly, D(A+K) = D(A) as sets and

‖x‖A+K = ‖x‖+ ‖(A+K)x‖ ≤ ‖x‖+ ‖Ax‖+ ‖K‖‖x‖A ≤(1 + ‖K‖)‖x‖A.

Now assume that there is no constant C > 0 with ‖x‖A ≤ C‖x‖A+K . Then we find asequence (xn) in D(A) with ‖xn‖A = 1 and ‖xn‖A+K → 0. Since K is compact [D(A)]→X we find a convergent subsequence of (Kxn) and may assume Kxn → y in X. By‖xn‖A+K → 0 we have ‖xn‖ → 0 and ‖(A+K)xn‖ → 0. Hence

Axn = (A+K)xn −Kxn → 0− y = y in X.

Since A is closed we infer y = 0, hence also Axn → 0 in X, i.e. ‖xn‖A → 0, a contradiction.The assertion on the essential spectrum follows from 11.8.

41

We give somehow typical examples.

11.17. Example: Let X = f ∈ C[0,∞) : limx→∞ f(x) = 0 with sup-norm ‖ · ‖∞, andA = d

dx, D(A) = f ∈ X ∩ C1[0,∞) : f ′ ∈ X. We know that σess(A) = iR.

Let m ∈ C[0,∞) such that m(t) = 0 for t ≥ a where a > 0. Then f 7→ mf is compact from[D(A)] to X (by Arzela-Ascoli). Hence, if we define Bf := f ′+mf for f ∈ D(B) := D(A),then σess(B) = iR.

By an approximation argument, the same holds for m ∈ X.

11.18. Example: In X = L2(Rd) we consider A = −∆, defined as in 10.20 on D(A) =W 2,2(Rd) = H2(Rd) by

−∆f =d∑j=1

∂2j f = F−1(ξ 7→ 4π2|ξ|2f(ξ)).

We know that A is closed, the graph norm is equivalent to ‖ · ‖H2 , and σ(A) = [0,∞): Forλ ∈ C \ [0,∞) we have λ ∈ ρ(A) and

R(λ,A)f = (λ+ ∆)−1f = F−1(ξ 7→ (λ− 4π2|ξ|2)−1︸ ︷︷ ︸∈L∞

f(ξ)), f ∈ L2(Rd).

Now we show that σess(A) = [0,∞): For λ ≥ 0 we have N(λ−A) = 0 since (λ−A)f = 0implies (λ − 4π2|ξ|2)f(ξ) = 0 for a.e. ξ ∈ Rd and then f = 0 a.e. and f = 0 a.e. asξ : |ξ| =

√λ is a null-set. Moreover,

R(λ− A) = F−1((λ− 4π2| · |2)FH2(Rd)) = F−1((λ− 4π2| · |2)(1 + 4π2| · |2)−1L2(Rd))

contains the dense set

F−1(f ∈ L2 : ∃ε > 0 : f(ξ) = 0 if |ξ| ≥ 1ε

or ||ξ| −√λ| < ε ) ).

So R(λ− A) is dense in L2(Rd) but not closed in L2(Rd) (since (λ− A)−1 is unbounded).Hence σ(A) = σess(A) = [0,∞).

Now we let mj ∈ L∞(Rd) for j = 0, 1, . . . , d with mj(x) = 0 for |x| ≥ a where a > 0 is fixed.

We define Bf :=∑d

j=1 mj∂jf +m0f for f ∈ H2(Rd). We show that B : [D(A)]→ L2(Rd)

is compact, i.e. B ∈ K(H2(Rd), L2(Rd)).

To this end we choose χ ∈ C∞c (Rd) such that χ = 1 on B(0, a+1) and suppχ ⊆ B(0, a+2)and observe Bf =

∑dj=1mjχ∂jf +m0χf . The operators f 7→ χ∂jf , f 7→ χf are bounded

H2(Rd)→ H10 (B(0, a+ 2)), the embedding H1

0 (B(0, a+ 2))→ L2(Rd) is compact, and themultiplication operators f 7→ mjf are bounded on L2(Rd).

We conclude that B : H2(Rd) → L2(Rd) is compact as claimed. By 11.16 we thus haveσess(−∆ +B) = σess(−∆) = [0,∞).

42

12 Operators in Hilbert space

When nothing else is said, H denotes a complex Hilbert space with inner product (·|·).Recall the following properties of an inner (or scalar) product:

• (·|·) is linear in the first component,

• (y|x) = (x|y) for all x, y ∈ H,

• these two properties imply that (·|·) is antilinear in the second component, i.e. (x|αy+z) = α(x|y) + (x|z) for al x, y, z ∈ H, α ∈ C.

• (x|x) ≥ 0 and (x|x) = 0 ⇐⇒ x = 0 for all x ∈ H.

These properties imply that x 7→ ‖x‖ :=√

(x|x) defines a norm on H and that theCauchy-Schwarz inequality

|(x|y)| ≤ ‖x‖‖y‖

holds for all x, y ∈ H.

A space H equipped with such an inner product (·|·) is called a Hilbert space if it is complete

for the norm ‖‖ associated with the inner product.

Let H ′ := L(H,C) denote the dual space of H. Then the map

JH : H 7→ H ′, y 7→ (·|y)

is bijective, isometric, and antilinear. Let H∗ denote the anti-dual space of H, i.e. the linearspace of all continuous antilinear functionals H → C. Then the map

JH : H 7→ H∗, x 7→ (x|·)

is bijective, isometric, and linear. end Thu06.06.19

12.1. Definition: Let H1, H2 be Hilbert spaces and T ∈ L(H1, H2). The adjoint operatorT ∗ ∈ L(H2, H1) of T is defined by

(x|T ∗y)H1 = (Tx|y)H2 for all x ∈ H1, y ∈ H2,

i.e. T ∗ = J−1H1T ′JH2 where T ′ ∈ L(H ′2, H

′1) denotes the dual operator of T . We clearly have

‖T ∗‖L(H2,H1) = ‖T‖L(H1,H2).

In the same way we can define the adjoint operator A∗ as an operator from H2 to H1 fora densely defined operator A : H1 ⊇ D(A)→ H2.

An operator S ∈ L(H) is called self-adjoint if S∗ = S, normal if SS∗ = S∗S, and unitaryif SS∗ = S∗S = I.

43

Rules: (T ∗)∗ = T , (S + T )∗ = S∗ + T ∗, (αT )∗ = αT ∗, (ST )∗ = T ∗S∗.

Remark: For S ∈ L(H) we have

S self-adjoint ⇐⇒ ∀x, y ∈ H : (Sx|y) = (x|Sy)

S normal ⇐⇒ ∀x, y ∈ H : (Sx|Sy) = (S∗x|S∗y)

S unitary ⇐⇒ S is bijective and isometric.

12.2. Lemma: Let S ∈ L(H).

(a) ‖S∗S‖ = ‖SS∗‖ = ‖S‖2 and S∗S, SS∗ are self-adjoint.

(b) If S is normal, then

‖S‖ = r(S) = max|λ| : λ ∈ σ(S).

Proof. (a) We have

‖Sx‖2 = |(Sx|Sx)| = |(x|S∗Sx)| ≤ ‖x‖‖S∗Sx‖,

hence‖S‖2 ≤ ‖S∗S‖ ≤ ‖S∗‖‖S‖ = ‖S‖2.

(b) We calculate r(S). Using (a) repeatedly and normality of S once we have

‖S2‖2 = ‖S2(S2)∗‖ = ‖(SS∗)2‖ = ‖(SS∗)(SS∗)∗‖ = ‖SS∗‖2 = ‖S‖4.

Hence ‖S2‖ = ‖S‖2. Iteration yields ‖S2k‖ = ‖S‖2k for any k ∈ N, from which r(S) = ‖S‖follows.

12.3. Definition und Lemma: Let T ∈ L(H) and

W (T ) := (Tx|x) : ‖x‖ = 1

be the numerical range of T . Then σ(T ) ⊆ W (T ).

Proof. Let λ 6∈ W (T ). Then d := d(λ,W (T )) > 0 and for ‖x‖ = 1:

d ≤ |λ− (Tx|x)| = |((λ− T )x|x)| ≤ ‖(λ− T )x‖ · ‖x‖ = ‖(λ− T )x‖.

Hence λ− T is injective, (λ− T )−1 : R(λ− T )→ H is bounded, and R(λ− T ) is closed.3

Now R(λ − T ) is dense if and only if (λ − T )∗ = λ − T ∗ is injective. Since W (T ∗) = µ :µ ∈ W (T ), we have d(λ,W (T ∗)) = d(λ,W (T )) = d > 0. The previous argument showsthat λ− T ∗ is injective.

3see Exercise 29, FA

44

12.4. Corollary: If S ∈ L(H) is self-adjoint then W (S) ⊆ R and

σ(S) ⊆ [m,M ] ⊆ [−‖S‖, ‖S‖],

where m := inf(Sx|x) : ‖x‖ = 1 and M := sup(Sx|x) : ‖x‖ = 1. Moreover, m,M ∈σ(S) and M = ‖S‖ or m = −‖S‖.

Proof. For x ∈ H with ‖x‖ = 1 we have

(Sx|x) = (x|Sx) = (Sx|x),

i.e. (Sx|x) ∈ R, and|(Sx|x)| ≤ ‖Sx‖‖x‖ ≤ ‖S‖,

which implies [m,M ] ⊆ [−‖S‖, ‖S‖]. The inclusion σ(S) ⊆ [m,M ] follows from 12.3.By 12.2 we have r(S) = ‖S‖. By W (S − m) = W (S) − m, σ(S − m) = σ(S) − m,r(S −m) = ‖S −m‖ we infer ‖S −m‖ = M −m ∈ σ(S −m), i.e. M ∈ σ(S). In the sameway, W (S −M) = W (S)−M etc and −‖S −M‖ = m−M ∈ σ(S −M), i.e. m ∈ σ(S).Then r(S) = ‖S‖ implies M = ‖S‖ or m = −‖S‖.

The following is a first spectral theorem for bounded self-adjoint operators and establishes afunctional calculus for functions that are continuous on the spectrum. We write pj(λ) = λj

for j ∈ N0, so p0 = 1σ(S) and p1 = idσ(S).

12.5. Theorem: Let S ∈ L(H) be self-adjoint. There exists a unique continuous linearmap Φ : C(σ(S))→ L(H) such that

Φ(p0) = I, Φ(p1) = S, Φ(f · g) = Φ(f)Φ(g), f, g ∈ C(σ(S)),

[Φ is multiplicative, i.e. an algebra homomorphism.]

For any f ∈ C(σ(S)) the operator Φ(f) is normal,

Φ(f)∗ = Φ(f),

and Φ(f) is self-adjoint if and only if f is real-valued.

Moreover, Φ is an isometry, i.e. ‖Φ(f)‖L(H) = ‖f‖∞,σ(S) for all f ∈ C(σ(S)).

Proof. For polynomials p(λ) =∑n

j=0 ajλj we define Φ(p) :=

∑nj=0 ajS

j, and let P thealgebra of all polynomial functions p : σ(S) → C. Then Φ : P → L(H) is an algebrahomomorphism, Φ(p)∗ = Φ(p) and Φ(p) is normal for each p ∈ P. Moreover, Φ(p) isself-adjoint if p is real-valued. Clearly, Φ(p0) = I and Φ(p1) = S. end Tue

11.06.19By Weierstraß, P is dense in C(σ(S)) for the sup-norm ‖ · ‖∞. So we only have to checkthat Φ : P→ L(H) is an isometry. For p ∈ P we have

‖Φ(p)‖2 = ‖Φ(p)∗Φ(p)‖ = ‖Φ(p)Φ(p)‖ = ‖Φ(pp)‖ = r(Φ(pp))

= max|λ| : λ ∈ σ(Φ(pp)) = max|pp(µ)| : µ ∈ σ(S) = ‖p‖2∞,σ(S),

45

where we used the polynomial spectral mapping result from the Exercise 23. If Φ(p) isself-adjoint, then we obtain Φ(p) = Φ(p)∗ = Φ(p), from which p = p, i.e. real-valuedness ofp, follows by injectivity of Φ.

The properties of Φ are preserved when approximating continuous functions by polynomi-als.

As an application of 12.5 we show “optimality” of self-adjoint operators with respect toresolvent bounds.

12.6. Corollary: Let S ∈ L(H) be self-adjoint. Then, for any λ ∈ ρ(S),

R(λ, S) = Φ((λ− (·))−1) and ‖R(λ, S)‖ =1

d(λ, σ(S)).

Proof. We have

(λ− S)Φ((λ− (·))−1) = Φ(λ− (·))Φ((λ− (·))−1) = Φ(1σ(S)) = I,

and Φ((λ − (·))−1)(λ − S) = I is proved similarly. Recall that Φ is isometric and observe‖(λ− (·))−1‖∞,σ(S) = d(λ, σ(S))−1.

12.7. Remark: Let S ∈ L(H) be self-adjoint. Then the functional calculus Φ from 12.5is an extension of the Dunford functional calculus: if F is holomorphic on a complexneighborhood of σ(S) then

Φ(F |σ(S)) =1

2πi

∫Γ

F (λ)R(λ, S) dλ,

where Γ is suitably chosen (→ 9.3): By 12.6 we have R(λ, S) = Φ((λ − (·))−1), and Γ →σ(S), λ 7→ (λ− (·))−1, is continuous. Hence we can interchange Φ with the integral:

1

2πi

∫Γ

F (λ)R(λ, S) dλ =1

2πi

∫Γ

F (λ)Φ((λ−(·))−1) dλ = Φ( 1

2πi

∫Γ

F (λ)

λ− (·)dλ)

= Φ(F |σ(S)).

12.8. Definition: We call a self-adjoint operator S ∈ L(H) positive and write S ≥ 0 if(Sx|x) ≥ 0 for all x ∈ H. If T ∈ L(H) is another self-adjoint operator then S ≥ T meansS − T ≥ 0, i.e. (Sx|x) ≥ (Tx|x) for all x ∈ H.

12.9. Lemma: Let S ∈ L(H) be self-adjoint and Φ : C(σ(S))→ L(H) be the functionalcalculus for S from 12.5. If f, g ∈ C(σ(S)) are real-valued and f ≥ g then Φ(f) ≥ Φ(g) inthe sense of 12.8.

46

Proof. It suffices to prove the case g = 0, i.e. f ≥ 0. Then√f is well defined,

√f ∈

C(σ(S)), and Φ(f) = Φ(√f√f) = Φ(

√f)Φ(

√f), where Φ(

√f) is selfadjoint. This implies,

for x ∈ H,(Φ(f)x|x) = (Φ(

√f)Φ(

√f)x|x) = (Φ(

√f)x|Φ(

√f)x) ≥ 0.

Before turning to the study of self-adjoint compact operators we give a lemma on generalself-adjoint operators. Recall

• x ⊥ y :⇐⇒ (x|y) = 0 for x, y ∈ H,

• M ⊥ N :⇐⇒ ∀x ∈M, y ∈ N : (x|y) = 0 for linear subspaces M,N ⊂ H,

• M⊥ := x ∈ H : x ⊥M = x ∈ H : ∀y ∈M : x ⊥ y for linear subspaces M of H.

If M is a linear subspace of H then M⊥ is always closed, M⊥

= M⊥, and (M⊥)⊥ = M .One has always H = M ⊕M⊥ where M ⊥M⊥.

12.10. Lemma: Let S ∈ L(H) be self-adjoint. Then:

(a) For λ, µ ∈ σ(S) with λ 6= µ: N(λ− S) ⊥ N(µ− S).

(b) For λ ∈ σ(S): N((λ− S)2) = N(λ− S).

(c) The space H is the orthogonal direct sum of N(S) and R(S), i.e. H = N(S) ⊕ R(S)and N(S) ⊥ R(S).

Proof. (a) For x ∈ N(λ− S) and y ∈ N(µ− S) we have

λ(x|y) = (λx|y) = (Sx|y) = (x|Sy) = µ(x|y),

and x ⊥ y, since λ 6= µ.

(b) Let x ∈ N((λ− S)2). Then

‖(λ− S)x‖2 = ((λ− S)x|(λ− S)x) = (x| (λ− S)2x︸ ︷︷ ︸=0

) = 0,

i.e. (λ− S)x = 0, and x ∈ N(λ− S). The reverse inclusion is clear.

(c) Let x ∈ N(S) and Sy ∈ R(S). Then (x|Sy) = (Sx|y) = 0, hence N(S) ⊥ R(S). If, onthe other hand, x ∈ H such that x ⊥ R(S) then

‖Sx‖2 = (Sx|Sx) = (x| S2x︸︷︷︸∈R(S)

) = 0,

i.e. Sx = 0 and x ∈ N(S). Hence N(S) = R(S)⊥

.

47

12.11. Theorem: Let S ∈ L(H) be compact and self-adjoint and dimH = ∞. Thereexists a real sequence (λn)n∈N0 with λn → 0 and an orthonormal sequence (en)n∈N0 in Hsuch that

S =∞∑n=0

λ(·|en)en,

where the series converges in operator norm.

Proof. We apply Theorem 11.9 to S. By 12.10(b) we know that p = 1 for every λ ∈ σ(S) \0. We also know that σ(S) consists of a null sequence. For any λ ∈ σ(S) \ 0 we choosea finite orthonormal basis of N(λ− S). We obtain thus an orthonormal sequence (en)n∈Nwhere we assume that the corresponding eigenvalues are ordered such that |λ1| ≥ |λ2| ≥ . . ..Here N = 0, . . . , n0 is finite (if σ(S) \ 0 is finite) or N = N0.

We let P :=∑

n∈N(·|en)en. Then P is the orthogonal projection onto H1 :=

span en : n ∈ N. Letting H0 := H⊥1 we have, for x ∈ H0 and n ∈ N ,

(Sx|en) = (x|Sen) = λn(x|en) = 0.

i.e. Sx ∈ H0. This means that S0 := S|H0 ∈ L(H0). Clearly, S0 is self-adjoint and compact.By 11.9, σ(S0) \ 0 = ∅, hence S0 = 0 by 12.2(b), and H0 ⊂ N(S). On the other hand,N(S) is by 12.10(a) orthogonal to each N(λn − S), n ∈ N . Hence N(S) ⊂ H0, and weobtain H0 = N(S), and by 12.10(c) also H1 = R(S).

But thenSx = SPx =

∑n∈N

λn(x|en)en for all x ∈ H.

If N = 0, . . . , n0 is finite, we set λn = 0 for n > n0, and choose an orthonormal sequence(en)n>n0 in N(S).

For x ∈ H and k ∈ N we have, by Pythagoras and Bessel’s inequality,

‖Sx−k∑

n=0

λn(x|en)en‖2 =∑n>k

|λn(x|en)|2 ≤ ‖x‖2(supn>k|λn|)2,

which proves convergence of the series in operator norm, since (λn) is a null sequence.end Thu13.06.19

12.12. Theorem: Let G be another Hilbert space with dimG = ∞ and T ∈ K(H,G).Then there exists a decreasing null sequence (sn)n∈N0 in [0,∞) and orthonormal systems(en)n∈N0 in H and (fn)n∈N0 in G such that

T =∞∑n=0

sn(·|en)fn,

where the series converges in operator norm.

48

Proof. The operator T ∗T ∈ L(H) is compact and self-adjoint. Moreover, T ∗T ≥ 0 inthe sense of 12.8, and σ(T ∗T ) ⊆ [0, ‖T‖2] by 12.4. By 12.11 we obtain a decreasing nullsequence (sn)n∈N0 and an orthonormal system (en)n∈N0 in H such that

T ∗T =∞∑n=0

s2n(·|en)en.

For n ∈ N0 with sn > 0 we let fn := s−1n Ten. For n,m ∈ N0 with snsm > 0 we then have

(fn|fm) = (snsm)−1(Ten|Tem) = (snsm)−1(T ∗Ten|em) =s2n

snsm(en|em) = δnm.

If N = n ∈ N0 : sn > 0 is finite then we extend (fn)n∈N to an orthonormal sequence(fn)n∈N0 in G. If y ⊥ en for all n ∈ N then

‖Ty‖2 = (T ∗Ty|y) = 0

by the representation of T ∗T . Hence, for any x ∈ H,

Tx = T(x−

∑n∈N

(x|en)en︸ ︷︷ ︸∈N(T )

)+ T

(∑n∈N

(x|en)en

)

=∑n∈N

(x|en)Ten =∑n∈N

sn(x|en)fn =∞∑n=0

sn(x|en)fn.

The proof for convergence in operator norm is the same as in 12.11.

12.13. Corollary: Let G be another Hilbert space and

F(H,G) := T ∈ L(H,G) : dimR(T ) <∞

denote the space of finite rank operators from H to G. Then F(H,G) is dense in K(H,G)with respect to operator norm.

Remark: This result is in general false in the Banach space context (by an example ofEnflo 1973). The rank of an operator in the dimension of its range.

12.14. Definition and Remark: A representation of T ∈ K(H,G) as a series withthe properties of 12.12 is called a Schmidt representation of T . The sequence (sn)n∈N0 isuniquely determined by T (as the decreasing eigenvalue sequence of T ∗T ), but in generalnot the orthonormal systems (en) and (fn). The sequence (sn)n∈N0 is called the sequenceof singular values of the operator T and denoted by (sn(T ))n∈N0 . The following gives acharacterization of the singular values of an operator as approximation numbers :

For T ∈ K(H,G) and n ∈ N0 one has

sn(T ) = inf‖T − U‖ : U ∈ L(H,G), dimR(U) ≤ n =: αn(T ).

Here αn(T ) measures how good T can be approximated by operators of rank of at most n.

49

Proof. Let∑∞

j=0 sj(·|ej)fj be a Schmidt representation of T . Then, for n ∈ N0 and x ∈ H,we have by Bessel’s inequality

‖Tx−n−1∑j=0

sj(x|ej)fj‖2 ≤∞∑j=n

s2j |(x|ej)|2 ≤ s2

n‖x‖2.

This implies that αn(T ) ≤ sn.

Now let U ∈ L(H,G) with dimR(U) ≤ n. The restriction of U to the (n+ 1)-dimensionalspace line0, . . . , en has non-trivial kernel, so we find y =

∑nj=0 ξjej with ‖y‖ = 1 and

Uy = 0. By Pythagoras we thus obtain

‖T − U‖2 ≥ ‖(T − U)y‖2 = ‖Ty‖2 = ‖n∑j=0

sjξjfj‖2 =n∑j=0

s2j |ξj|2 ≥ s2

n

n∑j=0

|ξj|2 = s2n.

Hence αn(T ) ≥ sn.

Remark: For 1 ≤ p <∞ the Schatten p-class is defined by

Sp(H,G) := T ∈ K(H,G) : (sn(T ))n∈N0 ∈ lp

and νp(T ) := (∑∞

n=0 sn(T )p)1/p for T ∈ Sp(H,G). Elements of S2(H,G) are called Hilbert-Schmidt opertors and elements of S1(H,G) are called nuclear operators or, for G = H, aresaid to be of trace class.

In many respects, the spaces Sp(H,G) may be viewed as “non-commutative” analogs ofthe spaces lp (cf., e.g., §16 in Meise/Vogt “Introduction to Functional Analysis”).

Remark: If H and G are separable infinite-dimensional Hilbert spaces, then the sequences(en)n∈N0 and (fn)n∈N0 in 12.11 and 12.12 can be chosen to be orthonormal bases of H andG, respectively. This can be seen from the proof.

A related observation is that, for a compact operator T ∈ K(H,G), the space R(T ) isalways separable. Moreover, if T ∈ K(H) is self-adjoint and injective, then the space H isseparable (since the space H0 in the proof is trivial).

50

13 The spectral theorem for self-adjoint operators

For a self-adjoint operator S ∈ L(H), we want to extend the functional calculus of Theorem12.5 to a functional calculus for bounded Borel measurable functions on σ(S).

13.1. Definition: For any compact subset K ⊆ R we let

Bb(K) := f : K → C : f is Borel measurable and bounded ,

equipped with the norm ‖f‖∞ := supt∈K |f(t)| (notice, that Bb(K) is a space of functions,not a space of co-classes of functions!). Then Bb(K) is a Banach space.

If (fn) is a sequence of functions M → C and f : M → C is a function, then we writefn → f bounded pointwise, if fn → f pointwise and supn ‖fn‖∞ <∞. Observe that Bb(K)is closed under bounded pointwise convergence.

We call a map Ψ : Bb(K) → L(H) σ-continuous, if fn → f bounded pointwise implies(Ψ(fn)x|y)→ (Ψ(f)x|y) for all x, y ∈ H.

13.2. Lemma: Let K ⊆ R be compact. Then Bb(K) is the smallest subset M of CK

satisfying

(1) C(K) ⊆M ,

(2) M is closed under bounded pointwise convergence.end Tue18.06.19

Proof. Let M0 be the smallest subset M of CK such that (1) and (2) hold, i.e. the inters-section of all subsets M of CK with (1) and (2). Since Bb(K) satisfies (1) and (2), M0 iswell-defined and we have M0 ⊆ Bb(K).

For λ ∈ C \ 0, 1λM0 satisfies (1) and (2), hence M0 ⊆ 1

λM0, i.e. λM0 ⊆M0.

If f ∈ C(K), then M0−f satisfies (1) and (2), thus M0 ⊆M0−f , i.e. f+M0 ⊆M0. HenceC(K) + M0 ⊆ M0. Now let S := f : f + M0 ⊆ M0. We just have shown C(K) ⊆ S. If(fn) is a sequence in S with fn → f bounded pointwise and if g ∈ M0, then (fn + g) is asequence in M0 (by fn ∈ S) and fn + g → f + g bounded pointwise. By (2) for M0 we havef + g ∈M0. Since g ∈M0 was arbitrary, we obtain f ∈ S, and have shown (2) for S. Thisyields M0 ⊆ S, i.e. M0 +M0 ⊆M0.

We thus have shown that M0 is a complex vector space. Moreover, M := g : |g| ∈ M0satisfies (1) and (2), hence M0 ⊆ M , i.e. f ∈ M0 implies |f | ∈ M0. Since M0 is a vectorspace we obtain

maxf, g,minf, g =1

2((f + g)± |f − g|) ∈M0

for real-valued f, g ∈M0.

51

Now we let F := A ⊆ K : 1A ∈ M0. Then F is a σ-algebra in K: ∅, K ∈ F is clear by0, 1K ∈ C(K) ⊆ M0. If A ∈ F, then K \ A ∈ F, since M0 is a vector space. If (Aj)j∈Nis a sequence in F, then

⋃nj=1Aj ∈ F, since the max of a finite family of functions in M0

belongs to M0. Finally⋃j∈NAj ∈ F follows from (2) for M0.

Also by (2), F contains all sets (−∞, b] ∩ K, b ∈ R, hence all Borel subsets of K. Sinceeach f ∈ Bb(K) can be uniformly on K approximated by a sequence (fn) of simple Borelfunctions, we obtain Bb(K) ⊆M0.

13.3. Lemma: Let K ⊆ R be compact. If Φ,Ψ : Bb(K) → L(H) are σ-continuous andcoincide on C(K) then Φ = Ψ.

Proof. Let

M := f ∈ Bb(K) : Φ(f) = Ψ(f) = f ∈ Bb(K) : ∀x, y ∈ H : (Φ(f)x|y) = (Ψ(f)x|y).

Then M satisfies (1), (2) from 13.2, hence M = Bb(K), i.e. Φ = Ψ.

The following proposition is essential for our proof of the Spectral Theorem. We postponethe proof of this proposition for now.

13.4. Proposition: Let K ⊆ R be compact and φ ∈ C(K)′. Then there exists a unique

σ-continuous extension φ ∈ Bb(K)′ of φ. Moreover, we have ‖φ‖ = ‖φ‖.

We shall also use the following lemma on the relation of continuous sesquilinear forms andlinear operators.

13.5. Lemma: Let β : H ×H → C be sesquilinear (i.e. linear in the first and antilinearin the second component) and M ≥ 0 such that

|β(x, y)| ≤M‖x‖‖y‖ for all x, y ∈ H.

Then there exists a unique linear operator B ∈ L(H) such that

β(x, y) = (Bx|y) for all x, y ∈ H,

and one has ‖B‖ ≤M .

Proof. For any x ∈ H, we have β(x, ·) ∈ H ′ and define Bx := J−1H (β(x, ·)). Then B is

linear (since it is a composition of two antilinear maps), and ‖Bx‖ ≤M‖x‖ for all x ∈ H,i.e. B ∈ L(H) and ‖B‖ ≤M . For x, y ∈ H we have

(Bx|y) = (y|Bx) = (JH(Bx))(y) = β(x, ·)(y) = β(x, y).

Uniqueness of B is clear.

52

We now present the Spectral Theorem for bounded self-adjoint operators.

13.6. Theorem (Functional calculus for bounded Borel measurable functions):Let S ∈ L(H) be self adjoint. Then there exists a unique σ-continuous extension Ψ :Bb(σ(S))→ L(H) of the functional calculus Φ : C(σ(S))→ L(H) from 12.5. The map Ψis linear and multiplicative, i.e. an algebra homomorphism, and satisfies ‖Ψ‖ = ‖Φ‖ = 1.Moreover, we have

(i) for each f ∈ Bb(σ(S)) the operator Ψ(f) is normal and Ψ(f) = Ψ(f)∗,

(ii) for real-valued f ∈ Bb(σ(S)) the operator Ψ(f) is self-adjoint,

(iii) for f ≥ 0 we have Ψ(f) ≥ 0 in the sense of 12.8,

(iv) if fn → f bounded pointwise then Ψ(fn)x→ Ψ(f)x for all x ∈ H.

Warning: We have ‖Φ(f)‖ = ‖f‖∞ for f ∈ C(σ(S)) and ‖Ψ(g)‖ ≤ ‖g‖∞ for all g ∈Bb(σ(S)), but it may happen that ‖Ψ(g)‖ < ‖g‖∞ for some g ∈ Bb(σ(S)): for λ ∈ σ(S)one has Ψ(1λ) = 0 if and only if λ 6∈ σp(S). (→ exercises).

Proof. Uniqueness follows from 13.3. For the existence proof, we put, for x, y ∈ H,

φx,y : C(σ(K))→ C, f 7→ φx,y(f) := (Φ(f)x|y).

Then φx,y ∈ C(σ(S))′ and ‖φx,y‖ ≤ ‖x‖‖y‖. Moreover, the map H × H 7→ C(σ(S))′ issesquilinear.

By 13.4 we find for each pair (x, y) ∈ H × H a unique σ-continuous extension φx,y :

Bb(σ(S)) → C and ‖φx,y‖ = ‖φx,y‖ ≤ ‖x‖‖y‖. We show that (x, y) 7→ φx,y is sesquilinear:

For α ∈ C and x, y, z ∈ H the σ-continuous funkcionals φαx+y,z and αφx,z + φy,z or φx,αy+z

und αφx,y + µx,z, respectively, coincide on C(σ(S)). By 13.3 they coincide on Bb(σ(S)).

For each f ∈ Bb(σ(S)) we define

βf : H ×H → C, (x, y) 7→ βf (x, y) := φx,y(f).

Then|βf (x, y)| = |φx,y(f)| ≤ ‖f‖∞‖x‖‖y‖ for all x, y ∈ H, f ∈ Bb(σ(S)).

Moreover, each βf is sesquilinear. By 13.5 we find, for each f ∈ Bb(σ(S)), a unique operatoroperator Ψ(f) ∈ L(H) such that

φx,y(f) = βf (x, y) = (Ψ(f)x|y) for all x, y ∈ H.

We have ‖Ψ(f)‖ ≤ ‖f‖∞. Clearly Ψ(f) = Φ(f) for all f ∈ C(σ(S)). By construction,Ψ : f 7→ Ψ(f) auf Bb(σ(S)) is σ-continuous.

53

We show that f 7→ Ψ(f) is linear: For α ∈ C and f, g ∈ Bb(σ(S)) we have

((αΨ(f) + Ψ(g))x|y) = α(Ψ(f)x|y) + (Ψ(g)x|y) = αφx,y(f) + φx,y(g) = φx,y(αf + g)

= (Ψ(αf + g)x|y),

for all x, y ∈ H. By 13.5 (uniqueness) we thus obtain αΨ(f) + Ψ(g) = Ψ(αf + g), i.e.Ψ : f 7→ Ψ(f) is linear.

For f ∈ C(σ(S)) the map g 7→ Ψ(fg)−Ψ(f)Ψ(g) is σ-continuous and vanishes on C(σ(S)),by 13.3 it thus vanishes on Bb(σ(S)). We obtain Ψ(fg) = Ψ(f)Ψ(g) for all g ∈ Bb(σ(S)).For fixed g ∈ Bb(σ(S)) the σ-continuous map f 7→ Ψ(fg)−Ψ(f)Ψ(g) vanishes on C(σ(S)),hence on Bb(σ(S)) by 13.3. We obtain Ψ(fg) = Ψ(f)Ψ(g) for all f, g ∈ Bb(σ(S)), i.e. Ψ ismultiplicative.

Now we prove (i). Since f 7→ Ψ(f) − Ψ(f)∗ is σ-continuous on Bb(σ(S)) and vanishes onC(σ(S)), it vanishes by 13.3 on Bb(K). Hence we have Ψ(f) = Ψ(f)∗ for all f ∈ Bb(σ(S)).For each f ∈ Bb(σ(S)) we then have

Ψ(f)Ψ(f)∗ = Ψ(f)Ψ(f) = Ψ(|f |2) = Ψ(f)Ψ(f) = Ψ(f)∗Ψ(f),

so Ψ(f) is normal.

We prove (ii). For real-valued f we have

Ψ(f)∗ = Ψ(f) = Ψ(f),

i.e. Ψ(f) is self-adjoint. For the proof of (iii) let f ≥ 0, then g :=√f ∈ Bb(σ(S)) and for

x ∈ H we have

(Ψ(f)x|x) = (Ψ(g2)x|x) = (Ψ(g)Ψ(g)x|x) = (Ψ(g)x|Ψ(g)x) ≥ 0,

where we used that Ψ(g) is self-adjoint. For the proof of (iv) it suffices to study the casef = 0. So let fn → 0 bounded pointwise and x ∈ H. We then have

‖Ψ(fn)x‖2 = (Ψ(fn)∗Ψ(fn)x|x) = (Ψ(|fn|2)x|x) −→ 0,

by σ-continuity since |fn|2 → 0 bounded pointwise.

We still have to prove Proposition 13.4.

Proof of Proposition 13.4. Uniqueness follows from 13.3. We assume K 6= ∅ and proveexistence. We first restrict to the case of intervals. Letting a := minK and b := maxKwe have K ⊆ [a, b] and [a, b] \ K is open (in [a, b] but also in R!). Hence [a, b] \ K isan at most countable disjoint union of open intervals Ij = (aj, bj) with aj, bj ∈ K. Let Rdenote the restriction map R : C[a, b]→ C(K), f 7→ f |K and define the extension operator

J : Bb(K)→ Bb[a, b] by Jf(x) = f(x) for x ∈ K and Jf(x) :=x−ajbj−aj f(aj) +

bj−xbj−aj f(bj) for

x ∈ Ij. Then ‖Jf‖∞ = ‖f‖∞ for all f , J(C(K)) ⊆ C[a, b] and RJ = IC(K). If φ ∈ C(K)′

54

and φR has a σ-continuous extension ψ ∈ Bb[a, b]′ then φ := ψJ ∈ Bb(K)′ is a σ-continuous

extension of φ.

We may thus assume that K = [a, b] and φ ∈ C[a, b]′. Now we use FA 3.10 and find a

function g ∈ BV0[a, b] with ‖g‖BV = ‖φ‖ such that φ(f) =∫ baf(t) dg(t) as a Stieltjes

integral for any f ∈ C[a, b], where we may assume that g is continuous from the right.Any real-valued BV -function is the difference of two monotone increasing functions, so wefind monotone increasing and right continuous functions gj : [a, b] →, j = 1, . . . , 4, withgj(0) = 0 and

g(t) = g1(t)− g2(t) + i(g3(t)− g4(t)), t ∈ [a, b].

We extend each gj to a function on R by letting gj(t) = 0 for t < a and gj(t) = gj(b) fort > b. Then µj((c, d]) := gj(d)− gj(c) induces a positive and finite Borel measure µj on R,for which we have∫

f dµ1 −∫f dµ2 + i

(∫f dµ3 −

∫f dµ4

)=

∫ b

a

f(t) dg(t), f ∈ C[a, b].

For each j, the map f 7→∫f dµj ∈ Bb[a, b]

′ is a σ-continuous extension of f 7→∫ bag(t) dgj(t). Hence

f 7→∫f dµ1 −

∫f dµ2 + i

(∫f dµ3 −

∫f dµ4

)∈ Bb[a, b]

′

is a σ-continuous extension of φ. One also has |∫f dµ| ≤ ‖f‖∞|µ|([a, b]) and |µ|([a, b]) =

‖g‖BV , but we do not go into details here.4

end Thu27.06.19

13.7. Definition: Let A be a linear operator in H, i.e. A : H ⊇ D(A)→ H is linear.

(i) A is called symmetric if (Ax|y) = (x|Ay) for all x, y ∈ D(A).

(ii) If A is densely defined then A is called self-adjoint if A = A∗.

Recall that, in the situation of (ii), the adjoint operator A∗ of A is given by

x ∈ D(A∗) and A∗x = y ⇐⇒ ∀z ∈ D(A) : (Az|x) = (z|y).

Remark: If A is densely defined in H then A is symmetric if and only if A ⊆ A∗. Here,A ⊆ A∗ means that x ∈ D(A) implies x ∈ D(A∗) and A∗x = Ax.

In particular, any self-adjoint operator is symmetric. Moreover, any self-adjoint operatoris closed.

Remark: If A is symmetric then (Ax|x) ∈ R for all x ∈ D(A), since

(Ax|x) = (x|Ax) = (Ax|x).

4We refer to Rudin: Real and Complex Analysis.

55

13.8. Lemma: If A is densely defined and symmetric in H, then A is closable and itsclosure A is symmetric.

Proof. By the remark above we have A ⊆ A∗, and A∗ is always closed. Hence also A ⊆A∗ = (A)∗ (for the last identity see below).

Recall: In the situation of 13.8, we have by 8.17 and reflexivity of H:

A is closable ⇐⇒ A∗ is densely defined.

Actually, the proof shows that, in this case, A = (A∗)∗. Applied to the closed operator A∗

in place of A, we get(A)∗ = ((A∗)∗)∗ = A∗.

13.9. Lemma: Let A be symmetric and closed. Then, for any z ∈ C \ R, the operatorz − A is injective and R(z − A) is closed. If R(z − A) = H then ‖R(z, A)‖ ≤ 1/|Im z|.

Proof. Since A is symmetric, we have for x ∈ D(A) and ξ + iη ∈ C with η 6= 0 by thepreceding remark

‖(ξ + iη − A)x‖2 = (ξ2 + η2)‖x‖2 − 2Re ((ξ + iη)x|Ax) + ‖Ax‖2

= (ξ2 + η2)‖x‖2 − 2ξ(x|Ax) + ‖Ax‖2

= η2‖x‖2 + ‖(ξ − A)x‖2 ≥ η2‖x‖2.

Since ξ + iη − A is closed, the assertion follows.

13.10. Proposition: If A is a self-adjoint operator in H then σ(A) ⊂ R, and ‖R(z, A)‖ ≤|Im z|−1 for any z ∈ C \ R.

Proof. We have (z−A)∗ = z−A for any z ∈ C. For z ∈ C \R we have by 13.9 that z−A,z − A are injective and that R(z − A) is closed. But

R(z − A)⊥ = y ∈ H : ∀x ∈ D(A) : ((z − A)x|y) = 0 = N((z − A)∗) = N(z − A) = 0,

i.e. R(z − A) = H, and we obtain z ∈ ρ(A). The resolvent estimate is from 13.9.

13.11. Proposition: For any densely defined operator in A the following are equivalent:

(i) A is closable and A is self-adjoint.

(ii) A is symmetric and σ(A) ⊂ R.

(iii) A is symmetric and there exists z ∈ C \R such that z−A∗ and z−A∗ are injective.

56

Proof. (i) =⇒ (ii): If A is self-adjoint, then A and hence also A are symmetric. σ(A) ⊂ Rholds by 13.10.

(ii) =⇒ (iii): By 13.8, A is symmetric. Moreover, (A)∗ = A∗. For z ∈ C \ R we thus havez, z ∈ σ(A∗), and z − A∗, z − A∗ are injective.

(iii) =⇒ (i): By 13.8, A is closable, and we have A ⊂ A∗ = (A)∗. By 13.9 (and its proof)the assumption implies z, z ∈ ρ(A), and this in turn implies z, z ∈ ρ(A∗). Hence we haveρ(A) ∩ ρ(A∗) 6= ∅ which by A ⊂ A∗ is only possible if A = A∗.

Remark: Here we have used the following: If B,C are clsoed operators in a Banach spaceX such that B ⊂ C and there exists λ ∈ ρ(B) ∩ ρ(C), then B = C.

For the proof observe that λ − C : D(C) → X is bijective so that λ − B : D(B) → X

cannot be surjective if D(B)⊂6= D(C).

13.12. Definition: A densely defined symmetric operator A in H is called essentiallyself-adjoint if its closure A is self-adjoint in H.

Remark: By 13.11, a densely defined and symmetric operator A is essentially self-adjointif and only if the operators i−A∗ and −i−A∗ are both injective. We will get back to thatlater on.

We aim at an extension of the functional calculus from Theorem 13.6 to unbounded self-adjoint operators A. This will be done by an orthogonal decompopsition of H with respectto an auxiliary bounded self-adjoint operator B associated to A.

13.13. Lemma: Let A be self-ajoint in H and set

B :=1

2i(R(i, A)∗ −R(i, A)), C :=

−1

2(R(i, A) +R(i, A)∗).

Then B,C ∈ L(H) are self-adjoint, BA ⊂ AB = C, B is injective and 0 ≤ B ≤ I (in thesense of 12.8).

Proof. Self-adjointness of B and C is clear. By R(i, A)∗ = R(−i, A) we obtain easilyBA ⊆ AB. Moreover, since AR(λ,A) = λR(λ,A)− I, we have

AB =1

2i(AR(−i, A)− AR(i, A)) =

1

2i(−iR(−i, A)− iR(i, A)) = C.

By 13.9, ‖R(±i, A)‖ ≤ 1, and we obtain B ≤ I. For x ∈ H and y = R(i, A)x we have

(Bx|x) =1

2i((x|R(i, A)x)− (R(i, A)x|x)) = Im (x|R(i, A)x) = Im ((i−A)y|y) = (y|y) ≥ 0.

Hence B ≥ 0. On the other hand, Bx = 0 implies (i−A)x = y = 0 and x = 0 by injectivityof i− A. Hence B is injective.

57

Remark: By the resolvent equation 8.6(a) we have

B =1

2i(R(−i, A)−R(i, A)) =

1

2i((2i)R(i, A)R(−i, A)) = R(i, A)R(i, A)∗,

and the properties of B also follow from this representation.

We shall use the functional calculus Ψ from 13.6 for the operator B and write f(B) := Ψ(f)for f ∈ Bb(σ(B)). If f is defined on a superset of σ(B) then we write f(B) := (f |σ(B))(B).Observe that σ(B) ⊆ [0, 1] by 13.13 and 12.3 and that 0 6∈ σp(B) by 13.13.

13.14. Proposition: Let A be a self-adjoint operator in H, and let B, C be as in 13.13.For any n ∈ N, define functions θn, sn : R → R by θn := 1( 1

n+1, 1n

] and sn(t) := 1tθn(t), and

let Pn := θn(B). Then:

(a) For each n ∈ N, Pn is an orthogonal projection in H and

PnA ⊂ APn = sn(B)C ∈ L(H).

(b) With Hn := R(Pn) we have Hn ⊆ D(A), A(Hn) ⊆ Hn and Hn ⊥ Hk for all n, k ∈ Nwith n 6= k. Moreover, An := A|Hn ∈ L(Hn) is self-adjoint in the Hilbert space Hn.

(c) We have

x =∑n∈N

Pnx for all x ∈ H (convergence in H)

and

D(A) = x ∈ H :∑n∈N

‖APnx‖2 <∞

Ax =∑n∈N

APnx for all x ∈ D(A) (convergence in H).

(d) For each n ∈ N we have

σ(An) ⊆ σ(A) ∩([−√n,−√n− 1] ∪ [

√n− 1,

√n]).

Proof. (a) By tsn(t) = θn(t), t ∈ R, we get Bsn(B) = θn(B) = Pn. Hence, by 13.13,

APn = ABsn(B) = Csn(B)(∗)= sn(B)C ∈ L(H)

(we say more on the equality (∗) below in the proof of (d)). Moreover, again by 13.13,

PnA = Bsn(B)A = sn(B)BA ⊆ sn(B)AB = sn(B)C = APn.

By 13.6, we have P 2n = Pn and P ∗n = Pn.

58

(b) Each Hn is a closed subspace of H and thus itself a Hilbert space. By (a), APn ∈ L(H),so Hn ⊂ D(A) and, for x ∈ Hn:

PnAx = APnx = Ax, i.e. Ax ∈ Hn.

Hence A(Hn) ⊆ Hn. For x ∈ Hn, y ∈ Hk and n 6= k we have by 13.6:

(x|y) = (Pnx|Pky) = (PkPn︸ ︷︷ ︸=0

x|y) = 0,

and thus Hn ⊥ Hk. Finally, An ∈ L(Hn) is self-adjoint in Hn, since it inherits symmetryfrom A.

(c) We let P0 := 10(B) and H0 := P0(H) = R(P0). P0 is an orthogonal projection andBP0 = (0 · 10)(B) = 0 by 13.6, so H0 ⊆ N(B) = 0 by 13.13. Hence P0 = 0.

Using 13.6, we thus obtain

IH = 1[0,1](B) = 1(0,1](B) + P0 = 1(0,1](B).

We havem∑n=1

θn = 1( 1m+1

,1] → 1(0,1] bounded pointwise as m→∞.

By 13.6 we obtain, for any x ∈ H,

‖m∑n=1

Pnx− x‖2 = ‖1(0, 1m+1

](B)x‖2 = (1(0, 1m+1

](B)x|1(0, 1m+1

](B)x)

= (1(0, 1m+1

](B)x|x)→ 0 (m→∞),

since 1(0, 1m+1

] → 0 bounded pointwise as m → ∞. Moreover, by orthogonality of the

summands we have ‖x‖2 =∑

n∈N ‖Pnx‖2. For x ∈ D(A) we thus have

Ax =∑n∈N

PnAx =∑n∈N

APnx and∑n∈N

‖APnx‖2 = ‖Ax‖2 <∞.

Now let x ∈ H such that∑

n∈N ‖APnx‖2 <∞. Then∑

n∈NAPnx converges in H, and by(b) we have that

xm :=m∑n=1

Pnx ∈ D(A), Axm =m∑n=1

APn

for each m ∈ N, and xm → x in H, Axm → y :=∑

n∈NAPnx in H as m → ∞. Byclosedness of A we obtain x ∈ D(A) and Ax =

∑n∈NAPnx.

(d) Let n ∈ N. Then Hn is invariant under B, since BPn = Bθn(B) = θn(B)B by 13.6.We set Bn := B|Hn . Since B commutes with resolvents of A, the space Hn is also invariant

59

under resolvents of A.5 Hence, for any λ ∈ ρ(A), we have λ ∈ ρ(An) and

R(λ,An) = R(λ,A)|Hn

(in particular, we have σ(An) ⊆ σ(A)): Indeed, for λ ∈ ρ(A) the operator λ−An, which isa restriction of λ−A, is clearly injective. Moreover, for y ∈ Hn and x := R(λ,A) we havex ∈ Hn and (λ − An)x = (λ − A)R(λ,A)y = y, and λ − An : Hn → Hn is also surjective.Hence we have, for any n ∈ N,

Bn = R(i, An)R(−i, An), B−1n = (i− An)(−i− An) = 1 + A2

n.

Since σ(Bn) ⊆ [ 1n+1

, 1n] by 13.6 we thus have σ((1+A2

n)−1) ⊆ [ 1n+1

, 1n]. By spectral mapping

this implies σ(A2n) ⊆ [n− 1, n] and finally

σ(An) ⊆ [−√n,−√n− 1] ∪ [

√n− 1,

√n],

so (d) is proved.

Remark: But by 13.14 we have

σ(A) =⋃n∈N

σ(An) =⋃n∈N

σ(An)

where the last equality is due to the fact that by (d), for any Cauchy sequence in⋃n σ(An),

there exists n0 ∈ N and a convergent subsequence contained in σ(An0). From this we obtain

σ(A) ∩ ((−√n,−√n− 1) ∪ (

√n− 1,

√n)) = σ(An) ∩ ((−

√n,−√n− 1) ∪ (

√n− 1,

√n))

for each n ∈ N.

In 13.14, we have established an orthogonal decomposition of the space H into closedsubspaces Hn, n ∈ N, such that Hn ⊆ D(A), An := A|Hn ∈ L(Hn) and An is self-adjoint inHn and corresponds to a bounded part of the spectrum of A. We now apply 13.6 to eachof the parts An and thus define a functional calculus for A for bounded Borel measurablefunctions on the spectrum σ(A).

13.15. Theorem: Let A be a self-adjoint operator in H and, for n ∈ N, let Pn, Hn, Anbe as in 13.14. For f ∈ Bb(σ(A)) we define

f(A)x :=∑n∈N

f(An)Pnx, x ∈ H.

5The following argument also proves (∗) in the proof of (a) above: Let T ∈ L(H) such that TB = BT .Then Tp(B) = p(B)T for all polynomials p. By Weierstraß, one can approximate continuous functionson σ(B) by polynomials, and this gives Tf(B) = f(B)T for all f ∈ C(σ(B)). Finally, application of 13.2shows Tf(B) = f(B)T for all f ∈ Bb(σ(B)), where 13.2(2) is shown with the help of σ-continuity of themap Ψ from 13.6.

Here and in (∗) this is applied to f = θn and T = R(λ,A), T = C, respectively.

60

Then f(A) ∈ L(H) is well-defined, ‖f(A)‖ ≤ ‖f‖∞, and the map Ψ : B(σ(A))→ L(H) isan algebra homomorphism with Ψ(1σ(A)) = I, Ψ((z − (·))−1) = R(z, A) for z ∈ C \ R.

For f ∈ Bb(σ(A)) one has Ψ(f)∗ = Ψ(f). In particular, each f(A) is normal, f(A) isself-adjoint for real-valued f , and f(A) ≥ 0 if f ≥ 0.

If fm → f bounded pointwise as m→∞ then fm(A)x→ f(A)x for any x ∈ H.

Proof. Since f(An) ∈ L(Hn) with ‖f(An)‖ ≤ ‖f‖∞,σ(An) ≤ ‖f‖∞,σ(A) we have∑n∈N

‖f(An)Pnx‖2 ≤ ‖f‖2∞,σ(A)

∑n∈N

‖Pnx‖2 = ‖f‖2∞,σ(A)‖x‖2, x ∈ H.

Hence∑

n f(An)Pnx converges and

‖f(A)x‖ =(∑n∈N

‖f(An)Pnx‖2)1/2

≤ ‖f‖∞,σ(A)‖x‖.

This implies that Ψ : Bb(σ(A))→ L(H) is linear and continuous with norm≤ 1. Ψ(1σ(A)) =IH follows from 13.14(c). For x ∈ H and z ∈ C \ R we have (cp. 13.14(d) and its proof)

R(z, A)x =∑n∈N

R(z, A)Pnx =∑n∈N

R(z, An)Pnx.

By 12.6 we have R(z, An) = (z − (·))−1(An) for any n ∈ N, so R(z, A) = Ψ((z − (·))−1) isproved.

Notice that, for each m ∈ N,

Pmf(A)x =∑n∈N

Pmf(An)Pnx = f(Am)Pmx.

Hence, for f, g ∈ Bb(σ(A)) and x ∈ H,

(gf)(A) =∑n∈N

(gf)(An)Pnx =∑n∈N

g(An)f(An)Pnx =∑n∈N

g(An)Pnf(A)x = g(A)(f(A)x),

and Ψ is multiplicative. By orthogonality of the summands we have, for f ∈ Bb(σ(A)) andx, y ∈ H,

(f(A)x|y) =∑n∈N

(Pnf(A)x|Pny) =∑n∈N

(f(An)Pnx|Pny)Hn

=∑n∈N

(Pnx|f(An)∗Pn)Hn =∑n∈N

(Pnx|f(An)Pn(y))Hn

= (x|f(A)y),

which implies that f(A)∗ = f(A).

61

The assertions on normality, self-adjointness and positivity follow as before. Let (fm)m∈Nbe a sequence in Bb(σ(A)) that converges bounded pointwise to a function f . By linearityof Ψ we may assume that f = 0. We have for the functional calculus from 13.6 thatfm(An)x→ 0 as m→∞ for all n ∈ N and x ∈ Hn. Hence, for fixed x ∈ H, we obtain

fm(An)Pnx→ 0 for each n ∈ N.

Letting M := supm∈N ‖fm‖∞ we have, for any N ∈ N,

‖fm(A)x‖2 =∑n∈N

‖fm(An)Pnx‖2 ≤N∑n=1

‖fm(An)Pnx‖2 +M∑n>N

‖Pnx‖2.

For a given ε > 0 we now choose N such that M∑

n>N ‖Pnx‖2 ≤ ε/2, and then m0 such

that∑N

n=1 ‖fm(A)Pnx‖2 ≤ ε/2 for all m ≥ m0. Thus fm(A)x→ 0 in H is proved.

Remark: One can show that Ψ : Bb(σ(A))→ L(H) with the stated properties is unique.

Moreover, one can actually see that Pn = (θn ψ)(A) where ψ(z) = 11+z2

, so Pn = ηn(A)where ηn = 1(−

√n,−√n−1]∪[

√n−1,

√n).

We briefly mention another aspect of the Spectral Theorem.

13.16. Spectral measures: Let A be self-adjoint in H and let Ψ denote the functionalcalculus from 13.15. For all x, y ∈ H

µx,y : B(R)→ C, M 7→ µx,y(M) := (Ψ(1M)x|y) = (1M(A)x|y),

is a compex Borel measure on R. For x = y, µx,x is a positive finite measure with

µx,x(R) = (1R(A)x|x) = (x|x) = ‖x‖2,

i.e., for x ∈ H with ‖x‖ = 1, µx,x is a probability maesure on R.

For each f ∈ Bb(σ(A)) we have the repreesentation6

(f(A)x|y) =

∫f(λ) dµx,y(λ).

One wries this as

f(A) =

∫f(λ) dE(λ),

where the spectral measure (or spectral resolution E of A is given by

E : B(R)→ L(H), M 7→ E(M) := Ψ(1M) = 1M(A).

6If A is bounded this is similar to what we have done in the construction of the functional calculus in13.6.

62

Note that E is not σ-additive w.r.t. operator norm. By 13.15 however, we have that B(R)→H, M 7→ E(M)x, is σ-additive for each x ∈ H. For f ∈ Bb(R) and x ∈ H we then have

‖f(A)x‖2 = (f(A)x|f(A)x) = (|f |2(A)x|x) =

∫|f |2 d(E(·)x|x).

This opens a way to define f(A) as (in general) unbounded operator in H for each Borel-measurable function f : σ(A)→ C with domain

D(f(A)) = x ∈ H :

∫|f |2 d(E(·)x|x) <∞,

see, e.g., Rudin: Functional Analysis.

The following is another version of the Spectral Theorem (here we omit the proof, ee, e.g.,Davies: Spectral Theory and Differential Operators).

13.17. Spectral Theorem (multiplicator version): Leti A be a self-adjoint operatorin the separable Hilbert space H. then there exists a σ-finite measure space (Ω,A, µ), ameasurable function m : Ω→ σ(A) and a unitary operator U : H → L2(µ) such that

D(A) = x ∈ H : m · Ux ∈ L2(µ), Ax = U−1(m · Ux).

The functional calculus for the operator A is then obtained from the functional calculusfor the multiplication operator f 7→ m · f in L2(µ).

We come back to the question of (essential) self-adjointness for symmetric operators. Firstwe present an example.

Example: Let A = −∆ with D(A) = C∞c (Rd) in H = L2(Rd). Clearly, A is densely definedand symmetric. We determine the adjoint operator A∗. For f, g ∈ L2(Rd) we have

f ∈ D(A∗) and A∗f = g ⇐⇒ ∀ ϕ ∈ C∞c (Rd) : (−∆ϕ|f) = (f |g)

⇐⇒ ∀ ϕ ∈ C∞c (Rd) :

∫4π2|ξ|2ϕ(ξ)f(ξ) dξ =

∫ϕ(ξ)g(ξ) dξ.

Since f , g ∈ L2(Rd) we obtain equality of the integrals for all ϕ ∈ S (Rd). By C∞c (Rd) ⊂S (Rd) and the corollary in FA 5.2 we then obtain

f ∈ D(A∗) and A∗f = g ⇐⇒ 4π2|ξ|2f(ξ) = g(ξ) for a.e. ξ ∈ Rd

⇐⇒ f ∈ H2(Rd) and −∆f = g,

where for the last equivalence we used f, g ∈ L2(Rd) and the definitions of H2(Rd) and the(weak) Laplace operator. Thus we have A∗ = −∆ with D(A∗) = H2(Rd) = W 2,2(Rd).

Finally, A is essentially self-adjoint, since the graph norm of A is equivalent to ‖ ·‖W 2,2 andD(A) = C∞c (Rd) is dense in (W 2,2(Rd), ‖ · ‖W 2,2) (see FA 5.4).

For the general case we state the following in the situation of 13.9.

63

13.18. Lemma: Let A be symmetric and closed. If λ, µ ∈ C \R such that Imλ · Imµ > 0then

codimR(λ− A) = codimR(µ− A).

Proof. Let λ ∈ C \ R. We show the assertion for µ ∈ C \ R with |λ − µ| < |Imλ|. Thelemma follows by taking λ = ±in and n → ∞. We let X := [D(A)]. By 13.9, λ − A isinjective and Y := R(λ− A) is closed. Let Z := X × Y ⊥ and consider

z − A : X × Y ⊥ → H, (x, y) 7→ (z − A)x+ y, z ∈ C \ R.

For z = λ, this operator is bijective with inverse Rh := ((λ−A)−1Ph, (I − P )h), where Pdenotes the orthogonal projection from H onto Y . If |µ− λ| < |Imλ|, then

µ− A = λ− A+ (µ− λ)π1 = (IH + (µ− λ)π1R)λ− A,

where π1 : X × Y ⊥ → H, (x, y) 7→ x, and ‖π1R‖ = ‖(λ− A)−1P‖L(H) ≤ 1/|Imλ| by 13.9.Hence

(µ− A)−1 = R∞∑k=0

(−π1R)k(µ− λ)k

and µ− A is bijective X × Y ⊥ → H. In particular, R(µ − A) + Y ⊥ = H. Moreover, if

x ∈ D(A) and (µ − A)x = −y ∈ Y ⊥ then µ− A(x, y) = 0, so x = 0 = y, which impliesR(µ−A)∩Y ⊥ = 0. This means that Y ⊥ is a complement both for R(λ−A) and R(µ−A),so codimR(λ− A) = codimR(µ− A).

Remark: The proof is similar to what we have done for Fredholm operators. The numbersn±(A) := codimR(±i − A) are called defect indices, and it can be shown that A has aself-adjoint extension if and only if n+(A) = n−(A).

If A is in addition positive, i.e. (Ax|x) ≥ 0 for all x ∈ D(A), then one can show in a similarway that n+(A) = n−(A) = codim (µ+A) for any µ > 0, and A has self-adjoint extensions.We will present how to get one in the next section. end Tue

09.07.19

64

14 Functional calculus for sectorial operators

A very nice way to obtain self-adjoint (and more general) operators in a Hilbert space isvia sesquilinear forms. We have seen this in 13.5 for the bounded case, but here we wantto define unbounded operators. We start with a motivation.

14.1. The Dirichlet form: We recall the Divergence Theorem:

Let Ω ⊆ Rd be a bounded domain with ∂Ω ∈ C1 and U ⊇ Ω be open. Then,for F ∈ C1(U), ∫

Ω

divF dx =

∫∂Ω

ν · F dσ,

where ν : ∂Ω→ Rd denotes the outer unit normal.

Letting F = v∇u we obtain

Corollary: Let Ω ⊆ Rd be a bounded domain with ∂Ω ∈ C1 and let U ⊇ Ω be open.Then, for u ∈ C2(U) and v ∈ C1(U):∫

Ω

∇u · ∇v dx = −∫

Ω

(∆u)v dx+

∫∂Ω

∂u

∂νv dσ. (∗)

Now let Ω ⊆ Rd be an arbitrary domain. For u, v ∈ W 1,2(Ω) we define

a(u, v) :=

∫Ω

∇u · ∇v dx.

Then a : W 1,2(Ω) ×W 1,2(Ω) → C is sesquilinear, a(u, v) = a(v, u) for all u, v ∈ W 1,2(Ω),and

Re a(u, u) =

∫Ω

|∇u|2 dx ≥ 0.

Notice that

(u|v)W 1,2 =

∫Ω

∇u · ∇v + uv dx = a(u, v) + (u|v)L2

‖u‖2W 1,2 =

∫Ω

|∇u|2 + |u|2 dx = a(u, u) + ‖u‖2L2 .

In particular, a(·, ·) + (·|·)L2 is a scalar product on W 1,2(Ω) and W 1,2(Ω) is complete w.r.t.this scalar product. This also holds for the closed linear subspace

W 1,20 (Ω) = C∞c (Ω)

‖·‖W1,2

of W 1,2(Ω). Since C∞c (Ω) is dense in L2(Ω), also W 1,20 (Ω) and W 1,2(Ω) are dense in L2(Ω).

We shall define the Laplace operator with Dirichlet and with Neumann boundary conditionsin arbitrary domains Ω ⊆ Rd via the Dirichlet form a with domain W 1,2

0 (Ω) and W 1,2(Ω),respectively. We first remark that, for a bounded domain Ω with C1-boundary and an opensuperset U ⊇ Ω, we have:

65

• If u ∈ C2(U) with u = 0 on ∂Ω and v ∈ C∞c (Ω) then

a(u, v) =

∫Ω

(−∆u)v dx = (−∆u|v)L2 .

• If u ∈ C2(U) with ∂u∂ν

= 0 on ∂Ω and v ∈ C1(U) then

a(u, v) =

∫Ω

(−∆u)v dx = (−∆u|v)L2 .

14.2. Sesquilinear forms: Let H be a Hilbert space and V ⊆ H be a dense linearsubspace. A sesquilinear form a : V × V → C is called

• symmetric if a(u, v) = a(v, u) for all u, v ∈ V ,

• accretive, if Re a(u, u) ≥ 0 for all u ∈ V ,

• sectorial, if there exists ω ∈ [0, π/2) such that a(u, u) ∈ Σω for all u ∈ V , whereΣ0 := [0,∞) and

Σω := z ∈ C \ 0 : |arg z| ≤ ω ∪ 0 for ω ∈ (0, π/2).

In this case a is called sectorial of angle ω.

If a is sectorial of angle ω then a is accretive and

|Im a(u, u)| ≤ tanω Re a(u, u), u ∈ V.

The adjoint form a∗ : V × V → C of a is defined by a∗(u, v) = a(v, u). We clearly have(a∗)∗ = a; a = a∗ ⇔ a is symmetric; a is accretive⇔ a∗ is accretive; a is sectorial of angleω ⇔ a∗ is sectorial of angle ω.

For each sesquilinear form a the sesquilinear forms

Re a :=1

2

(a + a∗

), Im a :=

1

2i

(a− a∗

)are symmetric and we have a = Re a + i Im a.

WARNING: We have (Re a)(u, u) = Re(a(u, u)

)for all u ∈ V , but (Re a)(u, v) is not

real in general! The same holds for Im a.

Estimates: If a is sectorial of angle ω then we have for all u, v ∈ V :

|(Re a)(u, v)| ≤ Re a(u, u)1/2 Re a(v, v)1/2,

|(Im a)(u, v)| ≤ (tanω) Re a(u, u)1/2 Re a(v, v)1/2,

|a(u, v)| ≤ (1 + tanω) Re a(u, u)1/2 Re a(v, v)1/2.

66

Proof. The first inequality is Cauchy-Schwarz, and the third inequality is implied by thefirst and the second.

For the proof of the second inequality we may assume that (Im a)(u, v) is real (otherwisewe multiply u with a suitable γ ∈ C with |γ| = 1). Then we have

(Im a)(u, v) =1

4

((Im a)(u+ v, u+ v)− (Im a)(u− v, u− v)

),

and by sectoriality of a thus

|(Im a)(u, v)| ≤ tanω

4

((Re a)(u+ v, u+ v) + (Re a)(u− v, u− v)

)=

tanω

2

(Re a(u, u) + Re a(v, v)

).

For each α > 0 we hence have

|(Im a)(u, v)| = |(Im a)(αu, α−1v)|

≤ tanω

2

(α2 Re a(u, u) + α−2 Re a(v, v)

).

If Re a(v, v)Re a(u, u) = 0 then the assertion follows letting α → 0+ or α → ∞. If

Re a(v, v)Re a(u, u) 6= 0 then the assertion follows with α =(Re a(v, v)/Re a(u, u)

)1/4.

Remark: If a : V × V → C is sectorial of angle ω and δ > 0 then

(u|v)V := (Re a)(u, v) + δ(u|v)H

defines a scalar product on V (usually one takes δ = 1, all these scalar products areequivalent). In this case the sesquilinear form a i called closed (on V ) if V is a Hilbertspace w.r.t. (·|·)V . Obviously, a is closed if and only if a∗ is closed.

Example: In 14.1 the Dirichlet form a is symmetric, sectorial of angle 0 and closed, bothon V = W 1,2(Ω) and on V = W 1,2

0 (Ω). Thus we can apply the following proposition inboth situations. end Thu

11.07.1914.3. Proposition: Let H be a Hilbert space and V ⊆ H a dense linear subspace. Leta : V × V → C be a sesquilinear which is sectorial of angle ω and closed. Define the linearoperator A in H by the following for all u, f ∈ H:

u ∈ D(A) and Au = f :⇐⇒ u ∈ V and ∀v ∈ V : a(u, v) = (f |v).

Then A is a closed linear operator in H, D(A) is dense in (V, (·|·)V ), we have σ(A) ⊆ Σω

and‖R(λ,A)‖L(H) ≤ 1/d(λ,Σω) for all λ ∈ C \ Σω.

Moreover, we have (Ax|x)H ∈ Σω for all x ∈ D(A). If we define in this way an operatorfor the adjoint sesquilinear form a∗ then we obtain the adjoint operator A∗. In particular,A is self-adjoint in H if a is symmetric.

67

Remark: Note that (by denseness of V in H) the rule above really defines a linear operatorA in H. This operator is called the operator associated with a, sometimes denoted by A ∼ a.

Proof. We denote by V ∗ the antidual space of (V, (·|·)V ), i.e. the linear space of all contin-uous antilinear functionals φ : V → C. We study the linear operator

B : V → V ∗, u 7→ a(u, ·) + (u|·)H ,

which is continuous by the estimates in 9.2. For u ∈ V with ‖u‖V = 1 we have

‖Bu‖V ∗ = sup‖v‖V =1

|a(u, v) + (u|v)H | ≥ |a(u, u) + (u|u)H | ≥ Re a(u, u) + ‖u‖2H = ‖u‖2

V = 1,

and B is injective and R(B) is closed in V ∗. Moreover, R(B) is dense in V ∗, otherwisethere were (by reflexivity of V and Hahn-Banach) a v ∈ V with B(u)(v) = 0 for all u ∈ V ,which contradicts the estimate we just proved. Hence B : V → V ∗ is an isomorphism andhas a continuous inverse R : V ∗ → V with norm ≤ 1.7

We identify H with its antidual space H∗. Since V is dense and continuously embeddedin H, the injection H∗ → V ∗ is continuous. Hence we have continuous embeddings V →H = H∗ → V ∗. Here we identify h ∈ H with (h|·)H

∣∣∣V∈ V ∗.

Now we let R := R|H . Then R ∈ L(H) and R is injective (since R is injective). We defineA := R−1 − IH . Then A is a closed linear operator in H and (A + 1)−1 = R ∈ L(H), i.e.−1 ∈ ρ(A). Moreover, we have for u, f ∈ H:

u ∈ D(A), Au = f ⇐⇒ R(f + u) = u ⇐⇒ u ∈ V and R(f + u) = u

⇐⇒ u ∈ V and Bu = u+ f

⇐⇒ u ∈ V and ∀v ∈ V : a(u, v) + (u|v)H = (u|v)H + (f |v)H

⇐⇒ u ∈ V and ∀v ∈ V : a(u, v) = (f |v)H .

D(A) is dense in V : We have D(A) = R(R) = R(H). Since R : V ∗ → V is an isomorphism,it suffices to show that H = H∗ is dense in V ∗. But otherwise there would exist v ∈ Vwith v 6= 0 and (h|v)H = 0 for all h ∈ H, which implies ‖v‖H = 0, a contradiction.

Now let the operator B be associated with the sesquilinear form a∗ via

v ∈ D(B) and Bv = g :⇐⇒ v ∈ V and ∀w ∈ V : a∗(v, w) = (g|w).

Then we have for u ∈ D(A) and v ∈ D(B): u, v‖V ‖ and

(Au|v) = a(u, v) = a∗(v, u) = (Bv|u) = (u|Bv).

This implies v ∈ D(A∗) and A∗v = Bv, i.e. B ⊆ A∗. By −1 ∈ ρ(B) and −1 ∈ ρ(A∗) (whichfollows from −1 ∈ ρ(A)) we obtain B = A∗. If a is symmetric then a∗ = a and we obtainA = A∗, i.e. A is self-adjoint.

7This also follows from Lax-Milgram.

68

For all u ∈ D(A) we obviously have (Au|u)H = a(u, u) ∈ Σω. For λ ∈ C\Σω and u ∈ D(A)with ‖u‖ = 1 we thus have

‖(λ− A)u‖ ≥ |((λ− A)u|u)| = |λ− (Au|u)| ≥ d(λ,Σω).

Hence λ−A is injective and R(λ−A) is closed in H. But then also (λ−A)∗ = λ−A∗ isinjective, which implies that R(λ−A) is dense inH. We conclude λ ∈ ρ(A) and ‖R(λ,A)‖ ≤1/d(λ,Σω) as asserted.

Examples: (1) Let Ω ⊆ Rd be an arbitrary domain. Then the Dirichlet Laplace operator−∆D on Ω is the operator associated with the Dirichlet form a from 14.1 on VD = W 1,2

0 (Ω),and the Neumann Laplace operator −∆N on Ω is the operator assciated with the Dirichletform a from 14.1 on VN = W 1,2(Ω). Note that we have VD ⊆ VN but D(−∆D) 6⊆ D(−∆N)in general, since then −1 ∈ ρ(−∆D) ∩ ρ(−∆N) implied −∆D = −∆N (which is false, e.g.,for bounded domains with C1-boundary). end Tue

16.07.19(2) Let Ω ⊆ Rd be a bounded domain with ∂Ω ∈ C1 and let a : Ω → Cd×d be bounded

and measurable such that ξta(x)ξ ∈ Σω for a.e. x ∈ Ω, all ξ ∈ Cd and some ω ∈ [0, π/2).

Assume there exists η > 0 such that

Re ξta(x)ξ ≥ η|ξ|2 for a.e. x ∈ Ω and all ξ ∈ Cd.

Then the sesquilinear form

a(u, v) =

∫Ω

(a(x)∇u) · ∇v dx, u, v ∈ V = W 1,2(Ω),

is sectorial of angle ω and closed in H = L2(Ω). The operator associated with a is formallygiven by Au = −div (a(·)∇u) with boundary conditions ν · a(·)∇u = 0 on ∂Ω (conormalboundary condition).

14.4. Closable forms: Let H be a Hilbert space and V ⊆ H be a dense linear subspace.Let a : V × V → C be a sectorial form that is not closed. We can extend a uniquely toa continuous sesquilinear form a on the completion V of (V, (·|·)V ). This extended form is

again sectorial of the ame angle. In order to apply Proposition 9.3 we have to realize V asa linear subspace of H. This is pssible if and only if the following holds:

For each ‖·‖V -Cauchy sequence (un) in V with ‖un‖H → 0 one has ‖un‖V → 0.

Sectorial forms satisfying this property are called closable. The condition is equivalent tothe continuous extension of the embedding J : V → H, u 7→ u, onto V being injective. Inthis case, a is clled the closure of a.

Examples: (1) Let A be densely defined and symmetric in H with (Ax|x) ≥ 0 for allx ∈ D(A). Then a(u, v) := (Au|v) defines a symmetric sesquilinear form on V := D(A)

69

which is sectorial of angle 0. This form is closable (see exercise) and the closure of this form

is associated with the so-called Friedrichs extension A of A. Note that A is self-adjoint inH with (Au|u) ≥ 0 for all u ∈ D(A).

(2) Let H = L2(0, 1), V = C[0, 1] and a : V × V → C given by a(u, v) = u(0)v(0). Then ais symmetric and sectorial, but not closed. Consider, e.g., un(t) = (1− nt)1[0,1/n](t). Then

(un) is Cauchy w.r.t. ‖u‖V =√|u(0)|2 + ‖u‖2

L2 with ‖un‖L2 → 0, but ‖un‖V → 1.

14.5. Square roots: Let A be self-adjoint in H with (Ax|x) ≥ 0 for all x ∈ D(A) and letPn, Hn, and An be as in 13.14. We define the operator A1/2 via

D(A1/2) = x ∈ H :∑n

‖A1/2n Pnx‖2 <∞,

A1/2x =∑n

A1/2n Pnx for x ∈ D(A1/2).

Then D(A) ⊆ D(A1/2) and D(A) is dense in D(A1/2) w.r.t. the graph norm of A1/2.Moreover, A1/2 is self-adjoint in H, A1/2 ≥ 0, and A1/2A1/2 = A, i.e.

D(A) = x ∈ D(A1/2) : A1/2x ∈ D(A1/2), Ax = A1/2A1/2x for x ∈ D(A).

Sketch of proof. 8 For the operators An ∈ L(Hn), we use 13.6, in particular A1/2n is self-

adjoint in Hn and A1/2n A

1/2n = An. For x ∈ D(A) we have by Cauchy-Schwarz∑

n

‖A1/2n Pnx‖2 =

∑n

(AnPn|Pnx) ≤(∑

n

‖AnPnx‖2)1/2(∑

n

‖Pnx‖2)1/2

= ‖Ax‖‖x‖,

hence we have D(A) ⊆ D(A1/2). The set of all x ∈ H with Pnx 6= 0 just for finitely manyn is a subset of D(A) and dense in D(A1/2) w.r.t. to the graph norm of A1/2. Symmetryof A1/2 is clear. By −1−A = (i±A1/2)(i∓A1/2) the operator i±A1/2 is surjective. Thisimplies self-adjointness by 13.9 und 13.11. Checking the algebraic properties is easy (seealso the mentioned exercise).

Now let V be dense in H and a : V × V → C be symmetric, sectorial and closed, andA ∼ a. Then

D(A1/2) = V and ‖A1/2u‖H = a(u, u) for u ∈ V .

Proof. For u ∈ D(A) ⊆ D(A1/2) ∩ V we have

a(u, u) = (Au|u) = (A1/2u|A1/2u) = ‖A1/2u‖2H ,

and D(A) is dense both in(D(A1/2),

(‖u‖2

H +‖A1/2u‖2H

)1/2)and (by 14.3) in

(V,(a(u, u)+

‖u‖2H

)1/2). This implies the assertion.

8In an exercise, f(A) had been constructed more generally for Borel measurble and locally boundedf : σ(A)→ C. Here, f =

√· is locally bounded on σ(A) ⊆ [0,∞).

70

Square Root Problem (Kato ’61/’62): Let V be dense in H, let a : V × V → Cbe sectorial and closed, and A ∼ a, B ∼ Re a. Do we have D(B1/2) = D(A1/2) and‖B1/2u‖ ∼ ‖A1/2u‖? Here, B1/2 is defined via the holomorphic functional calculus (seebelow).

The answer is “No”, in general (McIntosh). The answer is “Yes” for elliptic operators indivergence form A = −div (a(·)∇) on Rd (McIntosh et al. ’01).

14.6. Operatos with compact resolvents: Let V be dense in H, let a : V × V → C besymmetric, sectorial and closed, and A ∼ a. The following are equivalent

(i) A has compact resolvents,

(ii) the embedding V → H is compact,

(iii) A1/2 has compact resolvents.

In this case we can order the eigenvalues of A (with multiplicity) by size 0 ≤ λ1 ≤ λ2 ≤ . . .and we have the following “mini-max” formula for each n ∈ N:

λn = infλ(L) : L ⊆ V, dimL = n,

where λ(L) = maxa(u, u) : u ∈ L, ‖u‖ = 1.end Thu18.07.19

Proof. We find a corresponding orthonormal sequence (en) of eigenvectors. Then Au =∑j λj(u|ej)ej, A1/2u =

∑j λ

1/2j (u|ej)ej, and a(u, u) =

∑j λj|(u|ej)|2. We clearly have

λ1 = infa(u, u) : u ∈ V . Now let n ≥ 2 and put Mn := line1, . . . , en. Then λ(Mn) = λn.Let, on the other hand, let L be a linear subspace of V of dimension n and let Pu :=∑n−1

j=1 (u|ej)ej be the orthogonal projection in H onto Mn−1. Since Mn has dimension >n − 1, we find u0 ∈ L with ‖u0‖ = 1 and Pu0 = 0, i.e. with u0 ⊥ ej for j = 1, . . . , n − 1.We then have

a(u0, u0) =∞∑j=n

λj|(u0|ej)|2 ≥ λn

∞∑j=n

|(u0|ej)|2 = λn‖u0‖2 = λn.

The assertion is proved.

More can be found in Section 4.5 in Davies: Spectral Theory and Differential Operators.

Examples: (1) Let Ω ⊆ Rd be a bounded domain, then the embedding W 1,20 (Ω) → L2(Ω)

is compact (see FA). Hence the formula applies to the Dirichlet Laplace operator on Ω.

(2) Let Ω ⊆ Rd be a bounded domain with ∂Ω ∈ C1, then the embedding W 1,2(Ω) → L2(Ω)is compact. Hence the formula applies to the Neumann Laplace operator on Ω.

71

14.7. Sectorial operators: Let X be a Banach space and ω ∈ [0, π). A linear operator Ain X is called sectorial of angle ω if σ(A) ⊆ Σω and for each θ ∈ (ω, π) there exists Mθ > 0such that

‖R(λ,A)‖ ≤ Mθ

|λ|for all λ ∈ C \ Σθ.

Warning: The definiion of this notion is not uniform in the literature!

Example: Let V be dense in the Hilbert space H, let a : V × V → C be sectorial of angleω ∈ [0, π/2) and closed, and let A ∼ a. Then A is sectorial of angle ω. In particular, aself-adjoint operator A with A ≥ 0 is sectorial of angle 0.

14.8. Holomorphic functional calculus for sectorial operators: Let ω ∈ [0, π), Abe sectorial of angle ω in the Banach space X.

Remark: If D(A) is dense in X then, for all x ∈ X, we have λ(λ+A)−1x→ x as λ→∞.If R(A) is dense in X then, for all x ∈ X we have A(λ + A)−1x→ x as λ→ 0+ and A isinjective. Here, λ > 0 is real.

Proof. We have M := supλ>0 ‖λ(λ+A)−1‖ <∞ and supλ>0 ‖A(λ+A)−1‖ ≤M +1. Henceit is sufficient to show convergence on a dense subspace. For x ∈ D(A) we have

λ(λ+ A)−1x− x = −A(λ+ A)−1x = −λ−1 λ(λ+ A)−1Ax︸ ︷︷ ︸bounded

→ 0 (λ→∞),

and for x = Ay ∈ R(A) we have

A(λ+ A)−1x− x = −λ(λ+ A)−1x = −λA(λ+ A)−1y︸ ︷︷ ︸bounded

→ 0 (λ→ 0+).

The assertions are proved.

Assumption: From now on we suppose that A has dense domain and range.

For θ ∈ (0, π) let Σ0θ denote the interior of Σθ, and let H∞(Σ0

θ) denote the space ofall holomorphic functions ϕ : Σ0

θ → C such that there exist ε, C > 0 with |ϕ(z)| ≤C min|z|ε, |z|−ε for all z ∈ Σ0

θ.

Example: For each θ ∈ (0, π) we have ϕ(z) = z(1+z)2

= 11+z− 1

(1+z)2∈ H∞(Σ0

θ).

For θ ∈ (ω, π) and ϕ ∈ H∞(Σ0θ) we define

ϕ(A) :=1

2πi

∫Γσ

ϕ(λ)R(λ,A) dλ,

where Γσ = ∂Σσ with σ ∈ (ω, θ) is parametrized by γσ(t) = |t|e−iσsgn (t), t ∈ R, and theintegral is an absolutely convergent improper Riemannian integral with values in L(X) (ora Bochner integral). Notice that, for λ on Γσ,

‖ϕ(λ)R(λ,A)‖ ≤ CMσ|λ|−1 min|λ|ε, |λ|−ε = CMσ min|λ|ε−1, |λ|−1−ε.

72

By Cauchy’s integral theorem the definition does not depend on σ ∈ (ω, θ): intersect, forω < σ < σ < θ and 0 < r < R, the set Σσ \ Σσ with B(0, R) \ B(0, r) and integrateϕ(λ)R(λ,A) along the boundary. By Cauchy this integral is = 0. The estimates of theintegrand on |λ| = R and |λ| = r are . rε−1 and . R−ε−1, respectively, and the length ofthe intervals are ∼ r and ∼ R, respectively. The integrals over these parts of the boundarythus tend to 0 as r → 0+ and R→∞, respectively.

The map H∞0 (Σ0θ) → L(X), ϕ 7→ ϕ(A), is linear and multiplicative. The proof of multi-

plicativity is very similar to what we have done in 9.1(a), see also 9.3. We only have to usethat by Cauchy, for any g ∈ H∞0 (Σ0

θ),

1

2πi

∫Γσ

g(λ)

λ− µdµ =

g(λ) , λ ∈ Σ0

σ

0 , λ 6∈ Σσ.

Clearly, ϕ(A) commutes with resolvents of A. end Tue23.07.19A first extension: For

F ∈ H∞0 (Σ0θ) + span

1

1 + z, 1

=: E(Σ0θ),

i.e. for F (z) = ϕ(z) + a1+z

+ b with ϕ ∈ H∞0 (Σ0θ), we define

F (A) := ϕ(A) + a(1 + A)−1 + bI ∈ L(X).

Then E(Σ0θ) → L(X) is linear and multiplicative. Linearity is clear. In order to prove

multiplicativity let G = ψ + c1+z

+ d ∈ E(Σ0θ). We study the product with the function F

from above. There are nine terms but we only have to look closer at the following three:aψ

1+z, cϕ

1+z, and ac

(1+z)2.

Lemma 1: For ϕ ∈ E(Σ0θ) we have

(1 + A)−1ϕ(A) =1

2πi

∫Γσ

ϕ(λ)

1 + λ

R(λ,A)dλ.

For the proof recall (1+A)−1R(λ,A) = (1+λ)−1(R(λ,A)−R(−1, A)) and apply Cauchy’sTheorem for sectors.

Lemma 2: If F ∈ H∞0 (Σ0θ) + span(1 + z)−1 is holomorphic on a neighborhood of B(0, δ)

then

F (A) =1

2πi

∫Γσ,δ

F (λ)R(λ,A) dλ,

where Γσ,δ = ∂(Γσ∪B(0, δ)) is parametrized by arc length and Γσ lies to the left. For δ < 1we have furthermore

(1 + A)−1F (A) =1

2πi

∫Γσ,δ

F (λ)

1 + λR(λ,A) dλ.

73

Proof. By Cauchy the assertion is clear for ϕ ∈ H∞0 . So we just have to show it forF (z) = (1 + z)−1, i.e. we have to show

(1 + A)−1 =1

2πi

∫Γσ,δ

R(λ,A)

1 + λdλ.

We integrate, for R > 1, along the boundary of (C \ (Σθ ∪ B(0, δ)) ∩ B(0, R). By Cauchythis inegral equals −R(−1, A) = (1 + A)−1. For |λ| = R the integrand satisfies a bound. R−2, where the length of the interval is ∼ R. The integral over this part of the boundaryis therefore . R−1 and vanishes for R→∞. The proof of the second formula is similar tothe proof of Lemma 1.

Via Lemma 1 and Lemma 2 we obtain multiplicativity. We also obtain for ϕ(z) = z(1 +z)−2 = (1 + z)−1 − (1 + z)−2 that ϕ(A) = A(1 + A)−2. This operator is injective.

Example: Suppose that θ < π/2. For t > 0 we have F (z) = e−tz ∈ H∞0 (Σ0θ) + span(1 +

z)−1. We obtain e−tAe−sA = e−(t+s)A for all t > 0. Taking δ = 1/t we obtain the normestimate

‖e−tA‖ ≤ Mσ

π

∫ ∞1/t

e− cos(σ)tr dr

r+Mσt

∫ 2π

0

e−Re(teiα/t)/t dα,

where the right hand side is independent of t. Hence we have supt>0 ‖e−tA‖ <∞. One canshow that z 7→ e−zA is analytic on a suitable sector depending on θ (→ bounded analyticsemigroups, evolution equations).

Moreover, one can obtain for each ϕ ∈ H∞0 (Σ0θ) by similar techniques supt>0 ‖ϕ(tA)‖ <∞.

This applies (in case θ < π/2) to functions ϕ(z) = zke−z with k ∈ N.

We shall extend the holomorphic functional calculus to certain functions F outside E(Σ0θ),

but we allow F (A) to be unbounded. We follow an algebraic approach (cp. M. Haase: TheFuctional Calculus for Sectorial Operators).

14.9. A functional calculus of unbounded operators: In the situation of 14.8 let F :Σ0θ → C be a holomorphic function such that there exists g ∈ H∞0 (Σ0

θ) with Fg ∈ H∞0 (Σ0θ)

and g(A) injective. Such a g is called a regularizer for F . We then define

F (A) := g(A)−1(Fg)(A),

i.e. for x, y ∈ X we have x ∈ D(F (A)) and F (A)x = y if and only if (Fg)(A)x = g(A)y.Note that F (A) is a closed linear operator.

Example: If there exists C > 0 such that |F (z)| ≤ C max|z|−n, |z|n, then z 7→ zm(1 +z)−2m is a regularizer for F for every m > n.

F (A) is well-defined: Let h be another regularizer for F . Then also gh is a regularizer forF and (Fg)(A)x = g(A)y is equivalent to h(A)(Fg)(A)x = h(A)g(A)y. On the left hand

74

side we have (Fgh)(A)x = g(A)(Fh)(A)x and on the right ahnd side we have g(A)h(A)y.Hence the condition is equivalent to (Fh)(A)x = h(A)y.

F 7→ F (A) is “linear”: If g is a regularizer for F and h is a regularizer for G then gh is aregularizer for F and for G, and linearity is easily shown.

F 7→ F (A) is “multiplicative”: We have F (A)G(A) ⊆ (FG)(A) with D(F (A)G(A)) =D(G(A)) ∩D((FG)(A)).

Proof. Let g be a regularizer for F and h be a regularizer for G. Then gh is a regularizerfor FG. First let x ∈ D(G(A)) with G(A)x = y ∈ D(F (A)) and F (A)y = z. Then

(FGgh)(A)x = (Fg)(A)(Gh)(A)x = (Fg)(A)h(A)y = h(A)(Fg)(A)y = h(A)g(A)z = (gh)(A)z,

hence x ∈ D((FG)(A)) and (FG)(A)x = z. Now let x ∈ D(G(A)) ∩ D((FG)(A)) andy = G(A)x, z = (FG)(A)x. We have to show y ∈ D(F (A)) and F (A)y = z. But gh is aregularizer for F and

(gh)(A)z = (FGgh)(A)x = (Fg)(A)(Gh)(A)x = (Fg)(A)h(A)y = (Fgh)(A)y,

which implies the assertion.

14.10. Fractional powers: For α ∈ R we let Fα(z) := zα. Then Fα has regularizers andAα := Fα(A) is well-defined. For all α, β ∈ R we then have

AαAβ = Aα+β on D(AαAβ) = D(Aβ) ∩D(Aα+β).

14.11. Bounded H∞-calculus: Each F ∈ H∞(Σ0θ) has regularisizers and F (A) is a well-

defined closed operator in X. One says that A has a bounded H∞(Σ0θ)-calculus if there

exists C > 0 such that

‖F (A)‖ ≤ C‖F‖∞,Σ0θ

for all F ∈ H∞(Σ0θ).

One can show that this is the case if and only if there exists C > 0 such that

‖ϕ(A)‖ ≤ ‖ϕ‖∞,Σ0θ

for all ϕ ∈ H∞0 (Σ0θ).

The reason behind is the so-called “convergence lemma”. More details on this and thebounded H∞-functional calculus can be found in Chapter 5 of

M. Haase: The Functional Calculus for Sectorial Operators, Birkhauser 2006,

and Section 9 of

P. Kunstmann, L. Weis: Maximal Lp-Regularity for Parabolic Equations,Fourier Multiplier Theorems and H∞-Functional Calculus, in Functional Ana-lytic Methods for Evolution Equations, Springer Lecture Notes 1855, Springer2004, pp. 65–312.

75