Seminar Linear Operators Boundary Triplets and Self ... · Seminar Linear Operators Boundary...

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Seminar Linear Operators Boundary Triplets and Self-adjoint Extensions. Positive Self-adjoint Extensions Nguyen 314119 September 22, 2013 Introduction Studying the constructions and properties of self-adjoint extensions of a densely defined lower semi-bounded symmetric operator T on some Hilbert space H is one of the most important topics in [2]. By using a boundary triplet (K, Γ 0 , Γ 1 ) of the adjoint T , all self-adjoint extensions of T can be parametrized as restric- tions T B of T on D(T B )= {x D(T ) : (Γ 0 x, Γ 1 x) ∈ B} , where B is a self-adjoint relation on K, or equivalently, as restrictions T B of T on D(T B )= {x D(T ):Γ 0 x D(B) and BΓ 0 x = P B Γ 1 x} , where B is a self-adjoint operator on some closed subspace K B of K, and P B is the orthogonal projection of K onto K B .([2, Section 14.2, esp. Theorem 14.10]) Moreover, by Theorem 1.21 of Krein-Naimark, one can represent T B and its domain by Gamma-Fields, Weyl functions, and the operator T 0 := T {0}⊕K T B | B={0}⊕K . This representation leads to an interesting fact, which is the main result of Section 2 of this text: if the operator T 0 is the Friedrich extension T F of T , one can express the form associated to an extenstion T B in terms of the form associated to T F and of the positive self-adjoint relation B− M (λ). Section 3 considers the case where the greatest lower bound m T of T is positive. In this case, we obtain the decomposition of the form of T B in terms of the form of the Friedrich extension T F and of B. One can think the part of T F as the ”hard” 1 one of T B , which is ”fixed”, and the part of B as the ”soft” one, which may be change if one changes B. This decomposition is very useful to compare positive self-adjoint extensions of T (Theorem 3.1). Considering 1 According to [2, Remark in p.292], Krein called T F the hard extension of T . 1

Transcript of Seminar Linear Operators Boundary Triplets and Self ... · Seminar Linear Operators Boundary...

Page 1: Seminar Linear Operators Boundary Triplets and Self ... · Seminar Linear Operators Boundary Triplets and Self-adjoint Extensions. Positive Self-adjoint Extensions Nguyen 314119 September

Seminar Linear Operators

Boundary Triplets and Self-adjoint Extensions.

Positive Self-adjoint Extensions

Nguyen 314119

September 22, 2013

Introduction

Studying the constructions and properties of self-adjoint extensions of a denselydefined lower semi-bounded symmetric operator T on some Hilbert space H isone of the most important topics in [2]. By using a boundary triplet (K,Γ0,Γ1)of the adjoint T ∗, all self-adjoint extensions of T can be parametrized as restric-tions TB of T ∗ on

D(TB) = x ∈ D(T ∗) : (Γ0x,Γ1x) ∈ B ,

where B is a self-adjoint relation on K, or equivalently, as restrictions TB of T ∗

onD(TB) = x ∈ D(T ∗) : Γ0x ∈ D(B) and BΓ0x = PBΓ1x ,

where B is a self-adjoint operator on some closed subspace KB of K, and PB isthe orthogonal projection of K onto KB .([2, Section 14.2, esp. Theorem 14.10])

Moreover, by Theorem 1.21 of Krein-Naimark, one can represent TB and itsdomain by Gamma-Fields, Weyl functions, and the operator T0 := T0⊕K ≡

TB|B=0⊕K.This representation leads to an interesting fact, which is the main result of

Section 2 of this text: if the operator T0 is the Friedrich extension TF of T ,one can express the form associated to an extenstion TB in terms of the formassociated to TF and of the positive self-adjoint relation B −M(λ).

Section 3 considers the case where the greatest lower bound mT of T ispositive. In this case, we obtain the decomposition of the form of TB in termsof the form of the Friedrich extension TF and of B. One can think the part ofTF as the ”hard”1 one of TB , which is ”fixed”, and the part of B as the ”soft”one, which may be change if one changes B. This decomposition is very usefulto compare positive self-adjoint extensions of T (Theorem 3.1). Considering

1According to [2, Remark in p.292], Krein called TF the hard extension of T .

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an extreme case of the operator B, we obtain a representation of the Krein-von Neumann extensions, which is the smallest extension among all positiveself-adjoint extensions of T .

The last section introduces two examples for the Krein-von Neumann exten-sion, where T is an differential operator and the Krein-von Neumann extensionTN is expressed as a differential operator with boundary conditions. The maintool to solve theses examples is the decomposition of TN stated in Section 3.

The materials, which we use in this text, but do not belongs to [2, Section14.7-8], are recalled in the preliminary, Section 1.

1 Preliminary

In this section, we recall some important knowledge about symmetric and self-adjoint operators on Hilbert spaces, which we use in next Sections.

1.1 Some basic definitions: lower semi-boundedness, self-adjoint operators and their forms, Friedrichs exten-sion

Definition 1.1. Assume A is a self-adjoint operator on some Hilbert space H.Let EA denote its spectral measure by the well-known spectral theorem (forself-adjoint operators). We define the (sesquilinear) form of A by

D[A] := D(|A|1/2) =

x ∈ H :

R|λ|d EA(λ)x, x < ∞

A[x, y] :=

Rλ EA(λ)x, y , for x, y ∈ D[A].

Remark 1. In order to understand this definition, one should be familiar withthe spectral theorem and the functional calculus. We recall only very essentialfacts. By the spectral theorem and functional calculus, given a self-adjointoperator A on some Hilbert space H, we can define for a Borel function f a newoperator f(A) by

D(f(A)) =

x ∈ H :

R|f(λ)|2d EA(λ)x, x < ∞

f(A)x, y :=

Rf(λ)d EA(λ)x, y , for x ∈ D(A), y ∈ H.

(see [2, eq. (5.10) and (5.11)].) Now it rises a question of whether we get the”origin” domain of A if we use the above formula for f(λ) := λ. The answer isYES (see the paragraph after [2, eq. (5.11)]), and therefore, the domain of Acan be represent as

D(A) =

x ∈ H :

R|λ|

2d EA(λ)x, x < ∞

.

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Moreover, if we choose f(λ) := |λ|1/2, then, by the above representation ofD(A), we have D(A) ⊆ D(|A|1/2) ≡ D[A]. Then, for x, y ∈ D[A],

|A[x, y]| ≡

Rλ EA(λ)x, y

Rλd EA(λ)x, x

1/2

Rλd EA(λ)x, x

1/2

< ∞

by [2, Lemma 4.8. (ii)]. Therefore A[x, y] is well-defined for x, y ∈ D[A].

Definition 1.2. lower semi-bounded operators Let A : H ⊇ D(A) → H be alinear operator on some Hilbert space H. Then A is called bounded from belowif there is a constant m ∈ R such that

Ax, x ≥ m x2, for x ∈ D(A).

Further, mA := supAx, x x

−2 : x ∈ D(A), x = 0

denotes the greatest

lower bound of A.

Definition 1.3. Let A be a lower semi-bounded self-adjoint operator on someHilbert space H. Then the form norm of A can be defined by

xA:= A[x] + (1−mA) x , ∀x ∈ D[A].

We can replace mA by another lower bound m ≤ mA in order to get an equiv-alent norm. See Proposition 1.6.

Proposition 1.4. Let A be self-adjoint on H. Then D(A) is the set of x ∈ D[A]for which there exists a vector u ∈ H such that A[x, y] = u, y for y ∈ D[A]. Ifthis holds, then u = Ax, and hence

A[x, y] = Ax, y for x ∈ D(A), y ∈ D[A].

Proof. [2, Proposition 10.4]

Definition 1.5. order relations Let A and B be symmetric operators on someHilbert space H. We write A B if D(A) ⊆ D(B) and Ax, x ≥ Bx, x forx ∈ D(A). If A and B are self-adjoint, we write A ≥ B if D[A] ⊆ D[B] andA[x] ≥ B[x] for all x ∈ D[A].

Proposition 1.6. Let A ≥ mI be a lower semi-bounded self-adjoint operator.

1. For x, y ∈ D[A] = D((A−mI)1/2), we have

A[x, y] =(A−mI)1/2x, (A−mI)1/2y

+m x, y .

2. D(A) is dense in D[A] with respect to the form norm of A.

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3. If B is a linear operator such that D(B) ⊆ D[A] and A[x, y] = Bx, y forx ∈ D(B) and y ∈ D(A), then B ⊆ A.

4. For λ ≤ mA, the norm(A− λI)1/2·

is equivalent to the form norm ofA, which is stronger than the norm induced from H. This means that inthe definition of the form norm (Definition 1.3), we can replace mA byanother lower bound m of A.

Proof. We prove only the last assertion. For the others, see [2, Proposition10.5]. To do this, we show

min(1;mA−λ) uA≤ (A−λI) u ≤ max(1;mA−λ) u

A, for u ∈ D[A].

Let u ∈ D[A]. Then

(A− λI)[u] = A[u]− λ u2

= (A[u]−mA u2) + (mA − λ) u2

≤ max(1,mA − λ)A[u]−mA u

2 + u2

= max(1,mA − λ) u2A,

and

(A− λI)[u] = (A[u]−mA u1) + (mA − λ) u2

≥ min(1,mA − λ)A[u]−mA u

2 + u2

= min(1,mA − λ) u2A.

Clearly, the form norm is stronger than the norm induced from H, because

u2A= A[u]−mA u

2 + u2≥ u

2, for u ∈ D[A].

Proposition 1.7. Let A and B be lower semi-bounded self-adjoint operators onsome Hilbert space H, and let λ ∈ R, λ < min(mA,mB). Then λ ∈ ρ(A)∩ρ(B),and the relation A ≥ B holds if and only if (B − λI)−1 ≥ (A− λI)−1

.

Proof. This is [2, Corollary 10.13]

Essentials about Friedrichs and Krein-von Neumann extensions Werecall only the essential facts about the well-known Friedrichs and Krein-vonNeumann extensions. Let T be a densely defined lower semi-bounded symmetricoperator. We recall that the Friedrich extension of T , which is denoted by TF ,is the largest with respect to the relation ≥ among all lower semi-bounded self-adjoint extension of T . Moreover, the greatest lower bound of TF is equal tothat of T , i.e. mTF = mT . Moreover, D(T ) is dense in D(TF ) with respectto the form norm of TF defined in Definition 1.3, where A is replaced by TF .

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More about this topic, see [2, 10.4]. It may be useful to write again the idea toconstruct TF . One can show that the form sT defined by sT [x, y] := Tx, y forx, y ∈ D(sT ) := D(T ) is closable with the (closed!) closure sT (see [2, Lemma10.16], and also [2, Section 10.1] for basic facts of forms, closed and closableforms), and TF is defined as

D(TF ) = x ∈ D(sT ) : there is ux ∈ H with sT [x, y] = ux, y for y ∈ D(sT ) ,

and TFx := ux, which is uniquely determined because D(sT ) ⊇ D(T ) is densein H. Because sT is closed, TF is self-adjoint by the Representation Theoremof semi-bounded forms ([2, Theorem 10.17]).

The Krein-von Neumann extension TN is the topic of [2, Chapter 13]. If T isa densely defined POSITIVE symmetric operator, then the Krein-von Neumannextension TN of T is the smallest (with respect to ≥) among all positive self-adjoint extension of T on H. In this case, we have TF ≥ A ≥ TN for all positiveself-adjoint extension A of T on H.

1.2 Linear Relations on Hilbert spaces

Definition 1.8. Linear Relations on Hilbert spaces [2, Section 14.1]Let (Hi, ·, ·1),i = 1, 2 be Hilbert spaces. A linear subspace T ⊆ H1 ⊕ H2 iscalled a linear relation from H1 into H2. For such a linear relation, we definethe domain D(T ), the range R(T ), the kernel N (T ), and the multivalued partM(T ) by

D(T ) := x ∈ H1 : (x, y) ∈ T for some y ∈ H2 ,

R(T ) := y ∈ H2 : (x, y) ∈ T for some x ∈ H1 ,

N (T ) := x ∈ H1 : (x, 0) ∈ T ,

M(T ) := y ∈ H2 : (0, y) ∈ T .

The closure T is the closure of the linear subspace T with respect to the producttopology of H1 ⊕H2. If T = T , then T is called closed. Further, we define theinverse T

−1 and the adjoint T ∗ by

T−1 := (y, x) : (x, y) ∈ T ,

T∗ := (y, x) : v, y2 = u, x1 for all (u, v) ∈ T .

If T is (the graph of) a linear operator, then these notions introduced here co-incides with the corresponding notions of operators, if they exist. But, differentfrom linear operators, the adjoint and the closure of linear relations are alwaysdefined.

Moreover, if S and T be linear relations from H1 to H2, and let α ∈ C−0,we define

αT := (x,αy) : (x, y) ∈ T

S + T := (x, u+ v) : (x, u) ∈ S and (x, v) ∈ T .

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A linear relation T on a Hilbert space H is called symmetric if T ⊆ T∗, i.e.

x, v = y, u for all (x, y), (u, v) ∈ T , and self-adjoint, if T = T∗.

Let T be a symmetric relation. For α ∈ R, we write T ≥ αI, if y, x ≥

α x, x for all (x, y) ∈ T . In this case, T is called lower semi-bounded. Inparticular, we call T positive, if y, x ≥ 0 for (x, y) ∈ T .

By orthogonal projection, one can construct an self-adjoint operator B froma self-adjoint linear relation B on some Hilbert space H, such that this corre-spondence B ↔ B is one-to-one. This is the content of the following proposition,which we do not want to prove. Here, we denote S the set of all self-adjoint Bacting on a closed linear subspace HB of H, i.e. B : HB ⊇ D(B) → HB , andPB the orthogonal projection of H onto HB .

Proposition 1.9. There is a one-to-one correspondence between the sets ofoperators B ∈ S(H) and of self-adjoint relations B on H given by

B = G(B)⊕0⊕ (HB)

≡(x,Bx+ y) : x ∈ D(B), y ∈ (HB)

⊥,

where B is the operator part Bs, and (HB)⊥ is the multivalued part M(B) ofB. Moreover B ≥ 0 if and only if B ≥ 0.

Proof. This is [2, Proposition 14.2].

It leads to

Definition 1.10. Let B be a self-adjoint relation and B its operator part. Wedefine the form associated with B by

B[x, x] := B[x, x], for all x, x∈ D[B] := D[B].

Moreover, we have

Lemma 1.11. Let B be a linear relation on some Hilbert space H such thatB−1 is the graph of a positive self-adjoint operator on H. Then B is a positiveself-adjoint relation.

Proof. This is [2, Lemma 14.3].

1.3 Boundary Triplets of Adjoints of Symmetric Opera-tors

Definition 1.12. Let T be a densely defined symmetric operator on someHilbertspace H. Let (K, ·, ·

K) be a Hilbert space, and Γ0 : D(T ∗) → K and

Γ1 : D(T ∗) → K be linear mappings. Then (K,Γ0,Γ1) is a boundary triplet forthe adjoint T ∗, if the following conditions are satisfied:

i) [x, y]T∗ := T ∗x, y−x, T ∗

y = Γ1x,Γ0yK−Γ0x,Γ1yK for x, y ∈ D(T ∗),

ii) the mapping D(T ∗) x → (Γ0x,Γ1x) ∈ K ⊕K is surjective.

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Definition 1.13. Let (K,Γ0,Γ1) be a boundary triple for T∗ and B a linear

relation on K. We define the parametrized restriction2 TB of T ∗ to the domain

D(TB) := x ∈ D(T ∗) : (Γ0x,Γ1x) ∈ B

Recall that S(K) the set of self-adjoint operators B acting on a closed sub-space KB of K and PB is the orthogonal projection onto KB.

Definition 1.14. For B ∈ S(K), we define TB to be the parameterized restric-tion of T ∗ to the domain

D(TB) := x ∈ D(T ∗) : Γ0x ∈ D(B) and BΓ0x = PBΓ1x .

Comparing to above definitions, we see that if B is the operator part of B,then TB = TB.

The parametrization TB;B relation on K, or equivalently TB ;B ∈ S(K)gives us a characterization of the set of self-adjoint extension of T . This is thefollowing result which we do not want to prove.

Theorem 1.15. Suppose T is a densely defined symmetric operator on someHilbert space H, and (K,Γ0,Γ1) is a boundary triplet for T

∗. For any operatorS on H, the following are equivalent:

(i) S is a self-adjoint extension of T on H.

(ii) There is a self-adjoint relation B on K such that S = TB, or equivalently,there is an operator B ∈ S(K) such that S = TB .

Moreover, the relation B and the operator B are uniquely determined by S.

Proof. see [2, Theorem 14.10]

Corollary 1.16. If (K,Γ0,Γ1) is a boundary triplet for T∗, then there exist

self-adjoint extensions T0 and T1 of the symmetric operator T on H defined byD(T0) = N (Γ0) and D(T1) = N (Γ1).

Proof. This is [2, Corollary 14.8]. Choose B0 = 0⊕K and B1 = K⊕0, thenT0 := TB0 and T1 := TB1 are self-adjoint operators satisfying D(T0) = N (Γ0)and D(T1) = N (Γ1).

Now let T be a symmetric operator with a real regular point µ. Recall thatµ is called a regular point of T if there is a c = cµ > 0 such that (T − µI)x ≥

cµ x for all x ∈ D(T ). Then a basic result ([2, Proposition 3.16]) implies thatthere exists a self-adjoint extension A of T such that µ is in the resolvent ρ(A)of A. If we have a boundary triplet (K,Γ0,Γ1), we can ”parameterize” A byA = TB for a relation B. Now it rises the question of whether we can choose aboundary triplet such that A is exactly the operator T0 in Corollary 1.16. Theanswer is YES. Moreover, we have M(µ) = 0 and γ(µ) = IK K. This is theresult of [2, Example 14.6, Theorem 14.12, Example 14.12].

2I have not found a special name for this in the literature.

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Theorem 1.17. Let T be a densely defined symmetric operator on some Hilbertspace H. Suppose that A is a fixed self-adjoint extension of T on H and µ ∈

R ∩ ρ(A). Set Nµ := N (T ∗ − µI). For B ∈ S(Nµ), let

D(TB) :=x+Rµ(A)(Bu+ v) + u : x ∈ D(T ), u ∈ D(B), v ∈ Nµ ∩D(B)⊥

,

TB := T∗ D(TB), that is,

TB (x+Rµ(A)(Bu+ v) + u) := Tx+ (I + µRµ(A))(Bu+ v) + µu.

Then the operator TB is a self-adjoint extension of T on H. Each self-adjointextension of T is of the form TB with uniquely determined B ∈ S(Nµ)

Proof. [2, Example 14.6, Theorem 14.12]

1.4 Gamma Fields and Weyl Functions

Let T be a densely defined symmetric operator on H and (K,Γ0,Γ1) be a bound-ary triplet for T ∗, and T0 the self-adjoint extension of T in Corollary 1.16 suchthat D(T0) = N (Γ0). Set Nz := N (T ∗ − zI).

Lemma 1.18. 1. Γ0 and Γ1 are continuous mappings of (D(T ∗), ·T∗) into

K.

2. For each z ∈ ρ(T0), the mapping Γ0 is a continuous bijective mapping ofthe subspace Nz onto K with bounded inverse denote by γ(z).

Proof. This is [2, Lemma 14.13]

It leads to

Definition 1.19. We call the map ρ(T0) z → γ(z) ∈ B(K,H) the gammafield and the map ρ(T0) z → M(z) ∈ B(K) the Weyl function of the operatorT0 associated with the boundary triplet (K,Γ0,Γ1).

Proposition 1.20. Some properties of gamma fields and Weyl functions Letz ∈ ρ(T0), where T0 is defined in Corollary 1.16. Let γ(z),M(z) be the gammafield and Weyl function as in the above definition. Then

1. N (γ(λ)∗) = N⊥z

and γ(z)∗ is a bijection of Nz onto K.

2. For z ∈ ρ(T0), we have M(z)∗ = M(z).

1.5 The Krein Naimark Resolvent Formula

Theorem 1.21. Krein Naimark Resolvent Formula Let T be a densely definedsymmetric operator on H, and (K,Γ0,Γ1) a boundary triplet for T

∗. Suppose

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that B is a closed relation on K and z ∈ ρ(T0). Then the proper extension TB

of T is given by

D(TB) =

f = (T0 − zI)−1(y + v) + γ(z)u :u ∈ K, v ∈ Nz,

(u, γ(z)∗v) ∈ B −M(z),y ∈ HNz

(1)

TBf = TB

(T0 − zI)−1(y + v) + γ(z)u

= zf + y + v. (2)

If z ∈ ρ(TB)∩ρ(T0), the relation B−M(z) has an inverse (B−M(z))−1 ∈ B(K),and

(TB − zI)−1− (T0 − zI)−1 = γ(z)(B −M(z))−1

γ(z)∗. (3)

Proof. [2, Theorem 14.18]

2 Boundary Triplets and Semibounded Self-adjointOperators

In this section we assume that T is a densely defined lower semi-bounded sym-metric operator and (K,Γ0,Γ1) is a boundary triplet for T

∗. Recall that T0

denotes the self-adjoint operator defined in Corollary 1.16, and γ(z) and M(z)denote the gamma field and the Weyl function of T0.

Proposition 2.1. Suppose that T0 is the Friedrichs extension TF of T . Let B bea self-adjoint relation on K. If the self-adjoint operator TB is lower semibounded,so is the relation B. More precisely, if λ < mT and λ ≤ mTB , then B−M(λ) ≥0.

Proof. i) Fix λ< λ. First, assume we can show that B −M(λ) ≥ 0, which

is equivalent to

y, x − M(λ)x, x ≥ 0, for all x ∈ D(B).

Letting λ → λ, by the fact that M(·) is continous B(K)-valued function,

we obtainy, x − M(λ)x, x ≥ 0, for all x ∈ D(B),

which means B −M(λ) ≥ 0.

ii) We show now, that B−M(λ) ≥ 0. Recall that for the Friedrichs extension,we have mT = mTF . Since λ

< mT = mTF = mT0 and λ

< λ ≤ mTB , we

have λ ∈ ρ(TB) ∩ ρ(T0), by Proposition 1.7. By the resolvent formula (3)

in Theorem 1.21,

(TB − λI)−1

− (T0 − λI)−1 = γ(λ)(B −M(λ))−1

γ(λ)∗.

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Since TF is the largest lower semibounded self-adjoint extension of T , wehave T0 = TF ≥ TB, and taking the inversion (Proposition 1.7) leads to(TB − λ

I)−1 ≥ (T0 − λ

I)−1

. By the above resolvent formula, we obtain

γ(λ)(B −M(λ))−1γ(λ)∗ ≥ 0,

which is equivalent to

γ(λ)(B −M(λ))−1

γ(λ)∗x, x≥ 0,

and to (B −M(λ))−1

γ(λ)∗x, γ(λ)∗x≥ 0,

for all x ∈ K. Recall that γ(λ)∗ ∈ B(K) and γ(λ)∗K = K. Hence we canreplace γ(λ)∗x by y in the previous formula to get

(B −M(λ))−1

y, y≥ 0, for all y ∈ K.

By the inversion (using Lemma 1.11), B −M(λ) ≥ 0.

The following theorem is the main result of this section, it states that thedecomposition of the form of TB−λ in terms of the form of TF −λ and the formof the positiv linear relation B −M(λ). The main tool to prove this theorem isTheorem 1.21.

Recall that mT is the greatest lower bound of T , as in Definition 1.2.

Theorem 2.2. Let T be a densely defined lower semibounded symmetric oper-ator, and let (K,Γ0,Γ1), be a boundary triplet for T

∗ such that the operator T0

is the Friedrichs extension TF of T . Let λ ∈ R,λ < mT . Suppose that B is aself-adjoint relation on K such that B −M(λ) ≥ 0. Then TB − λI is a positiveself-adjoint operator, and

D[TB] = D[TF ] γ(λ)D[B −M(λ)]

(TB − λI)[x+ γ(λ)u, x + γ(λ)u] = (TF − λI)[x, x] + (B −M(λ))[u, u](4)

for x, x ∈ D[TF ] and u, u

∈ D[B −M(λ)].

Proof. The idea of this proof is to consider D(TB) instead of D[TB]. Recall thatwe have a representation of D(TB) by Theorem 1.21. From this, we obtain thesum

D(TB) = D(TF ) + γ(λ)D(B −M(λ)),

which looks like the first equation in (4), but we have here only (). Also, by thisrepresentation, we can show the second formula in (4), but only for x ∈ D(TF )and u ∈ D(B − M(λ)). The remaining can be treated by approximation. Onthe other hand, we see that the second formula of (4) looks like a

2 = b2 + c

2,which comes from the definition of the Gamma-field.

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1. In the first step, we show the ”orthogonality” (TB − λI)[x, γ(λ)u] = 0 forx ∈ D[TF ] and γ(λ)u ∈ D[TB], u ∈ K. By the properties of the Friedrichsextension in Subsection 1.1, there is a sequence (xn) from D(T ), whichconverges to x in the form norm of TF , and TF ≥ TB. The latter impliesthat D[TF ] ⊆ D[TB], and that the form norm of TB is weaker than theform norm of TF . Therefore, the sequence (xn) converges to x also in theform norm of TB.

Using that T ⊂ TB and γ(λ)u ∈ Nλ := N (T ∗ − λI), we derive

(TB − λI)[x, γ(λ)u] = limn→∞

(TB − λI)[xn, γ(λ)u]

= limn→∞

(TB − λI)xn, γ(λ)u

= limn→∞

(T − λI)xn, γ(λ)u = 0.

In the second equation, we used the fact that

A[x, y] = Ax, y , for all x ∈ D(A), y ∈ D[A],

where A is replaced by TB − λ.

2. Since B is self-adjoint, M(λ) = M(λ)∗ ∈ B(K) by Proposition 1.20, andB−M(λ) ≥ 0 by assumption, B−M(λ) is a positive self-adjoint relation.Let C denote its operator part, which is defined in Propsition 1.9. Then C

is a positive self-adjoint operator acting on some subspace KC of K, moreprecisely C : KC ⊇ D(C) → KC , and the form associated with B −M(λ)is defined by (B−M(λ))[u, u] = C[u, u] for u, u ∈ D[B−M(λ)] := D(C).

Clearly, λ < mT = mTF implies that λ ∈ ρ(TF ). Hence, by Theorem 1.21,where z is replaced by λ ∈ R and T0 = TF , the operator TB is describedby

D(TB) =

f = (TF − λI)−1(y + v) + γ(λ)u withu ∈ K, v ∈ Nλ,

(u, γ(λ)∗v) ∈ B −M(λ),y ∈ H Nλ

, (5)

TBf ≡ TB

(TF − λ)−1(y + v) + γ(λ)u

= λf + y + v. (6)

Now let f ∈ D(TB). By the above description, there exist v ∈ Nλ andy ∈ H Nλ such that

f = (TF − λI)−1(y + v) + γ(z)u =: x+ γ(λ)u, and

(u, γ(λ)∗v) ∈ B −M(λ).

The latter implies that u ∈ D(B − M(λ)) = D(C) ⊂ KC and Cu =PCγ(λ)∗v, where PC is the orthogonal projection of K onto KC . By (6),we have (TB − λI)f = y + v. Since γ(λ)u ∈ Nλ (recall that γ(λ) := (Γ0

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Nλ)−1), we have y ⊥ γ(λ)u. Using these facts, we compute

(TB − λI)f, f = y + v, x+ γ(λ)u

= y + v, x+ v, γ(λ)u

= (TF − λI)x, x+ γ(λ)∗v, uK

= (TF − λI)x, x+ PCγ(λ)∗v, u

K

= (TF − λI)x, x+ Cu, uK.

Therefore, since TF − λI ≥ 0 and C ≥ 0, it follows that TB − λI ≥ 0 and(TB − λI)1/2f

2=

(TF − λI)1/2x2+

C1/2u

2

K

. (7)

3. Now we prove the inclusion D[TF ] + γ(λ)D[C] ⊆ D[TB]. Because TF isthe largest self-adjoint extension of T , TF ≥ TB, and so D[TF ] ⊆ D[TB].Hence it suffices to show γ(λ)D[C] ⊆ D[TB].

Let u ∈ D(C), where D(C) is dense in D[C] with respect to the form normof C. By Proposition 1.20, γ(λ)∗ is a bijection ofNλ onto K, so there existsa vector v ∈ Nλ such that γ(λ)∗ = Cu. Since C is the operator part ofB −M(λ), we have

(u, γ(λ)∗v) = (u,Cu) ∈ B −M(λ).

Hence, by choosing y = 0 in (5) and (6), we obtain

f := (TF − λI)−1v + γ(λ)u =: x+ γ(λ)u ∈ D(TB)

and (TB − λI)f = v. Since

x ∈ D(TF ) ⊆ D[TF ] ⊆ D[TB]

f ∈ D(TB) ⊆ D[TB],

we have γ(λ)u = f − x ∈ D[TB],, so (TB − λI)[x, γ(λ)u] = 0 as shownin the first step. Further, v = (TF − λI)x, and γ(λ) is bounded, so theadjoint γ(λ)∗ is everywhere defined. From these facts, we obtain

(TB − λI)[f ] = TBf, f

= v, x+ γ(λ)u

= (TF − λI)x, x+ γ(λ)∗v, uK

= (TF − λI)[x] + Cu, uK,

(TB − λI)[f ] = (TB − λI)[x+ γ(λ)u]

= (TB − λI)[x] + (TB − λI)[γ(λ)u] + 2Re (TB − λI)[x, γ(λ)u]

= (TB − λI)[x] + (TB − λI)[γ(λ)u].

Comparing both formulas yields Cu, uK= (TB − λI)[γ(λ)u]. Therefore,

C1/2u

K

=(TB − λI)1/2γ(λ)u

, for u ∈ D(C). (8)

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Now we show that the above formula can be extended to all u ∈ D[C].Let u ∈ D[C]. Since D(C) is dense in D[C] with respect to the formnorm of C, there exist a sequence (un) ⊆ D(C) with un − u → 0 andC[un − u] → 0. By the boundedness of γ(λ), we have γ(λ)un → γ(λ)u inH. Now replace u by un − um in (8), we see that

(TB − λI)1/2γ(λ)un

is a Cauchy sequence in H. Because (TB − λI)1/2 is closed, we concludethat γ(λ)u ∈ D[(TB − λI)1/2] = D[TB]. Therefore γ(λ)D[C] ⊆ D[TB].

4. We prove the converse inclusion D[TB] ⊆ γ(λ)D[C]. We use (7). BecauseTB,TF and C are self-adjoint and positive, the norms

(TB − λI)1/2· ,

(TF − λI)1/2· ,

C1/2·

are equivalent to the form norm of TB, TF , and C respectively. Now letf ∈ D[TB], then there is a Cauchy sequence (fn) ⊆ D(TB) which convergesto f with respect to the form norm of TB. Write fn = xn + γ(λ)un withxn ∈ D(TF ) and un ∈ D(C) as in previous steps. By (7), and the factsof equivalent norms, (xn) is a Cauchy sequence with respect to the formnorm of TF and therefore in H, and (un) is a Cauchy sequence withrespect to the form norm of C and therefore in K. Let x ∈ D[TF ] andu ∈ D[C] be the corresponding limits. Because γ(λ) is bounded, we haveγ(λ)un → γ(λ)u in H, and hence xn + γ(λ)un → x + γ(u) in H. Thismeans f = x+ γ(λ)u ∈ D[TF ] + γ(λ)D[C].

5. Putting things together, we have D[TB] = D[TF ] + γ(λ)D[C]. Moreover,using the Cauchy sequences in the preceeding paragraph, we can takethe limit as n → ∞ in (7), where f, x, u are replaced by (fn, xn, un), toconclude that this formula remains valid for f = x+γ(λ)u ∈ D[TB]. Since(B −M(λ))[u] = C[u] =

C1/2uKby Definition 1.10, this means that

(TB − λI)[x+ γ(λ)u] = (TF − λI)[x] + (B −M(λ))[u],

and using polarization, we have

(TB − λI)[x+ γ(λ)u, x + γ(λ)u] = (TF − λI)[x, x] + (B −M(λ))[u, u],

for x, x ∈ D[TF ] and u, u ∈ D[B −M(λ)].

6. Finally, we show that D[TF ] + γ(λ)D[C] is a direct sum. By the previousseparation f = x+ γ(λ), assume f = 0, then the r.h.s of (7) is 0. Becausethe norm

(TF − λI)1/2· is equivalent to the form norm of TF which is

stronger than the norm of H, we have x = 0, and so γ(λ)u = f − x = 0.

3 Positive Self-adjoint Extensions

In this section, suppose T is a densely defined symmetric operator on H withpositive greatest lower bound mT > 0, i.e.

Tx, x ≥ mT x2, for all x ∈ D(T ).

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Our aim is to apply the preceeding results to investigate the set of all positiveself-adjoint extensions of T .

Since 0 < mT = mTF , we have 0 ∈ ρ(TF ), by Proposition 1.7. Hence,Theorem 1.17 applies with µ = 0 and A = TF . Assume that B ∈ S(N (T ∗)),where S(X ) the set of self-adjoint operators on some Hilbert space X . Thenthe self-adjoint extensions of T on H are precisely the operators TB defined asfollows

D(TB) =

x+ (TF )−1(Bu+ v) + u :x ∈ D(T ), u ∈ D(B),v ∈ N (T ∗) ∩D(B)⊥

,

TB

x+ (TF )

−1(Bu+ v) + u= Tx+Bu+ v.

(9)

Let S(N (T ∗))+ denote the set of positive operators in S(N (T ∗)).

Remark 2. The following Theorem 3.1 expresses the decomposition of the ex-tension TB into terms of TF and B. Its proof is based on the above formulas ofD(TB) and TB , which is proved through the construction of a boundary triplet,and therefore, do not look intuitive, if one omits to read precisely its proof. Thisis the disadvantage of the introduction of [2]. The introduction of [1] is maybemore intuitive: one gets TB via its forms defined by

D[TB ] = D[TF ]D[B],

TB [y + u, y + u

] = TF [y, y] +B[u, u], for y, y ∈ D[TF ], u, u

∈ D[B],

as the formulas in Theorem 3.1 (of course, one has to show that this form isa closed quadratic form, which represents a self-adjoint operator!), and proveswith this definition the representation of TB at the beginning of this section.

Theorem 3.1. (a) For B ∈ S(N (T ∗)), we have TB ≥ 0 if and only if B ≥ 0.In this case the greatest lower bound mB and mTB satisfy

mTmB(mT +mB)−1

≤ mTB ≤ mB .

(b) If B ∈ S(N (T ∗))+, then D[TB ] = D[TF ]D[B], and

TB [y + u, y + u

] = TF [y, y] +B[u, u], for y, y

∈ D[TF ], u, u

∈ D[B].

(c) If B1, B2 ∈ S(N (T ∗))+, then B1 ≥ B2 is equivalent to TB1 ≥ TB2 .

Proof. i) First, suppose that B ≥ 0. Let f ∈ D(TB), then f is of the formf = y + u with y = x + (TF )−1(Bu + v) with x ∈ D(T ), u ∈ D(B), andv ∈ N (T ∗)∩D(B)⊥, and we have TBf = Tx+Bu+v = TF y. We can writethis because T ⊆ TF = TF , so x ∈ D(T ) ⊆ D(TF ), therefore y ∈ D(TF ).Further, mT = mTF , T

∗u = 0, and v, u = 0. Putting things together, we

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compute

TBf, f = TF y, y + u

= TF y, y+Tx+Bu+ v, u

= TF y, y+ Bu, u

≥ mT y2 +mB u

2

≥ mTmB(mT +mB)−1(y+ u)2

≥ mTmB(mT +mB)−1

y + u2

≥ mTmB(mT +mB)−1

f2.

In the third equation, we usedTx, u

= T ∗ ∗

x, u = x, T ∗u = 0, and in

the second inequality, we used

αa2 + βb

2≥ αβ(α+ β)−1(a+ b)2, for α > 0, a, b,β ≥ 0.

Therefore we proved the part B ≥ 0 ⇒ TB ≥ 0 and mTmB(mT +mB)−1 ≤

mTB .

ii) In order to prove the remaining assertion, we apply the result stated in theparagraph before Theorem 1.17 for A = TF (as a self-adjoint extension ofT ) and µ = 0. We stress that there exists a boundary triplet (K,Γ0,Γ1)such that A is the operator T0 in Corollary 1.16. We have M(0) = 0and γ(0) = IH|K. Because TF = T0, we can apply Proposition 2.1 andTheorem 2.2 in the last section. To obtain b) and the part TB ≥ 0 ⇒ B ≥ 0,we apply both results with λ = 0. Set y = y

= 0 and u = u ∈ D(B) in b),

we haveBu, u = TBu, u ≥ mTB u

2,

so it is obvious, that mTB ≤ mB .

Part c) is an immediate consequence of part b).

By the definition of the order relation ≥ (Definition 1.5) the set S(N (T ∗))+contains a largest operator and a smallest operator. These are the following twoextreme cases:

1) D(B) = 0. By definition of ≥, the operator B is larger than all otheroperators, then B is the largest operators in S(N (T ∗))+. Then by (9),and T ⊆ TF , we have D(TF ) ⊇ D(TB) = D(T ) + (TF )−1N (T ∗), andTB(x+(TF )−1

v) = Tx+v = TF (x+(TF )−1v) for x ∈ D(T ) and v ∈ N (T ∗).

Because TF ≥ TB by the paragraph about Friedrichs and Krein-von Neu-mann extensions stated in Subsection 1.1, we have D(TF ) ⊆ D(TB), and soin this case TB = TF . Hence, D(TF ) = D(T ) + (TF )−1N (T ∗).

2) B = 0, D(B) = N (T ∗). By definition of ≥, this operator is smaller thanall other operators in S(N (T ∗))+, it is the smallest element of S(N (T ∗))+.

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Therefore, by Corollary 13.15, this operator TB is the Krein-von Neumannextension TN of T . From (9) (with Bu = 0 and v = 0) and Theorem 3.1, wededuce the following formula for the Krein-von Neumann extension:

D(TN ) = D(T )N (T ∗),

TN (x+ u) = Tx for x ∈ D(T ), u ∈ N (T ∗),

D[TN ] = D[TF ]N (T ∗),

TN [y + u, y + u

] = TF [y, y] for y, y ∈ D[TF ], u, u

∈ N (T ∗).

(10)

The direct sums in the first and the third equation are shown in Theorem3.1.

Theorem 3.2. Let T be a densely defined positive symmetric operator on H

such that mT > 0. For any positive self-adjoint operator A on H, the followingstatements are equivalent:

a) A is an extension of T .

b) There is an operator B ∈ S(N (T ∗))+ such that A = TB.

c) TF ≥ A ≥ TN .

Proof. a) → b) and b) → c) are just simple applications of the previous results.To show a) → b), we use the parameterization A = TB in Theorem 1.17 b).Because A = TB is positive, by Theorem 3.1, the operator B is also positive.Further, b) → c) holds by Theorem 3.1 c) because TF and TN are the two aboveextreme cases.

Now we show c) → a).

i) The inequalities TF ≥ A ≥ TN mean that D[TN ] ⊇ D[A] ⊇ D[TF ] andtTN ≤ tA ≤ tTF . If y ∈ D[TF ], then TF [y] = TN [y] by the last equation of(10), and hence A[y] = TF [y] = TN [y], because tTN ≤ tA ≤ tTF . Therefore,by polarization,

A[x, y] = TF [x, y] for x, y ∈ D[TF ].

ii) Next, we verify that

A[x, u] = 0, for x ∈ D[TF ], u ∈ N (T ∗) ∩D[A].

Let λ ∈ C. Using the last equation of (10) and tTN ≤ tA, we compute

TF [x] = TN [x+ λu] ≤ A[x+ λu] = A[x+ λu, x+ λu]

= A[x, x] + 2Re (λA[x, u]) + |λ|2A[u, u]

= TF [x] + 2Re (λA[x, u]) + |λ|2A[u, u]

Thus, 0 ≤ 2ReλA[x, u]+ |λ|2A[u, u] for all λ, which implies, by λ → 0, thatA[x, u] = 0.

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iii) Finally, in order to show T ⊆ A, we need to show, by Proposition 1.4,that A[x, f ] = Tx, f for all x ∈ D(T ) and f ∈ D(A). Let x ∈ D(T )and f ∈ D(A). Then f ∈ D[TN ] = D[TF ] N (T ∗), so f = y + u withy ∈ D[TF ] and u ∈ N (T ∗). Since x ∈ D[TF ] and u = f −y ∈ D[A], we haveA[x, u] = 0 as shown in the last step. Further, T ∗

u = 0. We obtain

A[x, f ] = A[x, y + u] = A[x, y]

= TF [x, y] = TFx, y

= Tx, y = Tx, f − u = Tx, f .

4 Some examples

In the following examples, we note that T is always closed. Our aim is tocharacterize the domain D(TN ) of the Krein-von Neumann extension TN as asubspace of the Sobolev space H

2 which satisfies some boundary conditions.

Example 1. Let T := −d2

dx2 with D(T ) := H20 (0, 1) on the Hilbert space H =

L2(0, 1). with the adjoint operator T

∗ = d −d2

dx2 with D(T ∗) = H2(0, 1). (see

[2, Example 1.4]). By the Poincare Inequality, there is a c > 0 with

Tf, f = f2≥ c f

2, for f ∈ D(T ).

Therefore, T has a positive lower bound. Further T is symmetric. We will showthat Krein-von Neumann extension TN of T has the domain

D(TN ) =f ∈ H

2(0, 1) : f (1) = f(0) = f(1)− f(0)

.

(Because H2(0, 1) ⊆ C

1([0, 1]), each boundary value here is well-defined.) Letf ∈ D(TN ) = D(T ) +N (T ∗) (first formula of (10)). Here, by solving the ODEy = 0, we have N (T ∗) is the set of functions of the form x → c+dx, where c, d

are constant. Therefore f(x) = g(x)+c+dx for a function g ∈ D(T ) = H10 (0, 1).

Recall that

H10 (0, 1) =

g ∈ H

10 (0, 1) : g(0) = g(1) = g

(0) = g(1)

.

Therefore, by a little algebra, the function f satisfies the boundary conditionf(1) = f

(0) = f(1)− f(0). Conversely, assume f ∈ D(T ∗) = H2(0, 1) satisfies

this boundary condition, then we set c = f(0), d = f(1) and g(x) := f(x) −f(0)− (f(1)−f(0))x. It is easy to check that g(0) = g(1) = g

(0) = g(1). Then

g ∈ H20 (0, 1). Therefore, f ∈ D(T ) +N (T ∗) = D(TN ).

In the last example, one can solve N (T ∗) explicitly. Therefore it is easy tofind the boundary condition of TN by the decomposition (10). In general, it isnot trivial to formulate and solve the same problem for d-dimensional DirichletLaplacian operators as seen in the following

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Example 2. Dirichlet Laplacian on bounded domains of Rd Differential Opera-tors are not the topic of this text. We recall only essential points which we needfor the example. Let Ω ⊂ Rd with d ≥ 2 be bounded with smooth boundary∂Ω. Let H = L

2(Ω). We define the operator L = −∆ first for smooth functionswith compact supports

Lf = −∆f := −

d

1

∂2f

∂x2i

, for f ∈ D(L) = C∞

0 (Ω).

By partial integration, L is symmetric. Moreover, L is densely defined. There-fore L is closable. We define the minimal operator Lmin as the closure L ofL, and the Dirichlet Laplacian as the Friedrichs extension of Lmin. Now setT = Lmin. Then T is lower semibounded (by the Poincare Inequality) andsymmetric. We can show that

D(T ) = H20 (Ω) =

f ∈ H

2(Ω) : f |∂Ω =∂f

∂ν|∂Ω = 0

,

D(TF ) = H2(Ω) = H

2(Ω) ∩H10 (Ω) =

f ∈ H

2(Ω) : f |∂Ω = 0,

where f |∂Ω,∂f

∂ν|∂Ω can be defined as the trace operators. (see [2, Theorem 10.19,

Theorem D.7].)In order to formulate this example, we need the following decomposition

D(T ∗) = D(TF )N (T ∗), (11)

which is [2, Eq (14.19)] with A replaced by TF and µ replaced by 0.Let f ∈ H

2(Ω). Then f ∈ D(T ∗), because, by integration by part, we have

f, (−∆)g = (−∆)f, g , for g ∈ D(−∆) = C∞

0 (Ω),

hence f ∈ D((−∆)∗) = D((−∆)∗

) = D(T ∗). Now by the above decomposition,there exists uniquely an H(f) ∈ D(T ∗) such that f −H(f) ∈ D(TF ) ⊆ H

2(Ω).Because f ∈ H

2(Ω), we have H(f) ∈ H2(Ω). Moreover, by the boundary

condition of D(TF ), we have f −H(f) = 0 on ∂Ω, so f |∂Ω = H(f)|∂Ω.

Now we show that f ∈ D(TN ) if and only if ∂f

∂ν|∂Ω = ∂H(f)

∂ν|∂Ω. Suppose

f ∈ D(TN ), then by (10) f = g + h where g ∈ D(T ) and h ∈ N (T ∗). Theng ∈ D(TF ), and by the decomposition (11), we have h = H(f). Since g ∈ D(T ),

we have ∂g

∂ν|∂Ω = 0, and hence ∂f

∂ν|∂Ω = ∂h

∂ν|∂Ω = ∂H(f)

∂ν|∂Ω. Conversely, suppose

∂f

∂ν|∂Ω = ∂H(f)

∂ν|∂Ω. Set g := f −H(f). Then ∂g

∂ν|∂Ω = 0. On the other hand, by

the definition of H(f) and the boundary condition of D(TF ), we have g|∂Ω = 0.Therefore g ∈ H

20 (Ω) = D(T ), and hence f = g+ h ∈ D(T ) +N (T ∗) = D(TN ).

References

[1] Alberto Alonso and Barry Simon. The Birman- Krein- Vishik Theory ofSelf-adjoint Extensions of Semibounded Operators. J. Operator Theory,4:251–270, 1980.

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[2] Konrad Schmudgen. Unbounded Self-adjoint Operators on Hilbert Space.Springer, 2012.

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