4 Adjoint of the Dirac operator

RIESZ BASES OF EIGENFUNCTIONS OF 1D DIRAC OPERATOR WITH STRICTLY REGULAR BOUNDARY

CONDITIONS

by

HAT˙ICE G ¨OZEL

Submitted to the Graduate School of Engineering and Natural Sciences in partial fulfillment of

the requirements for the degree of Master of Science

Sabancı University Spring 2011

Hatice G¨c ozel 2011 All Rights Reserved

ABSTRACT

One dimensional Dirac operators Lbc(v)y = i1 0

0 −1

 dy

dx + v(x)y, y =y₁ y₂



, x ∈ [0, π]

considered with L²-potentials v(x) =

 0 P (x) Q(x) 0



, P, Q ∈ L²([0, π]) ,

and subject to regular boundary conditions bc have discrete spectrum. In this thesis, we study basic properties of Riesz bases, prove existence of Riesz bases consisting of root functions of Dirac operators L_bc subject to strictly regular bc, find adjoint operator (Lbc)^∗, find all self-adjoint bc, and calculate some special self-adjoint ex- tensions.

OZET¨

L_bc(v)y = i1 0 0 −1

 dy

dx + v(x)y, y =y₁ y₂



, x ∈ [0, π]

denklemiyle verilen,

 0 P (x) Q(x) 0



, P, Q ∈ L²([0, π])

L² potansiyeli ve reg¨uler sınır ¸sartlarıyla d¨u¸s¨un¨ulen, tek boyutlu Dirac operat¨or¨un¨un ayrık spektrumu vardır. Bu tezde, Riesz tabanının genel ¨ozelliklerini inceliyoruz, g¨u¸cl¨u reg¨uler sınır ¸sartlarıyla d¨u¸s¨un¨ulen Dirac operat¨or¨u L_bc’nin ¨ozvekt¨orlerinden olu¸san Riesz tabanının varlıˇgını ispatlıyoruz, e¸slenik operat¨or¨u (L_bc)^∗’ı buluyoruz,

¨

oze¸slenik sınır ¸sartlarını belirliyoruz ve bazı ¨ozel ¨oze¸slenik geni¸slemeleri hesaplıyoruz.

TABLE OF CONTENTS

ABSTRACT iv

OZET¨ v

1. INTRODUCTION 1

2. RIESZ BASIS 2

3. RIESZ BASIS OF ROOT FUNCTIONS OF

DIRAC OPERATOR 11

4. ADJOINT OF THE DIRAC OPERATOR 21

5. SELF-ADJOINT DIRAC OPERATORS AND

SELF-ADJOINT bc 27

6. SELF-ADJOINT EXTENSIONS 29

REFERENCES 34

1 Introduction

The differential expression

L(v)y = i1 0 0 −1

 dy

dx + v(x)y, y =y₁ y2

 , is known as one dimensionsal Dirac operator. The matrix

v(x) =

 0 P (x) Q(x) 0



is called Dirac potential. In this thesis, we consider Dirac operators L_bc on [0, π]

with L² potentials, that is P, Q ∈ L²([0, π]), and with domain Dom (L_bc(v)) =



y =y₁ y2



: y₁ and y₂ are absolutely continuous, y satisfies the boundary conditions bc, and y⁰₁, y₂⁰ ∈ L²([0, π])} .

If v ≡ 0, then L_bc(0) is denoted by L⁰_bc and called the free Dirac operator. A regular boundary condition bc is given by the linear system of equations

y₁(0) + by₁(π) + ay₂(0) = 0, dy₁(π) + cy₂(0) + y₂(π) = 0,

where bc − ad 6= 0. Moreover, bc is called strictly regular if (b − c)²+ 4ad 6= 0.

In the second section, we study the basic properties of Riesz bases. If (e_γ, γ ∈ Γ) is an orthonormal basis in a Hilbert space H and A : H → H is an automorphism, then the system (f_γ, γ ∈ Γ), f_γ = Ae_γ, is called a Riesz basis. Riesz bases are unconditional bases. Moreover, Bari-Markus theorem is proven which states that if (e_n, n ∈ N) is a Riesz basis in a Hilbert space H and (fn, n ∈ N) is a minimal system of vectors such that

∞

X

n=1

kf_n− e_nk² < ∞,

then (f_n, n ∈ N) is also a Riesz basis. Bari-Markus theorem will be used to show the existence of a Riesz basis consisting of root functions of the Dirac operator L_bc. In the third section, we study eigenvalues and eigenfunctions of Dirac operators.

Dirac operators subject to regular boundary conditions bc have discrete spectrum.

It is shown that for strictly regular bc, every eigenvalue of the free Dirac operator is simple and has the form λ⁰_k,α = τ_α+ k, where α = 1, 2 and k ∈ 2Z, and spectrum consists only of eigenvalues. For each strictly regular bc, there is an N ∈ 2N such that

Sp(L_bc) ⊂ R_N ∪ [

|n|>N

D_n¹ ∪ D_n²
,

where RN is a rectangle containing 2N eigenvalues of Lbc and each of the discs D_n^α = z : 

z − λ⁰_n,α

< ρ = ρ(bc) , α = 1, 2 and |n| > N , contains exactly one

simple eigenvalue of L_bc. Using this spectra localization of the operators L_bc and Bari-Markus theorem, it is shown that there is a Riesz basis which consists of eigen- functions and (at most finitely many) associated functions.

In the fourth section, we show that the adjoint operator of Dirac operator L_bc(v) subject to regular boundary conditions is Lbc^∗(v^∗), where boundary conditions bc^∗ given by the system

bg₁(0) + g₁(π) + dg₂(π) = 0 ag₁(0) + g₂(0) + cg₂(π) = 0, and

v^∗ = 0 Q P 0

 .

In the last two sections, we find the form of self-adjoint boundary conditions bc and self-adjoint Dirac operators. Furthermore, we give a characterization of self-adjoint extensions of an unbounded operator and we calculate some special self-adjoint extensions.

2 Riesz bases

In this section, we give basic facts about bases. We define Riesz bases and give some basic properties of Riesz bases. We consider only separable Hilbert spaces.

Definition 1. Let H be a Hilbert space. A system (e_n, n ∈ N) is called a basis in H if

x =

∞

X

n=1

c_ne_n, ∀x ∈ H (2.1)

where c_n’s are uniquely determined and the series converges in norm. If the se- ries converges unconditionally, then (e_n, n ∈ N) is called an unconditional basis.

Moreover a basis is called orthonormal if it is an orthonormal system in H which means

he_n, e_mi = δ_n,m, ∀n, m ∈ N.

It is a general fact that there are orthonormal bases in every Hilbert space H.

Definition 2. Two systems (e_n, n ∈ N) and (fm, m ∈ N) in H are called biorthog- onal if

he_n, f_mi = δ_n,m, ∀n, m ∈ N.

Theorem 3. Assume that (f_n, n ∈ N) is a basis in a Hilbert space H. Then there exists a unique family 

fe_n, n ∈ N

in H such that x =X

n

D x, ef_nE

f_n, ∀x ∈ H,

where (f_n, n ∈ N) and 

fe_n, n ∈ N

are biorthogonal.

Proof. Suppose (f_n, n ∈ N) is a basis in a Hilbert space H. Then

x =

∞

X

n=1

c_n(x)f_n,

where cn(.) are linear functions due to the uniqueness of expansion (??). Let P_n(x) = c_n(x)f_n

and let

S_N(x) =

N

X

k=1

c_k(x)f_k.

First we show that the projections S_N, N = 1, 2, · · · , are uniformly bounded. Now define

kxk₁ := sup

N

N

X

k=1

c_k(x)f_k

< ∞.

This norm is well-defined since P

n

c_n(x)f_n converges. The fact that k.k₁ is indeed a norm in H easily follows from the fact that k.k is a norm in H. We show that (H, k.k₁) is a Banach space. Let (x_n, n ∈ N) be a Cauchy sequence in (H, k.k1). Fix ε > 0. Then there is µ such that

kxn− xmk₁ = sup

N

N

X

k=1

[ck(xn) − ck(xm)] fk

< ε, for n, m ≥ µ,

which implies

N

X

k=1

[ck(xn) − ck(xm)] fk

< ε, ∀N, ∀n, m ≥ µ.

Then for every N ∈ N and for every n, m ≥ µ

k[c_N(x_n) − c_N(x_m)] f_Nk =

N

X

k=1

[c_k(x_n) − c_k(x_m)] f_k−

N −1

X

k=1

[c_k(x_n) − c_k(x_m)] f_k

≤

N

X

k=1

[ck(xn) − ck(xm)] fk

+

N −1

X

k=1

[ck(xn) − ck(xm)] fk

< 2ε.

Therefore

|c_N(x_n− x_m)| < 2ε/ kf_Nk , for n, m ≥ µ.

So, for every N ∈ N, (cN(x_n)) is a Cauchy sequence of numbers and converges, say to c^∗_N. By definition of k.k₁,

kx_n− x_mk ≤ kx_n− x_mk₁ < ε, ∀n, m ≥ µ.

Therefore xn is a Cauchy sequence in (H, k.k) and kxn− xk → 0 for some x in H.

Let m → ∞. Then by the last inequality, we obtain

kx_n− xk ≤ ε, ∀n ≥ µ. (2.2)

We know that

N

X

k=1

[c_k(x_n) − c_k(x_m)] f_k

< ε, ∀N, ∀n, m ≥ µ.

Let m → ∞. Then we obtain

N

X

k=1

[ck(xn) − c^∗_k] fk

≤ ε, ∀N, ∀n ≥ µ. (2.3)

Fix n > µ. Then we have for every N ∈ N,

N

X

k=1

c_k(x)f_k−

N

X

k=1

c^∗_kf_k

≤

N

X

k=1

c_k(x)f_k− x

+ kx − x_nk +

x_n−

N

X

k=1

c_k(x_n)f_k +

N

X

k=1

ck(xn)fk−

N

X

k=1

c^∗_kfk

. By (??) and (??), the second and the fourth term on the right hand side are less then ε. Since (f_k) is a basis, for large enough N , the first and the third terms are also less than ε. Thus it follows that for large enough N

N

X

k=1

c_k(x)f_k−

N

X

k=1

c^∗_kf_k

< 4ε.

Since

N

P

k=1

c_k(x)f_k → x, it follows that

N

P

k=1

c^∗_kf_k → x, that is x =

N

P

k=1

c^∗_kf_k. Hence c^∗_k = c_k(x) by uniqueness of c_k. Since

N

X

k=1

[c_k(x_n) − c^∗_k] f_k

≤ ε, ∀N, ∀n ≥ µ, we obtain that for every n ≥ µ,

kx_n− xk₁ = sup

N

N

X

k=1

[c_k(x_n) − c_k(x)] f_k

= sup

N

N

X

k=1

[c_k(x_n) − c^∗_k] f_k

≤ ε.

So kx_n− xk₁ → 0 as n → ∞. Thus (H, k.k₁) is a Banach space.

By definition of k.k₁, we have

kxk ≤ kxk₁, ∀x ∈ H, and this means that the identity operator

I : (H, k.k₁) → (H, k.k)

is a continuous one to one mapping. So by Open Mapping Theorem, it has a continuous inverse. So there is a constant c > 0 such that

kxk₁ ≤ c kxk , ∀x ∈ H.

Therefore

S_N(x) =

N

X

k=1

c_k(x)f_k

≤ c kxk , ∀N.

By this way, we have shown that S_N’s are uniformly bounded. Then kP_n(x)k = kS_n(x) − S_n−1(x)k ≤ kS_n(x)k + kS_n−1(x)k ≤ 2c kxk . So

kP_n(x)k = |c_n(x)| kf_nk ≤ 2c kxk , which implies

|c_n(x)| ≤ 2c kf_nkkxk .

Hence cn(x) is continuous for n = 1, 2, · · · . By Riesz Representation Theorem, each c_n(x) can be written in the form

c_n(x) = D x, ef_nE

,

where efn is a uniquely determined element in H. So every x ∈ H can be written as x =

∞

X

n=1

D x, ef_nE

f_n.

Since c_n(f_m) = δ_n,m for n, m = 1, 2, · · · , c_n’s and f_n’s are biorthogonal systems.

And c_n(f_m) = δ_n,m means that

c_n(f_m) = D

f_m, ef_nE

= δ_n,m.

Thus f_n’s and ef_n’s are biorthogonal. Hence we have shown that for any basis (f_n, n ∈ N) in H, there is a unique biorthogonal system 

fe_n, n ∈ N

in H such that

x =

∞

X

n=1

D x, ef_nE

f_n, ∀x ∈ H.

The bases that we consider are unconditional bases, so we don’t have convergence problems related to order of the elements. Thus for practical uses we use countable bases of the form (e_γ, γ ∈ Γ) where Γ is a countable set of indices, instead of bases of the form (e_n, n ∈ N).

Let H be a Hilbert space and let (e_γ, γ ∈ Γ) be an orthonormal basis in H. If A : H → H is an automorphism, then the system (fγ, γ ∈ Γ) given by

f_γ = Ae_γ, γ ∈ Γ (2.4)

is also a basis in H. For every x ∈ H, we have x = A A⁻¹x
= A X

γ

A⁻¹x, e_γ e_γ

!

=X

γ

x, A⁻¹
∗

e_γ f_γ =X

γ

D x, ef_γE

f_γ. Definition 4. A basis of the form (??) is called a Riesz basis.

So we have also showed that (fγ) is a basis with its biorthogonal system fe_γ = A^−1
∗

e_γ, γ ∈ Γ. (2.5)

Riesz bases are unconditional bases since orthonormal bases are unconditional bases.

Lemma 5. Let H₁ and H₂ be two Hilbert spaces and let A : H₁ → H₂ be an isomorphism. If (e_γ, γ ∈ Γ) is an orthonormal basis in H₁, then the system

f_γ = Ae_γ, γ ∈ Γ is a Riesz basis in H₂.

Proof. Assume H₁ and H₂ are two Hilbert spaces and A : H₁ → H₂ is an isomor- phism. Let (e_γ, γ ∈ Γ) be an orthonormal basis in H₁. Take any orthonormal basis in H₂, say (φ_γ, γ ∈ Γ). Then the operator B : H₂ → H₁ defined by

Bφ_γ = e_γ, ∀γ ∈ Γ.

Then B is clearly an isomorphism. Now take C : H2 → H2 given by C = A ◦ B.

Then C is an isomorphism and

Cφ_γ = f_γ, ∀γ ∈ Γ.

Thus (f_γ, γ ∈ Γ) is a Riesz basis in H₂.

Recall that `²(Γ) is the space consisting of the generalized sequences (x_γ, γ ∈ Γ) such that

X

γ

|xγ|² < ∞.

We consider `²(Γ) equipped with the inner product hx, yi =X

γ

x_γy_γ,

where x = (x_γ, γ ∈ Γ) and y = (y_γ, γ ∈ Γ).

Now we give a characterization of Riesz bases by the following theorem.

Theorem 6. Suppose that (fγ, γ ∈ Γ) is a basis in H and



feγ, γ ∈ Γ

 is its biorthogonal system. Then (f_γ, γ ∈ Γ) is a Riesz basis if and only if

c ≤ kf_γk ≤ C, ∀γ ∈ Γ (2.6)

and

m kxk² ≤X

γ

   D

x, ef_γE  

2

kf_γk² ≤ M kxk², (2.7)

for some positive constants c, C, m and M .

Proof. First let (f_γ, γ ∈ Γ) be a Riesz basis in H with its biorthogonal system



fe_γ, γ ∈ Γ

. Then there is an orthonormal basis (e_γ, γ ∈ Γ) in H and an automor- phism A such that

A(e_γ) = f_γ. Then we have

kf_γk = kAe_γk ≤ kAk and 1 = ke_γk =

A⁻¹f_γ ≤

A⁻¹ kf_γk , which gives us

1

kA⁻¹k ≤ kf_γk ≤ kAk . (2.8)

So we get (??) with c = _kA¹−1k and C = kAk. Also we have that P

γ

x, (A⁻¹)^∗eγ

2kfγk² = P

γ

|hA⁻¹x, eγi|²kfγk²

≤ kAk²P

γ

|hA⁻¹x, e_γi|² by (??)

= kAk²kA⁻¹xk²

≤ kAk²kA⁻¹k²kxk². and

P

γ

x, (A⁻¹)^∗e_γ 

2kf_γk² = P

γ

|hA⁻¹x, e_γi|²kf_γk²

≥ _kA−1¹ _k2

P

γ

|hA⁻¹x, e_γi|² by (??)

= ¹

kA⁻¹k² kA⁻¹xk²

≥ _kA−1¹ _k2 1

kAk² kxk². since

kxk =

A⁻¹Ax

≤ kAk A⁻¹x

which means

A⁻¹x

≥ kxk kAk. Combining these results, we get

1 kA⁻¹k²

1

kAk² kxk² ≤X

γ

   D

x, ef_γE  

2

kf_γk² ≤ kAk² A⁻¹

2kxk², (2.9)

which proves (??) with m = _kA−1k¹2kAk² and M = kAk²kA⁻¹k². Now let (fγ, γ ∈ Γ) be a basis in H and



feγ, γ ∈ Γ



be its biorthogonal system such that (??) and (??) holds. Since Γ is a countable set, we may think that Γ = {γ_i, i = 1, 2, · · · }. With this enumaration, consider the operator B : `²(Γ) → H given by

B ((x_γ)) =

∞

X

i=1

x_γif_γi. Let

S_N =

N

X

i=1

x_γif_γi. Then

D

S_N, ff_γiE

= x_γi, i = 1, 2, · · · , N.

So by using (??) and (??), we get that k kSNk² ≤X

γ

|xγi|² ≤ K kSNk²,

where k = m/C² and K = M/c². By the same argument, we get that k kS_{N +M} − S_Nk² ≤

N +M

X

i=N

|x_γi|² → 0 as N → ∞

since (xγ) ∈ `². So (SN) is a Cauchy sequence in H. Since H is complete, (SN) is convergent to some s in H. Thus the series

∞

P

i=1

x_γif_γi converges which shows that B is well-defined.

Next we prove that B is continuous. Set x = B ((xγ)) =

∞

X

i=1

xγifγi.

Then by (??) and (??)

k kxk² = k kB ((x_γ))k² ≤

∞

X

i=1

|x_γi|² = k(x_γ)k²_`2 ≤ K kB ((x_γ))k².

So B is continuous since

kB ((x_γ))k ≤ 1

√

kk(x_γ)k_`2, and B⁻¹ is continuous since

kB ((x_γ))k ≥ 1

√K k(x_γ)k_`2. Thus B is an isomorphism.

Let (e_γ, γ ∈ Γ) be the orthonormal basis in `² given by e_γ(α) = δ_γ,α.

Then by definition of B,

Beγ = fγ, ∀γ ∈ Γ.

Thus (f_γ) is a Riesz basis by the previous lemma.

Definition 7. A system (f_n, n ∈ N) is called minimal if f_j ∈ span {f/ _k}_k6=j, ∀j ∈ N.

Theorem 8. (Bari-Markus Theorem) Let (en, n ∈ N) be a Riesz basis in a Hilbert space H and let (f_n, n ∈ N) be a minimal system of vectors such that

∞

X

n=1

kf_n− e_nk² < ∞.

Then (f_n, n ∈ N) is also a Riesz basis.

Proof. It sufficies to show that there is an isomorphism A such that A (e_n) = f_n. Since (e_n, n ∈ N) is a Riesz basis, there is an isomorphism B and orthonormal basis (φ_n, n ∈ N) such that B (φn) = e_n. So A ◦ B will be an isomorphism such that B ◦ A (φ_n) = f_n.

For x =

∞

P

n=1

x_ne_n, set

T x :=

∞

X

n=1

x_n(e_n− f_n) . The operator T is bounded since

kT xk =

∞

P

n=1

x_n(e_n− f_n)

≤

∞

P

n=1

|x_n| ke_n− f_nk

≤

 _∞ P

n=1

|x_n|²

1/2_∞ P

n=1

ke_n− f_nk²

1/2

≤

√ M c² s kxk ,

where c and M are coming from the inequalities (??),(??) and s² =

∞

P

n=1

kf_n− e_nk². Let (T_k, k ∈ N) be the sequence of finite rank operators given by

Tkx =

k

X

n=1

xn(en− fn), for x =

∞

X

n=1

xnen.

We have kT_k− T k → 0 as k → ∞, since kT_kx − T xk =

∞

P

n=k+1

x_n(e_n− f_n)

≤

∞

P

n=k+1

|x_n| ke_n− f_nk

≤

 _∞ P

n=k+1

|x_n|²

1/2 _∞ P

n=k+1

ke_n− f_nk²

1/2

≤ kxk

 _∞ P

n=k+1

ke_n− f_nk²

1/2

, and

∞

P

n=1

ke_n− f_nk² is a convergent series. Since T_k are finite-dimensional operators, it follows that T is a compact operator.

Now consider the operator A = 1 − T . A is invertible if 1 is not in the spectrum of T . Let x ∈ ker A, that is

(1 − T )x = 0, for x =

∞

X

n=1

x_ne_n.

Then it follows

∞

X

n=1

x_nf_n= 0,

which implies (since (f_n, n ∈ N) is a minimal system) that xn = 0 for every n, hence x = 0. So 1 is not an eigenvalue of the operator T . Recall that spectrum of a compact operator contains only eigenvalues. Since T is a compact operator and 1 is not an eigenvalue of T , 1 is not in the spectrum of T . Then A is invertible, so it is an isomorphism. Hence (f_n, n ∈ N) is also a Riesz basis.

3 Riesz Basis of Root Functions of Dirac Operator

In this section, we study the spectrum of free Dirac operator L⁰_bcsubject to strictly regular boundary conditions bc. We show that the eigenfunctions of L⁰_bc form a Riesz basis in L²([0, π], C²). Moreover, we show the existence of Riesz basis of root functions of Dirac operator L_bc subject to strictly regular boundary conditions bc.

The differential expression

L(v)y = i1 0 0 −1

 dy dx+

 0 P (x) Q(x) 0



y. (3.1)

is known as the one dimensional Dirac operator. The matrix v =

 0 P (x) Q(x) 0

 is called the Dirac potential. In case v ≡ 0, we write L⁰ = L(0) and call L⁰ free Dirac operator. A general boundary condition for the Dirac operator is given by a system of two linear equations

a₁y₁(0) + b₁y₁(π) + a₂y₂(0) + b₂y₂(π) = 0

c1y1(0) + d1y1(π) + c2y2(0) + d2y2(π) = 0. (3.2) Of course, equivalent systems of the form (??) define one and the same bc. Each boundary condition is determined by the matrix of the coefficients of (??)

a₁ b₁ a₂ b₂ c₁ d₁ c₂ d₂



. (3.3)

But if we multiply this matrix from the left by a 2x2 invertible matrix, we get another matrix that determines the same bc.

We may assign to every boundary condition bc of the form (??) a corresponding operator L⁰_bc as follows. Let

Dom (L⁰_bc) =



y =y₁ y₂



: y₁ and y₂ are absolutely continuous, y satisfies the boundary conditions bc, and y₁⁰, y⁰₂ ∈ L²([0, π])} , and let

L⁰_bc(y) = i1 0 0 −1

 y⁰₁ y⁰₂

 . Theorem 9. L⁰_bc is a closed densely defined operator.

Proof. Let f_n

gn

!

, L⁰ f_n gn

!!

−→

f g

! , h₁

h2

!!

in L²([0, π], C²).

This means that fn

g_n

!

, if_n⁰

−ig⁰_n

!!

−→

f g

! , h1

h₂

!!

in L²([0, π], C²).

So



f_n, if_n⁰ _k.k

−→ (f, h₁) and 

g_n, −ig⁰_n _k.k

−→ (g, h₂) . Since fn’s are measurable functions and fn

−→ f , fk.k n converges to f in measure.

Then there is a subsequence f_nk such that f_nk(x) −→ f (x). Thus there exists c ∈^a.e.

[0, π] such that f_n(c) → f (c). Now define H1(x) := 1

i

x

Z

c

h1(t)dt.

Then

|f_n(x) − f_n(c) − H₁(x)|² =    

x

R

c

f_n⁰(t) − ¹_ih₁(t)
dt    

2

≤

_π R

0

f_n⁰(t) −¹_ih₁(t) dt

2

≤

_π R

0

f_n⁰(t) −¹_ih₁(t) 

2dt

 _π R

0

1²dt



=

f_n⁰ −¹_ih₁

2π.

By this inequality, we get that

f_n(x) − f_n(c)−→ H^unif ₁(x), since f_n⁰ −→^{k.k h}_i¹. And also f_n(x) − f_n(c)−→ f (x) − f (c), so^a.e.

H₁(x) = f (x) − f (c) a.e.

Since we identify functions that are equal almost everywhere, we may think that the previous equality holds for every x ∈ [0, π]. So

h₁(x) = iH₁⁰(x) = if⁰(x) a.e. (3.4) We can similarly define H₂(x) := −¹_i

x

R

c

h₂(t)dt and get that

h₂(x) = −ig⁰(x). (3.5)

Since we identify functions that are equal almost everywhere and f_n’s are absolutely continuous functions, we may think that f_n(x) → f (x), ∀x ∈ [0, π]. So f satisfies bc. Similarly g also satisfies bc. So by (??) and (??)

f g

!

∈ dom(L⁰_bc) and L⁰ f g

!

= h₁ h₂

! .

Thus we have shown that the graph of L⁰_bc is closed, which means that L⁰_bcis a closed operator.