Self-adjoint Operators

Self-adjointness is the most important notion on unbounded operators in this book.

The main results about self-adjoint operators are the spectral theorem proved in Sect. 5.2 and the corresponding functional calculus based on it. A large effort is made in this book to prove that certain symmetric operators are self-adjoint or to extend symmetric operators to self-adjoint ones.

Definition 3.5 A densely defined symmetric operatorT on a Hilbert spaceHis called self-adjoint if T =T^∗ and essentially self-adjoint, briefly e.s.a., if T is self-adjoint, or equivalently, ifT =T^∗.

Let us state some simple consequences that will be often used without mention.

A self-adjoint operatorT is symmetric and closed, sinceT^∗is always closed.

LetT be a densely defined symmetric operator. Since thenT ⊆T^∗,T is self-adjoint if and only ifD(T )=D(T^∗). Likewise,T is essentially self-adjoint if and only ifD(T )=D(T^∗).

Any self-adjoint operatorT onHis maximal symmetric, that is, ifSis a symmet-ric operator onHsuch thatT ⊆S, thenT =S. Indeed,T ⊆Simplies thatS^∗⊆T^∗. Combined withS⊆S^∗andT^∗=T, this yieldsS⊆T, so thatT =S.

Some self-adjointness criteria follow easily from the next result. The nice direct sum decomposition (3.10) of the domainD(T^∗)is called von Neumann’s formula.

Proposition 3.7 LetT be a densely defined symmetric operator. Then D

T^∗

=D(T )˙+N

T^∗−λI

˙+N

T^∗−λI

forλ∈C\R, (3.10) dimD

T^∗

/D(T )=d₊(T )+d₋(T ). (3.11) Proof Let us abbreviateNλ:=N(T^∗−λI )andN_λ:=N(T^∗−λI ). The inclusion D(T )+Nλ+N_λ⊆D(T^∗)is obvious. We prove thatD(T^∗)⊆D(T )+Nλ+N_λ.

Letx∈D(T^∗). By Corollary 2.2 we have

H=R(T−λI )⊕N_λ. (3.12)

We apply (3.12) to the vector (T^∗−λI )x∈H. Then there exist x0∈D(T ) and x₋ ∈N_λsuch that(T^∗−λI )x=(T−λI )x₀+x₋ . Sinceλ /∈R, we can setx₋:=

(λ−λ)⁻¹x₋. The preceding can be rewritten as(T^∗−λI )(x−x0−x₋)=0, that is,x₊:=x−x0−x₋is inNλ. Hence,x=x0+x₊+x₋∈D(T )+Nλ+N_λ.

To prove that the sum in (3.10) is a direct sum, we assume thatx0+x₊+x₋=0 forx₀∈D(T ),x₊∈Nλ, andx₋∈N_λ. Then

T^∗−λI

(x₀+x₊+x₋)=(T−λI )x₀+(λ−λ)x₋=0.

The vector(λ−λ )x₋=(T−λI )x₀is inR(T−λI )∩N_λ. Therefore, it follows from (3.12) thatx₋=0 and(T−λI )x0=0. SinceT is symmetric andλis not real, the latter yieldsx₀=0 by Lemma3.4. Sincex₀=x₋=0, we getx₊=0.

The preceding proves (3.10). (3.11) is an immediate consequence of (3.10).

3.2 Self-adjoint Operators 43 Recall that a densely defined symmetric operatorT is essentially self-adjoint if and only ifD(T )=D(T^∗). By (3.10) (or (3.11)) the latter is satisfied if and only if d₊(T )=d₋(T )=0. That is, using formulas (3.4)–(3.7), we obtain the following:

Proposition 3.8 LetT be a densely defined symmetric operator onH, and letλ₊ andλ₋be complex numbers such that Imλ₊>0 and Imλ₋<0.

Then the operator T is essentially self-adjoint if and onlyd₊(T )=0 (equiv-alently,N(T^∗−λ₊I )= {0}, or equivalently,R(T −λ₊I )=H) andd₋(T )=0 (equivalently,N(T^∗−λ₋I )= {0}, or equivalently,R(T−λ₋I )=H).

For lower semibounded operators, we have the following stronger result.

Proposition 3.9 LetT be a densely defined symmetric operator such that there is a real number inπ(T ); in particular, ifT is lower semibounded, the latter is fulfilled, andC\ [m_T,∞)⊆π(T ). ThenT is essentially self-adjoint if and only ifd_λ(T )=0 (equivalently,N(T^∗−λI )= {0}, or equivalently,R(T−λI )=H) for one, hence all,λ∈π(T ).

Proof Sinced_λ(T )=d_±(T )by Proposition3.3andC\[m_T,∞)⊆π(T )by Propo-sition3.2, the assertions follow at once from Proposition3.8.

The next proposition characterizes the numbers of the resolvent set of a self-adjoint operator. Condition (ii) therein is often useful to detect the spectrum of the operator.

Proposition 3.10 LetT be a self-adjoint operator on a Hilbert spaceH. For any complex numberλ, the following conditions are equivalent:

(i) λ∈ρ(T ).

(ii) λ∈π(T ), that is, there exists a constantc_λ>0 such that(T−λI )x ≥c_λx for allx∈D(T ).

(iii) R(T −λI )=H.

Moreover, ifλ∈C\R, thenλ∈ρ(T )andR_λ(T ) ≤ |Imλ|⁻¹. Proof SinceT is self-adjoint, by Corollary 2.2 we have

R(T−λI )^⊥=N(T−λI ). (3.13) First suppose thatλ∈C\R. Then λ∈π(T ), and henceR(T −λI )is closed by Proposition 2.1(iv). Further, N(T −λI )= {0} and N(T −λI )= {0} by Lemma 3.4(i). Therefore, it follows from (3.13) that R(T −λI )=H. Hence, λ∈ρ(T )by Proposition 2.7(i), and all three conditions (i)–(iii) are satisfied.

Lety∈H. Thenx:=R_λ(T )y∈D(T )andy=(T−λI )x. Insertingxinto (3.3) yieldsy ≥ |Imλ|R_λ(T )y. That is,R_λ(T ) ≤ |Imλ|⁻¹.

From now on assume thatλ∈R.

(i)→(ii) is obvious, sinceρ(T )⊆π(T ).

(ii)→(iii): Sinceλ∈π(T )by (ii),N(T−λI )= {0}, soR(T−λI )is dense in Hby (3.13). BecauseR(T−λI )is closed by Proposition 2.1(iv),R(T−λI )=H.

(iii)→(i): SinceR(T −λI )=Hby (iii), we haveN(T −λI )= {0}again by (3.13). Therefore, it follows from Proposition 2.7(i) thatλ∈ρ(T ).

Now we give some self-adjointness criteria that do not assume that the symmetric operator is densely defined. They are essentially based on the following proposition.

Proposition 3.11 LetT be a symmetric operator onH. If there exists a complex numberλsuch thatR(T −λI )=HandR(T−λI )is dense inH, thenT is self-adjoint, andλandλare inρ(T ).

Proof We first show thatD(T )is dense inH. Lety∈D(T )^⊥. SinceR(T−λI )= H, there exists a vectoru∈D(T )such that y=(T −λI )u. Therefore, we have 0= y, x = (T −λI )u, x = u, (T −λI )xforx∈D(T ), sou∈R(T −λI )^⊥. SinceR(T −λI )is dense,u=0 and hencey=0. Thus,D(T )is dense.

Hence,T^∗is well defined. Letw∈D(T^∗). Applying once more the assumption R(T −λI )=H, there is av∈D(T )such that(T^∗−λI )w=(T−λI )v. Then

(T −λI )x, w

= x,

T^∗−λI w

=

x, (T−λI )v

=

(T −λI )x, v for x ∈D(T ). Because R(T −λI ) is dense, w=v and so w∈D(T ), that is, D(T^∗)⊆D(T ). SinceT is symmetric, this implies thatT is self-adjoint.

SinceT is self-adjoint andR(T −λI )=H,λ∈ρ(T )by Proposition3.10.

In particular, the preceding result implies again Proposition3.8.

Note that the assumptionR(T−λI )=Hin Proposition3.11is crucial. It cannot be replaced by the density ofR(T −λI )inH; see Exercise 12.

The caseλ=0 in Proposition3.11and Corollary 1.9 yields the following.

Corollary 3.12 IfT is a symmetric operator onHsuch thatR(T )=H, thenT is self-adjoint, and its inverseT⁻¹is a bounded self-adjoint operator onH.

Proposition 3.13 LetT be a closed symmetric operator onH, and letλ₊, λ₋∈C, where Imλ₊>0 and Imλ₋ <0. The operator T is self-adjoint if and only if d₊(T )=0 (equivalently, λ₋ ∈ρ(T ), or equivalently, R(T −λ₋I )=H) and d₋(T )=0 (equivalently,λ₊∈ρ(T ), or equivalently,R(T −λ₊I )=H).

Proof From Proposition3.10, a self-adjoint operator satisfies all these conditions.

To prove the converse, we first note thatR(T−λI )is closed for anyλ∈C\R by Proposition 2.1(iv), becauseλ∈π(T )andT is closed. Therefore,d_±(T )=0 if and only ifR(T −λ_±I )=H, or equivalently,λ_±∈ρ(T )by Proposition3.10.

Thus, it suffices to assume thatd₊(T )=d₋(T )=0. But thenR(T±iI )=H, soT

is self-adjoint by Proposition3.11.

By the preceding proposition a closed symmetric operatorT is self-adjoint if and only ifC\R⊆ρ(T ). We restate this fact by the following corollary.

3.2 Self-adjoint Operators 45 Corollary 3.14 A closed symmetric linear operatorT onHis self-adjoint if and only ifσ (T )⊆R.

Since the deficiency indices are constant on connected subsets ofπ(T ), Proposi-tion3.13implies the following result.

Proposition 3.15 LetT be a closed symmetric operator. Suppose thatπ(T ) con-tains a real number. ThenT is self-adjoint if and only ifd_λ(T )=0 (equivalently, λ∈ρ(T ), or equivalently,R(T −λI )=H) for one, hence all,λ∈π(T ).

In the following examples we construct self-adjoint extensions of symmetric op-erators by adding appropriate elements to the domains and using Proposition3.6.

Example 3.4 (Examples 1.4, 1.5, and3.2continued) Leta, b∈R,a < b. Recall from Examples 1.4 and3.2thatT = −i_dx^d onD(T )=H₀¹(a, b)is a closed sym-metric operator with deficiency indices(1,1). In Example 1.5 it was shown that for anyz=e^iϕ∈T, the operatorS_z= −i_dx^d with boundary conditionf (b)=zf (a)is self-adjoint and hence a self-adjoint extension ofT.

We rederive this result by applying Proposition3.6to the subspaceEz:=C·uϕ, whereu_ϕ(x):=exp(iϕ(b−a)⁻¹x). Using formula (1.13), one verifies thatD(T_Ez) is a symmetric subspace. Thus,T_Ez is a closed symmetric operator with deficiency indices(0,0)by Proposition3.6and hence self-adjoint by Proposition3.13. Since uϕ(b)=zuϕ(a), we haveD(T_Ez)⊆D(Sz)andT_Ez⊆Sz. Hence,T_Ez=Sz. ◦ Example 3.5 (Example3.3continued) As in Example 3.3, we let T = −_dx^d22 on D(T )= {f ∈H²(R):f(0)=0}. By Lemma 1.12, applied to f, the functional f→f(0)is continuous onH²(R), soD(T )is a closed subspace ofH²(R). Since the graph norm ofT is equivalent to the norm ofH²(R),D(T )is complete in the graph norm ofT. Therefore,T is closed.

ForB∈R, we define the operatorTB= −_dx^d22 on the domain D(TB)=

f∈H²(−∞,0)⊕H²(0,+∞):

f(+0)=f(−0), f (+0)−f (−0)=Bf(0) . Statement T_Bis a self-adjoint extension ofT.

Proof Using the integration-by-parts formula (3.8), one easily verifies thatT_B is symmetric. From Example3.3we know that the symmetric operatorT has defi-ciency indices(1,1). Clearly,D(T )⊆D(TB), andTBis an extension ofT.

Choose numbersa, b >0 andα, β∈Csuch thatα(2+aB)+β(2+bB)=0 and aα+bβ =0. Define the function f on R by f (x)=αe^−ax+βe^−bx and f (−x)= −f (−x)ifx >0. Thenf ∈D(T_B)andf /∈D(T ). PutE:=C·f. Since T_E⊆TB andTB is symmetric,D(T_E)is a symmetric subspace ofD(T^∗). There-fore, by Proposition3.6,T_E is a closed symmetric operator with deficiency indices (0,0). Hence,T_E is self-adjoint by Proposition3.13andT_E=T_B.

IfB<0, then−4B⁻²is an eigenvalue of T_B. An eigenvector is the functionf witha= −2B⁻¹,β=0, that is,f (x)=e^2B−1xandf (−x)= −f (x),x >0. ◦ The problem of describing self-adjoint extensions (if there are any) of a general symmetric operator will be studied in detail in Chaps. 13 and 14. In general, there is no self-adjoint extension in the same Hilbert space. Here we prove only two simple results. The second shows by an easy construction that a densely defined symmetric operator has always a self-adjoint extension on a larger Hilbert space.

Proposition 3.16 IfT is a densely defined closed symmetric operator onHandμ is a real number inπ(T ), then there exists a self-adjoint extensionAofT onH such thatμ∈ρ(A).

Proof Upon replacingT by T +μI, we can assume without loss of generality thatμ=0. LetP denote the projection of HontoN(T^∗). Sinceμ=0∈π(T ), we haveR(T^∗)=Hby Proposition3.2(iii). Hence, there exists a linear subspace D0ofD(T^∗)such thatT^∗D0=N(T^∗). DefineD(A):=D(T )+(I−P )D0 and A:=T^∗D(A). Obviously,Ais an extension ofT.

We show that Ais symmetric. Let u_k=x_k+y_k, wherex_k∈D(T )and y_k ∈ (I−P )D0,k=1,2. From the relationT^∗(I−P )D0=T^∗D0=N(T^∗)⊥(I−P )D0

we obtainT^∗y_j, y_k =0 forj, k=1,2. Therefore, Au1, u2 =

T x1+T^∗y1, x2+y2

= T x1, x2 +

T^∗y1, y2

+ T x1, y2 +

T^∗y1, x2

= x1, T x₂ +

x₁, T^∗y₂

+ y₁, T x₂

= x₁, T x₂ +

y₁, T^∗y₂ +

x₁, T^∗y₂

+ y₁, T x₂

=

x1+y1, T x2+T^∗y2

= u1, Au2, that is,Ais symmetric.

Since 0∈π(T )andT is closed, we haveH=R(T )⊕N(T^∗)by Corollary 2.2.

By construction,A(I−P )D0=T^∗(I−P )D0=N(T^∗), and soR(A)=H. There-fore, by Corollary3.12,Ais self-adjoint, and 0∈ρ(A).

Proposition 3.17 Let T be a densely defined symmetric operator on a Hilbert spaceH. Then there exists a self-adjoint operatorAon a Hilbert spaceG which containsHas a subspace such thatT ⊆AandD(T )=D(A)∩H.

Proof LetT_dbe the operator on the “diagonal” subspaceHd= {(x, x):x∈H}of the Hilbert spaceG:=H⊕H defined by Td(x, x)=(T x, T x),x ∈D(T ), with domainD(T_d):= {(x, x):x∈D(T )}. The mapUgiven byU (x):=^√1₂(x, x)is an isometry ofHontoHdsuch thatU T U⁻¹=T_d. Hence, the operatorsT andT_dare unitarily equivalent, so it suffices to prove the assertion forT_d.

3.2 Self-adjoint Operators 47

Consider the following block matrix on the Hilbert spaceG=H⊕H:

A=

0 T T^∗ 0

.

That is,Ais defined byA(x, y)=(T y, T^∗x)for(x, y)∈D(A)=D(T^∗)⊕D(T ).

A straightforward verification shows thatAis self-adjoint operator onG. Obviously,

Td⊆AandD(A)∩Hd=D(Td).

We close this section by proving two simple technical results. They will be used in Sects.3.4, 5.2, 7.1, and 7.3.

Proposition 3.18 LetT be a densely defined closed linear operator from a Hilbert spaceH1into a Hilbert spaceH2. Then:

(i) I +T^∗T is a bijective mapping of H1. Its inverse C:=(I +T^∗T )⁻¹ is a bounded self-adjoint operator onH1such that 0≤C≤I.

(ii) T^∗T is a positive self-adjoint operator onH1, andD(T^∗T )is a core forT. Proof (i): Recall that G(T^∗)=V (G(T ))^⊥ by Lemma 1.10, where V (x, y) = (−y, x), x∈H1,y ∈H2. Hence, H2⊕H1=G(T^∗)⊕V (G(T )). Therefore, for eachu∈H1, there exist vectorsx∈D(T )andy∈D(T^∗)such that

(0, u)= y, T^∗y

+V (x, T x)=

y−T x, T^∗y+x ,

so y=T x and u=x +T^∗y=x +T^∗T x=(I +T^∗T )x. That is, I +T^∗T is surjective. The operatorI+T^∗T is injective, because

I+T^∗T x²=

x+T^∗T x, x+T^∗T x

= x²+T^∗T x²+2T x² (3.14) forx∈D(T^∗T ). Thus,I+T^∗T is bijective.

Letu=(I+T^∗T )x, wherex∈D(T^∗T ). Thenx=Cu, and by (3.14), Cu = x ≤I+T^∗T

x= u.

Therefore,Cis a bounded operator onH1. From (3.14) we also derive Cu, u = x, u = x²+ T x²≤I+T^∗T

x²= u².

Therefore, C is symmetric and hence self-adjoint, since C is bounded. Further, 0≤C≤I.

(ii): SinceC =(I+T^∗T )⁻¹ is self-adjoint as just proved, so are its inverse I+T^∗T by Theorem 1.8(iv) and henceT^∗T. We haveT^∗T x, x = T x²≥0 for x∈D(T^∗T ), that is, the operatorT^∗T is positive.

ThatD(T^∗T )is a core forT means thatD(T^∗T )is dense in the Hilbert space (D(T ),·,·T), where ·,·T is defined by (1.3). If y ∈D(T ) is orthogonal to D(T^∗T )in(D(T ),·,·T), then

0= y, xT = y, x + T y, T x = y,

I+T^∗T x

forx∈D(T^∗T ). Hence,y=0, sinceR(I+T^∗T )=H1. This proves thatD(T^∗T )

is a core forT.