Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors, and Self-adjointness of Symmetric Operatorsand Self-adjointness of Symmetric Operators

In this section we develop criteria for the self-adjointness of symmetric operators and the strong commutativity of self-adjoint operators by means of analytic vectors, quasi-analytic vectors, and Stieltjes vectors. These are classes of vectorsx which are defined by growth conditions on the sequence(Tⁿx)_n∈N0.

Let us begin with some basics on quasi-analytic functions. Let(m_n)_n∈N0 be a positive sequence, andJ ⊆Rbe an open interval. We denote byC{m_n}the set of functionsf ∈C^∞(J)for which there exists a constantK_f >0 such that

f⁽ⁿ⁾(t )≤K_fⁿm_n for allt∈J, n∈N0. (7.20) A linear subspaceC ofC^∞(J)is called a quasi-analytic class, provided that the following holds: iff ∈C and there is a pointt0∈J such thatf⁽ⁿ⁾(t0)=0 for all n∈N0, thenf (t )≡0 onJ.

The quasi-analytic classes among the linear subspacesC{mn}are characterized by the following famous Denjoy–Carleman theorem.

Proposition 7.6 C{mn}is a quasi-analytic class if and only if ∞

n=1

inf

k≥nm^{1/ k}_{k −}1

= ∞. (7.21)

A proof of Proposition7.6can be found in [Hr, Theorem 1.3.8]. In the case of log convex sequences a proof is given in [Ru1, Theorem 19.11].

For the results obtained in this section, it suffices to have the following:

Corollary 7.7 Let(m_n)_n∈N0 be a positive sequence such that ∞

n=1

m⁻_n^1/n= ∞. (7.22)

Suppose thatf ∈C^∞(J)and there is a constantKf >0 such that (7.20) is sat-isfied. If there exists at0∈J such thatf⁽ⁿ⁾(t0)=0 for alln∈N0, thenf (t )≡0 onJ.

Proof Since obviously m^1/n_n ≥infk≥nm^{1/ k}_k , (7.22) implies (7.21), soC{mn}is a quasi-analytic class by Proposition7.6. This gives the assertion.

Example 7.2(m_n=n!)_n∈N Sincen! ≤nⁿ, the sequence(n!)_n∈N0 satisfies (7.21) and (7.22), soC{n!} is a quasi-analytic class. But any function f ∈C{n!}has a holomorphic extension to some strip{z:Rez∈J, |Imz|< δ},δ >0 (see, e.g., [Ru1, Theorem 19.9]). Using the latter fact, the assertion of Corollary7.7is well

known in this special case. ◦

7.4 Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors 145 Example 7.3 (log convex sequences) Let(m_n)_n∈N0be a positive sequence such that m0=1 and

m²_n≤m_n−1m_n+1 forn∈N. (7.23) Condition (7.23) means that(logm_n)_n∈N0 is a convex sequence. It is well known and easily checked that thenm^1/n_n ≤m^{1/ k}_k forn≤k, som^1/n_n =infk≥nm^{1/ k}_k . There-fore, by Proposition7.6, in this caseC{m_n}is a quasi-analytic class if and only if

nm⁻_n^1/n= ∞, that is, if (7.22) is satisfied. ◦ Next, we define and investigate various classes of vectors for a linear operatorT on a Hilbert spaceH. A vectorx∈His called aC^∞-vector ofT ifx belongs to

D^∞(T ):=_∞

n=1D Tⁿ

.

The reason for this terminology stems from the following fact which follows easily from Proposition 6.1: IfT is self-adjoint, then a vector x∈Hbelongs to D^∞(T )if and only ift→e^{it T}xis aC^∞-map ofRinto the Hilbert spaceH.

Definition 7.1 Letx∈D^∞(T ). We write

x∈D^b(T )andxis called bounded forT if there is a constantBx>0 such that

Tⁿx≤B_xⁿ forn∈N, (7.24)

x∈D^a(T )andxis called analytic forT if there exists a constantC_x>0 such that Tⁿx≤C_xⁿn! forn∈N0, (7.25) x∈D^qa(T )andxis called quasi-analytic forT if

∞ n=1

Tⁿx^−1/n= ∞, (7.26)

x∈D^s(T )andxis called a Stieltjes vector forT if ∞

n=1

Tⁿx^−1/(2n)= ∞. (7.27)

Obviously, eigenvectors ofT are bounded vectors. The setsD^a(T )andD^b(T ) are linear subspaces ofD^∞(T ), and we have

D^b(T )⊆D^a(T )⊆D^qa(T )⊆D^s(T ). (7.28) The second inclusion in (7.28) follows from the inequalityTⁿx^−1/n≥(Cxn)⁻¹, sincen! ≤nⁿ. A vectorxis inD^qa(T )(resp.D^s(T )) if and only ifλxis forλ=0, but the sum of two vectors fromD^qa(T )(resp.D^s(T )) is in general not inD^qa(T ) (resp.D^s(T )).

Example 7.4 (Example7.3continued) Suppose that the operatorT is symmetric andx is a unit vector ofD^∞(T ). Setmn:=Tⁿx. Then the assumptions of Exam-ple7.3are fulfilled, sincem₀=1 and

m²_n=Tⁿx²=

Tⁿx, Tⁿx

=

Tⁿ⁻¹x, Tⁿ⁺¹x

≤m_n−1m_n+1 forn∈N. Hence, as stated in Example7.3,C{Tⁿx}is a quasi-analytic class if and only if

nTⁿx^−1/n= ∞, that is,x∈D^qa(T ).

In Example 7.3 it was also noted that T²ⁿ⁺¹x^1/(2n+1)≥ T²ⁿx^1/(2n) for n∈N. This in turn yields that

kT^kx^{−1/ k}≤2

nT²ⁿx^−1/(2n). Therefore, x∈D^qa(T )if and only if

nT²ⁿx^−1/(2n)= ∞. ◦

Example 7.5 (Examples of analytic vectors) LetAbe a self-adjoint operator onH, and letα >0. Clearly,e^−α|A|ande^−αA2 are bounded operators onH. Suppose that u, v∈H.

Statement x:=e^−α|A|uandy:=e^−αA2vare analytic vectors forA.

Proof By Stirling’s formula [RA, p. 45] there is a null sequence(ε_n)such that n! =√

2π n^n+1/2e⁻ⁿ(1+εn) forn∈N. (7.29) Hence, there is a constantc >0 such that n²ⁿ≤ce²ⁿ(n!)² for n∈N. By some elementary arguments one shows thatλ²ⁿe^−2α|λ|≤α⁻²ⁿn²ⁿforλ∈R. Thus,

Rλ²ⁿe^−2α|λ|d

E_A(λ)u, u

≤

Rα⁻²ⁿn²ⁿd

E_A(λ)u, u

≤cu²α⁻²ⁿe²ⁿ(n!)². Hence,u∈D(Aⁿe^−α|A|)by (4.26), that is, x∈D(Aⁿ). Since the first integral is equal toAⁿe^−α|A|u²= Aⁿx²by (4.33), the preceding shows thatx∈D^a(A).

Sinceλ²ⁿe^−2αλ2 ≤λ²ⁿe^−2α|λ| for λ∈R,|λ| ≥1, a slight modification of this

reasoning yieldsy∈D^a(A). ◦

Proposition 7.8 Suppose that the operatorT is self-adjoint andx∈D^a(T ). For anyz∈C,|z|< C_x⁻¹, whereCxis given by (7.25), we havex∈D(e^zT)and

e^zTx= lim

n→∞

n k=0

z^k

k!T^kx. (7.30)

Proof Fixz∈C, |z|< C_x⁻¹. Let E be the spectral measure of T, andE_x(·)the measureE(·)x, x. Letn∈N. ThenTn:=T E([−n, n])is a bounded self-adjoint operator onHwith spectrum contained in[−n, n]. Since the series_∞

k=0z^k k!λ^k con-verges to the functione^zλuniformly on[−n, n], we havee^zTnx=_∞

k=0z^k

k!T_n^kx(see Proposition 4.12(v)). Using properties of the functional calculus, we derive

7.4 Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors 147 _n

−n

e^zλ2dE_x(λ) 1/2

=e^zTE

[−n, n]x=e^zTnx= ^∞

k=0

z^k k!T_n^kx

≤^∞

k=0

|z|^k

k! T_n^kx≤^∞

k=0

|z|^k k! T^kx

≤ ∞ k=0

|zCx|^k=

1− |zCx|₋1

. Lettingn→ ∞, we get

R|e^zλ|²dE_x(λ) <∞. Therefore,x∈D(e^zT)by (4.26).

Replacingzby|z|and by−|z|in the preceding, we conclude that the functions e^2|z|λande^−2|z|λareL¹(R, Ex). Hence,e^2|zλ|is inL¹(R, Ex). The sequence

ϕ_n(λ):=

e^zλ− n k=0

z^k k!λ^k

2

, n∈N,

converges to zero onRand has the integrable majorante^2|zλ|, because ϕn(λ)=

k≥n+1

z^k k!λ^k

²≤e^2|zλ|.

Therefore, it follows from Lebesgue’s dominated convergence theorem (Theo-rem B.1) that

e^zTx−

n k=0

z^k k!T^kx

2

=

R

e^zλ−

n k=0

z^k k!λ^k

2

dE_x(λ)=

Rϕ_n(λ) dE_x(λ)→0 asn→ ∞. This proves formula (7.30).

Corollary 7.9 Suppose thatT is a self-adjoint operator onH. For anyx∈H, the following statements are equivalent:

(i) x∈D^a(T ).

(ii) There is ac >0 such thatx∈D(e^zT)for allz∈C,|Rez|< c.

(iii) There is ac >0 such thatx∈D(e^c|T|).

Proof Clearly, by formula (4.26),x∈D^∞(T )if (ii) or (iii) holds. The implication (i)→(ii) was proved by Proposition7.8. The equivalence of (ii) and (iii) follows at once from (4.26). Suppose that (iii) is satisfied and setu=e^c|T|x. Thenx=e^−c|T|u,

sox∈D^a(T )by Example7.5.

The following corollary contains some explanation for the terminology “analytic vector”; another reason is given by Example7.8below.

Corollary 7.10 LetT be a self-adjoint operator onH,y∈H, andx∈D^a(T ). Then f (z):= e^izTx, yis a holomorphic function on the strip{z: |Imz|< C_x⁻¹}, where C_xis given by (7.25).

Proof Lets∈R. By formula (7.30), applied withzreplaced by i(z−s), the series f (z)=

e^izTx, y

=

e^(iz−is)Tx, e^−isTy

=^∞

n=0

(i(z−s))ⁿ n!

Tⁿx, e^−isTy

converges if|z−s|< C_x⁻¹. Hence,f is holomorphic on the set{|Imz|< C_x⁻¹}.

Before we turn to the main results of this section, we prove three simple lemmas.

Lemma 7.11 Let(r_n)_n∈Nbe a positive sequence and setm_n=αr_n+β forn∈N, whereα >0 andβ >0. If

nr_n^−1/n= ∞, then

nm⁻_n^1/n= ∞.

Proof LetMbe the set{n:αrn≤β}. Forn∈M, we havemn≤2βand m⁻_n^1/n≥(2β)^−1/n≥(1+2β)⁻¹,

so

nm⁻n^1/n= ∞when the setMis infinite. IfMis finite, then there is ak∈N such thatmn≤2αrnfor alln≥k, and so

n≥k

m⁻_n^1/n≥

n≥k

(2α)^−1/nr_n^−1/n≥

n≥k

(1+2α)⁻¹r_n^−1/n= ∞.

Lemma 7.12 IfT if a symmetric operator, thenD^s(T )⊆D^s(T+zI )forz∈C.

Proof Letx∈D^s(T ). Since λx∈D^s(T ) forλ∈C, we can assume without loss of generality thatT x=0 andx =1. As noted in Examples7.3and7.4, we then haveT^lx^{1/ l}≤ Tⁿx^1/nforl≤n. Using this inequality, we derive forn∈N,

(T+zI )ⁿx=

n l=0

n l

z^n−lT^lx

≤

n l=0

n l

|z|^n−lT^lx

≤ n

l=0

n l

|z|^n−lTⁿx^l/n=Tⁿx^1/n+ |z|n

=Tⁿx1+ |z|Tⁿx^−1/nn

≤ Tⁿx

1+ |z|T x⁻¹n

. Sincex∈D^s(T ), we have

nTⁿx^−1/(2n)= ∞. By the preceding inequality this implies that

n(T+zI )ⁿx^−1/(2n)= ∞, that is,x∈D^s(T+zI ).

Lemma 7.13 IfT is a self-adjoint operator onH, thenD^b(T )is dense inH. Proof LetEbe the spectral measure ofT. Ifa >0 andx∈E([−a, a])H, then

Tⁿx²= _a

−a

λ²ⁿd

E(λ)x, x

≤a²ⁿx²≤

a(1+ x)2n

forn∈N, sox∈D^b(T ). For anyy∈H,y=lim_a→∞E([−a, a])y. Therefore, the set

a>0E([−a, a])His contained inD^b(T )and dense inH.

7.4 Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors 149 The following three theorems are the main results in this section. They all contain sufficient conditions for the self-adjointness of symmetric operators in terms of the density of special classes of vectors. In the proof of Theorem 7.15we shall use Theorem 10.17 below. Theorems7.14and7.15are due to A.E. Nussbaum.

Theorem 7.14 IfT is a symmetric operator on a Hilbert spaceHsuch that the linear span ofD^qa(T )is dense inH, thenT is essentially self-adjoint.

Proof First, we note thatT is densely defined, because the linear span ofD^qa(T )is dense and obviously contained inD(T ).

We prove that N(T^∗−iI )= {0}. Assume to the contrary that there exists a nonzero vectory∈N(T^∗−iI ). Letx∈D^qa(T ),x=0. By Proposition 3.17 the symmetric operator T has a self-adjoint extension Ain a possibly larger Hilbert spaceG⊇H. Define a functionf onRbyf (t )= e^{it A}x, y −e^tx, y.

SinceT ⊆A, we haveD^∞(T )⊆D^∞(A), so x∈D^∞(A). Therefore, by a re-peated application of Proposition 6.1(iii), it follows thatf∈C^∞(R)and

f⁽ⁿ⁾(t )=

e^{it A}(iA)ⁿx, y

−e^tx, y fort∈R, n∈N. (7.31) SinceT^∗y=iy, we haveTⁿx, y = Tⁿ⁻¹x,iy = · · · = x,iⁿy, and hence

f⁽ⁿ⁾(0)=

(iA)ⁿx, y

− x, y =iⁿ Tⁿx, y

− x, y =0 forn∈N0. (7.32) Leta >0 andJ =(−a, a). Putα= y,β=e^axyandm_n=αTⁿx+βfor n∈N0. Sincex∈D^qa(T ), we have

nTⁿx^−1/n= ∞, and hence

nm⁻_n^1/n=

∞by Lemma7.11. From (7.31) we obtain

f⁽ⁿ⁾(t )≤Aⁿxy +e^axy =αTⁿx+β=m_n fort∈J, n∈N0, sof satisfies Eq. (7.20) withK_f =1. Further,f⁽ⁿ⁾(0)=0 for alln∈N0by (7.32).

By the preceding we have shown that all assumptions of Corollary7.7are fulfilled.

From Corollary7.7we conclude thatf (t )≡0 on(−a, a)and hence onR, because a >0 was arbitrary. That is, we havee^{it A}x, y =e^tx, yfor allt∈R.

But the functione^{it A}x, yis bounded on R, whilee^t is unbounded on R. So the latter equality is only possible when x, y =0. Thus, we have proved that D^qa(T )⊥y. Since the linear span ofD^qa(T )is dense inH, we gety=0, which is the desired contradiction. Therefore,N(T^∗−iI )=0.

A similar reasoning shows that N(T^∗+iI )=0. Hence, T is essentially

self-adjoint by Proposition 3.8.

Theorem 7.15 Suppose thatT is a semibounded symmetric operator onH. If the linear span ofD^s(T )is dense inH, thenT is essentially self-adjoint.

Proof The proof follows a similar pattern as the proof of Theorem 7.14. By Lemma7.12we can assume without loss of generality thatT ≥I. Then 0∈π(T ) by Lemma 3.3. Therefore, by Proposition 3.9, it suffices to prove thatN(T^∗)= {0}.

Assume to the contrary thatN(T^∗)= {0}and take a unit vector y∈N(T^∗). Let

x∈D^s(T ),x=0. By Theorem 10.17 (which we take for granted for this proof!) there exists a self-adjoint operatorA≥I onHwhich extendsT.

By Proposition 6.7,u(t ):=(cosA^1/2t )xsolves the Cauchy problemu(0)=x, u(0)=0 for the wave equationu(t )= −Au(t )andu(t )= −(A^1/2sinA^1/2t )xfor t∈R. Define a functionf onRbyf (t ):= u(t ), y − x, y.

Since x ∈D^∞(T )⊆D^∞(A), from the functional calculus of the self-adjoint operatorAwe easily derive that

u^(2k)(t )=(−A)^ku(t ) and u^(2k−1)(t )=(−A)^(k−1)u(t ) fork∈N. Therefore,u∈C^∞(R,H), and hence f ∈C^∞(R). Since y =1 andA≥I, it follows from the preceding and the formulas foruanduthat

f^(2k−1)(t )=(−A)^k−1u(t ), y≤A^k−1A^1/2x≤A^kx=T^kx, f^(2k)(t )=(−A)^ku(t ), y≤A^kx=T^kx

fork∈Nand|f (t )| ≤2xonR. That is, settingm₀=2x,m_2k= T^kx, and m_2k−1= T^kx, the preceding shows that|f⁽ⁿ⁾(t )| ≤m_nfor alln∈N. Clearly,

n

m⁻n^1/n≥

n

Tⁿx^−1/(2n)= ∞ by (7.27). Sinceu(0)=0,u(0)=xandT^∗y=0, we obtain

f^(2k−1)(0)=

(−A)^ku(0), y

=0, f (0)= u(0), y

− x, y =0, f^(2k)(0)=

(−A)^ku(0), y

=

(−T )^kx, y

=

−(−T )^k−1x, T^∗y

=0 for k∈N, that is, we have shown that f⁽ⁿ⁾(0)=0 for alln∈N0. Therefore, all assumptions of Corollary7.7are satisfied, so we conclude thatf (t )≡0 onR.

Sincef (t )=0 and cosA^1/2t is a bounded self-adjoint operator fort ∈R, we have

x, y = u(t ), y

=

cosA^1/2t x, y

= x,

cosA^1/2t y

.

Therefore,x, y−(cosA^1/2t )y =0 for all vectorsx∈D^s(T ). By assumption the linear span ofD^s(T )is dense inH. Hence, it follows thaty−(cosA^1/2t )y=0. If Eis the spectral measure ofA, from the functional calculus we derive

0=y−

cosA^1/2t y²=

_∞

1

1−cos

λ^1/2t²d

E(λ)y, y .

This implies thatλ^1/2t∈2π·Zfor anyλin the support of the measureE(·)y, y. Becauset∈Rwas arbitrary, this only possible when this measure is zero. Since y =1, we arrive at a contradiction. This completes the proof.

The following result is the famous Nelson theorem.

Theorem 7.16 LetT be a symmetric operator onH. If the spaceD^a(T )of analytic vectors forT is dense inH, thenT is essentially self-adjoint. IfT is closed, thenT is self-adjoint if and only ifD^a(T )is dense inH.

7.4 Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors 151 Proof SinceD^a(T )is a linear space andD^b(T )⊆D^a(T )⊆D^qa(T )by (7.28), the assertions follow immediately from Theorem7.14and Lemma7.13.

Nelson’s and Nussbaum’s theorems are powerful tools for proving self-adjoint-ness by means of growth estimates ofTⁿx. We illustrate this by an example.

Example 7.6 (Jacobi operators) Let{en:n∈N0}be an orthonormal basis of a Hilbert spaceH. Given a complex sequence(a_n)_n∈N and a real sequence(b_n)_n∈N, we define the linear operatorT with domainD(T )=Lin{e_n:n∈N0}by

T e_n=a_n−1e_n−1+b_ne_n+a_ne_n+1, n∈N0, (7.33) wheree₋1:=0. It is easily seen thatT is symmetric.

Operators of the form (7.33) are called Jacobi operators. They play a crucial role in the study of the moment problem in Sect. 16.1, see, e.g., (16.7).

Statement Suppose that there exist two positive real constantsαandβ such that

|ak| ≤αn+β and|bk| ≤αn+β for allk≤n,k, n∈N. Then we haveD^a(T )= D(T ), and the operatorT is essentially self-adjoint.

Proof The main part of the proof is to show that each vectore_k is analytic forT. From formula (7.33) we concludeTⁿe_k is a sum of at most 3ⁿ summands of the formγmem, wherem≤k+nandγm is a product ofnfactorsaj,aj, orbj with j < k+n. Therefore, by the above assumption,Tⁿe_k ≤3ⁿ(α(n+k)+β)ⁿ.

Setc:=k+βα⁻¹. Clearly,c0:=sup_n(ⁿ⁺_n^c)ⁿ<∞. As noted in Example7.5, it follows from Stirling’s formula that there is a constantc1>0 such thatnⁿ≤c1eⁿn! forn∈N. From the preceding we obtain

Tⁿek≤3ⁿ

α(n+k)+β_n

=(3α)ⁿ(n+c)ⁿ≤(3α)ⁿc0nⁿ≤c0c1(3αe)ⁿn!

forn∈N0. This implies thatek∈D^a(T ). Hence,D(T )=D^a(T ), becauseD^a(T ) is a linear space. By Nelson’s Theorem7.16,T is essentially self-adjoint. ◦ Example 7.7 (Annihilation and creation operators) Recall from (7.11) that the an-nihilation operatorAand the creation operatorA⁺act on the standard orthonormal basis{en:n∈N0}ofH=l²(N0)by

Ae_n=√

n e_n−1 and A⁺e_n=√

n+1e_n+1, n∈N0, wheree₋₁:=0. SinceA^∗=A⁺, setting

P0= 1

√2 i

A−A⁺

and Q0= 1

√2

A+A⁺ , P0andQ0are symmetric operators onl²(N0), and we have

P₀e_n= 1

√2 i(√

n e_n−1−√

n+1e_n+1), Q₀e_n= 1

√2(√

n e_n−1+√

n+1e_n+1).

Since the assumptions of the statement in Example7.6are satisfied, the restrictions of P₀ and Q₀ to D0:=Lin{e_n:n∈N0} are essentially self-adjoint. It is easily checked that the operatorsP0andQ0satisfy the canonical commutation relation

P0Q0−Q0P0= −iID0. (7.34) The operatorP = −i_dx^d and the multiplication operatorQ=M_x by the variablex onL²(R)satisfy this relation as well. It is well known that the Hermite functions

h_n(x):=(−1)ⁿ 2ⁿn!√

π₋1/2

e^x2/2 dⁿ

dxⁿe^−x2, n∈N0,

form an orthonormal basis ofL²(R). Hence, there is a unitary operatorUofL²(R) ontol²(N0)defined byU h_n=e_n. Then it can be shown thatU P U^∗x=P₀x and

U QU^∗x=Q0xfor allx∈D0. ◦

Example 7.8 (Analytic vectors for the self-adjoint operatorT = −i_dx^d onL²(R)) LetF denote the Fourier transform, andM_xthe multiplication operator by the in-dependent variable x on L²(R). For this example, we will take for granted that T =F⁻¹M_xF. (This follows from formula (8.2) in Sect. 8.1).

Statement f ∈L²(R)is an analytic vector forT if and only iff is the restriction to Rof a holomorphic functionF on a strip{z: |Imz|< c}for somec >0 satisfying

|supy|<c

R

F (x+iy)²dx <∞. (7.35) Proof Since T and Mx are unitarily equivalent, f ∈D^a(T ) if and only if fˆ= F(f )is inD^a(M_x). Obviously,D^a(M_x)=D^a(|M_x|)=D^a(M_|x|). Thus, by Corol-lary7.9,f ∈D^a(T )if and only iffˆis inD(e^cM|x|), or equivalently,e^c|x|f (x)ˆ is in L²(R)for somec >0. By a classical theorem of Paley–Wiener [Kz, Theorem 7.1]

the latter is equivalent to the existence of a functionF as stated above.

From this statement it follows that the null vector is the only analytic vector for T contained inC₀^∞(R). Nevertheless, the restriction ofT toC₀^∞(R)is essentially self-adjoint, sinceC₀^∞(R)is a core forT as shown in Example 6.1. However, by Theorem7.16this cannot happen if the symmetric operatorT is closed.

Let(εn)_n∈Nbe a positive null sequence. Then the function g(z):=^∞

n=1

2⁻ⁿ 1 z−n−εni

is holomorphic on the setC\{n+ε_ni:n∈N}. In particular,gis analytic onR. One easily checks thatg∈D^∞(T ). Butgis not an analytic vector forT, because

condi-tion (7.35) is not fulfilled. ◦

Our next main theorem states that if two commuting symmetric operators have sufficiently many joint (!) quasi-analytic vectors, their closures are strongly com-muting self-adjoint operators. For this, we need a preliminary lemma.

7.4 Analytic Vectors, Quasi-analytic Vectors, Stieltjes Vectors 153 IfT is a linear operator with domainDandTD⊆D, we denote by{T}^cthe set of all densely defined linear operatorsS satisfyingD⊆D(S)∩D(S^∗),SD⊆D, andT Sx=ST xfor allx∈D.

Lemma 7.17 LetT be a symmetric operator with domainDsuch thatTD⊆D.

For any operatorS∈ {T}^c, we haveSD^qa(T )⊆D^qa(T ). In particular,D^qa(T )is invariant under the operatorT.

Proof Letx∈D. Using the properties ofT andS∈ {T}^c, we obtain TⁿSx²=

T²ⁿSx, Sx

=

ST²ⁿx, Sx

=

T²ⁿx, S^∗Sx

≤T²ⁿxS^∗Sx. Letx∈D^qa(T ). Without loss of generality we can assume thatx =1. Then we

have

nT²ⁿx^−1/(2n) = ∞ by Example7.4. The preceding inequality implies thatTⁿSx^1/n≤ T²ⁿx^1/(2n)(1+ S^∗Sx). Hence,

n

TⁿSx^−1/n≥

n

T²ⁿx^−1/(2n)

1+S^∗Sx⁻¹= ∞,

so thatSx∈D^qa(T ).

Theorem 7.18 LetA and B be symmetric operators acting on the same dense domainDof Hsuch thatAD⊆D,BD⊆D, andABx=BAxforx∈D. If the linear spanD_Qof

Q:= Sx:S∈ {A}^c∩ {B}^c, x∈D^qa(A)∩D^qa(B)

is dense inH(in particular, ifD^a(A)∩D^a(B)is dense inH), then the closuresA andBare strongly commuting self-adjoint operators onH.

Proof By the definition of the set{T}^cwe haveQ⊆D. From Lemma7.17it follows thatQis contained inD^qa(A)andD^qa(B). Therefore,AandBare self-adjoint by Theorem7.14.

LetA0andB₀denote the restrictions ofAandB toD_QandE_Q:=(A−iI )D_Q, respectively. SinceQ⊆D^qa(A₀)has a dense linear span, Theorem7.14applies also toA0, soA0is essentially self-adjoint. Hence,E_Q=(A0−iI )D_Q=Lin(A−iI )Q is dense inH. But(A−iI )Q⊆D^qa(B) by Lemma7.17, because Q⊆D^qa(B) and(A−iI )∈ {B}^c. Thus,(A−iI )Q⊆D^qa(B0), sinceB0=BE_Q. Hence, The-orem7.14applies toB0and the set(A−iI )Qas well and implies thatB0is essen-tially self-adjoint. SinceB₀⊆Band henceB₀⊆B, we obtainB₀=B.

Since A and B commute on D, we have B0(A−iI )x =(A−iI )Bx for x∈D_Q. Settingy=(A−iI )x, the latter yields(A−iI )⁻¹B0y=B(A−iI )⁻¹yfor y∈E_Q=D(B₀). From this relation we derive(A−iI )⁻¹B₀ y=B( A−iI )⁻¹y fory∈D(B0). SinceB0=Bas noted above, we get(A−iI )⁻¹B⊆B(A−iI )⁻¹. Hence, by Proposition 5.27, the self-adjoint operatorsAandBstrongly commute.

Completion of the proof of Proposition 5.20 Suppose thatvis a generating vector forA. Puty=e^−A2v. LetD0denote the linear span of vectorsAⁿy,n∈N0, and let H0be its closure inH. To complete the proof, it suffices to show thatH0=H.

First, we prove thatv∈H0. Setpn(t ):=_n

k=0t^2k

k! andϕn(t ):=pn(t )e^−t2. The sequence(ϕn(t ))n∈Nconverges monotonically to 1 onR. From the functional calcu-lus we therefore conclude thatϕ_n(A)v=p_n(A)y→vinH. Hence,v∈H0, since p_n(A)y∈D0.

Obviously,A0:=AD0 is a symmetric operator on the Hilbert spaceH0 and A0D0⊆D0. By Example 7.5, we havey ∈D^a(A). SinceAD^a(A)⊆D^a(A), the latter yields D0⊆D^a(A), so that D0=D^a(A0). Since A0 has a dense set D0

of analytic vectors, A0 is self-adjoint by Theorem 7.16. Hence, Proposition 1.17 applies, andA decomposes asA=A₀⊕A₁ with respect to the orthogonal sum H=H0⊕H^⊥₀. Clearly,E_A

0(M)=E_A(M)H0. Fromv∈H0we getE_A(M)v= E_A

0(M)v∈H0. Sincevis a generating vector forA, the span of vectorsE_A(M)v

is dense inH. Therefore,H=H0.

No documento Graduate Texts in Mathematics 265 (páginas 165-175)