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Existence of dynamics

ドキュメント内 HUSCAP Journals (ページ 32-42)

In order to prove that there exists dynamics of these quantum fields, we have only to see that Theorem 2.7 can be applied to our case by checking that Assumptions 2.1, 2.2 and 2.3 are valid. We see in what follows that this is indeed the case withA=INb,H0 =Hfr,H1= Hint,D=Fb,0.

Hereafter, we omit trivial tensor product like⊗I orI⊗when no confusion may occur and just write, for instance,Helinstead ofHelIand so forth.

Lemma 6.6. (I) Nbis self-adjoint and non-negative.

(II) Nband Hfrare strongly commuting.

(III) The interaction Hamiltonian Hintand its adjoint Hintare relatively bounded with respect to Nb1/2. (IV) For all L≥0andΨ∈R(ENb([0,L])), HintΨand Hint Ψbelong to R(ENb([0,L+1])).

Proof. The statements (I) and (II) are well known facts.

We prove (III). As in the proof of Lemma 6.4, it can be seen thatHintisNb1/2- bounded. We prove that Hint is alsoH1/2b - bounded. Take arbitraryΨ∈ Fb,0. By Lemma 6.5, we obtain

Hint Ψ∥=∥HintηΨ∥

Mint∥(Nb+1)12ηΨ∥

Mint∥(Nb+1)12Ψ∥. (6.78)

SinceFb,0is a core ofNb1/2, we have the assertion (III).

Finally, we prove (IV). Notice that the spectrum of the photon number operatorNbis a purely discrete set{0,1,2, . . .}, and that for allL≥0 and [L], the largest integer satisfying [L]≤L,

R(ENb([0,L]))=R(ENb([0,[L]]))=F[L]. (6.79) SupposeΨbelongs toR(ENb([0,L])). Then, it is clear thatΨbelongs toF[L]+1=R(ENb([0,L+1])), since the interaction Hamiltonian creates at most one photon. By Lemma 6.5, and the fact that ηandNb are strongly commuting and thusηpreserves photon number, it immediately follows thatHint Ψ ∈ F[L]+1 =

R(ENb([0,L+1])). □

From Lemma 6.6, we can apply the general theory constructed in the earlier sections to obtain:

Theorem 6.1. For eachΨ∈ Fb,0, the series U(t,t)Ψ:= Ψ +(−i)

t

t

1Hint1)Ψ +(−i)2

t

t

1

τ1

t

2Hint1)Hint2)Ψ +· · · (6.80) converges absolutely, where each of integrals is strong integral. Furthermore, U(t,t) has properties stated in Theorems 2.1, 2.2, 2.3 and 2.5, with H1replaced by Hint, H0by Hfrand D byFb,0.

As we have already seen in the general theory, onceU(t,t) is constructed, we immediately obtain a time evolution which is generated byW(t) :=e−itHfrU(t,0). Here, recall that we omit the overline to mean the operator closure. In the present application to physics, it should be made sure that this time evolution is physically acceptable, that is, the probability amplitude is conserved

⟨Ψ|Φ⟩=⟨W(t)Ψ|W(t)Φ⟩,

for suitable vectorsΨ,Φ. The following theorem is concerned with this aspect:

Theorem 6.2. The time evolution operator

W(t) :=eitHfrU(t,0) (6.81)

satisfies

W(t)W(t)1. (6.82)

In particular,

W(t)Ψ|W(t)Φ⟩=⟨Ψ|Φ⟩ (6.83)

for allΨ,Φ∈D(W(t)).

To prove Theorem 6.2, we prepare some facts.

Lemma 6.7. For eachΨ∈ Fb,0,

ηU(t,t)ηΨ =U(t,t)Ψ (6.84)

holds.

Proof. From Lemma 6.5, the operator identitiesηHint η=HintandηHfrη=Hfrhold, which imply

ηHint(t)η=Hint(t). (6.85)

We claim that (6.85) implies the identity

ηUn(t,t)ηΨ =Un(t,t)Ψ (6.86) for alln=0,1,2, . . . andΨ∈ Fb,0. In fact, by Lemma 3.8, one finds

ηUn(t,t)ηΨ =(−i)n

t

t

1. . .

τn1

t

nHint1). . .Hintn)Ψ

=(−i)n

t

t

n. . .

τ2

t

1Hintn). . .Hint1)Ψ

=Un(t,t)Ψ. (6.87)

Hence (6.86) is true for alln, and by summing up overn=0,1,2, . . ., we obtain (6.84). □ Proposition 6.2. For all t,t∈R, U(t,t)is invertible and the operator equality

U(t,t)=U(t,t)1 (6.88)

holds.

Proof. Fixt,t. Firstly, we prove thatU(t,t) is injective. Note that if for allΨ∈ Fb,0,⟨Ψ|Φ⟩= 0, then it follows thatΦ =0. From Lemma 6.7, we have the operator relation

ηU(t,t)⊂U(t,t)η, (6.89)

sinceFb,0 is a core ofU(t,t), and sinceU(t,t)is closed. Now, suppose thatΦsatisfiesU(t,t)Φ = 0.

LetΨ∈ Fb,0be arbitrary. It follows that

0=⟨U(t,t)Ψ|U(t,t)Φ⟩

=⟨U(t,t)Ψ, ηU(t,t)Φ⟩

=⟨U(t,t)Ψ,U(t,t)ηΦ⟩

=⟨

U(t,t)U(t,t)Ψ|Φ⟩

=⟨Φ|Ψ⟩, (6.90)

where we have used (6.89) in the third equality, and Theorem 2.3 (2.9) in the last equality. SinceΨ∈ Fb,0

is arbitrarily taken, we findΦ =0. This proves thatU(t,t) is injective.

Secondly, we prove

U(t,t)⊂U(t,t)1. (6.91)

Take arbitraryΨ∈ Fb,0. Then, we have from Theorem 2.3, (2.9) U(t,t)U(t,t)Ψ = Ψ, which impliesΨ∈D(U(t,t)−1) and

U(t,t)Ψ =U(t,t)1Ψ, Ψ∈ Fb,0. But sinceFb,0is a core ofU(t,t), the relation (6.91) follows.

Finally, we prove

U(t,t)⊃U(t,t)1. (6.92)

LetΨ∈D(U(t,t)−1)=R(U(t,t)). Then, there is someΦ∈D(U(t,t)) withΨ = U(t,t)Φ. SinceFb,0is a core ofU(t,t), we can choose a sequencen}n ⊂ Fb,0satisfying

Φn→Φ, U(t,t)ΦnU(t,t)Φ = Ψ, (6.93) asntends to infinity. Therefore, we have

U(t,t)U(t,t)Φn= Φn →Φ, n→ ∞. SinceU(t,t) is closed, we conclude thatΨ =U(t,t)Φ∈D(U(t,t)) and

U(t,t)Ψ =U(t,t)U(t,t)Φ = Φ =U(t,t)1Ψ.

This proves (6.92). □

Proposition 6.3. For all t∈R, the operator W(t)is injective and

W(t)1 =W(t) (6.94)

holds as an operator equality.

Proof. Sincee−itHfris unitary andU(t,0) is injective by Lemma 6.2, one findsW(t) is injective.

We prove (6.94). Fixt∈R. For allΨ∈D(W(t))=D(U(t,0)), we have by Theorem 2.3 (2.9) W(t)W(t)Ψ =eitHfrU(t,0)eitHfrU(t,0)Ψ

=U(0,t)U(t,0)Ψ

= Ψ. (6.95)

This meansΨ∈D(W(t)1) and

W(t)Ψ =W(t)−1Ψ.

Since Fb,0 is a core ofW(t), we conclude W(t)W(t)−1. The proof of the inverse inclusion is very

similar to the proof of (6.92), and we omit it. □

Proof of Theorem 6.2. LetΨ∈ Fb,0. Then, we have

ηW(t)ηΨ =W(t)Ψ. (6.96)

Thus, by taking closure, it follows that

ηW(t)η⊃W(t). (6.97)

But, we knowW(t)=W(t)−1from Proposition 6.3. This proves (6.82).

Moreover, sinceW(t)Φ∈D(W(t)1)⊂D(ηW(t)η), we obtain

W(t)Ψ|W(t)Φ⟩=⟨W(t)Ψ, ηW(t)Φ⟩

=⟨

Ψ,W(t)ηW(t)Φ⟩

=⟨

Ψ|W(t)−1W(t)Φ⟩

=⟨Ψ|Φ⟩. (6.98)

This completes the proof. □

We next consider the Heisenberg equations of motion for quantum fieldsAµandψl. Lemma 6.8. (I) Aµ(0,f)and Aµ(0, f)are Nb1/2- bounded and closed.

(II) ψl(0,g)andψl(0,g)are Nb1/2- bounded and closed.

(III) For all L ≥ 0, Ψ ∈ R(ENb([0,L])) implies that Aµ(0,f, Aµ(0,f)Ψ, ψl(0,g)Ψ, andψl(0,g)Ψ belong to R(ENb([0,L+1])).

Proof. The assertion (I) follows from Lemma 6.2. The closedness is obvious.

The assertion (II) is an immediate consequence of the fact that Dirac fields are bounded operators.

The claim (III) immediately follows from the fact that photon field operators Aµ(0, f) andAµ(0,f) create at most one photon and Dirac fieldsψl(0,g) andψl(0,g) create no photon.

Put fort∈Randµ=0,1,2,3,l=1,2,3,4,

A0,µ(t,f)Ψ:=eitHfrAµ(0,f)e−itHfrΨ, f ∈ S(R3), (6.99) ψ0,l(t,g)Ψ:=eitHfrψl(0,g)e−itHfrΨ, g∈ S(R3). (6.100) Lemma 6.9. (I) For eachΨ∈D(Nb1/2)and each f,g ∈ S(R3), A0,µ(t,fandψ0,l(t,g)Ψare strongly

continuously differentiable in t∈R.

(II) The strong derivatives of A0,µ(t,fandψ0,l(t,g)Ψ, which are denoted by A0,µ(t,fandψ0,l(t,g)Ψ, are closable for all t∈R.

(III) The operators A0,µ(t,fandψ0,l(t,g)Ψare A1/2-bounded uniformly in t ∈R. That is, there exist constants c0,c1,d0,d1 ≥0such that for all t∈Randξ∈D(A1/2),

A0,µ(t,f)Ψ∥ ≤c0Nb1/2Ψ∥+c1∥Ψ∥, (6.101)

∥ψ0,l(t,g)Ψ∥ ≤d0Nb1/2Ψ∥+d1∥Ψ∥. (6.102) Proof. (I) It follows from the general theorem [1, Lemma 4-49] thatt7→aµ(eitωf)Ψandt7→aµ(eitωf)Ψ

are strongly differentiable and the strong derivatives become d

dtaµ(eitωf)Ψ =aµ(iωeitωf)Ψ, (6.103) d

dtaµ(eitωf)Ψ =aµ(iωeitωf)Ψ, (6.104)

forΨ ∈ D(Nb1/2) and f ∈ S(R3). From Lemma 6.2 (i) and (6.103), (6.104), the continuity of the mappings

t7→ d

dtaµ(eitωf)Ψ, t7→ d

dtaµ(eitωf)Ψ

are obvious. For the fermion creation and annihilation operators, the operator valued functions t7→B(eitEMG), t7→B(eitEMG), GD(EM) (6.105) are continuously differentiable in the operator norm and the derivatives are

d

dtB(eitEMG)= B(iEMeitEMG), (6.106) d

dtB(eitEMG)= B(iEMeitEMG). (6.107) In fact, one finds from the well known estimates for the fermion creation and annihilation operators (6.35) that

B(ei(t+h)EMG)B(eitEMG)

hB(iEMeitEMG)

ei(t+h)EMGeitEMG

hiEMeitEMG H

el

→0, (6.108)

ash→0, and the mappings

t7→B(iEMeitEMG), t7→B(iEMeitEMG)

are continuous in the operator norm. On the other hand, from these facts, the assertion (I) immedi-ately follows.

(II) From (6.103) and (6.104), we have A0,µ(t,f)Ψ =aµ

iωeitωbf

√2ω

Ψ +a

iωeitωbf

√2ω

Ψ, Ψ∈D(Nb1/2). (6.109) It is clear from this expression that the adjoint operator of A0,µ(t,f) is defined on dense subspace Fb,0and therefore it is closable. From (6.106) and (6.107), the closability ofψ0,l(t,g) is obvious.

(III) This statement clearly follows from Lemma 6.2 (i) and (6.35).

□ Finally, we have arrived at the existence of strong solutions for the Heisenberg equations of motion for quantum fields, by combining all the results obtained so far. We can define from Lemma 6.6, Lemma 6.8 and Theorem 2.7 the following filed operators:

D(Aµ(t, f)) :=Fb,0, Aµ(t,f)Ψ:=W(t)Aµ(0,f)W(t)Ψ, Ψ∈ Fb,0, f ∈ S(R3), (6.110) D(ψl(t,g)) :=Fb,0, ψl(t,g)Ψ:=W(t)ψl(0,g)W(t)Ψ, Ψ∈ Fb,0, g∈ S(R3), (6.111) and one concludes from Lemma 6.9:

Theorem 6.3. For all f,g∈ S(R3),µ=0,1,2,3and l=1,2,3,4, the operator valued functionsR∋t7→

Aµ(t,f)andR∋t7→ψl(t,g)are strong solutions of the differential equations d

dtAµ(t,f)Ψ =[iHQED,Aµ(t,f)]Ψ, Ψ∈D(Hfr)∩ Fb,0, (6.112) d

dtψl(t,g)Ψ =[iHQED, ψl(t,g)]Ψ, Ψ∈D(Hfr)∩ Fb,0. (6.113) We remark that from Theorem 6.2 the above quantum fields are also written as

Aµ(t,f)Ψ =W(t)Aµ(0,f)W(t)Ψ, Ψ∈ Fb,0, (6.114) ψl(t,g)Ψ =W(t)ψl(0,g)W(t)Ψ, Ψ∈ Fb,0. (6.115)

A A little about the function t 7→ U(t, t

In Section 3, we have constructed the solution of the differential equation d

dtϕ(t)=−iH1(t)ϕ(t), ϕ(0)=ξ, (A.1)

via the evolution operatorU(t,t),

ϕ(t)=U(t,t)ξ.

In this appendix, we investigate the properties ofϕ(t) = U(t,t)ξ (ξ ∈ D) as a function oft ∈R. Let us denote byIa closed interval inR, fixtI and set

K:=(A+1)1/2,

for short. Throughout the appendix, we assume only Assumption 2.1, which is sufficient to ensure that there is a solution of (A.1).

Lemma A.1. Suppose that Assumption 2.1 holds. Let u : ID be a strongly continuous mapping satisfying

u(t)VL, tI (A.2)

for some constant L≥0. Then, the mapping

It7→H1(t)u(t)∈ H (A.3)

is strongly continuous. Moreover, for all tI, the strong Riemann integral

t

t

ds H1(s)u(s). (A.4)

belongs to VL+b. Proof. Lets,tI.

||H1(t)u(t)−H1(s)u(s)|| ≤ ||(H1(t)−H1(s))u(t)||+||H1(s)(u(t)−u(s))||

≤ ||(H1(t)−H1(s))K−1·Ku(t)||+||H1(s)K−1|| ||K(u(t)u(s))||. (A.5) Firstly, note that the mapping

t7→eitH0H1K−1e−itH0 (A.6)

is strongly continuous, sinceH1K−1 is a bounded operator. Thus, the first term of (A.5) tends to zero as st.

Secondly, note also that there is someL≥0 such that

u(t)u(s)VL, s,tI Therefore, the second term of (A.5) also vanishes asstends tot.

From Assumption 2.1,H1(t)u(t)∈VL+b(t∈I). SinceVL+bis closed, one finds

t

t

ds H1(s)u(s)VL+b.

From Lemma A.1, it follows that under Assumption 2.1 we may define a linear transformationJ in the linear space

C :={u:ID|uis strongly continuous and there is anLsuch thatu(t)VLfor alltI} (A.7) by

(Ju)(t) :=

t

t

ds H1(s)u(s), tI, (A.8)

wheretI is fixed ([t,t]Iift <t; [t,t]⊂I ift<t).

Lemma A.2. Under assumption 2.1, for all n=1,2, . . . and u∈ C, the estimate

||(Jnu)(t)|| ≤ |tt|n

n! Cn(L+(n−1)b+1)1/2(L+(n−2)b+1)1/2. . .(L+1)1/2sup

s∈I ||u(s)|| (A.9) holds, where L≥0is a constant depending on u satisfying

u(t)VL, tI.

Moreover,

(Jnu)(t)VL+nb, tI. (A.10)

Proof. We prove by induction. Whenn=1, the assertion is clear.

Suppose the lemma is true for somen. Then, by noting that (Jnu)(t)VL+nb, tI, we have

||(J◦Jnu)(t)|| ≤

t

t

ds||H1(s)(Jnu)(s)||

t

t

ds||H1K1|| ||K(Jnu)(s)||

C(L+nb+1)1/2

t

t

ds||(Jnu)(s)||

C(L+nb+1)1/2

t

t

ds|st|n

n! Cn(L+(n−1)b+1)1/2. . .(L+1)1/2sup

s∈I ||u(s)||

= |tt|n+1

(n+1)! Cn+1(L+nb+1)1/2(L+(n−1)b+1)1/2. . .(L+1)1/2sup

sI ||u(s)||, (A.11) and, by Lemma A.1,

(Jn+1u)(t)VL+(n+1)b, tI. (A.12)

Hence, by induction, the assertion follows. □

Denote the approximated solution byϕn: ϕn(t)=

n

k=0

Sn(t,t)ξ=

n

k=0

((−iJ)nξ)(t), ξ∈D,

where in the last expression,ξis identified with a constant function which belongs toC. As mentioned above, ϕn belongs to C for alln ∈ N. But, the limit function ϕdoes not have to belong toC. We identify the set of functions to whichϕbelongs.

Definition A.1. Let T be a self-adjoint operator. We say that a function u : ID(T) is T -uniformly integrable if the following two conditions are satisfied:

(1) t7→T u(t)is strongly continuous.

(2) limΛ→∞suptI||T ET([Λ,∞))u(t)||=0.

Forα≥0, we denote

Cα:={u:ID(Kα)|uisKα- uniformly integrable}. (A.13) The set Cα becomes a vector space by naturally defined addition and scalar multiplication. For each u∈ Cα, we introduce

||u||α,:=sup

tI ||Kαu(t)||. (A.14)

This is finite, sinceKαu(·) is strongly continuous. It is clear that|| · ||α,∞is a norm ofCα. Further, we have Proposition A.1. The normed space(Cα,|| · ||α,)is a Banach space.

Proof. We show only the completeness. Take a Cauchy sequence in (Cα,|| · ||α,∞),{un}n. Then, for all ϵ >0, there is anN ∈Nsuch thatnmNimplies that

||Kα(un(t)−um(t))||< ϵ, tI. (A.15)

Since{Kαun(t)}nis Cauchy inH, there exists somev(t)∈ Hsuch that

nlim→∞Kαun(t)=v(t).

On the other hand, sinceKα≥1, one finds forn,mN

||un(t)−um(t)|| ≤ ||K−α|| ||Kα(un(t)−um(t))||

≤ ||Kα(un(t)−um(t))||< ϵ. (A.16)

Therefore,{un(t)}nis also Cauchy and there is someu(t)∈ Hsuch that

nlim→∞un(t)=u(t).

Hence, we haveu(t)D(Kα) andKαu(t) = v(t) (tI). Takem → ∞in (A.15). Then, one has for all nN,

||Kα(un(t)−u(t))|| ≤ϵ, tI. (A.17) This means

||unu||α, ≤ϵ for allnN.

We showu∈ Cα, that is,uisKα- uniformly integrable. Note thatKαu(·) — a uniform limit ofKαun(·)

— is strongly continuous. Take an arbitraryϵ > 0. We can choose N ∈ Nin such a way that nN implies that

sup

t∈I ||Kα(un(t)−u(t))||< ϵ.

Then for an arbitraryΛ≥0, one has

||KαEKα([Λ,∞))u(t)|| ≤ ||KαEKα([Λ,∞))(un(t)−u(t))||+||KαEKα([Λ,∞))un(t)||

< ϵ+||KαEKα([Λ,∞))un(t)||. (A.18)

Taking the limit in whichΛ→ ∞, we have sup

t∈IT

||KαEKα([Λ,∞))u(t)||< ϵ+sup

t∈IT

||KαEKα([Λ,∞))un(t)|| →ϵ, Λ→ ∞. (A.19) Sinceϵ >0 is arbitrary, one has

Λ→∞lim sup

tIT

||KαEKα([Λ,∞))u(t)||=0,

namely,u∈ Cα. □

The next lemma shows the relation betweenCαandC.

Lemma A.3. For allα≥0,C ⊂ Cαas normed spaces. Moreover,Cis dense inCαwith respect to the

|| · ||α,∞norm topology.

Proof. First, we prove thatC ⊂ Cα. Letu ∈ C. To show thatuisKα-uniformly integrable, note that there is anL≥0 such that

u(t)VL, tI, by definition ofC. One has

||Kα(u(t)−u(s))|| ≤(L+1)α/2||u(t)u(s)||, which shows thatKαu(·) is strongly continuous. LetΛ>(L+1)α/2. Then,

||KαEKα([Λ,∞))u(t)||=||KαEA([Λ2/α−1,∞))u(t)||=0. (A.20) Hence,uisKα-uniformly integrable, that is,u∈ Cα.

Next, we prove thatCis dense inCα. Take anyu∈ Cα. Define{un}n ⊂ Cby un(t) :=EA([0,n))u(t), n=1,2, . . . .

Then it follows that

||unu||α, =sup

t∈IT

||Kα(EA([0,n))−1)u(t)||

=sup

tIT

||KαEA([n,∞))u(t)||

=sup

t∈IT

||KαEKα([(1+n)α/2,∞))u(t)|| →0, n→ ∞. (A.21)

This proves thatCis dense. □

This lemma tells us thatCαis the completion ofCwith respect to the norm|| · ||α,. Lemma A.4. Let0≤α≤β. ThenCβ ⊂ Cαand the inclusion mapping

ι:Cβ → Cα

is continuous.

Proof. Letu∈ Cβ. Then,Kα−βis bounded and thus the mapping t7→Kαu(t)= Kα−βKβu(t) is strongly continuous. Further, we have

sup

t∈IT

||KαEKα([Λ,∞))u(t)|| ≤ ||Kα−β||sup

t∈IT

||KβEKα([Λ,∞))u(t)||

=||Kαβ||sup

tIT

||KβEKβ([Λβ/α,∞))u(t)|| →0 Λ→ ∞. (A.22) These imply thatu∈ Cα. HenceCβCα.

The second assertion immediately follows from the fact that

||Kαu(t)|| ≤ ||Kαβ|| ||Kβu(t)||, u∈ Cβ.

□ Proposition A.2. The approximated solutionn}n⊂ Cis a Cauchy sequence in|| · ||α,for allα≥0.

Proof. Fixα≥0. ChooseLforξ∈Din such a way that ξ ∈VL holds. Since

ϕn=

n

k=0

(−iJ)kξ, and from Lemma A.2, we obtainϕn(t)∈VL+nb. Therefore, we have

||ϕn+1−ϕn||α,∞=sup

tIT

||Kαn+1(t)−ϕn(t))||

≤sup

t∈I

(L+nb+1)α/2||ϕn+1(t)−ϕn(t)||

≤ |I|n+1

(n+1)!Cn(L+nb+1)α/2(L+nb+1)1/2. . .(L+1)1/2||ξ||, (A.23) where|I|denotes the Lebesgue measure of the intervalI. By d’Alembert’s ratio test, one obtains

n=0

||ϕn+1−ϕn||α,<∞. Hence, for anyϵ >0, there is anN∈N, such thatnmNimplies

||ϕn−ϕm||α,

n−1

k=m

||ϕk+1−ϕk||α, < ϵ.

□ Theorem A.1. For anyξ ∈ D, the solutionϕ, which is the limit ofϕn = Sn(·,t)ξ, belongs to∩

α0Cα. That is, U(·,tis Kα-uniformly integrable for allα≥0.

Proof. Fixα≥0. By Proposition A.1 and Proposition A.2, there is an elementuαCαsatisfying

||ϕn−ϕα||α,∞→0.

Takeβ >0. Without loss of generality, we may assumeα≤β. Then, we obtain

||ϕα−ϕβ||α,≤ ||ϕα−ϕn||α,+||ϕn−ϕβ||α,

≤ ||ϕα−ϕn||α,∞+||Kαβ|| ||ϕn−ϕβ||β,∞ →0, n→ ∞, (A.24) which shows thatϕα = ϕβ. Ifα = 0, then||ϕn−ϕ||0, → 0. Therefore, we conclude that for allα≥ 0,

ϕ=ϕα∈ Cα. □

ドキュメント内 HUSCAP Journals (ページ 32-42)

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