Schrodinger equation and heat equation with translation invariance on the infinite-dimensional vector space ; Self-adjoint Laplace operator with translation invariance on infinite-dimensional space

Heat equation and Schr¨odinger equation with translation invariance on the infinite-dimensional vector space R∞ Hiroki Yagisita (Kyoto Sangyo University) The standard Laplacian −4Rnin L2(Rn) is self-adjoint and translation invariant on the finite-dimensional linear space Rn. In this paper, we define a translation invariant operator −4R∞on R∞as a non-negative self-adjoint operator in some non-separable Hilbert space L2(R∞). The set L2(R∞) is a translation invariant subset of the set CM(R∞) of all complex measures on the product measurable space R∞. Furthermore, we show that for any f∈L2(Rn) and any u∈L2(R∞), the separations of variables e△R∞t(f⊗u) = (e△Rntf)⊗(e△R∞tu) (t∈[0,+∞)) and e√−1△R∞t(f⊗u) = (e√−1△Rntf)⊗ (e√−1△R∞tu) (t∈(−∞,+∞)) hold. This clearly shows that −4R∞is an analog of −4Rn. The starting point for the discussion in this paper is to naturally introduce a translation invariant structure of Hilbert space into CM(R∞). When |u|denotes the total variation of u∈CM(R∞) and u6= 0 holds, 1 ∥u∥2 CM(R∞)|u|is a probability measure on R∞.L2(R∞) is a closed linear subspace of CM(R∞). The inner product of L2(R∞) is defined as that of CM(R∞). [key word] Born and Heisenberg probabilistic interpretation, Schr¨odinger representation of CCR, Fisher information metric, information geometry. [data availability statement] It has no associated data. [statement and declaration] The author has no competing interests to declare that are relevant to the content of this article. No funds, grants, or other support was received. [affiliation] Department of Mathematics, Faculty of Science, Kyoto Sangyo University, Motoyama, Kamigamo, Kita-ku, Kyoto, 603-8555, Japan. [orcid] 0000-0002-9544-4434 1 1 Introduction In this paper, we define a translation invariant operator −4R∞on the infinite-dimensional vector space R∞as a non-negative self-adjoint operator and we examine evolution equations ut=4R∞uand ut=√−14R∞uto show that −4R∞is an analog of the standard Laplacian −4Rnin L2(Rn). Our construction of −4R∞is achieved by combining well basic results widely used in finite-dimensional analysis (e.g., [2, 3, 5, 7, 8, 12]), although it is an infinite-dimensional object. On the other hand, our method cannot be immediately applied to constructing an analog on a smooth domain of R∞, so we look forward to further research in the future. There is no known previous research that has given an analog on an infinite-dimensional linear space that is self-adjoint and translation invariant. In this sense, there does not seem to be any clear related literature. On the other hand, the amount of previous research on things that seem in some sense to be analogs on infinitedimensional linear spaces that are self-adjoint or translation invariant seems to be vast. We find it difficult to provide an unbiased citation. Let the set of all measurable sets of Rbe the topological σ-algebra of R(i.e., the smallest σ-algebra on Rcontaining all open sets of R). Let R∞ denote the countable product measurable space ∏n∈NR. In Sections 2, 3 and 4, we define some Hilbert space L2(R∞) which is a subset of the set of all complex measures on R∞and a non-negative self-adjoint operator −4R∞in L2(R∞). In Section 5, we set the stage for the following sections. In Section 6, we examine evolution equations ut=4R∞uand ut=√−14R∞uto show that −4R∞is an analog of the standard Laplacian −4Rnin L2(Rn). In Section 7, we show that L2(R∞) is not separable. In Section 8, we show that L2(R∞) and 4R∞are translation invariant on R∞. More specifically, it is as follows. There does not exist a σ-finite measure µon R∞with translation invariance that satisfies µ([0,1)∞) = 1 (i.e., ideal Lebesgue measure on R∞). Therefore, the space of square-integrable functions on R∞seems indefinable. However, on the other hand, according to Born and Heisenberg probabilistic interpretation of quantum mechanical wavefunction, when Ω is a measurable space, for a measure µon Ω and a function f∈L2(µ) that satisfy kfkL2(µ)6= 0, the probabilistic interpretation of the state vector ffor posi2 tion measurement is the probability measure 1 kfk2 L2(µ)|f|2dµ on Ω. This probability measure is the normalization of the total variation of the complex measure f|f|dµ on Ω. So, it raises the following question. For a measurable space Ω, let CM(Ω) denote the set of all complex measures on Ω, into CM(Ω), is it possible to introduce the structure of Hilbert space such that for any measure µon Ω and any f1, f2∈L2(µ), hf1|f1|dµ, f2|f2|dµiCM(Ω) =∫Ω f1f2dµ holds ? As a matter of fact, given restricting µto be σ-finite, this can be easily done consistently using Radon-Nikodym theorem, as we will do in Section 2. That is, for any measurable space Ω, the structure of Hilbert space such that for any σ-finite measure µon Ω and any f1, f2∈L2(µ), hf1|f1|dµ, f2|f2|dµiCM(Ω) =∫Ωf1f2dµ holds is introduced into CM(Ω). In particular, when Ω is a measurable space, for any σ-finite measure µon Ω, the isometry f7→ f|f|dµ from L2(µ) into CM(Ω) is determined. So, we can think that the well-known Hilbert space L2(Rn) is a closed linear subspace of Hilbert space CM(Rn). Since the sum in CM(Ω) that fits the inner product is defined in Definition 4 and it is simple, readers who have doubts here should take a look at Definition 4. For a vector a∈R∞, let τadenote the translation on CM(R∞) by the vector a. Let e1:= (1,0,0,···)∈R∞, e2:= (0,1,0,···)∈R∞,···. Let L2 k(R∞) denote the set of all u∈CM(R∞) such that limh↓+0 kτheku−ukCM(R∞)= 0 holds. Let H1 k(R∞) denote the set of all u∈CM(R∞) such that there uniquely exists v∈CM(R∞) such that limh↓+0 kτheku−u h−vkCM(R∞)= 0 holds. For u∈H1 k(R∞), let ∂u ∂xkdenote −v. In Section 3, we show that τais a unitary operator in CM(R∞). So, by Stone theorem, √−1∂ ∂xkis a self-adjoint operator in L2 k(R∞). 3 In Section 4, from the very simple quadratic form (Hermitian form) ∑ k∈Nh√−1∂u1 ∂xk ,√−1∂u2 ∂xkiCM(R∞), we define a closed linear subspace L2(R∞) of CM(R∞) and a non-negative self-adjoint operator −4R∞in L2(R∞). First, we introduce Sobolev type space H1(R∞) in CM(R∞) by the inner product hu1, u2iH1(R∞):= hu1, u2iCM(R∞)+∑ k∈Nh√−1∂u1 ∂xk ,√−1∂u2 ∂xkiCM(R∞). The closure of H1(R∞) in CM(R∞) is denoted by L2(R∞). We show that H1(R∞) is Hilbert space. Therefore, by |hf, φiCM(R∞)| ≤ kfkCM(R∞)kφkH1(R∞) and Riesz theorem, for any f∈L2(R∞), there uniquely exists u∈H1(R∞) such that for any φ∈H1(R∞), hf, φiCM(R∞)=hu, φiH1(R∞) holds. This map f7→ ufrom L2(R∞) to H1(R∞) is denoted by (1−4R∞)−1. We show that (1 − 4R∞)−1is self-ajoint in L2(R∞) and injective. So, the operator 4R∞in L2(R∞) defined by 4R∞u:= u−((1 −4R∞)−1)−1u is self-ajoint. We show that −4R∞is non-negative. In Section 5, when Ω1and Ω2are measurable spaces, u1is a complex measure on Ω1and u2is a complex measure on Ω2, it is shown that the product u1·u2can be naturally defined as a complex measure on Ω1×Ω2 and ku1·u2kCM(Ω1×Ω2)=ku1kCM(Ω1)ku2kCM(Ω2)holds. Since the product is defined in Definition 26 and it is simple, readers who have doubts here should take a look at Definition 26. For f∈L2(Rn) and u∈CM(R∞), let f⊗u∈CM(R∞) be defined as (f⊗u)(x1, x2,···) := (f(x1, x2,··· , xn)|f(x1, x2,··· , xn)|dx1dx2···dxn)·u(xn+1, xn+2,···). In Section 6, we show that for any f0∈L2(Rn) and any u0∈L2(R∞), the formulas for separation of variables e△R∞t(f0⊗u0) = (e△Rntf0)⊗(e△R∞tu0) (t∈[0,+∞)), 4 e√−1△R∞t(f0⊗u0) = (e√−1△Rntf0)⊗(e√−1△R∞tu0) (t∈(−∞,+∞)) hold. This clearly shows that 4R∞is an analog of the standard Laplacian 4Rn. In order to prove this, we had to resort to technical ingenuity. The proof path for the separations of variables is not very clear even to the author himself, so we hope that readers will improve it. In Section 7, we show that L2(R∞) has an uncountable orthogonal system (i.e., L2(R∞) is not separable). In Section 8, just to make sure, we confirm that L2(R∞) and 4R∞are translation invariant on R∞. Section 6, Section 7 and Section 8 can each be read independently. In addition, as appropriate, we once again state the proof of some fundamental facts that should hold, because there does not seem to be much accessible well-known literature that explicitly states the proof (and the author is the kind of person who worries about such things). Readers who are not too concerned can feel free to ignore such statements. On the one hand, in a familiar way, it uses well-known basic theorems such as Radon-Nikodym theorem without any specific reference to them. 5 2 Square root of density Throughout this paper, we may use the following three simple facts (Lemma 1, Lemma 2 and Remark) without specific mention. In a familiar way, we also use Radon-Nikodym theorem without any specific reference. Lemma 1: Let ηbe the map from Cto Cdefined for z∈C, as η(z) := z|z|. Let ζbe the map from Cto Cdefined for w∈C, as ζ(w) := w|w|−1 2 when w6= 0 holds and ζ(w) := 0 when w= 0 holds. Then, ηand ζare continuous. ζ◦ηand η◦ζare the identity map. □ Lemma 2: Let Ωbe a measurable space. Let {µn}∞ n=1 be a sequence of σ-finite measures on Ω. Let {un}∞ n=1 be a sequence of complex measures on Ω. Then, there exists a finite measure νon Ωsuch that for any n∈N,µn and unare absolutely continuous with respect to ν. Proof: There exists {En,m}∞ n,m=1 such that for any n, Ω = t∞ m=1En,m holds and for any m,µn(En,m)<+∞holds. As |un|is the total variation of un, let ν(E) := ∑ n(1 2n(|un|(E) 1 + |un|(Ω) +∑ m(1 2m µn(E∩En,m) 1 + µn(En,m)))). ■ Remark: Let Ωbe a measurable space. Let µbe a measure on Ω. Let ρbe a[0,+∞)-valued measurable function on Ω. Let fbe a complex measurable function on Ω. Then, the followings hold. (1) Suppose that fis non-negative. Then, f(ρdµ) = (fρ)dµ holds. (2) Suppose that f∈L1(ρdµ)or fρ ∈L1(µ)holds. Then, f∈L1(ρdµ), fρ ∈L1(µ)and f(ρdµ) = (fρ)dµ hold. Proof: Although it is a natural result, we include the proof just in case. (1) There exists a monotonic non-decreasing sequence {gn}nof non-negative simple measurable functions such that for any x, limn|gn(x)−f(x)|= 0 holds. Then, for any measurable set E, (f(ρdµ))(E) = ∫Ef(ρdµ) = limn(∫Egn(ρdµ)) = limn(∫E(gnρ)dµ) = ∫E(fρ)dµ = ((fρ)dµ)(E) holds. (2) From (1), f∈L1(ρdµ) and fρ ∈L1(µ) hold. So, there exist h1, h2, h3, h4∈ L1(ρdµ) such that h1, h2, h3and h4are non-negative and f= (h1−h2) + √−1(h3−h4) holds. Then, from (1), for any measurable set E, (f(ρdµ))(E) = ∫Ef(ρdµ) = (∫Eh1(ρdµ)−∫Eh2(ρdµ)) + √−1(∫Eh3(ρdµ)−∫Eh4(ρdµ)) = (∫E(h1ρ)dµ − 6 ∫E(h2ρ)dµ) + √−1(∫E(h3ρ)dµ −∫E(h4ρ)dµ) = ∫E(fρ)dµ = ((fρ)dµ)(E) holds. ■ Definition 3 (CM(Ω)): Let Ωbe a measurable space. Then, let CM(Ω) denote the set of all complex measures on Ω.□ Definition 4 (sum): Let Ωbe a measurable space. Let u1, u2∈CM(Ω). Then, there uniquely exists v∈CM(Ω) such that the following holds. There exist a σ-finite measure µon Ωand f1, f2∈L2(µ)such that u1=f1|f1|dµ, u2=f2|f2|dµ and v= (f1+f2)|f1+f2|dµ hold. Let u1+u2∈CM(Ω) be defined as u1+u2:= v. Proof: From Lemma 1 and Lemma 2, existence is easy. We show uniqueness. Suppose that (v1, µ1, f1,1, f2,1) and (v2, µ2, f1,2, f2,2) each satisfy the condition of the definition. Then, from Lemma 2, there exist a finite measure νand [0,+∞)-valued measurable functions ρ1, ρ2such that µ1=ρ1dν and µ2=ρ2dν hold. So, u1=f1,1|f1,1|dµ1= (f1,1√ρ1)|f1,1√ρ1|dν and u1=f1,2|f1,2|dµ2= (f1,2√ρ2)|f1,2√ρ2|dν hold. From Lemma 1, νa.e., f1,1√ρ1=f1,2√ρ2holds. Similarly, ν-a.e., f2,1√ρ1=f2,2√ρ2holds. v1= (f1,1+f2,1)|f1,1+f2,1|dµ1= (f1,1√ρ1+f2,1√ρ1)|f1,1√ρ1+f2,1√ρ1|dν = (f1,2√ρ2+f2,2√ρ2)|f1,2√ρ2+f2,2√ρ2|dν = (f1,2+f2,2)|f1,2+f2,2|dµ2=v2 holds. ■ Definition 5 (scalar multiple): Let Ωbe a measurable space. Let c∈Cand u∈CM(Ω). Then, there uniquely exists v∈CM(Ω) such that the following holds. There exist a σ-finite measure µon Ωand f∈L2(µ) such that u=f|f|dµ and v= (cf)|cf|dµ hold. Let cu ∈CM(Ω) be defined as cu := v. Proof: From Lemma 1, existence is easy. Similar to Definition 4, uniqueness can be confirmed. ■ Definition 6 (inner product complex measure): Let Ωbe a measurable space. Let u1, u2∈CM(Ω). Then, there uniquely exists v∈CM(Ω) such that the following holds. There exist a σ-finite measure µon Ωand f1, f2∈L2(µ)such that u1=f1|f1|dµ,u2=f2|f2|dµ and v=f1f2dµ hold. Let hhu1, u2ii ∈ CM(Ω) be defined as hhu1, u2ii := v. Proof: Similar to Definition 4, existence and uniqueness can be confirmed. ■ Definition 7 (inner product): Let Ωbe a measurable space. Let u1, u2∈CM(Ω). Then, let hu1, u2iCM(Ω) ∈Cdenote hhu1, u2ii(Ω).□ Proposition 8: Let Ωbe a measurable space. Then, CM(Ω) is Hilbert space. Proof: Similar to Definition 4, it can be confirmed that CM(Ω) is an 7 inner product space. Let {un}nbe Cauchy sequence in CM(Ω). We show that there exists v∈CM(Ω) such that limnkun−vkCM(Ω) = 0 holds. From Lemma 1 and Lemma 2, there exist a finite measure νand a sequence {fn}nin L2(ν) such that for any n,un=fn|fn|dν holds. So, because of kfk−flkL2(ν)= kuk−ulkCM(Ω),{fn}nis Cauchy sequence in L2(ν). There exists g∈L2(ν) such that limnkfn−gkL2(ν)= 0 holds. Then, because of kun−g|g|dνkCM(Ω) = kfn−gkL2(ν), limnkun−g|g|dνkCM(Ω) = 0 holds. ■ Remark: (1) Let Mbe a C∞-manifold. H¨ormander ([6]) defined densities of order 1 pon M(§2.4) and further defined an inner product for densities of order 1 2on M(§4.2). Let (·,·)Mbe the inner product defined by H¨ormander. Then, (·,·)Mcan be understood as a special case of the inner product of CM(M). For simplicity, assume that Mis compact. Let ω be a volume element of M. Then, the positive square root √ωis a C∞- density of order 1 2. For any C∞-density uof order 1 2, there uniquely exists a complex-valued C∞-function fsuch that u=f√ωholds. For any complex-valued C∞-functions f1and f2, (f1√ω, f2√ω)M=∫Mf1f2dω = hf1|f1|dω, f2|f2|dωiCM(M)holds. (2) Let Ω be a measurable space. As Ay, Jost, Lˆe and Schwachh¨ofer (Remark 3.5 of [1]) noted, Neveu ([11]) indicated that some real Hilbert space S1 2(Ω) is naturally defined. On the other hand, the set RM(Ω) of all real measures on Ω is a real Hilbert space. It can be seen that S1 2(Ω) and RM(Ω) are naturally isomorphic as real Hilbert spaces. Therefore, not only is a statistical manifold on Ω a submanifold of RM(Ω), but the inner product of RM(Ω) induces Fisher information metric. □ 8 3 Self-adjoint operator √−1∂ ∂xk As Ris the measurable space whose set of all measurable sets is the topological σ-algebra, R∞denotes the product measurable space ∏n∈NR. Lemma 9: Let α, β ∈R∞. Let Ebe a subset of R∞. Let Eα:= {x∈ R∞|x−α∈E}and Eβ:= {x∈R∞|x−β∈E}. Then, if Eαis a measurable set of R∞, then Eβis a measurable set of R∞. Proof: Let Mbe the set of all subsets Dof R∞such that {x∈R∞|x− (β−α)∈D}is a measurable set of R∞. Then, Mis a σ-algebra on R∞. Furthermore, for any k∈Nand any measurable set Bof R, (∏n∈{k}B)× (∏n∈N\{k}R)∈ M holds. So, if Dis a measurable set of R∞, then D∈ M holds. In particular, Eα∈ M holds. Therefore, since on the other hand, Eβ={x∈R∞|x−(β−α)∈Eα}holds, Eβis a measurable set of R∞.■ Lemma 10: Let a∈R∞and T∈CM(R∞). Let Tabe the map from the set of all subsets Eof R∞such that {x∈R∞|x+a∈E}is a measurable set of R∞to Cdefined as Ta(E) := T({x∈R∞|x+a∈E}). Then, Ta∈CM(R∞)holds. Proof: From Lemma 9, it is easy. ■ Definition 11 (translation): Let a∈R∞. Then, let a map τafrom CM(R∞)to CM(R∞)be defined as (τaT)(E) := T({x∈R∞|x+a∈E}). □ Lemma 12: Let a∈R∞. Let µbe a measure on R∞. Let µabe the map from the set of all subsets Eof R∞such that {x∈R∞|x+a∈E}is a measurable set of R∞to [0,+∞]defined as µa(E) := µ({x∈R∞|x+a∈E}). Then, µais a measure on R∞. Let fbe a complex measurable function on R∞. Let fabe the map from R∞to Cdefined as fa(x) := f(x−a). Then, fais a complex measurable function on R∞. If f∈L1(µ)holds, then fa∈L1(µa)and τa(fdµ) = fadµahold. 9 (2) The map (u1, u2)7→ u1·u2from CM(Ω1)×CM(Ω2)to CM(Ω1×Ω2) is bi-linear. Proof: (1) There exist finite measures µ1, µ2and f1, g1∈L2(µ1), f2, g2∈ L2(µ2) such that u1=f1|f1|dµ1, v1=g1|g1|dµ1, u2=f2|f2|dµ2and v2= g2|g2|dµ2hold. Then, from hhu1, v1ii =f1g1dµ1,hhu2, v2ii =f2g2dµ2and hhu1·u2, v1·v2ii =f1g1f2g2dµ1dµ2, it follows. (2) It is easy. ■ We may use (2) of the following remark without specific mention, as it is a natural result. Remark: (1) Let Ωbe a set. Let Abe an algebra on Ω. Let Bbe the smallest σ-algebra on Ωsuch that A⊂Bholds. Let u1and u2be real measures on the measurable space whose set of all measurable sets is B. Suppose that for any E∈ A,u1(E) = u2(E)holds. Then, u1=u2holds. (2) Let Ω1and Ω2be measurable spaces. Let u1, u2∈CM(Ω1×Ω2). Suppose that for any measurable set E1of Ω1and any measurable set E2of Ω2,u1(E1×E2) = u2(E1×E2)holds. Then, u1=u2holds. Proof: (1) Let |u1|be the total variation of u1. Let |u2|be the total variation of u2. For a measurable set E, let µ(E) := |u1|(E) + |u2|(E). Then, there exist [−1,+1]-valued measurable functions f1and f2such that u1=f1dµ and u2=f2dµ hold. Let Ef1<f2:= {x∈Ω|f1(x)< f2(x)}and Ef2<f1:= {x∈Ω|f2(x)< f1(x)}. Let ε > 0. Then, from basic facts about Carath´eodory outer measure (especially, Hopf extension theorem), there exists a sequence {En}∞ n=1 in Asuch that Ef1<f2⊂ ∪nEnand ∑nµ(En)< µ(Ef1<f2) + ε 2hold. Let E′ n:= En\∪n−1 k=1Ek. Then, because ∑nµ(E′ n\Ef1<f2) = µ(∪n(E′ n\Ef1<f2)) = µ((∪nEn)\Ef1<f2) = µ(∪nEn)−µ(Ef1<f2)<ε 2and E′ n∈ Ahold, ∫Ef1<f2|f1− f2|dµ =∑n∫E′ n∩Ef1<f2(f2−f1)dµ =∑n(∫E′ n(f2−f1)dµ −∫E′ n\Ef1<f2(f2− f1)dµ) = ∑n∫E′ n\Ef1<f2(f1−f2)dµ ≤∑n∫E′ n\Ef1<f22dµ < ε holds. Therefore, ∫Ef1<f2|f1−f2|dµ = 0 holds. Similarly, ∫Ef2<f1|f1−f2|dµ = 0 holds. (2) Let Pbe the set of all product sets of a measurable set of Ω1and a measurable set of Ω2. Let Abe the smallest algebra on Ω1×Ω2such that P ⊂ A holds. Then, Ais the set of all finite disjoint unions of elements of P. So, for any E∈ A,u1(E) = u2(E) holds. Let Bbe the smallest σ-algebra on Ω1×Ω2such that A ⊂ B holds. Then, Bis the set of all measurable sets of Ω1×Ω2. Therefore, from (1), Re(u1) = Re(u2) and Im(u1) = Im(u2) hold. ■ 16 6 Embedding finite-dimensional evolution {e4RNt}t∈[0,+∞)and {e√−14RNt}t∈(−∞,+∞) Let N∈N. As Ris the measurable space whose set of all measurable sets is the topological σ-algebra, let RNdenote the product measurable space ∏N n=1 R. Let dx denote the (ordinary) measure ∏N n=1 dxnon RN. Let 4RN denote the (ordinary) Laplacian in L2(RN). In Section 3, we defined {∂ ∂xk}k∈N. However, in RN×R∞, the variables x1, x2,··· , xNconflict. For notational consistency, we introduce the following definition. Definition 29 (N-shift): Let u∈CM(R∞). Then, let u+ Ndenote the map from the set of all subsets Eof ∏∞ n=N+1 Rsuch that {{xn}n∈N∈ R∞|{xn−N}∞ n=N+1 ∈E}is a measurable set of R∞to Cdefined as u+ N(E) := u({{xn}n∈N∈R∞|{xn−N}∞ n=N+1 ∈E}). □ Lemma 30: For any u∈CM(R∞),u+ N∈CM(∏∞ n=N+1 R)holds. The map u7→ u+ Nfrom CM(R∞)to CM(∏∞ n=N+1 R)is a unitary operator. Proof: It is easy. ■ Definition 31 (⊗): Let f∈L2(RN)and u∈CM(R∞). Then, let f⊗u∈CM(R∞)defined as f⊗u:= (f|f|dx)·(u+ N). □ Lemma 32: Let f1, f2∈L2(RN)and u1, u2∈CM(R∞). Then, hf1⊗u1, f2⊗u2iCM(R∞)=hf1, f2iL2(RN)hu1, u2iCM(R∞) holds. Proof: From Proposition 28 (1) and Lemma 30, it is easy. ■ Lemma 33: The map (f, u)7→ f⊗ufrom L2(RN)×CM(R∞)to CM(R∞)is bi-linear. Proof: From Proposition 28 (2) and Lemma 30, it is easy. ■ Lemma 34: Let a∈RN. Let fbe a complex measurable function on RN. Let fabe the map from RNto Cdefined as fa(x) := f(x−a). 17 Then, fais a complex measurable function on RN. Let Ebe a measurable set of RN. Let Ea:= {x∈RN|x−a∈E}. Then, Eais a measurable set of RN. If f∈L1(RN)holds, then fa∈L1(RN) and ∫Eafadx =∫Efdx hold. Proof: It is well known. ■ Lemma 35: Let {an}n∈N∈R∞. Let f∈L2(RN)and u∈CM(R∞). Let f{an}N n=1 be the map from RNto Cdefined as f{an}N n=1 (x) := f(x−{an}N n=1). Then, τ{an}n∈N(f⊗u) = f{an}N n=1 ⊗(τ{aN+n}n∈Nu) holds. Proof: Although it is a natural result, we include the proof just in case. Let Fbe a measurable set of RN. Let Gbe a measurable set of ∏∞ n=N+1 R. Then, from {x∈R∞|x+{an}∞ n=1 ∈F×G}={x∈RN|x+{an}N n=1 ∈F}× {x∈∏∞ n=N+1R|x+{an}∞ n=N+1 ∈G}and Lemma 34, (τ{an}∞ n=1 (f⊗u))(F× G) = (f⊗u)({x∈R∞|x+{an}∞ n=1 ∈F×G}) = (∫x+{an}N n=1∈Ff|f|dx)(u+ N({x∈ ∏∞ n=N+1R|x+{an}∞ n=N+1 ∈G})) = (∫Ff{an}N n=1 |f{an}N n=1 |dx)((τ{aN+n}∞ n=1 u)+ N(G)) holds. So, from Remark (2) at the end of Section 5, τ{an}∞ n=1 (f⊗u) = (f{an}N n=1 |f{an}N n=1 |dx)·((τ{aN+n}∞ n=1 u)+ N) holds. ■ Lemma 36: Let k∈N,u∈H1 k(R∞)and f∈L2(RN). Then, f⊗u∈ H1 N+k(R∞)and ∂(f⊗u) ∂xN+k =f⊗∂u ∂xk hold. Proof: There uniquely exists ek∈R∞such that ek,k = 1 holds and for any n∈N\ {k},ek,n = 0 holds. There uniquely exists eN+k∈R∞ such that eN+k,N+k= 1 holds and for any n∈N\ {N+k},eN+k,n = 0 holds. Then, from Lemma 35, Lemma 33 and Lemma 32, kτheN+k(f⊗u)−f⊗u h+ f⊗∂u ∂xkkCM(R∞)=kf⊗(τheku)−f⊗u h+f⊗∂u ∂xkkCM(R∞)=kf⊗(τheku−u h+ ∂u ∂xk)kCM(R∞)=kfkL2(RN)kτheku−u h+∂u ∂xkkCM(R∞)holds. ■ Lemma 37: Let k∈ {1,2,··· , N}. Then, there uniquely exists eN k∈RN such that eN k,k = 1 holds and for any n∈ {1,2,··· , N}\{k},eN k,n = 0 holds. 18 Let f, g ∈L2(RN). Suppose that lim h↓+0 kf(x−heN k)−f(x) h+g(x)kL2(RN)= 0 holds. Let u∈CM(R∞). Then, f⊗u∈H1 k(R∞)and ∂(f⊗u) ∂xk =g⊗u hold. Proof: There uniquely exists ek∈R∞such that ek,k = 1 holds and for any n∈N\ {k},ek,n = 0 holds. For a∈RN, let fa(x) := f(x−a). Then, from Lemma 35, Lemma 33 and Lemma 32, kτhek(f⊗u)−f⊗u h+g⊗ ukCM(R∞)=kfheN k⊗u−f⊗u h+g⊗ukCM(R∞)=k(fheN k−f h+g)⊗ukCM(R∞)= kfheN k−f h+gkL2(RN)kukCM(R∞)holds. ■ Lemma 38: (1) Let f∈H1(RN)and u∈H1(R∞). Then, f⊗u∈ H1(R∞)holds. (2) Let f∈L2(RN)and u∈L2(R∞). Then, f⊗u∈L2(R∞)holds. Proof: (1) From Lemma 37 and Lemma 36, f⊗u∈ ∩k∈NH1 k(R∞) holds and for any k∈ {N+1, N+2,···},∂(f⊗u) ∂xk=f⊗∂u ∂xk−Nholds. So, from Lemma 32, ∑∞ k=N+1 k∂(f⊗u) ∂xkk2 CM(R∞)=kfk2 L2(RN)(∑k∈Nk∂u ∂xkk2 CM(R∞))≤ kfk2 L2(RN) kuk2 H1(R∞)holds. (2) There exists a sequence {gn}nin H1(RN) such that limnkgn−fkL2(RN)= 0 holds. There exists a sequence {vn}nin H1(R∞) such that limnkvn− ukL2(R∞)= 0 holds. Then, from (1), for any n,gn⊗vn∈H1(R∞) holds. On the other hand, from Lemma 33 and Lemma 32, limnkgn⊗vn−f⊗ ukCM(R∞)= 0 holds. ■ In order to examine (4RNf)⊗uand f⊗(4R∞u), we introduce ⊗- contraction (Nand ∞). Definition 39 (N,∞): Let v∈CM(R∞). (1) Let f∈L2(RN). Then, there uniquely exists w∈CM(R∞)such that for any u∈CM(R∞), hf⊗u, viCM(R∞)=hu, wiCM(R∞) holds. Let fNv∈CM(R∞)be defined as fNv:= w. 19 (2) Let u∈CM(R∞). Then, there uniquely exists g∈L2(RN)such that for any f∈L2(RN), hf⊗u, viCM(R∞)=hf, giL2(RN) holds. Let u∞v∈L2(RN)be defined as u∞v:= g. Proof: Existence and uniqueness are easy (by Riesz Theorem). ■ Lemma 40: (1) Let f∈L2(RN)and v∈CM(R∞). Then, kfN vkCM(R∞)≤ kfkL2(RN)kvkCM(R∞)holds. (2) Let u, v ∈CM(R∞). Then, ku∞vkL2(RN)≤ kukCM(R∞)kvkCM(R∞) holds. Proof: It is easy. ■ Proposition 41: (1) Let k∈N,v∈H1 N+k(R∞)and f∈L2(RN). Then, fNv∈H1 k(R∞)and ∂(fNv) ∂xk =fN ∂v ∂xN+k hold. (2) Let k∈ {1,2,··· , N},v∈H1 k(R∞)and u∈CM(R∞). Let ∂ ∂xN,k be the (ordinary) generalized partial differential operator in L2(RN). Then, u∞vis an element of the domain of ∂ ∂xN,k and ∂(u∞v) ∂xN,k =u∞ ∂v ∂xk holds. Proof: (1) From Lemma 36 and Theorem 16, because for any u∈ H1 k(R∞), hu, fN∂v ∂xN+kiCM(R∞)=hf⊗u, ∂v ∂xN+kiCM(R∞)=−hf⊗∂u ∂xk, viCM(R∞)= −h ∂u ∂xk, f NviCM(R∞)holds, fN∂v ∂xN+k=−(∂ ∂xk)∗(fNv) = ∂(f⋄Nv) ∂xkholds. (2) In the same way as (1), from Lemma 37 and Theorem 16, it follows. ■ Lemma 42: (1) Let φ∈H1(R∞)and f∈L2(RN). Then, fNφ∈ H1(R∞)holds. (2) Let φ∈H1(R∞)and u∈CM(R∞). Then, u∞φ∈H1(RN)holds. (3) Let f∈L2(RN). Let ube an element of the domain of 4R∞. Then, f⊗(4R∞u)∈L2(R∞),f⊗u∈ ∩k∈NH1 N+k(R∞)and ∑k∈Nk∂(f⊗u) ∂xN+kk2 L2 N+k(R∞)< +∞hold. For any φ∈H1(R∞), −hf⊗(4R∞u), φiL2(R∞)=∑ k∈Nh∂(f⊗u) ∂xN+k ,∂φ ∂xN+kiL2 N+k(R∞) 20 holds. (4) Let u∈L2(R∞)and f∈H2(RN). Then, (4RNf)⊗u∈L2(R∞)and f⊗u∈ ∩N k=1H1 k(R∞)hold. For any φ∈H1(R∞), −h(4RNf)⊗u, φiL2(R∞)= N ∑ k=1h∂(f⊗u) ∂xk ,∂φ ∂xkiL2 k(R∞) holds. Proof: (1) From Proposition 41 (1) and Lemma 40 (1), it follows. (2) From Proposition 41 (2), it follows. (3) From Lemma 38 (2), f⊗(4R∞u)∈L2(R∞) holds. From Theorem 24 (1), u∈H1(R∞) holds. So, from Lemma 36, f⊗u∈ ∩k∈NH1 N+k(R∞) and ∑k∈Nk∂(f⊗u) ∂xN+kk2 L2 N+k(R∞)<+∞hold. From (1), Proposition 41 (1) and Lemma 36, hf⊗u, φiCM(R∞)−hf⊗(4R∞u), φiL2(R∞)=h(1 −4R∞)u, f N φiL2(R∞)=hu, fNφiH1(R∞)=hu, fNφiCM(R∞)+∑k∈Nh∂u ∂xk, fN∂φ ∂xN+kiCM(R∞)= hf⊗u, φiCM(R∞)+∑k∈Nh∂(f⊗u) ∂xN+k,∂φ ∂xN+kiL2 N+k(R∞)holds. (4) Similar to (3), from (2), Lemma 37, Lemma 38 (2) and Proposition 41 (2), it is shown. ■ Lemma 43: Suppose that −∞ < T0< T1<+∞holds. Let f∈ C1([T0, T1]; L2(RN)) and u∈C1([T0, T1]; CM(R∞)). Then, f⊗u∈C1([T0, T1] ;CM(R∞)) and d dt(f⊗u) = ( d dtf)⊗u+f⊗(d dtu) hold. Proof: From Lemma 32 and Lemma 33, it follows. ■ Theorem 44: Let f0∈L2(RN)and u0∈L2(R∞). Then, the followings hold. (1) Let t∈[0,+∞). Then, e△R∞t(f0⊗u0) = (e△RNtf0)⊗(e△R∞tu0) holds. (2) Let t∈(−∞,+∞). Then, e√−1△R∞t(f0⊗u0) = (e√−1△RNtf0)⊗(e√−1△R∞tu0) holds. 21 Proof: About (2), we show it. About (1), it can be similarly shown. Let f(t) := e√−1△RNtf0and u(t) := e√−1△R∞tu0. First, we show that if f0∈H2(RN) holds and u0is an element of the domain of 4R∞, then e√−1△R∞t(f0⊗u0) = f(t)⊗u(t) holds. From Lemma 43 and Lemma 38 (2), f⊗u∈C1((−∞,+∞); L2(R∞)), d dt(f⊗u) = √−1((4RNf)⊗u+f⊗(4R∞u)) hold. On the other hand, because from Lemma 38 (1), f⊗u∈H1(R∞) holds, from Lemma 42 (3) and Lemma 42 (4), for any φ∈H1(R∞), hf⊗u, φiL2(R∞)−h(4RNf)⊗u+f⊗(4R∞u), φiL2(R∞)=hf⊗u, φiH1(R∞) holds. So, f⊗u−((4RNf)⊗u+f⊗(4R∞u)) = (1 −4R∞)(f⊗u) holds. (4RNf)⊗u+f⊗(4R∞u) = 4R∞(f⊗u) holds. Therefore, if f0∈H2(RN) holds and u0is an element of the domain of 4R∞, then e√−1△R∞t(f0⊗u0) = f(t)⊗u(t) holds. There exist a sequence {g0,n}nin H2(RN) such that limnkg0,n−f0kL2(RN)= 0 holds. There exist a sequence {v0,n}nin the domain of 4R∞such that limnkv0,n −u0kL2(R∞)= 0 holds. Let gn(t) := e√−1△RNtg0,n and vn(t) := e√−1△R∞tv0,n. Then, limnkgn(t)−f(t)kL2(RN)= 0 and limnkvn(t)−u(t)kL2(R∞)= 0 hold. From Lemma 32, Lemma 33 and Lemma 38 (2), lim nkgn(t)⊗vn(t)−f(t)⊗u(t)kL2(R∞)= 0 holds. On the other hand, from limnkg0,n ⊗v0,n −f0⊗u0kL2(R∞)= 0, lim nke√−1△R∞t(g0,n ⊗v0,n)−e√−1△R∞t(f0⊗u0)kL2(R∞)= 0 holds. So, because of e√−1△R∞t(g0,n ⊗v0,n) = gn(t)⊗vn(t), it follows. ■ 22 7 Inseparability of L2(R∞) Lemma 45: Let {fn}n∈Nbe a sequence of [0,+∞)-valued measurable functions on R. Suppose that for any n∈N,kfnkL2(R)= 1 holds. Then, the followings hold. (1) Let m∈N. Then, (∏ n∈{m} fn(xn)2dxn)( ∏ n∈N\{m} fn(xn)2dxn) = ∏ n∈N fn(xn)2dxn holds. (2) Let {an}n∈N∈R∞. Then, τ{an}n∈N(∏ n∈N fn(xn)2dxn) = ∏ n∈N fn(xn−an)2dxn holds. Proof: Although it is a natural result, we include the proof just in case. Let N∈N. Let {En}N n=1 be a family of measurable sets of R. (1) ((∏n∈{m}fn(xn)2dxn)(∏n∈N\{m}fn(xn)2dxn))((∏N n=1 En)×(∏∞ n=N+1 R)) = ∏N n=1(∫Enfn(xn)2dxn) holds. (2) For n∈ {1,2,··· , N}, let Fn:= {xn∈R|xn+an∈En}. Then, (τ{an}n∈N(∏n∈Nfn(xn)2dxn))((∏N n=1 En)×(∏∞ n=N+1 R)) = (∏n∈Nfn(xn)2dxn) ((∏N n=1 Fn)×(∏∞ n=N+1 R)) = ∏N n=1(∫Fnfn(xn)2dxn) = ∏N n=1(∫Enfn(xn− an)2dxn) holds. ■ Lemma 46: Let {fn}n∈Nand {gn}n∈Nbe sequences of [0,+∞)-valued measurable functions on R. Suppose that for any n∈N,kfnkL2(R)= kgnkL2(R)= 1 holds. Suppose that there exists m∈Nsuch that hfm, gmiL2(R)= 0holds. Then, h∏n∈Nfn(xn)2dxn,∏n∈Ngn(xn)2dxniCM(R∞)= 0 holds. Proof: From Lemma 45 (1) and Proposition 28 (1), it follows. ■ Lemma 47: Let {fn}n∈Nbe a sequence of [0,+∞)-valued measurable functions on R. Suppose that for any n∈N,kfnkL2(R)= 1 holds. Let m∈N. Suppose that fm∈H1(R)holds. Then, ∏n∈Nfn(xn)2dxn∈H1 m(R∞)and ∂ ∂xm(∏n∈Nfn(xn)2dxn) = (∏n∈{m}f′ n(xn)|f′ n(xn)|dxn)·(∏n∈N\{m}fn(xn)2dxn) hold. Proof: There uniquely exists em∈R∞such that em,m = 1 holds and for any n∈N\ {m},em,n = 0 holds. From Lemma 45, for h > 0, τhem(∏n∈Nfn(xn)2dxn) = (∏n∈{m}fn(xn−h)2dxn)(∏n∈N\{m}fn(xn)2dxn) holds. 23 So, because from Proposition 28 (2), τhem(∏n∈Nfn(xn)2dxn)−∏n∈Nfn(xn)2dxn h+ (∏n∈{m}f′ n(xn)|f′ n(xn)|dxn)·(∏n∈N\{m}fn(xn)2dxn) = (∏n∈{m}(fn(xn−h)−fn(xn) h+ f′ n(xn))|fn(xn−h)−fn(xn) h+f′ n(xn)|dxn)·(∏n∈N\{m}fn(xn)2dxn) holds, from Proposition 28 (1), it follows. ■ Proposition 48: There exists an orthonormal system {uτ}τ∈∏n∈N{0,1}of L2(R∞). Proof: There exists f∈C∞(R) such that ∫R|f|2dx = 1, for any x∈R, f(x)≥0 holds and for any x∈R\(0,1), f(x) = 0 holds. There exists a sequence {Ln}n∈Nin (0,+∞) such that ∑ n∈N 1 L4 n <+∞ holds. For τ∈∏n∈N{0,1}, let uτ:= ∏ n∈N (1 Ln f(xn L2 n−τ(n)))2dxn. Then, from Lemma 46, {uτ}τ∈∏n∈N{0,1}is an orthonormal system of CM(R∞). On the other hand, because from Lemma 47 and Proposition 28 (1), for any τ∈∏n∈N{0,1}and any n∈N, k∂uτ ∂xnk2 CM(R∞)=1 L4 nkf′k2 L2(R) holds, for any τ∈∏n∈N{0,1},uτ∈H1(R∞) holds. So, {uτ}τ∈∏n∈N{0,1}is an orthonormal system of L2(R∞). ■ 24 8 Translation invariance of 4R∞ Lemma 49: Let a∈R∞and k∈N. Then, the followings hold. (1) Let u∈H1 k(R∞). Then, τau∈H1 k(R∞)and ∂(τau) ∂xk=τa(∂u ∂xk)hold. (2) Let u1, u2∈H1 k(R∞). Then, h∂(τau1) ∂xk,∂(τau2) ∂xkiCM(R∞)=h∂u1 ∂xk,∂u2 ∂xkiCM(R∞) holds. Proof: (1) From τbτc=τb+cand Proposition 13, it follows. (2) From (1) and Proposition 13, it follows. ■ Lemma 50: Let a∈R∞. Then, the followings hold. (1) Let u∈H1(R∞). Then, τau∈H1(R∞)holds. (2) Let u1, u2∈H1(R∞). Then, hτau1, τau2iH1(R∞)=hu1, u2iH1(R∞) holds. Proof: (1) From Lemma 49 (1) and Proposition 13, it follows. (2) From Lemma 49 (2) and Proposition 13, it follows. ■ Lemma 51: Let a∈R∞and f1, f2∈L2(R∞). Then, τaf1, τaf2∈ L2(R∞)and hτaf1, τaf2iL2(R∞)=hf1, f2iL2(R∞)hold. Proof: From Lemma 50 (1) and Proposition 13, it follows. ■ Proposition 52: Let a∈R∞. Let ube an element of the domain of 4R∞. Then, τauis an element of the domain of 4R∞and 4R∞(τau) = τa(4R∞u) holds. Proof: From Lemma 50, Lemma 51 and Proposition 13, for any φ∈ H1(R∞), hτau−τa(4R∞u), φiL2(R∞)=h(1−4R∞)u, τ−aφiL2(R∞)=hu, τ−aφiH1(R∞) =hτau, φiH1(R∞)holds. So, τau= (1 −4R∞)−1(τau−τa(4R∞u)) holds. ■ Remark: (1) Gross ([4]) considered an analog on infinite-dimensional real Hilbert space of the Laplacian. Of course, R∞is not Hilbert space. It is translation invariant. However, it is not a self-adjoint operator. (2) For p∈[1,+∞), let Lpbe Ornstein-Uhlenbeck operator in Lp(µ) with respect to Wiener measure µ. Of course, it may be more appropriate to think of Lpas an analog of Ornstein-Uhlenbeck operator on Rn, rather than as an analog of Laplace operator on Rn.Lpis a generator of a contraction semigroup and one of the main characters in Malliavin calculus (e.g., [10]). L2is a self-adjoint operator corresponding to infinite-dimensional Dirichlet form (e.g., [9]). However, Lpis not translation invariant. □ 25