Solution of a cauchy problem for a diffusion equation in a Hilbert space by a Feynman formula

Abstract

The Cauchy problem for a class of diffusion equations in a Hilbert space is studied. It is proved that the Cauchy problem in well posed in the class of uniform limits of infinitely smooth bounded cylindrical functions on the Hilbert space, and the solution is presented in the form of the so-called Feynman formula, i.e., a limit of multiple integrals against a gaussian measure as the multiplicity tends to infinity. It is also proved that the solution of the Cauchy problem depends continuously on the diffusion coefficient. A process reducing an approximate solution of an infinite-dimensional diffusion equation to finding a multiple integral of a real function of finitely many real variables is indicated.

Full text

ISSN 1061-9208, Russian Journal of Mathematical Physics, Vol. 19, No. 3, 2012, pp. 360–372. c
Pleiades Publishing, Ltd., 2012.
Solution of a Cauchy Problem
for a Diffusion Equation in a Hilbert Space
by a Feynman Formula
I. D. Remizov
Received May 20, 2012
Abstract. The Cauchy problem for a class of diffusion equations in a Hilbert space is studied.
It is proved that the Cauchy problem in well posed in the class of uniform limits of infinitely
smooth bounded cylindrical functions on the Hilbert space, and the solution is presented
in the form of the so-called Feynman formula, i.e., a limit of multiple integrals against a
gaussian measure as the multiplicity tends to infinity. It is also proved that the solution of
the Cauchy problem depends continuously on the diffusion coefficient. A process reducing
an approximate solution of an infinite-dimensional diffusion equation to finding a multiple
integral of a real function of finitely many real variables is indicated.
DOI 10.1134/S1061920812030089
1. INTRODUCTION
The main result of the present paper claims that the Cauchy problem
u
t(t, x)=g(x)tr(Au
xx(t, x)) ; t0,x∈H,
u(0,x)=u0(x); x∈H. (1)
has a solution (which is unique in some class of function) given by
u(t, x)= lim
n→∞
HH
...
H
ntimes
u0(y1)μy2
2t
ng(y2)A(dy1)μy3
2t
ng(y3)A(dy2)...μ
yn
2t
ng(yn)A(dyn−1)μx
2t
ng(x)A(dyn),(2)
and this solution continuously depends (in the uniform norm) on the functions gand u0.HereH
stands for a real separable Hilbert space, Ais a linear positive trace-class selfadjoint operator on
H,u:[0,+∞)×H→Ris the desired function, u0:H→Rand g:H→Rare given functions
(with g(x)g0≡const >0 for every x∈H), μx
(2t/n)g(x)Ais the Gaussian measure on Hwith the
mean xand the correlation operator (2t/n)g(x)A. For details, see Theorem 4.
Relations of the form (2) in which some function is represented as a limit of a multiple integral
as the multiplicity tends to infinity are referred to as Feynman formulas (see [21, 22]). The term
“Feynman formula” was introduced in this context by Smolyanov [26]. For a survey of current
results concerning Feynman formulas, see [27].
The paper is organized as follows. Section 2 contains the main notation, definitions, and auxiliary
facts. In Sec. 3 we discuss properties of the operators used in the setting and the solution of
problem (1), and the main theorem is proved in Sec. 4. In the final section we discuss the size of
classes in which we seek solutions of the Cauchy problem.
2. NOTATION, DEFINITIONS, AND PRELIMINARIES
Let Hdenote a separable Hilbert space over Rwith the inner product ·,·. Denote by Cb(H, R)
the Banach space of all bounded continuous functions H→Requipped with the uniform norm
f=sup
x∈H|f(x)|. Recall that a function f:H→Ris said to be cylindrical if there are vectors
e1,...,e
nin Hand a function fn:Rn→Rsuch that f(x)=fn(x, e1,,...,x, en,) for every
x∈H.LetD=C∞
b,c(H, R) denote the space of all bounded cylindrical functions H→Rwhich are
360

SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 361
infinitely Fr´echet differentiable at any point of Hand whose Fr´echet derivatives of any order are
bounded (and continuous).
If f:H→R,thenf(x) stands for the second-order Fr´echet derivative of fevaluated at x.and
A:H→Ha linear trace-class1selfadjoint positive nondegenerate operator with the domain H.
Introduce the differential operator ΔAby the rule (ΔAf)(x):=trAf(x).Let X=C∞
b,c(H, R)be
the closure of Din Cb(H, R). For any function g∈Xbounded away from zero (i.e., g(x)g0≡
const >0), introduce the operator gΔAby the rule (gΔAf)(x):=g(x)(ΔAf)(x)=g(x)trAf(x).
For a function u:[0,+∞)×H→Rwe pose the Cauchy problem for the diffusion equation (1).
The abstract Cauchy problem for a function U:[0,+∞)→Xof the form
(d/dt)U(t)=gΔAU(t); t0,
U(0) = u0,(3)
where gΔAstands for the closure of the unbounded differential operator gΔAand (gΔAf)(x)=
g(x)tr(Af(x)) is closely related to (1). Following [23], by a strong solution of problem (3) we mean
a function U:[0,+∞)→Xsuch that
U∈C1([0,+∞),X),U(t)∈D1;t0,(d/dt)U(t)=gΔAU(t); t0,U(0) = u0,(4)
where D1={f∈X:∃(fj)⊂D: limj→∞ fj=f,∃limj→∞ tr A(fj)}stands for the space of
functions f∈Xuniformly approximable by cylindrical functions fj∈Dfor which the sequence
tr A(fj) converges uniformly. By a mild solution of problem (3) we mean a function U:[0,+∞)→
Xsuch that
U∈C([0,+∞),X),t
0
U(s)ds ∈D1;t0,U(t)=gΔAt
0
U(s)ds +u0;t0.(5)
Problem (1) is reduced to (3) as follows. A function u:(t, x)→ u(t, x)of(t, x) is treated as a
function u:t→ [x→ u(t, x)] of twith the range in some function space of the argument x.Then
u(t, x)=(U(t))(x), t0,x ∈H. The terms “strong solution” and “mild solution” of problem (1)
are now transferred from solutions of problem (3) using this correspondence.
The following theorem is used below.
Theorem 1 (Chernoff product formula, see, e.g., [23, Th. III,5.2]).Let Xbe a real Banach
space and let Lb(X, X)be the space of all bounded linear operators on Xequipped with the operator
norm. Let a function S:[0,+∞)→Lb(X, X)be given such that S0=I,whereIstands for the
identity operator, let (St)mbe bounded for some constant Mfor any t0and m∈N,and
let the strong limit limt↓0t−1(Stϕ−ϕ)=:Lϕ exist for any ϕ∈D⊂X, where Dand (λ0I−
L)(D)are dense subspaces of Xfor some λ0>0. Then the closure Lof Lis an infinitesimal
generator of some bounded strongly continuous operator semigroup (Tt)t0given by the formula
Ttϕ= limn→∞(St/n )nϕ, where the limit exists for any ϕ∈Xand is uniform with respect to
t∈[0,t
0]for any t0>0.
Let x∈Hand B:H→Ha linear operator. By the symbol μx
Bwe denote the Gaussian
probability measure on a Borel sigma algebra in Hwith the mean xand the correlation operator B.
Let (Stϕ)(x):=Hϕ(x+y)μ2tg(x)A(dy)fort>0andS0ϕ:= ϕfor ϕ∈X.
The objective of the paper is to prove, using Theorem 1, that the solution of the Cauchy problem
(1) is given by the Feynman formula (2). We develop and strengthen results of [15] obtained there
for the simplest diffusion equation with a variable coefficient at the higher derivative, namely, we
consider here an infinite-dimensional case and prove the existence of a solving semigroup. We also
discuss the reduction of an approximate evaluation of integrals over an infinite-dimensional space
1For the definition and properties of trace-class operators in a Hilbert space, see, e.g., G. J. Murphy, C∗-Algebras
and Operator Theory (Academic Press, Inc., Boston, MA, 1990), §2.4.
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

362 REMIZOV
in the Feynman formula to evaluation of integrals over finite-dimensional spaces and the size of
function classes in use.
Recall some notation, definitions, and facts needed below. Unless otherwise stated, the sym-
bol Xstands for an abstract real Banach space. We use the apparatus of strongly continuous
one-parameter semigroups of bounded linear operators in Banach spaces and their infinitesimal
generators; for the definitions and the main properties, we use the book [23]. For a closed linear
operator L, we refer to the problem
(d/dt)F(t)=LF (t); t0,
F(0) = F0,(6)
for a function F:[0,+∞)→Xas the abstract Cauchy problem associated with the closed linear
operator L:X⊃Dom(L)→Xand the vector F0∈X.AfunctionF:[0,+∞)→Xis said to be
aclassical solution of the abstract Cauchy problem (6) if the function Fhas continuous derivative
F:[0,+∞)→Xin the strong operator topology for every t0, F(t)∈Dom(L) for every t0,
and (6) holds; a strongly continuous function F:[0,+∞)→Xis referred to as a mild solution of
(6) if t
0F(s)ds ∈Dom(L)andF(t)=Lt
0F(s)ds +F0for every t0. Recall that, if an operator
(L, Dom(L)) is the infinitesimal generator of a strongly continuous semigroup (Ts)s0, then, for
every F0∈Dom(L), there is a unique classical solution of (6) given by the formula F(t)=T(t)F0
and, for every F0∈X, there is a unique mild solution of (6) given by the formula F(t)=T(t)F0
(see [23, II, Proposition 6.2]).
A strongly continuous operator semigroup (Tt)t0is said to be contractive (or a contraction
semigroup) if Tt1 for every t0. A linear operator L:X⊃Dom(L)→Xon a Banach
space Xis said to be dissipative if Lx −λxλxfor every λ>0andanyx∈Dom(L). Recall
[23, II, Proposition 3.14] that a linear dissipative operator L:X⊃Dom(L)→Xon a Banach
space Xwhose domain Dom(L)isdenseinXadmits the closure, L:X⊃Dom(L)→X,which
is also dissipative. By the Lumer–Phillips theorem (see, e.g., [23, Th. II.3.15]), the closure Lof a
dissipative operator (L, Dom(L)) on a Banach space Xis an infinitesimal generator of a contraction
semigroup if and only if the image of the operator λI −Lis dense in Xfor some (and hence for
any) λ>0.
Let (eLnt)t0be strongly continuous operator semigroups on a Banach space Xwith infinitesimal
generators (Ln,Dom(Ln)) satisfying the stability condition eLntMewt for constants M1,
w∈R,andallt0andn∈N. Let there be a densely defined operator (L, Dom(L)) on Xsuch
that Lnx→Lx for all xinacoreDof Lsuch that the image of the operator (λ0I−L)isdense
in Xfor some λ0>0.Then the semigroups (eLnt)t0,n ∈Nconverge strongly (and uniformly
with respect to t∈[0,t
0] for any t0>0) to a strongly continuous semigroup (etL)t0with the
infinitesimal generator L[23, Th. III.4.9].
We also need some fundamentals concerning Fr´echet differentiation in Banach and Hilbert spaces
(for details, see [8]. In particular, the second Fr´echet derivative of a function f:H→Ris regarded
as a mapping f :H→Lb(H, H). For n+1 times Fr´chet differentiable function f,wehavethe
Taylor expansion
f(x+h)=f(x)+(1/1!)f(x)h+(1/2!)f (x)(h, h)+···+(1/n!)f(n)(x)(h,...,h)+Rn(x, h),(7)
where Rn(x, h)∈conv{(n+1)!
−1f(n+1)(x+θh)(h,...,h):θ∈(0,1)}⊂R, and therefore
|Rn(x, h)|(hn+1/(n+1)!) sup
z∈[x,x+h]f(n+1)(z).(8)
The following statement is used below.
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SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 363
Proposition 1. Le t fbe a cylindrical real-valued function on Hand let (ek)k∈Nbe an or-
thonormal basis in Hsuch that fdoes not depend on the coefficients f,ekwith k>nfor
some n∈N. The the function fhas a derivative in a direction hif and only if the function
fnhas a derivative in the direction (h, e1,...,h, en)∈Rn,andf(x)h=h, (∂1fn(x, e1,...,
x, en), ..., ∂
nfn(x, e1,...,x, en),0,0,0,...),where the symbol ∂jfnstands for the partial
derivative with respect to the jth argument of the function fn:Rn→R.Ifthefunctionfhas
Fr´echet derivative at x,then
f(x)=(∂1fn(x, e1,...,x, en), ..., ∂
nfn(x, e1,...,x, en),0,0,0,...)T,(9)
and fhas Fr´echet derivative on Hif and only if fnhas Fr´echet derivative on Rn.IfA:H→H
is a trace-class operator, then
tr Af(x)=
n

s=1
n

k=1Aes,e
k∂k∂sfn(x, e1,...,x, en).
The proof is straightforward.
We also need some definitions and facts concerning Gaussian probability measures on a Hilbert
space [SS]. Recall the definitions of measurable sets and of measures on H.AsetB⊂His said
to be cylindrical if there are an n-dimensional linear subspace Hn⊂Hand a Borel set Bn⊂B
such that B=P−1(Bn), where P:H→Hnstands for the orthogonal projection. For a fixed
n-dimensional subspace Hn⊂H, the family {P−1(Bn):Bnis a Borel set in Hn}is a σ-algebra
in H.IfHis infinite-dimensional, then the family of cylindrical sets in Hforms an algebra (rather
than a σ-algebra, because, e.g., if (ek)isabasisinH, then the countable intersection of the sets
{x:0<x, ek<1}is not a cylindrical set). We refer to this set algebra as the cylinder algebra and
to the smallest σ-algebra containing all cylinders as the cylindrical σ-algebra.SinceHis separable,
the cylindrical σ-algebra on Hcoincides with the Borel one.
By a cylindrical measure μon Hwe mean a real finitely additive set function defined on the
algebra of all cylindrical sets in Hand such that, for every finite-dimensional linear subspace
Hn⊂Hfor the orthogonal projection P:H→Hn, the restriction of the measure μto the σ-
algebra {P−1(Bn): Bnis a Borel subset of Hn}is a countably-additive measure.
For any finite-dimensional subspace Hn⊂H, the restriction of a cylindrical measure to the
σ-algebra {P−1(Bn):Bnis a Borel subset of Hn}is a countably additive measure. Therefore, for
the Borel cylindrical functions, the Lebesgue integral against a cylindrical measure has a definite
meaning.
By the Fourier transform of a cylindrical measure μon Hwe mean a function μ:H→C,
μ(z):=Heiz,yμ(dy).As is well known [9, p. 16], a cylindrical measure is defined by its Fourier
transform uniquely. For a real separable Hilbert space H, a cylindrical measure μx
Adefined on the
cylinder algebra is referred to as a nondegenerate Gaussian measure if the Fourier transform of μx
A
is of the form 
μx
A(z)=exp
iz,x−1
2Az, z,where x∈His the so-called expectation of μx
A
and A:H→His the so-called correlation operator of μx
A.Ifx= 0, then the measure is said to be
centered, and we write μ0
A=μA.
The correlation operators of the Gaussian measures used below are selfadjoint positive trace-class
operators. By [4, Ch. II, §2, 3◦]), if a function ϕ:H→Ris cylindrical, continuous, and bounded,
if e1,...,e
nis a full family of eigenvectors of an operator C,andc1,...,c
nare the corresponding
eigenvalues, then
H
ϕ(y)μA(dy)=(2π)−n/2(
n

i=1
ci)−1/2Rn
ϕn(z1,...,z
n)exp−
n

i=1
z2
i/(2ci)dz1...dz
n.(10)
By [4, Ch. II, §2, 1◦], if His a real separable Hilbert space, A:H→His a trace-class symmetric
positive-definite linear operator, μAis a centered Gaussian measure on Hwith the correlation
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

364 REMIZOV
operator A,andB:H→His a bounded linear operator, then
HBy,yμA(dy)=tr(AB),(11)
H
(By,y)2μA(dy) = (tr(AB))2+2tr(AB)2.(12)
Lemma 1. Let Hbe a real separable Hilbert space. Let A:H→Hbe a positive, trace-class, self-
adjoint linear operator. Let μAbe a centered Gaussian measure on Hwith the correlation operator
A.Letf:H→Rbe a continuous bounded function. Then
H
f(x)μtA(dx)=H
f(√tx)μA(dx).(13)
The proof uses the uniqueness of a measure having a given Fourier transform and the standard
theorem concerning the change of measure, in a Lebesgue integral, caused by a measurable mapping.
Finally, recall the following result (see [6, Th. 4.3.1 and 4.3.2, and Corollary 4.3.4].
Lemma 2. Let aij :Rn→R,i, j =1,...,n,beafunctioninC∞
b(Rn,R)(the space of bounded
real function on Rnhaving the bounded derivatives of all orders). Let the ellipticity condition
n
i,j=1 aij (x)ξiξjκξ2hold for some κ>0and every ξ=(ξ1,...,ξ
n)∈Rnand x∈Rn.
Let a constant λ>0and a function f∈C∞
b(Rn,R)be chosen. Then there is a unique function
u∈C∞
b(Rn,R)which is a solution of the equation
n

i=1
n

j=1
aij(x)∂2
∂xi∂xj
u(x)−λu(x)=f(x),(14)
and every function v∈C∞
b(Rn,R)satisfies the bound
sup
x∈Rn
n

i=1
n

j=1
aij(x)∂2
∂xi∂xj
v(x)−λv(x)
λsup
x∈Rn|v(x)|.(15)
3. FAMILIES OF OPERATORS AND THEIR PROPERTIES
Theorem 2. Let g∈Xbe such that g(x)go≡const >0for every x∈H,andletgbe a
uniform limit of smooth cylindrical functions in D.Lett>0and μ2tg(x)AacenteredGaussian
measure on Hwith the correlation operator 2tg(x)A.Fort0and ϕ∈Cb(H, R),write
(Stϕ)(x):=H
ϕ(x+y)μ2tg(x)A(dy)ast>0andS0ϕ:= ϕ. (16)
Then Stϕ∈Cb(H, R)for t0and ϕ∈Cb(H, R)St:Cb(H, R)→Cb(H, R)is a bounded linear
operator for any t0with unit norm. If g∈D(g∈X),thenD(X, respectively)is St-invariant
for any t0. The representation Hϕ(x+y)μ2tg(x)A(dy)=ϕ(x)+tg(x)tr(Aϕ(x)) + t2r(x, t),
where supx∈H|r(x, t)|1
6supz∈Hϕ(4)(z)(supz∈H|g(z)|)2((tr A)2+2trA2), holds for any ϕ∈D
and every t>0, and, finally, limt→0supx∈Ht−1(Stϕ)(x)−ϕ(x)−(gΔAϕ)(x)=0for every
function ϕ∈D.
Proof. The integral (16) exists in the Lebesgue sense because the function is bounded and
continuous and the (probability) measure is bounded. Let a number t>0andafunctionϕ∈
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 365
Cb(H, R) be chosen. By Lemma 1, Hϕ(x+y)μ2tg(x)A(dy)=Hϕ(x+2tg(x)y)μA(dy).It follows
from the bound
Stϕ=sup
x∈HH
ϕ(x+2tg(x)y)μA(dy)sup
x∈H|ϕ(x)|·μA(H)=ϕ·1 (17)
that the function Stϕis bounded. We claim that the function Stϕis continuous. Let xj→x.
Then ϕ(xj+2tg(xj)y)→ϕ(x+2tg(x)y) for any y∈H.Moreover,|ϕ(xj+2tg(xj)y)|
ϕ≡const and |ϕ(x+2tg(x)y)|ϕ≡const. Therefore, by the dominated convergence
theorem, limj→∞ Hϕ(xj+2tg(xj)y)μA=Hϕ(x+2tg(x)y)μA,i.e., (Stϕ)(xj)→(Stϕ)(x).
Thus, Stϕ∈Cb(H, R). The operator Stis obviously linear. If 1 is regarded as a constant function
on H,thenSt1 = 1, and hence St1.The bound (17) shows that St1.
To prove that D(X)isSt-invariant, choose a t>0. If g∈D, then the operator Sttakes a cylin-
drical function to a cylindrical one. Indeed, if ϕis cylindrical, then ϕ(x)=ϕn(x, e1,...,x, en)
for every x∈Hand for some n∈N, some function ϕn:Rn→R,andsomee1,...,e
nin H. Simi-
larly, since the function gis cylindrical by assumption, we have g(x)=gm(x, en+1,...,x, en+m)
for every x∈Hand for some m∈N, some function gm:Rm→R,andsomeen+1,...,e
n+min H.
By Lemma 1, H
ϕ(x+y)μ2tg(x)A(dy)=H
ϕ(x+2tg(x)y)μA(dy).(18)
Write
Φ(x1,...,x
n+m)=
H
ϕnx1+2tgm(xn+1,...,x
n+m)y, e1,
...,x
n+2tgm(xn+1,...,x
n+m)y, enμA(dy).
then (Stϕ)(x)=Φ(x, e1,...,x, en+m) for every x∈H, which means that Stϕis cylindrical.
We claim now that, if ϕis cylindrical and has bounded Fr´echet derivative of all orders, then so is
Stϕ. Let us apply Proposition 1. We claim first that Φ has Fr´echet derivatives of all orders. To this
end, we pass in the expression for Φ from the integral over Hto the integral over Rn(Since ϕis cylin-
drical). Introduce the following notation: Ψn:Hh→ (h, e1,...,h, en)∈Rnis a projection,
Hn= span(e1,...,e
n) is the related subspace of H,In:Hnh→ (h, e1,...,h, en)∈Rnis an
isomorphism, Pn:Hh→h, e1e1+···+h, enen∈Hnis the projection in H.ThenΨ
n=InPn
and ϕ(x)=ϕn(Ψnx). Moreover, write →
xn
1=(x1,...,x
n)∈Rnand →
xm
n+1 =(xn+1,...,x
m)∈Rm.
Since Ais nondegenerate and symmetric on H, it follows that PnAis nondegenerate and sym-
metric on Hn, and therefore it can be diagonalized in some basis. We may assume without loss
of generality that e1,...,e
nis this basis (because the change of a basis in Hnmodifies ϕnonly).
The matrix of the operator PnAon Hncoincides with the matrix of Cn=InPnAI−1
non Rn.Let
c1,...,c
n-be the eigenvalues of Cncorresponding to the eigenvectors Ψne1,...,Ψnen.Notethat
ci>0andgm→
xm
n+1g0≡const >0 for every →
xm
n+1 ∈Rm. By (10),
Φ→
xn
1,→
xm
n+1=H
ϕn→
xn
1+2tgm→
xm
n+1Ψn(y)μA(dy)
=(2π)−n/2n

i=1
ci−1/2Rn
ϕn→
xn
1+2tgm→
xm
n+1zexp −
n

i=1
z2
i/(2ci)dz.
Introduce a measure νon Rngiven by a density with respect to the Lebesgue measure as follows:
set
ν(A):=(2π)−n/2n

i=1
ci−1/2A
exp −
n

i=1
z2
i/(2ci)dz
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

366 REMIZOV
for every measure set A⊂Rn. It follows from this definition that
Φ→
xn
1,→
xm
n+1=Rn
ϕn→
xn
1+2tgm→
xm
n+1zν(dz).
The integrand is a composition of mappings with bounded continuous Fr´echet derivative, and
therefore has the same property. The Fr´echet derivative of the integrand is uniformly bounded,
while (Rn,ν) is a locally compact and countable at infinity normed linear space equipped with
a nonnegative Radon measure, and therefore we can apply Theorem 115 in [13] on the Fr´echet
differentiation under the sign of a Lebesgue integral. Repeating this argument for everyk∈N,we
conclude that, since the integrand has continuous bounded Fr´echet derivatives of order keverywhere
on Rn+m, it follows that the function Φ has the same property on Rn+m. By Proposition 1, the
function Stϕhas continuous bounded Fr´echet derivatives of order kon Hfor every k∈N,and
therefore Stϕ∈D.
Suppose now that ϕ∈X, i.e, ϕ∈Cb(H, R) and there is a sequence (ϕj)⊂Dsuch that ϕj→ϕ
uniformly. Let g∈X, i.e., g∈Cb(H, R) and there is a sequence (gj)⊂Dsuch that gj→g
uniformly. Choose a t>0. Denote the operator Stconstructed from gjby the symbol (Sj)t.As
was proved ab ove, (Sj)tϕj∈Dfor every j∈N. We claim that ((Sj)tϕj)(x)→(Stϕ)(x) uniformly
with respect to x∈H, and therefore Stϕ∈X.
Any function in Dhas bounded first derivative, and therefore is uniformly continuous by (7).
Therefore, all functions in Xare uniformly continuous, including ϕ.Sincea→ √ais uniformly
continuous, it follows that so is the function z→ ϕ(z+√2tzy). Further, for any chosen y,wehave
the convergence ϕjx+2tgj(x)y→ϕx+2tg(x)y, which is uniform with respect to x∈H.
There is an index j0such that ϕjx+2tgj(x)y−ϕx+2tg(x)yϕj+ϕ2ϕ+1
for every x∈H, y ∈Hand jj0. Therefore, the following sequence of numerical functions
is well defined, namely, Yj=y→ supx∈Hϕjx+2tgj(x)y−ϕx+2tg(x)y.As was
proved ab ove, Yj(y) converges to 0 pointwise. The functions Yjare jointly bounded, and therefore
HYj(y)μA(dy)→0 by the dominated convergence theorem. To sum up,

(Sj)tϕj−Stϕ
=sup
x∈HH
ϕjx+2tgj(x)y−ϕx+2tg(x)yμA(dy)
H
sup
x∈Hϕjx+2tgj(x)y−ϕx+2tg(x)yμA(dy)=H
Yj(y)μA(dy →0.
Choose some t>0andx∈Hand consider the integral Hϕ(x+y)μ2tg(x)A(dy).
We approximate the integrand by its Taylor polynomial of order three centered at x.Letus
stipulate that the remainder term R(x, y) is not small, since the vector yranges the entire space
H, and the vector xis chosen. However, since ϕis four times continuously Fr´echet differentiable
on H, it follows that the function R(x, y) such that
H
ϕ(x+y)μ2tg(x)A(dy)=Hϕ(x)+ϕ(x),y+1
2! ϕ(x)y, y
+1
3!ϕ(x)(y, y, y)+R(x, y)μ2tg(x)A(dy)
is defined on H×Heverywhere and, by (8), satisfies the bound
|R(x, y)|y4
(4)! sup
z∈[x,x−y]
ϕ(4)(z)
1
4!aϕy4,(19)
where we write aϕ=sup
z∈Hϕ(4)(z)and ag=sup
z∈H|g(z)|.
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 367
The sum can be integrated termwise, because every summand is dominated by a polynomial
with respect to the variable y, which the polynomials are integrable with respect to any Gaussian
measure with trace-class correlation operator ([4, p. 68]).
The integrals against the symmetric measure μ2tg(x)Aof functions ϕ(x),yand
(1/3!)ϕ(x)(y, y, y) that are odd with respect to yvanish. The number ϕ(x) does not depend
on yand μ2tg(x)Ais a probability measure, and therefore the integral of ϕ(x)isequaltoϕ(x). By
(11), H(1/2!)ϕ(x)y, yμ2tg(x)A(dy)=(1/2) tr(2tg(x)Aϕ(x)) = tg(x)tr(Aϕ(x)).Finally,
H
R(x, y)μ2tg(x)A(dy)
(18)
HR(x, 2tg(x)y)μA(dy)
(12)
H
1
4! aϕ2tg(x)4y4μA(dy)
(1/6)t2aϕa2
gHy4μA(dy)(19)
=1
6[aϕa2
g((tr A)2+2trA2)]t2.
This implies that limt→0supx∈Ht−1((Stϕ)(x)−ϕ(x))−(gΔAϕ)(x)= limt→0tsupx∈H|r(x, t)|=0.
Theorem 3. Let g∈X,letgbe bounded and bounded away from zero, and let gbe the uniform
limit of smooth cylindrical functions in D.ThengΔAϕ∈Dfor g∈Dand ϕ∈Dand gΔAϕ∈X
for g∈Xand ϕ∈D. Moreover, the operator λI −gΔA,whereIstands for the identity operator,
is surjective on Dfor any λ>0if g∈D(in particular, (λI −gΔA)(D)=Dis then dense in X),
and, if g∈X, then the operator (gΔA,D)is dissipative and closable, and the closure of (ΔA,D)
is (ΔA,D
1), which is also dissipative.
Proof. Let ϕ∈D, i.e., ϕ(x)=ϕn(x, e1,...,x, en) for some orthonormal family e1,...,e
n
in Hand for some function ϕn:Rn→Rwith bounded derivatives of all orders (see Proposition 1).
This proposition implies the equation
(ΔAϕ)(x)=trAϕ(x)=
n

s=1
n

k=1Aes,e
k∂k∂sϕn(x, e1,...,x, en).(20)
The function on the right-hand side of (20) is cylindrical, and the corresponding function on Rnis
a finite sun of functions each of which has bounded derivatives of all orders (together with ϕn). By
Proposition 1, ΔAϕ∈D.
Since Dis a function algebra, it contains products of its elements. If g∈X, there is a sequence
(gj)⊂Dsuch that g−gj→0; we have gΔAϕ−gjΔAϕΔAϕ·g−gj→0, and hence
gΔAϕ∈X.
Let g∈D.Chooseaλ>0, consider an arbitrary function ϕ∈D, and show that there is a
function f∈Dsolving the equation
λf(x)−g(x)trAf(x)=ϕ(x).(21)
Let vectors e1,...,e
nbe such that
ϕ(x)=ϕn(x, e1,...,x, en)andg(x)=gn(x, e1,...,x, en) (22)
for any x∈H,whereϕn:Rn→Rand gn:Rn→Rare smooth functions bounded together with
all their derivatives. We seek a solution of (21) in the form
f(x)=fn(x, e1,...,x, en),(23)
where fn:Rn→Ris a smooth function bounded together with all its derivatives.
By (20), (22), and (23), equation (21) for the function fof the above form is equivalent to the
following equation for the function fn:
gn(x1,...,x
n)
n

s=1
n

k=1Aes,e
k∂k∂sfn(x1,...,x
n)−λfn(x1,...,x
n)=−ϕn(x1,...,x
n),(24)
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

368 REMIZOV
where xjstands for x, ej.
Equation (24) is a finite-dimensional partial differential equation. Note that gn(x1,...,x
n)
g0≡const >0. Moreover, the quadratic form given by the n×nmatrix (Aes,e
k) is positive
definite because Ais positive. Therefore, the operator in (24) is elliptic, and equation (24) is of
the form (14). The functions gnand ϕnare bounded together with all their derivatives. Hence,
Lemma 2 can be applied to equation (24), and therefore equation (24) has a continuous solution
fnwhich is bounded together with all its derivatives. By Proposition 1, the function fdefined by
(23) is in D.
Thus, for any λ>0, the operator λI −gΔAis surjective on D, because the preimage of a
function ϕ∈Dcontains at least the function f(x)=fn(x, e1,...,x, en), where fn:Rn→Ris
a solution of equation (24).
Suppose now that g∈D. We claim that gΔAis dissipative. Let f∈Dand λ>0. As
above, fand gare cylindrical functions if and only if f(x)=fn(x, e1,...,x, en)andg(x)=
gn(x, e1,...,x, en), where fn:Rn→Rand gn:Rn→Rare smooth functions bounded together
with all their derivatives.
As above, the value of a function obtained by applying gΔAto fat a point x∈His equal to
the value of the function obtained by applying the above finite-dimensional second-order elliptic
differential operator to fnat the point (x, e1,...,x, en)∈Rn. The operator Ais positive
definite, and the function gsatisfies the inequality g(x)g0≡const >0, and therefore Lemma 2
can be applied to the finite-dimensional operator, which turns out to be dissipative, and thus the
operator gΔAis dissipative as well. Formally,
gΔAf−λf=sup
x∈H|g(x)tr(Af (x)) −λf(x)|
sup
(x1,...,xn)∈Rngn(x1,...,x
n)
n

s=1
n

k=1Aes,e
k∂k∂sfn(x1,...,x
n)−λfn(x1,...,x
n)
(15)
λsup
(x1,...,xn)∈Rn|fn(x1,...,x
n)|=λsup
x∈H|f(x)|=λf,
where xjstands for x, ej. The inequality
gΔAf−λfλf(25)
thus obtained means that gΔAis dissipative.
Next, by [23, Proposition II.3.14], it follows from the density of the domain of gΔAand its
dissipativity that gΔAis closable. By setting g(x)≡1, we see that the operator (ΔA,D)isalso
dissipative and closable. It follows from the definition of the space D1that the closure of (ΔA,D)
has the domain D1.
Let g∈X,andlet(gj)⊂Dbe a sequence such that g−gj→0. The images of the operators
(λI −gΔA)and(gΔA−λI) coincide. It suffices to prove that the image of (gΔA−λI)isdense
in D(then it is dense in Xas well). Choose a λ>0 and a function ψ∈D, which we intend to
approximate by values of the operator (gΔA−λI). Since the image of (gjΔA−λI)isD, for any
j∈Nthere is a function fj∈Dsuch that
gjΔAfj−λfj=ψ. (26)
We clai m t h a t , i f j→∞,then gΔAfj−λfj→ψ. To this end, we shall prove first that, for chosen
λ,ψ,andg,thenumbersetsupx∈Htr Af
j(x)j∈Nis bounded. Since g(x)g0≡const >0and
gj→g, there is an index j0such that
gj(x)g0/2forj>j
0.(27)
This gives the following bound which holds for j>j
0:
sup
x∈Htr Af
j(x)(26)
=sup
x∈H
|ψ(x)+λfj(x)|
gj(x)
(27)
2
g0sup
x∈H|ψ(x)|+λsup
x∈H|fj(x)|
(25)
2
g0
(ψ+gjΔAfj−λfj)(26)
=4
g0ψ.
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SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 369
Writing C=maxtr Af 
1,...,tr Af 
j0,4
g0ψ,weseethat
sup
j∈N
sup
x∈Htr Af
j(x)C≡const.(28)
It remains to show that, if j→∞,then gΔAfj−λfj→ψ. However, the following bound holds:
gΔAfj−λfj−ψ=gΔAfj−gjΔAfj+(gjΔAfj−λfj−ψ)
(26)
=
(g−gj)trAf 
j+0

g−gj·tr Af
j
(28)
g−gj·C,
where g−gj→0asj→∞.
Let g∈X,andlet(gj)⊂Dbe as above. Then, as was proved above, in accordance with (33),
the inequality gjΔAf−λfλϕholds for any function ϕ∈Dand every λ>0. Passing to the
limit as j→∞,weobtaingΔAϕ−λϕλϕ,which means that gΔAis a dissipative operator.
By [23, Proposition II.3.14], the densely defined (on D) dissipative operator (gΔA,D)inXadmits
the closure (gΔA,Dom(gΔA)), which is also dissipative.
We claim that Dom(gΔA)=D1. Indeed, let f∈Dom(gΔA). This means that there is a sequence
(fj)⊂Dsuch that fj→fand the sequence gΔAfjconverges. However, since the bound g0
g(x)gholds for any x∈H, it follows that the sequence gΔAfjconverges if and only if the
sequence ΔAfjconverges, which holds if and only if f∈Dom(ΔA). It immediately follows from the
definition of the space D1that Dom(ΔA)=D1. It remains to note that gΔAf=glimj→∞ ΔAfj=
limj→∞ gΔAfj=gΔAf, and therefore gΔA=gΔA.
4. FEYNMAN FORMULA SOLVES THE CAUCHY PROBLEM
Theorem 4. The Cauchy problem (1) has a unique strong solution for every u0∈D1,the
Cauchy problem (1) has a unique mild solution for every u0∈X, these solutions are norm continu-
ous (for the norm in X), uniformly continuous with respect to t∈[0,t
0]for any t0>0, continuously
depend on u0and g, satisfy the bound supx∈H|u(t, x)|supx∈H|u0(x)|for any t0,andare
given by
u(t, x)= lim
n→∞HH
...
HH
n
u0(y1)μy2
2t
ng(y2)A(dy1)μy3
2t
ng(y3)A(dy2)...μ
yn
2t
ng(yn)A(dyn−1)μx
2t
ng(x)A(dyn).
(29)
Proof. Let us verify the validity of the conditions of Theorem 1 for the operator family (St)t0
and the operator gΔA.First,S0=Iby the very definition of the family (St)t0. Second, by
Theorem 2, we have St=1foranyt0, and therefore (St)mSt·...·St=1·...·1=1
for any m∈N. Third, by the same theorem, the strong limit limt→0Stϕ−ϕ
t=gΔAϕexists in Xfor
every function ϕ∈D. Fourth, the space Dis automatically dense in its closure X.Moreover,by
Theorem 3, for every λ>0, the space (λI −gΔA)(D)isdenseinX. Therefore, by Theorem 1, the
closure of (gΔA,D) (coinciding with the operator (gΔA,D
1) by Theorem 3) generates a strongly
continuous semigroup etgΔAdefined for every ϕ∈Xby the rule etgΔAϕ= limn→∞ St/nnϕ, and
the limit is uniform with respect to t∈[0,t
0] for any chosen t0.Notethat,byTheorem3,the
Lumer–Phillips theorem [23, Th. II.3.15] can be applied to the semigroup etgΔA, and hence etgΔA
is a contraction semigroup.
Let us now introduce a function U:[0,+∞)→Xbytherule(U(t))(x)=u(t, x).If we set
(U0)(x)=u0(x),then problem (1) with u0∈D1becomes equivalent to problem (3) with u0∈D1
and problem (1) with u0∈Xto problem (3) with u0∈X. It follows from what was proved above,
from definitions after formula (6), and from Proposition II.6.2 in [23] that, for every U0∈D1,there
is a unique solution of the Cauchy problem (3) given by the formula U(t)=etgΔAu0, and, for every
RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 19 No. 3 2012

370 REMIZOV
U0∈X, there is a unique solution of the Cauchy problem (5) given by the formula U(t)=etgΔAu0.
Since the operator semigroup etgΔAis strongly continuous, it follows that, by Proposition I.1.2
in [23], the function t→ etgΔAϕ(x) is differentiable with respect to tat every point t0 uniformly
with respect to x∈Hfor every function ϕ∈D1.
Thus, the Cauchy problem (6) has a unique solution for every u0∈D1,whichisgivenby
the formula u(x, t)=(etgΔAu0)(x) = limn→∞ St/nnu0(x), and the Cauchy problem (7) has
a unique solution for every u0∈X, which is given by the formula u(x, t)=(etgΔAu0)(x)=
limn→∞ St/nnu0(x).
Let us explain how the formula u(x, t) = limn→∞ St/nnu0(x) leads to formula (37). The
change of variable Hψ(y)μA(dy)=Hψ(y−x)μx
A(dy) holds for every continuous bounded function
ψ:H→Rand every point x∈H. Applying this rule and replacing Aby 2tg(x)A,weobtain
(Stϕ)(x)=H
ϕ(x+y)μ2tg(x)A(dy)=H
ϕ(x+(y−x))μx
2tg(x)A(dy)=H
ϕ(y)μx
2tg(x)A(dy).
Therefore, the expression whose limit is taken in (37) for n=2is
St/22ϕ(x)=St/2St/2ϕ(x)=HH
ϕ(y1)μy2
2t
2g(y2)A(dy1)μx
2t
2g(x)A(dy2).
The expressions for n>2 can be obtained in a similar way. This proves formula (37).
Since the semigroup operators in the semigroup etgΔAare continuous, it follows that the solution
ustrongly continuously (in X) depends on u0. The bound supx∈H|u(t, x)|supx∈H|u0(x)|holds
because etgΔAis a contraction semigroup.
It remains to show that the solution continuously depends on g.Let(gj)⊂Xand gj→g.Then
we automatically have gj(x)const >0 beginning with some index j0. As was proved above, the
operators gjΔAare infinitesimal generators of the semigroups etgjΔA.SincegjΔAϕ→gΔAϕfor
any ϕ∈D, we can apply Theorem III.4.9 in [23] and conclude that the semigroups etgjΔAconverge
strongly (and uniformly with respect to t∈[0,t
0] for any chosen t0>0) to a strongly continuous
semigroup etgΔAwith the infinitesimal generator gΔA, i.e., for every function ϕ∈Xwe have
limj→∞ etgjΔAϕ=etgΔAϕuniformly with respect to t∈[0,t
0] for any chosen t0>0. Therefore,
the solution of the Cauchy problem (6) corresponding to gjconverges as j→∞to the solution of
the Cauchy problem (6) corresponding to guniformly with respect to x∈Hand uniformly with
respect to t∈[0,t
0] for any chosen t0>0.
Remark 1. The functions u0and gbelong to the space X, and therefore can be approximated
uniformly by cylindrical functions. Substituting cylindrical approximations (u0)jand gjinto the
Feynman formula (29) (instead of the functions u0and gthemselves), then the integrands become
cylindrical functions, and therefore the integral over the infinite-dimensional space Hcanthenbe
replaced by the integral over a finite-dimensional subspace of Hby (17). The passage to the limit
as j→+∞and n→+∞gives the function u(t, x) again. This means that we obtain a sequence
of cylindrical functions uniformly approximating the function u, and the functions in the sequence
can be obtained as integrals of finite multiplicity over finite-dimensional spaces.
5. HOW LARGE ARE FUNCTION CLASSES D1AND X?
Proposition 2. Let αk:R→Rbe a family of infinitely smooth functions that are uniformly
bounded together with their first and second derivatives, supp∈{0,1,2}supk∈Nsupt∈Rdpαk(t)/dtp
M≡const .For example, the functions αk(t)=sin(dk(t−tk)) and αk(t)=exp−dk(t−tk)2,where
dkand tkare constants, 0<d
k1, satisfy these conditions. Let a number series ∞
k=1 bkconverge
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SOLUTION OF A CAUCHY PROBLEM FOR A DIFFUSION EQUATION 371
absolutely. Let (ek)∞
k=1 be an orthonormal basis in H. Then the function f(x)=∞
k=1 bkαk(x, ek)
belongs to the class D1.
Proof. The sequence of cylindrical functions fj(x)=j
k=1 bkαk(x, ek) converges uniformly
to f. By Proposition 1, the sequence tr Af 
j(x)isoftheform
tr Af
j(x)=
j

k=1Aek,e
kbkα(x, ek),
and it converges as j→∞to the function ∞
k=1Aek,e
kbkα(x, ek), because

n2

k=n1
Aek,e
kbkα(x, ek)AM
n2

k=n1
|bk|,
and the series ∞
k=1 bkconverges absolutely.
Proposition 3. A nonconstant function in Xcannot have any limit at infinity. In particular,
the function x→ exp(−x2)belongs to Cb(H, R)but not to X.
Proof. Let f∈D. Then there is an n-dimensional subspace Hn⊂Hsuch that f(x)=f(Px)
for every x∈H,whereP:H→Hnis the orthogonal projection. Since f= const, there is a number
ε0>0andpointsx1,x
2∈Hnsuch that |f(x1)−f(x2)|>ε
0.Then|f(x1+y)−f(x2+y)|>ε
0
for every y∈ker P, and, in particular, for x1+yRand x2+yR, which contradicts the
existence of a limit of fat infinity.
Suppose now that f∈X. Then there is a sequence of functions (fj)⊂Dwhich converges
uniformly to f. There is an index jsuch that f−fj<ε0
8. Therefore, |fj(x1)−fj(x2)|>3ε0
4
and |fj(x1+y)−fj(x2+y)|>3ε0
4for every y∈ker Pand for two points x1,x
2of the space Hn
constructed from the function fj.Sincef−fj<ε0
8, it follows that |f(x1+y)−f(x2+y)|>ε
0/2,
which contradicts the existence of a limit at infinity.
Remark 2. Recall that, by the proof of Theorem 2, any function f∈Xis uniformly continuous.
Acknowledgments
The author expresses his gratitude to O. G. Smolyanov for setting the problem and for the interest
to the research, and also to Ya. M. Butko-Kinderknecht, D. V. Turaev, and N. N. Shamarov for
useful discussions and remarks concerning the manuscript.
The research was supported by the Russian Foundation for Basic Research under grant no. 10-
01-00724, by the Federal Targeted Program “Scientific and Scientific-Pedagogical Personnel of the
Innovative Russia” for 2009–2013, and also by a grant of the Government of the Russian Federation
according to the Resolution no. 220 “On Measures to Attract Leading Scientists to Russian Higher
Professional Education Institutions” by the agreement no. 11.G34.31.0054.
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