variable order subordination in the sense of bochner and pseudo-differential operators

Math. Nachr. 284, No. 8–9, 987 – 1002 (2011) / DOI 10.1002/mana.200810246

Variable order subordination in the sense of Bochnerand pseudo-differential operators

Kristian Evans∗ and Niels Jacob∗∗

Department of Mathematics, Swansea University, Swansea SA2 8PP, UK

Received 3 November 2008, revised 30 March 2009, accepted 1 May 2009Published online 9 May 2011

Key words Subordination in the sense of Bochner, pseudo-differential operators of variable order, Fellersemigroups

MSC (2010) 47006, 47007, 47G30, 35S05, 60J35

For a large class of pseudo-differential operators with a negative definite symbol q(x, ξ) in the sense of Hohand for a large family of x-dependent Bernstein functions f (x, ·) we prove that the pseudo-differential operatorwith symbol −f (x, q(x, ξ)) has an extension generating a Feller semigroup.

c© 2011 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

The main purpose of this note is to investigate subordination (in the sense of Bochner) of variable order.Bochner introduced a method called subordination which was used to obtain a new process from a given oneby a random time change. His original papers are [4] and [5]. We will however, study subordination using an an-alytic approach. There is a long history of constructing a functional calculus for generators of subordinate semi-groups, with the first general results obtained by Phillips [32]. Many calculi for this topic have been proposed,however, we only mention the paper of Berg, Boyadzhiev and de Laubenfels [3], Faraut [9], the monograph ofde Laubenfels [6] and the papers of Schilling [33] and [34]. We should note that Hirsch [13]–[14] had obtainedrelated results prior to this. The representation of fractional powers of generators is the best known result, this isdue to Balakrishnan [1], see also Yosida [38], Krasnosel’skii et al. [24] and Nollau [30]–[31]. At the root of ourwork is the result that for a continuous negative definite function ψ and a Bernstein function f , f ◦ ψ is also acontinuous negative definite function.

Subordination has also been studied on the level of pseudo-differential operators. In particular, in the casewhere −q(x,D) generates a Feller or sub-Markovian semigroup, in many cases −f(q(x, ξ)), f being a Bernsteinfunction, is also the symbol of a pseudo-differential operator generating a Feller or sub-Markovian semigrouprespectively.

This now leads us to the next step, i.e., subordination of variable order. By subordination of variable orderwe mean the case when we replace a fixed Bernstein function f by a family of Bernstein functions, f(x, ·)depending on x. Pseudo-differential operators of variable order of differentiation have already been studied byUnterberger and Bokobza [37], and in particular by Leopold [25], [26]. Feller semigroups obtained from the

symbol(1 + |ξ|2

)r(x)have been studied by Jacob and Leopold [21], where further work is due to Negoro [29],

in particular to Kikuchi and Negoro [22], [23]. It should also be noted that a Weyl-Hormander calculus can beused to consider operators of variable order of differentiation, see Baldus [2]. Moreover, Hoh [17] has shown thatwhen

f(x, q(x, ξ)) = (q(x, ξ))m (x)

∗ e-mail: [email protected], Phone: +44 1792 292905, Fax: +44 1792 295843∗∗ Corresponding author: e-mail: [email protected], Phone: +44 1792 295461, Fax: +44 1792 295843

c© 2011 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

988 K. Evans and N. Jacob: Variable order subordination

where 0 ≤ m(x) ≤ 1, then under certain conditions the pseudo-differential operator

−p(x,D)u = −(2π)−n2

∫Rn

eix·ξ (q(x, ξ))m (x) u(ξ) dξ

extends to the generator of a Feller semigroup. An approach using Dirichlet forms was recently proposed byUemura [35], [36].

The aim of this paper is to extend these ideas; we want to enlarge the class of examples and obtain a generalproof showing that the pseudo-differential −p(x,D) with symbol −f(x, q(x, ξ)) extends to the generator of aFeller semigroup, see our main result, Theorem 3.10. All our notations are standard and if not explained in thetext they can be found in [18]–[20].

2 Hoh’s symbolic calculus

Before starting with our main considerations we need to recollect some basic results from Hoh’s symbolic calcu-lus, see Hoh [15] or [16], compare also [19].

Definition 2.1 A continuous negative definite function ψ : Rn → R belongs to the class Λ if for all α ∈ N

n0

it satisfies ∣∣∂αξ (1 + ψ(ξ))

∣∣ ≤ c|α |(1 + ψ(ξ))2−ρ ( |α |)

2 , (2.1)

where ρ(k) = k ∧ 2 for k ∈ Nn0 .

Definition 2.2 A. Let m ∈ R and let ψ ∈ Λ. We then call a C∞-function q : Rn × R

n → C a symbol in theclass Sm,ψ

ρ (Rn ) if for all α, β ∈ Nn0 there are constants cα,β ≥ 0 such that

∣∣∂βx ∂α

ξ q(x, ξ)∣∣ ≤ cα,β (1 + ψ(ξ))

m −ρ ( |α |)2 (2.2)

holds for all x ∈ Rn and ξ ∈ R

n . We call m ∈ R the order of the symbol q(x, ξ).B. Let ψ ∈ Λ and suppose that for an arbitrarily often differentiable function q : R

n × Rn → C the estimate∣∣∂α

ξ ∂βx q(x, ξ)

∣∣ ≤ cα,β (1 + ψ(ξ))m2 (2.3)

holds for all α, β ∈ Nn0 and x, ξ ∈ R

n . In this case we call q a symbol of the class Sm,ψ0 (Rn ).

Note that Sm,ψρ (Rn ) ⊂ Sm,ψ

0 (Rn ). For q ∈ Sm,ψ0 (Rn ), hence also for q ∈ Sm,ψ

ρ (Rn ), we can define onS(Rn ) the pseudo-differential operator q(x,D) by

q(x,D)u(x) := (2π)−n2

∫Rn

eix·ξ q(x, ξ)u(ξ) dξ (2.4)

and we denote the classes of these operators by Ψm,ψρ (Rn ) and Ψm,ψ

0 (Rn ), respectively.

Theorem 2.3 If q ∈ Sm,ψ0 (Rn ) then q(x,D) maps S(Rn ) continuously into itself.

Let ψ : Rn → R be a fixed continuous negative definite function. For s ∈ R and u ∈ S(Rn ) (or u ∈ S′(Rn ))

we define the norm

‖u‖2ψ ,s =

∥∥(1 + ψ(D))s2 u∥∥2

0 =∫

Rn

(1 + ψ(s))s |u(ξ)|2dξ, (2.5)

where || · ||0 denotes the norm in L2(Rn ).The space Hψ,s(Rn ) is defined as

Hψ,s(Rn ) := {u ∈ S′(Rn ); ‖u‖ψ ,s < ∞}. (2.6)

The scale Hψ,s(Rn ), s ∈ Rn , and more general spaces have been systematically investigated in [10] and [11],

see also [19]. In particular we know that if for some ρ1 > 0 and c1 > 0 the estimate ψ(ξ) ≥ c1 |ξ|ρ1 holds for allξ ∈ R

n , |ξ| ≥ R, R ≥ 0, then the space Hψ,s(Rn ) is continuously embedded into C∞(Rn ) provided s > nρ1

.

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Math. Nachr. 284, No. 8–9 (2011) / www.mn-journal.com 989

The following result is due to Hoh [15] and is of much importance, compare also [19], Theorem 2.4.23.A.

Theorem 2.4 Let ψ ∈ Λ. For q1 ∈ Sm 1 ,ψρ (Rn ) and q2 ∈ Sm 2 ,ψ

ρ (Rn ) the symbol q of the operatorq(x,D) := q1(x,D) ◦ q2(x,D) is given by

q(x, ξ) = q1(x, ξ) · q2(x, ξ) +n∑

j=1

∂ξjq1(x, ξ)Dxj

q2(x, ξ) + qr1 (x, ξ) (2.7)

with qr1 ∈ Sm 1 +m 2 −2,ψ0 (Rn ).

Remark 2.5 An easy calculation yields q1 · q2 ∈ Sm 1 +m 2 ,ψρ (Rn ), ∂ξj

q1 ∈ Sm 1 −1,ψρ (Rn ) and

Dxjq2 ∈ Sm 2 ,ψ

ρ (Rn ). Hence the second term on the right-hand side in (2.7) belongs to Sm 1 +m 2 −1,ψρ (Rn ).

On S(Rn ) we may define the bilinear form

B(u, v) := (q(x,D)u, v)0 , q ∈ Sm,ψρ (Rn ). (2.8)

Now using Hoh’s symbolic calculus we obtain estimates for the bilinear form, compare again Hoh [15] or see[19].

Theorem 2.6 Let q ∈ Sm,ψρ (Rn ) be real valued and m > 0. It follows that

|B(u, v)| ≤ c ‖u‖ψ , m2‖v‖ψ , m

2(2.9)

holds for all u, v ∈ S(Rn ). Hence the bilinear form B has a continuous extension onto Hψ, m2 (Rn ). If in addition

for all x ∈ Rn

q(x, ξ) ≥ δ0(1 + ψ(ξ))m2 for |ξ| ≥ R (2.10)

with some δ0 > 0 and R ≥ 0, and

lim|ξ |→∞

ψ(ξ) = ∞ (2.11)

holds, then we have for all u ∈ Hψ, m2 (Rn ) the Garding inequality

ReB(u, u) ≥ δ0

2||u||2ψ , m

2− λ0 ||u||20 . (2.12)

3 Subordination of variable order

The method discussed in this section improves the ideas of [8]. Let f : Rn × (0,∞) → R be an arbitrarily often

differentiable function such that for y ∈ Rn fixed the function s → f(y, s) is a Bernstein function. We assume

that with some 0 < r1 ≤ 1 we have

supy∈Rn

f(y, s) ≤ c1sr1 for s ≥ γ0 (3.1)

as well as for some 0 < r0 such that r0 < r1 it holds

infy∈Rn

f(y, s) ≥ c2sr0 for s ≥ γ0 . (3.2)

In our applications we will consider symbols f(x, q(x, ξ)) where q(x, ξ) ≥ λ0(1 + ψ(ξ)) for some real-valuedcontinuous negative definite function ψ. Thus we can always confine ourselves to the case where γ0 ≥ 1. Considerthe negative definite symbol

p(x, ξ) = f(x, q(x, ξ)) (3.3)

where the symbol q(x, ξ) is comparable with a fixed continuous negative definite function ψ satisfyinglimξ→∞ ψ(ξ) = ∞ , i.e.,

0 < c3 ≤ q(x, ξ)ψ(ξ)

≤ c4 , (3.4)

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for all x ∈ Rn and ξ ∈ R

n . Note that the lower bounds imply ψ(ξ) > 0. Since ξ → ψ(ξ) − ψ(0) is also acontinuous negative definite function, the lower bound in (3.4) corresponds to an estimate c(1 + ψ(ξ)) ≤ q(x, ξ)for ψ being a continuous negative function which might have a zero. We find using Lemma 3.9.34.B in [18]

p(x, ξ) = f(x, q(x, ξ)) ≤ c1(1 + ψ(ξ))r1 (3.5)

and

p(x, ξ) ≥ c2(1 + ψ(ξ))r0 (3.6)

i.e., p(x, ξ) is bounded above and below by continuous negative definite functions.Under suitable assumptions on f , see below, the pseudo-differential operator

p(x,D)u(x) = (2π)−n2

∫Rn

eix·ξ p(x, ξ)u(ξ) dξ

= (2π)−n2

∫Rn

eix·ξ f(x, q(x, ξ))u(ξ) dξ (3.7)

has a symbol p ∈ S2r1 +2ε,ψρ (Rn ). The following section gives a detailed proof of this result.

3.1 Estimates for p(x, ξ)

Let f : Rn × (0,∞) → R be an arbitrarily often differentiable function such that for each x ∈ R

n the functionf(x, ·) : (0,∞) → R is a Bernstein function. For every Bernstein function h : (0,∞) → R the estimates

∣∣h(k)(s)∣∣ ≤ k!

skh(s), s > 0 and k ∈ N0 (3.8)

hold, compare [18], Lemma 3.9.34.D. Hence for f as above we find

∣∣f (k)(x, s)∣∣ ≤ k!

skf(x, s), s > 0, x ∈ R

n and k ∈ N0 , (3.9)

where

f (k)(x, s) =∂kf(x, s)

∂sk.

We assume now in addition:There exists η > 0 and δ0 > 0 such that for ε ∈ (0, η) and for all s ≥ δ0 it follows that

∣∣∂αx ∂k

s f(x, s)∣∣ ≤ cα,k,ε

1sk

f(x, s)sε (3.10)

holds for all x ∈ Rn and s ≥ δ0 with cα,k,ε independent of x and s.

Example 3.1 (Compare Hoh [17]) Consider f(x, s) = sm (x) for 0 < m ≤ m(x) ≤ M < 1. It follows that

∂ks sm (x) = Pk (m(x))

1sk

sm (x) (3.11)

where Pk (t) is a polynomial of degree less or equal to k. If we assume in addition that m(·) ∈ C∞(Rn ) and|∂αm(x)| ≤ mα for all α ∈ N

n0 we find using (3.11) that

∂αx ∂k

s sm (x) = ∂αx

(Pk (m(x))

1sk

sm (x))

=1sk

∑β≤α

(αβ

)∂α−β Pk (m(x))

∑β1 +···+βl ′=β

l′=1,...,|β |

c{βj }

l∏j=1

(∂βj (m(x) ln s)

)sm (x) .



Thus we arrive at

∣∣∂αx ∂k

s sm (x)∣∣ ≤ cα,k,m

Q(ln s)sk

sm (x)

provided s ≥ 1 (otherwise, if s ≥ δ0 > 0, we need to treat the terms involving ln s a bit differently, for examplewe may switch to | ln s|) where Q is a suitable polynomial. Since for ε > 0 we find a constant cε such thatQ(ln s) ≤ cεs

ε holds, we arrive at (3.10).

Example 3.2 Consider f(x, s) = sm (x )

2

(1− e−4s

m (x )2)

where m : Rn → R is arbitrarily often differentiable

such that 0 < m ≤ m(x) ≤ M < 2 and∣∣∂α

x m(x)∣∣ ≤ mα . A straight forward calculation, compare [7] shows

that for all α ∈ Nn0 and k ∈ N0 there exists ε > 0, 0 < ε < η, such that∣∣∣∣∂α

x ∂ks s

m (x )2

(1 − e−4s

m (x )2

)∣∣∣∣ ≤ cα,k,ε1sk

sm (x )+ ε

2

(1 − e−4s

m (x )2

). (3.12)

We now return to the general case. It should be noted here that due to their length calculations may have to besplit over many lines. To avoid any confusion if a calculation is written as∑

×∑

it means ∑∑and not (∑)(∑)

.

Eventually we need various controls on the symbol

(x, ξ) −→ f(x, q(x, ξ))

where q(x, ξ) comes from a certain symbol class which we will fix later. For this we use a formula to calculatehigher order derivatives of composed functions which is due to Fraenkel [12], compare also [18], p. 15.

Let u : Rm → C and vj : R

n → R, j = 1, . . . ,m, be smooth functions. Then for α ∈ Nn0 it holds with

v = (v1 , . . . vm )

∂αu(v(x)) = ∂αu(v1(x), . . . vm (x)) (3.13)

=∑

1≤|σ |≤|α |σ∈N

m0

(∂σu)(v(x))σ!

∑γ 1 +···+γ m =α

γ j ∈Nn0

Pγ 1 (σ1 , v1 ;x) · . . . · Pγ m (σm , vm ;x)

where for γ ∈ Nn0

Pγ (N, v;x) :=∑

ρ∈R(γ ,N )

N !ρ!

(∂β (1)v(x)

β(1)!

)ρ1

· . . . ·(

∂β (r)v(x)β(r)!

)ρr

(3.14)

with

R(γ,N) :=

⎧⎨⎩ρ ∈ N

r0

∣∣∣∣∣∣r∑

j=1

ρjβ(j) = γ and |ρ| = N

⎫⎬⎭ , (3.15)

Nn0,γ :=

{β ∈ N

n0 | 0 < β ≤ γ

}(3.16)



and with |Nn0,γ | = r an enumeration of N

n0,γ is given by β(1), . . . , β(r). In our concrete problem many reductions

happen.We consider first f : R

n × (0,∞) → R artificial as f : Rn × R

n × (0,∞) → R by setting f(x, s) =f(x1 , . . . , xn , 1, . . . , 1, s). Next we introduce the 2n + 1 functions

vj (x, ξ) =

⎧⎨⎩

xj , 1 ≤ j ≤ n,1, n + 1 ≤ j ≤ 2n,q(x, ξ), j = 2n + 1.

(3.17)

In the following multi-indices in N2n0 will be split as α =

(α(1) , α(2)

)where α(1) acts on the x-variables and

α(2) acts on the ξ variables. Our problem is to estimate

∂βx ∂α

ξ f(v1(x, ξ), . . . , v2n+1(x, ξ))

= ∂βx ∂α

ξ f(x1 , . . . , xn , 1, . . . , 1, q(x1 , . . . , xn , ξ1 , . . . ξn ))

=∑

1≤|σ |≤|α |+|β |σ∈N

2 n + 10

(∂σf)(v(x, ξ))σ!

×∑

γ 1 +···+γ 2 n + 1 =ω

γ j ∈N2 n0

Pγ 1 (σ1 , v1 ;x, ξ) · . . . · Pγ 2 n + 1 (σ2n+1 , v2n+1;x, ξ) (3.18)

where ω = (β, α). If γj =(δj1 , δ

j2

)then

N2n0,γ j =

{(ζ, τ) ∈ N

2n0 | |ζ| + |τ | > 0 and 0 < ζ ≤ δj

1 , 0 < τ ≤ δj2

}.

Let an enumeration of N2n0,γ j : η(1) = (ζ(1), τ(1)), . . . , η(rj ) = (ζ(rj ), τ(rj )) where rj = rj (γj ) be given.

Then we have with σ = (σ1 , . . . , σ2n+1)

R(γj , σj

)=

{ρ ∈ N

rj

0

∣∣∣∣∣rj∑

l=1

ρlη(l) = γj and |ρ| = σj

}

=

{ρ ∈ N

rj

0

∣∣∣∣∣rj∑

l=1

ρl(ζ(l), τ(l)) = γj and |ρ| = σj

}

=

{ρ ∈ N

rj

0

∣∣∣∣∣rj∑

l=1

ρlζ(l) = δj1 ,

rj∑l=1

ρlτ(l) = δj2 and |ρ| = σj

}

and

Pγ j (σj , vj , x, ξ) =∑

ρ∈R(γ j ,σj )

σj !ρ!

(∂

ζ (1)x ∂

τ (1)ξ vj (x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂

τ (rj )ξ vj (x, ξ)

ζ(rj )!τ(rj )!

)ρr j

where σ ∈ N2n+10 is such that σ = (σx, σξ , σ2n+1), σx, σξ ∈ N

n0 and σ2n+1 ∈ N0 .

When σξ �= 0 then ∂σf = 0 therefore∑1≤|σ |≤|α |+|β |

σ∈N2 n + 10

reduces to ∑1≤|σx |+σ2 n + 1 ≤|α |+|β |

σ∈N2 n + 10 , σξ =0

Consider Pγ j (σj , vj , x, ξ)



(i) 1 ≤ j ≤ n implies

∂ζ (k)x ∂

τ (k)ξ vj (x, ξ) = ∂ζ (k)

x ∂τ (k)ξ xj =

{1, ζ(k) = εj , τ(k) = 00, otherwise

therefore for 1 ≤ j ≤ n

Pγj(σj , vj , x, ξ) =

∑ρ∈R(γ j ,σj )

σj !ρ!

(∂

ζ (1)x ∂

τ (1)ξ vj (x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂


ζ(rj )!τ(rj )!

)ρr j

=∑

ρ∈{

ρ′∈Nr j0

∣∣∑ r jl = 1 ρ′

l ζ (l)=σj and ζ (l)=εj} σj !

ρ!

(∂

ζ (1)x ∂

τ (1)ξ vj (x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂


ζ(rj )!τ(rj )!

)ρr j

,

i.e., in this case Pγj(σj , vj , x, ξ) = cj,σ ;

(ii) n + 1 ≤ j ≤ 2n implies

∂ζ (k)x ∂

τ (k)ξ vj (x, ξ) = 0 whenever ζ(k) �= 0 or τ(k) �= 0, i.e., (ζ(k), τ(k)) �= 0,

i.e., for n + 1 ≤ j ≤ 2n

Pγj(σj , vj , x, ξ) =

∑ρ∈R(γ j ,σj )

σj !ρ!

(∂

ζ (1)x ∂

τ (1)ξ vj (x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂


ζ(rj )!τ(rj )!

)ρr j

= 0;

(iii) finally let j = 2n + 1 and set r = rj , i.e.,

R(γ2n+1 , σ2n+1

)=

{ρ ∈ N

r0

∣∣∣∣∣r∑

l=1

ρl(ζ(l), τ(l)) = γ2n+1 , |ρ| = σ2n+1

}

then

Pγ 2 n + 1 (σ2n+1 , v2n+1 , x, ξ) = Pγ 2 n + 1 (σ2n+1 , q(x, ξ))

=∑

ρ∈R(γ 2 n + 1 ,σ2 n + 1 )

σ2n+1!ρ!

(∂

ζ (1)x ∂

τ (1)ξ q(x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂

τ (rj )ξ q(x, ξ)

ζ(rj )!τ(rj )!

)ρr j

.

We observe that

(∂σf)(x1 , . . . xn , s) = 0 if σξ �= 0 ∈ Nn0 .



Thus we find using the previous calculations and (3.18)

∂βx ∂α

ξ f(x, q(x, ξ)) =∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))

×∑

γ 1 +···+γ 2 n + 1 =(β ,α)

γ j ∈N2 n0

Pγ 1 (σ1 , x1) · . . . · Pγ 2 n + 1 (σ2n+1 , q(x, ξ))

=∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))

×∑

δ 11 +···+δ 2 n + 1

1 =β

δ 12 +···+δ 2 n + 1

2 =α

δjk ∈N

n0

P(δ 11 ,δ 1

2 )(σ1 , x1) · . . . · P(δ 2 n + 11 ,δ 2 n + 1

2 )(σ2n+1 , q(x, ξ))

=∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δ 12 +···+δn

2 +δ 2 n + 12 =α

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

P(δ 11 ,δ 1

2 )(σ1 , x1) · . . . · P(δ 2 n + 11 ,δ 2 n + 1

2 )(σ2n+1 , q(x, ξ))

=∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δ 2 n + 12 =α

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

P(δ 11 ,δ 1

2 )(σ1 , x1) · . . . · P(δ 2 n + 11 ,δ 2 n + 1

2 )(σ2n+1 , q(x, ξ))

=∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δ 2 n + 12 =α

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

C(δ11 , . . . , δn

1 , δ12 , . . . , δn

2)P(δ 2 n + 1

1 ,α)(σ2n+1 , q(x, ξ))

=∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

(∂σf

σ!

)(x, q(x, ξ))



×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

C(δ11 , . . . , δn

1)

×∑

ρ∈R((δ 2 n + 11 ,α),σ2 n + 1 )

σ2n+1!ρ!

(∂

ζ (1)x ∂

τ (1)ξ q(x, ξ)

ζ(1)!τ(1)!

)ρ1

· . . . ·(

∂ζ (rj )x ∂

τ (rj )ξ q(x, ξ)

ζ(rj )!τ(rj )!

)ρr 2 n + 1

.

In order to estimate ∂βx ∂α

ξ f(x, q(x, ξ)) we assume further∣∣∂ζx∂τ

ξ q(x, ξ)∣∣ ≤ cζ τ (1 + ψ(ξ))

2−ρ ( |τ |)2 (3.19)

and the ellipticity condition

q(x, ξ) ≥ γ0(1 + ψ(ξ)). (3.20)

Taking (3.10) into account we find for every ε > 0 but sufficiently small that∣∣∂βx ∂α

ξ f(x, q(x, ξ))∣∣ ≤ C ′

α,β ,ε

∑1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

1q(x, ξ)σ2 n + 1

f(x, q(x, ξ))q(x, ξ)ε

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

×∑

ρ∈R((δ 2 n + 11 ,α),σ2 n + 1 )

(1 + ψ(ξ))(2−( 2∧τ ( 1 ) ) )

2 ρ1 · . . . · (1 + ψ(ξ))(2−( 2∧τ ( r ) ) )

2 ρr 2 n + 1

≤ C ′α,β ,ε

∑1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

1(1 + ψ(ξ))σ2 n + 1

f(x, q(x, ξ))q(x, ξ)ε

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

×∑

ρ∈R((δ 2 n + 11 ,α),σ2 n + 1 )

(1 + ψ(ξ))ρ1 +···+ρr 2 n + 1 (1 + ψ(ξ))−( (2∧τ ( 1 ) )2 ρ1 +···+ (2∧τ ( r ) )

2 ρr 2 n + 1 ).

Since for ρ ∈ R((

δ2n+11 , δ2n+1

2

), σ2n+1

), it follows that ρ1 + · · · + ρr2 n + 1 = σ2n+1 and we arrive at∣∣∂β

x ∂αξ f(x, q(x, ξ))

∣∣ ≤ Cα,β ,εf(x, q(x, ξ))q(x, ξ)ε∑

1≤|σx |+σ2 n + 1 ≤|α |+|β |σ=(σx ,0,σ2 n + 1 )∈N

2 n + 10

×∑

δ 11 +···+δn

1 +δ 2 n + 11 =β

δjk ∈N

n0 ,δ j

k =0, n+1≤j≤2n

δj1 ∈{0,εj }, 1≤j≤n, δ j

2 =0, 1≤j≤n

×∑

ρ∈R((δ 2 n + 11 ,α),σ2 n + 1 )

(1 + ψ(ξ))−12 ((2∧τ (1))ρ1 +···+(2∧τ (r))ρr ) .



Denote the last sum by UR((δ 2 n + 11 ,δ 2 n + 1

2 ),σ2 n + 1 ), i.e.,

U(ψ )R((δ 2 n + 1

1 ,δ 2 n + 12 ),σ2 n + 1 ) =

∑ρ∈R((γ 2 n + 1

1 ,γ 2 n + 12 )σ2 n + 1 )

(1 + ψ(ξ))−12 ((2∧τ (1))ρ1 +···+(2∧τ (r))ρr ) .

If |α| = 1 we get a contribution from R((

δ2n+11 , δ2n+1

2

), 1)

of the type |τ(l)| = 1, ρl = 1, and we have theestimate ∣∣∣∣U (ψ )

R((δ 2 n + 11 ,δ 2 n + 1

2 ),1)

∣∣∣∣ ≤ (1 + ψ(ξ))−12 = (1 + ψ(ξ))−

12 (2∧|α |) .

If |α| = 2 we get at least one contribution from R((

δ2n+11 , δ2n+1

2

), 2)

of the type |τ(l)| = 2, ρl = 1 or |τ(l)| =|τ(k)| = 1 and ρl = ρk = 1, l �= k, hence we get that the estimate∣∣∣∣U (ψ )

R((δ 2 n + 11 ,δ 2 n + 1

2 ),2)

∣∣∣∣ ≤ (1 + ψ(ξ))−1 = (1 + ψ(ξ))−12 (2∧|α |)

holds. Finally, for |α| ≥ 3 we find the estimate by analysing the possible terms in R((

γ2n+11 , γ2n+1

2

)σ2n+1

)for

σ2n+1 ≥ 3 ∣∣∣∣U (ψ )R((δ 2 n + 1

1 ,δ 2 n + 12 ),k)

∣∣∣∣ ≤ (1 + ψ(ξ))−12 (2∧|α |)

with k ≥ 3. Thus we have proved

Theorem 3.3 Suppose that f : Rn×(0,∞) → R is arbitrarily often differentiable and that f(x, ·) : (0,∞) →

R is a Bernstein function. Suppose further that (3.10) holds. Let q : Rn × R

n → R be an arbitrarily oftendifferentiable function satisfying with a fixed continuous negative definite function ψ : R

n → R conditions (3.19)and (3.20). Then for every ε > 0 sufficiently small and all α, β ∈ N

n0 it holds∣∣∂β

x ∂αξ f(x, q(x, ξ))

∣∣ ≤ Cα,β ,εf(x, q(x, ξ))q(x, ξ)ε(1 + ψ(ξ))−12 ρ(|α |) . (3.21)

Now, Theorem 3.3 implies together with Theorem 2.5.4 in [19] and (3.1) that p ∈ S2r1 +2ε,ψρ (Rn ) and that

p(x,D) maps the space Hψ,2r1 +2ε+s(Rn ) to the space Hψ,s(Rn ) i.e.,

||p(x,D)u||ψ ,s ≤ c ||u||ψ ,2r1 +2ε+s .

In particular p(x,D) is continuous from Hψ,s(Rn ) to Hψ,s−2r1 −2ε(Rn ) i.e.,

||p(x,D)u||ψ ,s−2r1 −2ε ≤ c′||u||ψ ,s .

We now consider the bilinear form

B(u, v) := (p(x,D)u, v)0 , u, v ∈ S(Rn ).

Since p ∈ S2r1 +2ε,ψρ (Rn ) we may apply Theorem 2.6 to get

|B(u, v)| ≤ κ||u||ψ ,r1 +ε ||v||ψ ,r1 +ε , (3.22)

for some κ > 0 and all u, v ∈ S(Rn ) i.e., B has a continuous extension onto Hψ,r1 +ε(Rn ) again denoted by B.Furthermore we have

Proposition 3.4 For u ∈ Hψ,2r1 +2ε(Rn ), r1 + ε − 12 < r0 , we have the Garding inequality

B(u, u) ≥ δ1 ||u||2ψ ,r0− λ0 ||u||20 , (3.23)

for some λ0 ≥ 0.



P r o o f. We have the lower bound

p(x, ξ) ≥ c(1 + ψ(ξ))r0 .

This implies

r(x, ξ) = p(x, ξ) − c(1 + ψ(ξ))r0 ≥ 0.

Now Theorem 2.5.5 in [19], which is due to Hoh, see [15], gives

(r(x,D)u, u)0 ≥ −k||u||22 r 1 + 2 ε−12

(3.24)

since (r(x,D)u, u)0 is real-valued, where

(r(x,D)u, u)0 = ReB(u, u) − c||u||2ψ ,r0. (3.25)

Putting (3.24) and (3.25) together we get

ReB(u, u) ≥ c ||u||2ψ ,r0− k||u||22 r 1 + 2 ε−1

2.

Under the assumption that

ψ(ξ) ≥ c0 |ξ|ρ0 (3.26)

and r1 + ε − 12 < r0 we get for every ε0 > 0

(1 + ψ(ξ))2 r 1 + 2 ε−1

2 ≤ ε20(1 + ψ(ξ))r0 + c2(ε0)

which leads to

||u||22 r 1 + 2 ε−12

≤ ε0 ||u||2ψ ,r0+ c(ε0)||u||20

implying the result.

We are now dealing with the space Hψ,r0 (Rn ) and the space Hψ,r1 +ε(Rn ) which is the smaller of thetwo. Since our estimates for B are in different spaces we seek to introduce an intermediate space, compareLouhivaara and Simader [27] and [28].

Firstly, we consider the symmetric part B of B i.e.,

B(u, v) =12(B(u, v) + B(v, u)),

on Hψ,r1 +ε(Rn ). Then

Bλ0 (u, v) := B(u, v) + λ0(u, v)0 , (3.27)

We have

|Bλ0 (u, v)| ≤ κ||u||ψ ,r1 +ε ||v||ψ ,r1 +ε

and

Bλ0 (u, u) ≥ γ||u||2ψ ,r0.

Since Bλ0 (u, v) is a scalar product on Hψ,r1 +ε(Rn ) we may consider the completion of Hψ,r1 +ε(Rn ) withrespect to Bλ0 (u, v). We denote this new intermediate space by Hpλ 0 (Rn ) . Clearly we have

Hψ,r1 +ε(Rn ) ↪→ Hpλ 0 (Rn ) ↪→ Hψ,r0 (Rn ) (3.28)

in the sense of continuous embeddings.



Lemma 3.5 The bilinear form Bλ0 is continuous on Hpλ 0 (Rn ), provided r1 + ε − 12 < r0 .

P r o o f. We find by using Corollary 2.4.23 in [19] that

12(pλ0 (x,D) + p∗λ0

(x,D)) =12(pλ0 (x,D) + pλ0 (x,D)) + r1(x,D)

= pλ0 (x,D) + r1(x,D)

where r1 ∈ S2r1 +2ε−1,ψρ (Rn ) and we used that p(x, ξ) is real-valued. Consider

|Bλ0 (u, v)| = |(pλ0 (x,D))u, v)0 |

≤ 12|((pλ0 (x,D) + p∗λ0

(x,D))u, v)0 | + |(r1(x,D)u, v)0 |

= |Bλ0 (u, v)| + |(r1(x,D)u, v)0 |.

We know that Bλ0 (u, v) is continuous on Hpλ 0 (Rn ) therefore our calculations are reduced to estimating|(r1(x,D)u, v)0 |.We know that r1 ∈ S2r1 +2ε−1,ψ

ρ (Rn ) which implies by Theorem 2.6 that

|(r1(x,D)u, v)0 | ≤ c ||u||ψ ,r1 +ε− 12||v||ψ ,r1 +ε− 1

2.

If r1 + ε − 12 < r0 we get

||u||ψ ,r1 +ε− 12≤ ||u||ψ ,r0 ≤ c||u||pλ 0

implying the result by (3.28).

By the Lax-Milgram theorem, for every g ∈ (Hpλ 0 )∗ ⊂ S′(Rn ) there exists a unique element u ∈ Hpλ 0

satisfying

Bλ0 (u, v) =⟨g, v⟩

(3.29)

for all v ∈ Hpλ 0 . We call u the variational solution to p(x,D)u + λ0u = g. From (3.28) we get

Hψ,−r0 (Rn ) = (Hψ,r0 (Rn ))∗ ↪→ (Hpλ 0 (Rn ))∗,

hence for g ∈ Hψ,−r0 (Rn ) there exists a unique u ∈ Hpλ 0 (Rn ) satisfying (3.29).

Proposition 3.6 For every g ∈ Hψ,−r0 (Rn ) there exists a unique u ∈ Hψ,r0 (Rn ) such that

pλ0 (x,D)u = p(x,D)u + λ0u = g (3.30)

holds as an equality in S′(Rn ).

P r o o f. Denote by u ∈ Hpλ 0 (Rn ) the unique solution to (3.30) for g ∈ Hψ,r0 (Rn ) given and take a sequence(uk )k∈N, uk ∈ S(Rn ) converging in Hpλ 0 (Rn ) to u.

It follows from

(pλ0 (x,D)uk , v)0 = Bλ0 (uk , v), v ∈ S(Rn )

and the continuity of pλ0 (x,D) from Hψ,s(Rn ) to Hψ,s−2r1 −2ε(Rn ) that for k → ∞⟨pλ0 (x,D)u, v

⟩= Bλ0 (u, v) =

⟨g, v⟩

for all v ∈ S(Rn ). Thus pλ0 (x,D)u = g in S′(Rn ) and the uniqueness follows from (3.23).

In order to get more regularity for variational solutions we have



Theorem 3.7 Assume (3.10), (3.19), (3.20) and (3.2), i.e., for some γ > 0

γ(1 + ψ(ξ))r0 ≤ f0(γ0(1 + ψ(ξ))) (3.31)

holds where

f0(s) := infx∈Rn

f(x, s). (3.32)

Then for every η > 0 sufficiently small the function p−1λ (x, ξ) belongs to the class S−2r0 +2η ,ψ

ρ (Rn ).

P r o o f. Let us assume for simplicity that δ0 = γ0 , compare (3.10). For λ > 0 let pλ (x, ξ) = λ+f(x, q(x, ξ)).From (2.27) in [18] we find with l = |α| + |β|∣∣∣∣∂α

ξ ∂βx

1pλ (x, ξ)

∣∣∣∣ ≤ 1pλ (x, ξ)

∑α1 +···+αl =α

β 1 +···+β l =β

c{αj ,β j }

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x pλ(x, ξ)pλ(x, ξ)

∣∣∣∣∣

=1

λ + f(x, q(x, ξ))

∑α1 +···+αl =α

β 1 +···+β l =β

c{αj ,β j }

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x pλ (x, ξ)pλ(x, ξ)

∣∣∣∣∣ .

From the definition of f1(s), (3.32), together with (3.20) we get

∣∣∣∣∂αξ ∂β

x

1pλ (x, ξ)

∣∣∣∣ ≤ 1f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

c{αj ,β j }

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x pλ(x, ξ)pλ (x, ξ)

∣∣∣∣∣

≤ cα,β1

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x (λ + f(x, q(x, ξ)))λ + f(x, q(x, ξ))

∣∣∣∣∣

= cα,β1

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x λ + ∂αj

ξ ∂β j

x f(x, q(x, ξ))λ + f(x, q(x, ξ))

∣∣∣∣∣ .

When α = β = 0 we find∣∣∣∣∂αξ ∂β

x

1pλ (x, ξ)

∣∣∣∣ ≤ c0,01

f1(γ0(1 + ψ(ξ))).

Moreover whenever αj = βj = 0, then∣∣∣∣∣∂αj

ξ ∂β j

x λ + ∂αj

ξ ∂β j

x f(x, q(x, ξ))λ + f(x, q(x, ξ))

∣∣∣∣∣ = 1.

Therefore we now only have to consider

∣∣∣∣∂αξ ∂β

x

1pλ (x, ξ)

∣∣∣∣ ≤ cα,β1

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

αj +β j >0

l∏j=1

∣∣∣∣∣∂αj

ξ ∂β j

x f(x, q(x, ξ))λ + f(x, q(x, ξ))

∣∣∣∣∣



Further, using (3.21) we get∣∣∣∣∂αξ ∂β

x

1pλ(x, ξ)

∣∣∣∣ ≤ 1f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

αj +β j >0

×l∏

j=1

cαj β j εf(x, q(x, ξ))q(x, ξ)ε(1 + ψ(ξ))−ρ (|α j |)

2

λ + f(x, q(x, ξ))

=1

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

αj +β j >0

[f(x, q(x, ξ))q(x, ξ)ε

λ + f(x, q(x, ξ)))

]l⎛⎝ l∏

j=1

cαj β j ε

⎞⎠

× (1 + ψ(ξ))−12 (ρ(|α1 |)+···+ρ(|αl |))

≤ γα,β ,ε1

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

αj +β j >0

[f(x, q(x, ξ))q(x, ξ)ε

λ + f(x, q(x, ξ))

]l

(1 + ψ(ξ))−12 ρ(|α |)

using the sub-additivity of ρ. Since∣∣∣ f (x,q(x,ξ))λ+f (x,q(x,ξ))

∣∣∣ ≤ 1 we find

∣∣∣∣∂αξ ∂β

x

1pλ(x, ξ)

∣∣∣∣ ≤ cα,β ,ε

f1(γ0(1 + ψ(ξ)))

∑α1 +···+αl =α

β 1 +···+β l =β

αj +β j >0

q(x, ξ)lε(1 + ψ(ξ))−12 ρ(|α |) . (3.33)

Since q(x, ξ) ≤ c0,0(1 + ψ(ξ)) it follows further that given η > 0 we can find cα,β ,η > 0 such that∣∣∣∣∂αξ ∂β

x

1pλ(x, ξ)

∣∣∣∣ ≤ cα,β ,η

f0(γ0(1 + ψ(ξ)))(1 + ψ(ξ))η− 1

2 ρ(|α |) . (3.34)

Taking into account (3.31) we eventually arrive at∣∣∣∣∂αξ ∂β

x

1pλ(x, ξ)

∣∣∣∣ ≤ cα,β ,η (1 + ψ(ξ))−2 r 0 + 2 η −ρ ( |α |)

2 , (3.35)

i.e., p−1λ ∈ S−2r0 +2η ,ψ

ρ (Rn ) proving the theorem.

We can now prove

Theorem 3.8 Let p(x, ξ) be given by (3.3) where we assume for q condition (3.4). For f it is supposed that(3.1) and (3.2) hold. Then we have for any ε > 0 and ˜η > 0 that p ∈ S2r1 +2ε,ψ

ρ (Rn ) and p−1λ ∈ S−2r0 +2η ,ψ

ρ (Rn )where we assume that r1 −r0 < 1

2 . Let u ∈ Hpλ 0 (Rn ) ⊂ Hψ,r0 (Rn ) be the solution to (3.30) for g ∈ Hψ,k (Rn ),k ≥ 0. Then it follows that u ∈ Hψ,k+2r0 −2η (Rn ).

P r o o f. The statements for p and p−1λ have already been proved, i.e., Theorems 3.3 and 3.7, respectively.

From Theorem 2.4 it follows that

p−1λ0

(x,D) ◦ pλ0 (x,D) = id + r(x,D) (3.36)

with r ∈ S2r1 +2ε−2r0 +2η−1,ψ0 (Rn ). Since pλ0 (x,D)u = g we deduce from (3.36) that

u = p−1λ0

(x,D) ◦ pλ0 (x,D)u − r(x,D)u

= p−1λ0

(x,D)g − r(x,D)u.



Now, p−1λ0

(x,D)g ∈ Hψ,2r0 −2η+k (Rn ) and r(x,D)u ∈ Hψ,3r0 −2η−2r1 −2ε+1(Rn ) implying that u ∈ Hψ,t(Rn )for t = (k + 2r0 − 2η) ∧ (3r0 − 2η − 2r1 − 2ε + 1) > r0 . With a finite number of iterations we arrive atu ∈ Hψ,k+2r0 −2η (Rn ).

Corollary 3.9 If k + 2r0 − 2η > nρ1

then u ∈ C∞(Rn ).

We finally arrive at

Theorem 3.10 Let f : Rn × (0,∞) → R be an arbitrarily often differentiable function such that for

y ∈ Rn fixed, the function s → f(y, s) is a Bernstein function. Moreover assume (3.1) and (3.2). In addi-

tion let ψ : Rn → R be a continuous negative definite function in the class Λ which satisfies in addition (3.26).

For an elliptic symbol q ∈ S2,ψρ (Rn ) satisfying (3.4) we define p(x, ξ) by (3.3). We know that for any ε > 0 and

η > 0 it holds p ∈ S2r1 +2ε,ψρ (Rn ) and 1

p+λ ∈ S−2r0 +2η ,ψρ (Rn ). If r1 − r0 < 1

2 then −p(x,D) extends to agenerator of a Feller semigroup on C∞(Rn ).

P r o o f. The statements for p and p−1λ have already been proved, i.e., Theorems 3.3 and 3.7, respectively. We

want to apply the Hille-Yosida-Ray theorem, Theorem 4.5.3 in [18]. We know that p(x,D) maps Hψ,2r1 +2ε+k (Rn )into Hψ,k (Rn ). Hence if k > n

2ρ1the operator

(− p(x,D),Hψ,2r1 +2ε+k (Rn )

)is densely defined on C∞(Rn )

with range in C∞(Rn ). That −p(x,D) satisfies the positive maximum principle on Hψ,2r1 +2ε+k (Rn ) followsfrom Theorem 2.6.1 in [19]. Now, for λ ≥ λ0 we know that for g ∈ Hψ,k+2r1 −2r0 +2η+2ε(Rn ) we have a uniquesolution to pλ (x,D)u = g belonging to Hψ,k+2r1 +2ε(Rn ) implying the theorem.

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variable order subordination in the sense of bochner and pseudo-differential operators

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