H ∞ -calculus, interpolation, and isomorphisms

Proof. (i) We have f1/f2∈H^∞(Ω)by assumption. Hence we get f1(T) =

f1

f₂ ·f2

(T)⊇

f1

f₂

(T)f2(T)

(cf. Theorem 1.19 (ii)) and (f₁/f₂)(T) ∈ L(X). So we obtain the claimed relation between the domains as well as the claimed estimate. On the other hand we have

f1(T) =

f2· f1

f₂

(T) =f2(T) f1

f₂

(T)

(cf. Theorem 1.19 (iii)), which yields the claimed relation between the ranges.

(ii) This follows immediately from (i).

Lemma 1.25. Let g, h∈HP(Ω) with g(z), h(z)6= 0for all z∈Ωandg⁻¹, h⁻¹∈H^∞(Ω). If we define f(z) :=g(z)·h(z),z∈Ω, we can conclude

f(T) =g(T)h(T) =h(T)g(T), which especially yieldsD(f(T)) =D(g(T)h(T)) =D(h(T)g(T)).

Proof. Without loss of generality we only showf(T) =g(T)h(T). From Theorem 1.19 we already know f(T)⊇g(T)h(T), f⁻¹(T) =g⁻¹(T)h⁻¹(T) =h⁻¹(T)g⁻¹(T),

which yieldD(g(T)h(T))⊆D(f(T)) =R(f⁻¹(T)) =R(g⁻¹(T)h⁻¹(T)) =R(h⁻¹(T)g⁻¹(T)). One can easily showR(h⁻¹(T)g⁻¹(T))⊆D(g(T)h(T)), which ends the proof.

1.2 H

^∞

-calculus, interpolation, and isomorphisms

Proposition 1.26 (H^∞-calculus and isomorphisms). Let X, Y be Banach spaces such that there exists an isomorphismΦ∈LIsom(Y, X). For eachk= 1, . . . , N let

Tk:D(Tk)⊆X →X be an arbitrary linear operator and define

Sk:D(Sk)⊆Y →Y, y7→Φ⁻¹TkΦy

whereD(S_k) := Φ⁻¹(D(T_k)). If we defineT:= (T₁, . . . , T_N)andS:= (S₁, . . . , S_N), we get the following assertions:

(i) IfT_k is sectorial (respectively, bisectorial), then the operator S_k is also sectorial (respectively, bi-sectorial) with ϕ_Sk =ϕ_Tk (respectively, ϕ^(bi)_S

k =ϕ^(bi)_T

k ). In particular, we have ρ(S_k) = ρ(T_k) and (λ−Sk)⁻¹= Φ⁻¹(λ−Tk)⁻¹Φfor allλ∈ρ(Tk). If Tis admissible, thenSis also admissible.

(ii) IfThas a bounded jointH^∞(Ω)-calculus, thenSalso has a bounded jointH^∞(Ω)-calculus and the representation

f(S) = Φ⁻¹f(T)Φ holds for allf ∈H_P(Ω).

(iii) IfThas anR-bounded jointH^∞(Ω)-calculus, thenSalso has anR-bounded jointH^∞(Ω)-calculus.

Proof. (i) Without loss of generality let Tk be a sectorial operator for all k = 1, . . . , N. Then we proves thatSadmits a bounded joint H^∞(Ω)-calculus.

Let f ∈ H_P(Ω) and m ∈ N0 such that Ψ^mf ∈ H₀^∞(Ω). Then we have Ψ(S) = Φ⁻¹Ψ(T)Φand (Ψ^mf)(S) = Φ⁻¹(Ψ^mf)(T)Φdue toΨ∈H₀^∞(Ω). We then derive

f(S) = Ψ(S)^−m(Ψ^mf)(S) = Φ⁻¹Ψ(T)^−mΦΦ⁻¹(Ψ^mf)(T)Φ

= Φ⁻¹f(T)Φ.

(iii) This easily follows from the representation in (ii) and Remark B.14 (iii).

Theorem 1.27 (H^∞-calculus and interpolation). Let F ∈ {(·,·)θ,p,[·,·]θ}, θ ∈ (0,1), p∈ (1,∞), and let{X0, X1} be an interpolation couple. Additionally, let

Tk:D(Tk)⊆X0→X0,

S_k:D(S_k)⊆X₁→X₁, k= 1, . . . , N

be linear operators with the compatibility conditions Tkx=Skxfor allx∈D(Tk)∩D(Sk)and

(λ−T_k)⁻¹x= (λ−S_k)⁻¹x, λ∈ρ(T_k)∩ρ(S_k), x∈X₀∩X₁. (1.4) Moreover we define fork= 1, . . . , N the interpolated operators

A_k:D(A_k)⊆ F({X0, X₁})→ F({X0, X₁}), D(A_k) :=F({D(Tk), D(S_k)})

with Akx:=Tkx0+Skx1 for x=x0+x1 ∈D(Ak) ,→ D(Tk) +D(Sk)with x0 ∈ D(Tk), x1 ∈ D(Sk).

Note that we equipD(Tk)andD(Sk)with the graph norm. Then we have the following results:

(i) Letk= 1, . . . , N. IfTk andSk are both sectorial (respectively, bisectorial), thenAk is also sectorial (respectively, bisectorial) with ϕA_k ≤max{ϕT_k, ϕS_k} (respectively, ϕ^(bi)_A

k ≤max{ϕ^(bi)_T

k , ϕ^(bi)_S

k }). If T andSare both admissible, then Ais also admissible.

(ii) LetT:= (T1, . . . , TN)andS:= (S1, . . . , SN)be tuples of sectorial and bisectorial operators such that T_k andS_k are of the same type. IfTandSadmit a boundedH^∞(Ω)-calculus, then the interpolated tuple A:= (A₁, . . . , A_N)also admits a boundedH^∞(Ω)-calculus and we have the representation

1.2:H^∞-calculus, interpolation, and isomorphisms 15 (iii) LetTandSbe the same operators as in (ii) but with anR-boundedH^∞(Ω)-calculus. IfX0, X1 are of class HT, then A also has an R-bounded H^∞(Ω)-calculus. In particular for N = 1 we obtain ϕ^R,∞_A

1 ≤max{ϕ^R,∞_T

1 , ϕ^R,∞_S

1 }(respectively, ϕ^R,∞,(bi)_A

1 ≤max{ϕ^R,∞,(bi)_T

1 , ϕ^R,∞,(bi)_S

1 }).

Proof. (i) We have to consider the relation between the resolvents of Tk, Sk and Ak. Without loss of generality we assume that Tk and Sk are sectorial. Let λ ∈ −Sπ−θ⁰ ⊆ ρ(Tk)∩ρ(Sk) with θ⁰ >max{ϕTk, ϕ_Sk}. Then we define the bounded operator

R^(k)_λ :X0+X1 → D(Tk) +D(Sk),

x0+x1 7→ (λ−Tk)⁻¹x0+ (λ−Sk)⁻¹x1. Note thatR^(k)_λ is well defined by (1.4). So we have

(R^(k)_λ )_|X0 = (λ−Tk)⁻¹, (R^(k)_λ )_|X1 = (λ−Sk)⁻¹, (R_λ^(k))_|F({X0,X1})∈L(F({X0, X1}), D(Ak)).

With this we can easily showλ∈ρ(A_k)and(λ−A_k)⁻¹= (R^(k)_λ )_|F({X0,X1}), which already yields

−S_π−θ⁰ ⊆ρ(A_k). Thus, we derive

k(λ−A_k)⁻¹k_{L(F({X0,X1}))} ≤ k(λ−T_k)⁻¹k^1−θ_L(X

0)k(λ−S_k)⁻¹k^θ_L(X

1), λ∈ −S_π−θ⁰

due to Remark A.4 (ii) and Remark A.7 (ii). We still have to show the density ofR(Ak)andD(Ak) inF({X0, X1}). Here we use the characterizations

R(Ak) = {x∈ F({X0, X1}) :_[F({X0,X1})] lim

n→∞−1/n(−1/n−Ak)⁻¹x= 0}, D(Ak) = {x∈ F({X0, X1}) :_[F({X0,X₁})] lim

n→∞−n(−n−Ak)⁻¹x=x}

given in Lemma 1.6. Letx∈X0∩X1 be arbitrary. Then we derive

−1/n(−1/n−Ak)⁻¹x = −1/n(−1/n−Tk)⁻¹x=−1/n(−1/n−Sk)⁻¹x,

−n(−n−A_k)⁻¹x = −n(−n−T_k)⁻¹x=−n(−n−S_k)⁻¹x.

Due toR(Tk) =D(Tk) =X0,R(Sk) =D(Sk) =X1, and Lemma 1.6 we have

[X₀] lim

n→∞−1/n(−1/n−Tk)⁻¹x= 0, [X₁] lim

n→∞−1/n(−1/n−Sk)⁻¹x= 0,

[X0] lim

n→∞−n(−n−Tk)⁻¹x=x, _[X1] lim

n→∞−n(−n−Sk)⁻¹x=x, which yields

0 =[X₀∩X1] lim

n→∞−1/n(−1/n−Ak)⁻¹x=_[F({X0,X₁})] lim

n→∞−1/n(−1/n−Ak)⁻¹x, x=_[X0∩X1] lim

n→∞−n(−n−Ak)⁻¹x=_[F({X0,X1})] lim

n→∞−n(−n−Ak)⁻¹x

due to X₀∩X₁ ,→ F({X₀, X₁}). So we have proved X₀∩X₁ ⊆ R(A_k), D(A_k) and therefore X0∩X1

,→ F({Xd 0, X1})(cf. Remark A.4 (i) and Remark A.7 (i)) yields R(A_k) =D(A_k) =F({X₀, X₁}).

Hence we have shown thatAk is sectorial withϕA_k ≤max{ϕT_k, ϕS_k}. The claimed admissibility ofAcan be shown easily by the representation of the resolvents byRλ and Remark 1.8.

(ii) Letf ∈H₀^∞(Ω) and define the operator

F:X0+X1 → X0+X1,

x0+x1 7→ f(T)x0+f(S)x1,

which is well-defined due to (1.4). For x∈ F({X0, X1})we obtain

(iii) Part (ii) already yields thatAadmits a boundedH^∞(Ω)-calculus. By assumption the families T1 := {h(T) :h∈H₀^∞(Ω),khk_∞≤1} ⊆L(X0),

T₂ := {h(S) :h∈H₀^∞(Ω),khk_∞≤1} ⊆L(X₁)

are both R-bounded. Thus, the representation of f(A) in part (ii) and Theorem B.24 yield the R-boundedness of

{h(A) :h∈H₀^∞(Ω),khk_∞≤1} ⊆L(F({X0, X1})).

ThereforeAadmits anR-boundedH^∞(Ω)-calculus.

The assertion on the angle in the case N = 1is obvious.

Remark 1.28. The theorem aboutH^∞-calculus and interpolation is essential for our purpose and there-fore we want to make some remarks for the special case ifX1 embeds into X0:

(i) The compatibility conditions for the resolvents in (1.4) are always fulfilled in case of X₁ ,→ X₀. This comes from the fact that we have (λ−S_k)⁻¹= (λ−T_k)⁻¹_|X

1 for all λ∈ρ(T_k)∩ρ(S_k)in this case.

1.2:H^∞-calculus, interpolation, and isomorphisms 17 (ii) For X1,→X0 the representation of f(A)in Theorem 1.27 (ii) can be given by

f(A) =f(T)_|F({X0,X1}), f ∈H^∞(Ω).

This can be easily seen by(λ−A_k)⁻¹= (λ−T_k)⁻¹_|F({X

0,X1}) forλ∈ρ(T_k)∩ρ(S_k).

The full strength of Theorem 1.27 will become much clearer in Chapter 5. There we develop a bounded H^∞-calculus of∇+= (∂t,∇)on Triebel-Lizorkin spaces. Except for Chapter 5 we always haveX1,→X0

in every situation.

Lemma 1.29(H^∞-calculus of a shifted operator). Let T: D(T)⊆X →X be a sectorial operator.

Then for allµ≥0the operatorS:=µ+T is also sectorial withϕ_S ≤ϕ_T. In this situation we also have

f(S) = [f(µ+·)](T) (1.5)

for all f ∈ HP(Sθ) with θ > ϕT. If T admits a bounded H^∞-calculus, then S also admits a bounded H^∞-calculus with ϕ^∞_S ≤ ϕ^∞_T . If T even admits an R-bounded H^∞-calculus, then S also admits an R-boundedH^∞-calculus withϕ^R,∞_S ≤ϕ^R,∞_T .

Proof. The sectoriality of S andϕ_S ≤ϕ_T are obvious. Letθ > ϕ_T andg(z) :=µ+z forz ∈S_θ. Then we haveg(T) =µ+T by Theorem 1.19 (iii). Now we directly obtain (1.5) by [KW04, Proposition 15.11], where the authors prove the transformation formula(f◦g)(T) =f(g(T))for a more general situation.

TriviallyS also admits a boundedH^∞-calculus since

kf(S)k_L(X)≤Ckf(µ+·)k_∞≤Ckfk_∞

for allf ∈H₀^∞(Sθ). The claimedR-boundedness is obvious bykf(µ+·)k_∞≤ kfk_∞and Remark B.14 (vi).

Lemma 1.30(H^∞-calculus for shifted operator tuple). LetTbe a tuple of operators with a bounded H^∞(Ω)-calculus andS:= (µ+T₁, T₂, . . . , T_N)forµ >0. IfT₁ is a sectorial operator, thenSalso admits a boundedH^∞(Ω)-calculus with

f(S) =fµ(T), fµ(z) :=f(µ+z1, z2, . . . , zN) for allf ∈HP(Ω).

Proof. (I) First, we prove the assertion for f ∈ H₀^∞(Ω). For all z⁰ ∈ Ω⁰ := QN

k=2Ω_k we define g(z⁰) :=f(µ+T1, z⁰). Then Lemma 1.29 and Lemma 1.22 yieldg(z⁰) =fµ(T1, z⁰)and

f(S) =g(T⁰) =fµ(T)

where T⁰ := (T2, . . . , TN). We obviously getkf(S)k_L(X) ≤Ckfµk_∞ ≤Ckfk_∞ for allf ∈H₀^∞(Ω) and thusSalso admits a bounded H^∞(Ω)-calculus.

(II) Letf ∈HP(Ω). Then we get

f(S) = Ψ(S)^−m(Ψ^mf)(S) = Ψµ(T)^−m(Ψ^m_µfµ)(T)

⊇ Ψ_µ(T)^−mΨ_µ(T)^mf_µ(T) =f_µ(T)

due to Theorem 1.19 and the result of part (iI). The converse inclusion can be obtained in the same way by

fµ(T) = Ψ(T)^−m(Ψ^mfµ)(T) = Ψ(T)^−m Ψ^m

Ψ^m_µ

· Ψ^m_µfµ

(T)

⊇ Ψ(T)^−m Ψ^m

Ψ^m_µ

(T) Ψ^m_µfµ

(T)

= Ψ(T)^−m Ψ^m

Ψ^m_µ

(T) (Ψ^mf) (S)

⊇ Ψ(T)^−mΨ(T)^mΨ^−m_µ (T) (Ψ^mf) (S)

= Ψ(S)^−m(Ψ^mf) (S) =f(S).

Proposition 1.31 (Compatibility ofH^∞-calculus on spaces of higher regularity). Let X, Y be Banach spaces with Y ,→X. Let

T_k:D(T_k)⊆X → X,

Sk:D(Sk)⊆Y → Y, k= 1, . . . , N

be sectorial or bisectorial operators (of the same type) withD(Sk)⊆D(Tk)(with respect toY ,→X) such that the tuplesT:= (T1, . . . , TN)andS:= (S1, . . . , SN)are admissible andTky=Sky for ally∈D(Sk) andk∈ {1, . . . , N}. Then we obtain

f(S)⊆f(T) for allf ∈H_P(Ω) whereΩis admissible for TandS.

Proof. (I) Letf ∈H₀^∞(Ω) andy∈Y. Then we have

f(S)y = 1

(2πi)^N

"

Z

Γ

f(z)

N

Y

k=1

(zk−Sk)⁻¹dz_[L(Y)]

#

y= 1

(2πi)^N Z

Γ

f(z)

N

Y

k=1

(zk−Sk)⁻¹ydz_[Y]

= 1

(2πi)^N Z

Γ

f(z)

N

Y

k=1

(z_k−T_k)⁻¹ydz_[X]= 1 (2πi)^N

"

Z

Γ

f(z)

N

Y

k=1

(z_k−T_k)⁻¹dz_[L(X)]

# y

= f(T)y

because of (z−Tk)⁻¹_|Y = (z−Sk)⁻¹for allz∈ρ(Tk)∩ρ(Sk)andY ,→X.

(II) Letf ∈H_P(Ω)andm∈N0 with(Ψ^mf)∈H₀^∞(Ω). Using (I) we obtain D(f(S)) = {y∈Y: (Ψ^mf)(S)y∈R(Ψ(S)^m)}

= {y∈Y: (Ψ^mf)(T)y∈R(Ψ(S)^m)}

⊆ {y∈Y: (Ψ^mf)(T)y∈R(Ψ(T)^m)}

= Y ∩D(f(T)).

This yields

D(f(S))⊆D(f(T)).

It is easy to prove thatΨ(T)^−mv= Ψ(S)^−mv for allv∈R(Ψ(S)^m). Fory∈D(f(S))we then get f(T)y = Ψ(T)^−m(Ψ^mf)(T)y= Ψ(T)^−m(Ψ^mf)(S)y

| {z }

∈R(Ψ(S)^m)

=f(S)y

by (I). So we have proved the assertion.

Chapter2

The joint time-space H ^∞ -calculus on spaces of higher regularity

In this chapter we want to establish a functional calculus for the operator tuple ∇+ := (∂t,∇) on Lp-spaces. The concept of the joint H^∞-calculus presented in Chapter 1 is a suitable framework for this purpose. One of our aims is to investigate the jointH^∞-calculus on spaces of higher regularity as Besov and Bessel-potential spaces. Partial results in this direction can be found in works of G. Dore and A. Venni, cf. [DV02c], where the authors consider the joint H^∞-calculus of ∇ onLp(Rⁿ). See also the work of R. Denk, J. Saal, and J. Seiler, cf. [DSS08], where the authors consider the jointH^∞-calculus of (∂_t,∆) on spaces of higher regularity. In particular, the ground spaces in [DSS08] can possess different scales in time and space, e.g.0B_p^s(R+, H_p^r(Rⁿ)).

In Chapter 5 we even expand the results of this chapter to Bessel-valued Triebel-Lizorkin spaces. In L_p-L_q theory these spaces are crucial for the understanding of traces in the space variable.

In Sections 2.1 and 2.2 we introduce the basic concepts of vector-valued Bessel-potential and Besov spaces. Here we essentially follow the same approaches as in [Ama09], where function spaces on the half space are defined by retractions and coretractions. H. Amann even considers anisotropic function spaces in [Ama09] but this is of less interest for our purposes.

2.1 Retractions and coretractions

In some situations we want to establish identities for function spaces where the functions are defined on a half space. To carry over the known results on function spaces where the functions are defined on the whole space we need the concept of retractions and coretractions. Especially these concepts are compatible with the real and complex interpolation functors, which is important for our purpose.

Definition 2.1 (Retraction and coretraction, cf. [Ama95, I.2.3], [Ama09, p. 4]). Let X and Y be locally convex spaces. A mappingr∈ L(X,Y) is called a retraction if there exists e ∈L(Y,X) such that(re)y=y for ally∈ Y. The mapping eis then called the corresponding coretraction.

Definition 2.2.LetX be a Banach space and letY be a locally convex space. For a mappingϕ∈L(X,Y) we define the image space

(ϕX,k · kϕX) equipped with the quotient norm

kukϕX:= inf

kfkX: f ∈ϕ⁻¹({u}) , u∈ϕ(X).

Remark 2.3. In the situation of Definition 2.2 it holds that (ϕX,k · k_ϕX) is a Banach space, which fulfillsϕX ,→ Y.

19

Lemma 2.4. Let X and Y be Banach spaces and let S be a locally convex space. Suppose there exists r∈L(X,S)∩L(Y,S). Then we have

r(X∩Y) = (rX)∩(rY) with equivalent norms.

Proof. We trivially have r(X ∩Y) = (rX)∩(rY) as an equality of sets, so we only have to show the equivalence of norms. Foru∈r(X∩Y)we derive

kuk_r(X∩Y) = inf{kfkX∩Y:f ∈X∩Y, rf =u}= inf{kfkX+kfkY: f ∈X∩Y, rf =u}

≥ inf{kfkX:f ∈X∩Y, rf =u}+ inf{kgkY:g∈X∩Y, rg=u}

≥ inf{kfk_X:f ∈X, rf =u}+ inf{kgk_Y: g∈Y, rg=u}

= kukrX+kukrY =kuk_(rX)∩(rY).

Using the bounded inverse theorem and Remark 2.3 we conclude the claimed equivalence of norms.

Lemma 2.5(Interpolation and retraction). LetA0, A1, Z0, Z1be Banach spaces andrk∈L(Ak, Zk), k= 0,1. Ifrkis a retraction inL(Ak, rkAk)fork= 0,1with corresponding coretractionek ∈L(rkAk, Ak) andr0x=r1xfor all x∈A0∩A1, then we have

r[A₀, A₁]_θ = [r₀A₀, r₁A₁]_θ, θ∈(0,1),

r(A0, A1)_θ,p = (r0A0, r1A1)_θ,p, θ∈(0,1), 1< p <∞ with equivalent norms wherer:A0+A1→r0A0+r1A1, a0+a17→r0a0+r1a1.

Proof. We only consider the complex interpolation functor. The proof for the real method is essentially the same. We have

r∈L([A0, A1]θ,[r0A0, r1A1]θ), which yields

r[A0, A1]θ,→[r0A0, r1A1]θ (2.1) according to Remark 2.3. Next, we apply [Tri78, Theorem 1.2.4] withBk:=rkAk,k= 0,1. This yields

Φ :=e_|[B0,B₁]_θ ∈LIsom [B0, B1]_θ, er([A0, A1]θ),k · k[A₀,A₁]_θ

(2.2)

and therefore e([B₀, B₁]_θ) =er([A₀, A₁]_θ)as an equality of sets. Due to (2.1) and (2.2) we also obtain [B₀, B₁]_θ =r([A₀, A₁]_θ) as an equality of sets. The bounded inverse theorem and (2.1) then yield the assertion.

Lemma 2.6 (Interpolation and isomorphism). Let A₀, A₁, B₀, B₁ be Banach spaces. If there exist r_k∈L_Isom(A_k, B_k),k= 0,1, withr₀x=r₁xfor allx∈A₀∩A₁, then we haveB_k=r_kA_k with equivalent norms and

r[A₀, A₁]_θ = [B₀, B₁]_θ, θ∈(0,1),

r(A0, A1)_θ,p = (B0, B1)_θ,p, θ∈(0,1), 1< p <∞ with equivalent norms wherer:A₀+A₁→r₀A₀+r₁A₁, a₀+a₁7→r₀a₀+r₁a₁.

Proof. We only consider the complex interpolation functor. The argumentation for the real method is exactly the same. Due tork∈LIsom(Ak, Bk)we have

kukr_kA_k=kr⁻¹_k ukA_k ≤CkukB_k. This yieldsr_kA_k =B_k. Lemma 2.5 then already shows

r[A₀, A₁]_θ= [B₀, B₁]_θ, θ∈(0,1).

2.2: Preliminaries on spaces of mixed scales 21

Im Dokument General parabolic mixed order systems in Lp and applications (Seite 23-31)