An Introduction to Interpolation Theory, 2007 [Lunardi_A.]

8/11/2019 An Introduction to Interpolation Theory, 2007 [Lunardi_A.]

1/125

An Introduction to Interpolation Theory

Dottorato di Ricerca in Matematica, consorzio

Milano-Insubria-Parma-Trieste

Alessandra Lunardi

February 2007


2/125

2


3/125

Contents

1 Real interpolation 71.1 The K-method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.2 The trace method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.3 Intermediate spaces and reiteration . . . . . . . . . . . . . . . . . . . . . . . 24

1.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.3.2 Applications. The theorems of Marcinkiewicz and Stampacchia . . . 311.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2 Complex interpolation 352.1 Definitions and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.1.2 The theorems of HausdorffYoung, RieszThorin, Stein . . . . . . . 452.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3 Interpolation and domains of operators 49

3.1 Operators with rays of minimal growth . . . . . . . . . . . . . . . . . . . . . 493.1.1 Two or more operators . . . . . . . . . . . . . . . . . . . . . . . . . . 543.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.2 The case where A generates a semigroup . . . . . . . . . . . . . . . . . . . . 583.2.1 Examples and applications. Schauder type theorems . . . . . . . . . 62

4 Powers of positive operators 674.1 Definitions and general properties . . . . . . . . . . . . . . . . . . . . . . . . 67

4.1.1 Powers of nonnegative operators . . . . . . . . . . . . . . . . . . . . 754.2 Operators with bounded imaginary powers . . . . . . . . . . . . . . . . . . . 76

4.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.2.2 The sum of two operators with bounded imaginary powers . . . . . 824.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4.3 M-accretive operators in Hilbert spaces . . . . . . . . . . . . . . . . . . . . . 884.3.1 Self-adjoint operators in Hilbert spaces . . . . . . . . . . . . . . . . . 93

5 Analytic semigroups and interpolation 995.1 Characterization of real interpolation spaces . . . . . . . . . . . . . . . . . . 1005.2 Generation of analytic semigroups by interpolation . . . . . . . . . . . . . . 1025.3 Regularity in abstract parabolic equations . . . . . . . . . . . . . . . . . . . 103

3


4/125

4

5.4 Applications to regularity in parabolic PDEs . . . . . . . . . . . . . . . . . 110

A The Bochner integral 113A.1 Integrals over measurable real sets . . . . . . . . . . . . . . . . . . . . . . . 113A.2 Lp and Sobole v s pac e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115A.3 WeightedLp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

B Vector-valued holomorphic functions 119

Index 125


5/125

Introduction

Let X, Y be two real or complex Banach spaces. By X = Y we mean that X and Yhave the same elements and equivalent norms. ByYXwe mean thatY is continuouslyembedded in X.

The couple of Banach spaces (X, Y) is said to be an interpolation coupleif bothXandYare continuously embedded in a Hausdorff topological vector spaceV. In this case theintersectionX Y is a linear subspace ofV, and it is a Banach space under the norm

v

XY= max

{v

X,

v

Y

}.

Also the sum X+ Y ={x+y : xX, y Y} is a linear subspace ofV. It is endowedwith the norm

vX+Y = inf xX, yY, x+y=v

xX+ yY.

As easily seen,X+ Y is isometric to the quotient space (X Y)/D, whereD ={(x, x) :xX Y}. SinceVis a Hausdorff space, then D is closed, and X+ Yis a Banach space.Moreover,xX xX+Y andyY yX+Y for all x X, y Y, so that both Xand Y are continuously embedded in X+ Y.

The spaceVis used only to guarantee thatX+ Y is a Banach space. It will disappearfrom the general theory.

If (X, Y) is an interpolation couple, anintermediate spaceis any Banach space Esuch

thatX YEX+ Y.

An interpolation spacebetweenXandY is any intermediate space such that for everyT L(X) L(Y) (that is, for every T L(X+Y) whose restriction to X belongs toL(X) and whose restriction to Y belongs to L(Y)), the restriction of T to E belongsto L(E). We could also require that there is a constant independent of T such thatTL(E)C(TL(X)+ TL(Y)), but often this property is neglected.

The general interpolation theory is not devoted to characterize all the interpolationspaces between X and Ybut rather to construct suitable families of interpolation spacesand to study their properties. The most known and useful families of interpolation spacesare the real interpolation spaces which will be treated in chapter 1, and the complex

interpolation spaceswhich will be treated in chapter 2.Interpolation theory has a wide range of applications. We shall emphasize applications

to partial differential operators and partial differential equations, referring to [36], [12] forapplications to other fields. In particular we shall give self-contained proofs of optimalregularity results in Holder and in fractional Sobolev spaces for linear elliptic and parabolicdifferential equations.

The domains of powers of positive operators in Banach spaces are not interpolationspaces in general. However in some interesting cases they coincide with suitable complex

5


6/125

6 Introduction

interpolation spaces. In any case the theory of powers of positive operators is very closeto interpolation theory, and there are important connections between them. Thereforein chapter 4 we give an elementary treatment of the powers of positive operators, withparticular attention to the imaginary powers.


7/125

Chapter 1

Real interpolation

Let (X, Y) be a real or complex interpolation couple.IfIis any interval contained in (0, +),Lp(I) is the Lebesgue space Lp with respect

to the measure dt/t in I. In particular, L (I) =L(I). See Appendix,2.

1.1 The K-method

Definition 1.1.1 For everyxX+ Y andt >0, setK(t,x,X,Y) = inf

x=a+b, aX, bY(aX+ tbY) . (1.1)

If there is no danger of confusion, we shall writeK(t, x)K(t,x,X,Y)instead ofK(t,x,X,Y).

Note thatK(1, x) =xX+Y, and for everyt >0 K(t, ) is a norm inX+Y, equivalentto the norm ofX+ Y. Now we define a family of Banach spaces by means of the functionK.

Definition 1.1.2 Let0< 0. By the transformation=t1, which preserves Lp(0, ), we get

(X, Y),p = (Y, X)1,p, 0<


8/125

8 Chapter 1

and

(X, Y) = (Y, X)1. (1.5)

So, pay attention to the order!Let us consider some particular cases.

(a) Let X=Y. Then X+ Y =X, and K(t, x)

min

{t, 1

}x

. Therefore

X(X, X),p, 0< 0, and limt+ tK(t, x) = 0. Therefore,only the behavior near t = 0 oftK(t, x) plays a role in the definition of (X, Y),pand of (X, Y). Indeed, one could replace the halfline (0, +) by any interval (0, a)in definition 1.1.2, obtaining equivalent norms.

The inclusion properties of the real interpolation spaces are stated below.

Proposition 1.1.3 For0< 0,

so that for every n N(takingt = 1/n) there are anX,bnY such that x = an+ bn,and

anX+1

nbnY 2

n x,.In particular,xbnX+Y =anX+Y anX 2x,n, so that the sequence{bn} goes to x in X+Y as n . This implies that (X, Y), is contained in Y.Arguing similarly (i.e., replacing 1/n by n and letting n ), or else recalling that(X, Y), = (Y, X)1,) we see that (X, Y), is contained also in X. Moreover, bydefinitionxX+Y =K(1, x). Therefore

xX+Y =K(1, x) x,, x(X, Y),,


9/125

Real interpolation 9

so that (X, Y), is continuously embedded in X+ Y.The inclusion (X, Y)(X, Y),is obvious because K(, x) is continuous (see exercise

1,1.1.1) so that tK(t, x) is bounded in every interval [a, b] with 0< a < b.Let us show that (X, Y),p is contained in (X, Y) and it is continuously embedded in

(X, Y), forp 0, recalling thatK(, x) is increasingwe get

tK(t, x) = (p)1/p +t

sp1ds1/p K(t, x)(p)1/p

+t

sp1K(s, x)pds

1/p, t >0.

(1.7)

The right hand side is bounded by (p)1/px,p. Therefore x(X, Y),, andx,(p)1/px,p. Changing with 1and XwithYwe obtain x,((1)p)1/px,p.Therefore,

x,[min{, 1 }p]1/px,p. (1.8)Moreover lettingt we get limt tK(t, x) = 0. To prove thatx(X, Y) we needalso that limt0 t

K(t, x) = 0. This can be seen as follows: since (X, Y),p = (Y, X)1,pthen

0 = limt+

t(1)K(t ,x,Y,X) = limt+

tK(t1, x , X , Y ) = lim0+

K(,x,X,Y).

Let us prove that (X, Y),p1 (X, Y),p2 for p1 < p2 < +. Forx (X, Y),p1 wehave

x,p2 = +

0tp2K(t, x)p2

dt

t

1/p2

=

+

0tp1K(t, x)p1(tK(t, x))p2p1

dt

t 1/p2

+

0tp1K(t, x)p1

dt

t

1/p2 supt>0 t

K(t, x)(p2p1)/p2

= (x,p1)p1/p2(x,)1p1/p2 ,and using (1.8) we find

x,p2[min{, 1 }p1]1/p11/p2x,p1. (1.9)Finally, from the inequalityK(t, x)min{1, t} xXY for everyxX Y it follows

immediately thatX Y is continuously embedded in (X, Y),p for 0<


10/125

10 Chapter 1

Proposition 1.1.4 IfYX, for0< 1< 2< 1 we have(X, Y)2,(X, Y)1,1. (1.10)

Therefore, (X, Y)2,p(X, Y)1,q for everyp, q[1, ].Proof. For x(X, Y)2, we have, using the inequalities K(t, x) xX for t1 andK(t, x)

t2

x

2, for 0< t

1,

x1,1= 10

t11K(t, x)dt +

+1

t11K(t, x)dt

10

t11x2,t2dt + +1

t11xXdt

12 1 x2,+

1

1xX,

(1.11)

and the statement follows since (X, Y)2,X+ Y =X because YX. Note that (1.10) is not true in general. See next example 1.1.10.

Proposition 1.1.5 For all (0, 1), p [1, ], (X, Y),p is a Banach space. For all(0, 1), (X, Y) is a Banach space, endowed with the norm of(X, Y),.Proof. Let{xn}nN be a Cauchy sequence in (X, Y),p. By the continuous embedding(X, Y),p X+ Y,{xn}nN is a Cauchy sequence in X+ Y too, so that it converges toan elementxX+ Y.

Let us estimatexn x,p. Fix >0, and letxn xm,p for n, mn. Sincey K(t, y) is a norm in X+ Y, for every n, m N and t > 0 we have K(t, xn x)K(t, xn xm) + K(t, xm x), so that

tK(t, xn x)tK(t, xn xm) + t max{t, 1}xm xX+Y. (1.12)Letp=. Then for every t >0 and n, mn

tK(t, xn x) + t max{t, 1}xm xX+Y.Letting m +, we find tK(t, xn x) for every t > 0. This implies thatx(X, Y), and that xnx in (X, Y),. Therefore (X, Y), is complete.

It is easy to see that (X, Y) is a closed subspace of (X, Y),. Since (X, Y), iscomplete, then also (X, Y) is complete.

Let now p


11/125


Letting first m and then 0 we get x(X, Y),p and xn x in (X, Y),p. So,(X, Y),p is complete.

The spaces (X, Y),p and (X, Y) are interpolation spaces, as a consequence of thefollowing important theorem.

Theorem 1.1.6 Let (X1, Y1), (X2, Y2) be interpolation couples. If T

L(X1, X2)

L(Y1, Y2), thenT L((X1, Y1),p, (X2, Y2),p)L((X1, Y1),(X2, Y2))for every(0, 1)andp[1, ]. Moreover,

TL((X1,Y1),p,(X2,Y2),p)(TL(X1,X2))1(TL(Y1,Y2)). (1.13)

Proof.IfTis the null operator, there is nothing to prove. IfT= 0, either TL(X1,X2)= 0orTL(Y1,Y2)= 0. Assume thatTL(X1,X2)= 0. Let x (X1, Y1),p: then for everyaX1, bY1 such that x = a + b and for everyt >0 we have

T aX2+ tT bY2 TL(X1,X2)aX1+ t TL(Y1,Y2)TL(X1,X2) bY1 ,so that, taking the infimum over all a,b as above, we get

K(t ,Tx,X2, Y2) TL(X1,X2)K

tTL(Y1,Y2)TL(X1,X2)

, x , X 1, Y1

. (1.14)

Settings= tTL(Y1,Y2)TL(X1,X2)

we get T x (X2, Y2),p, and

T x(X2,Y2),p TL(X1,X2)

TL(Y1,Y2)

T

L(X1,X2)

x(X1,Y1),p ,

and (1.13) follows. From (1.14) it follows also that

limt0 tK(t,x,X1, Y1) = limt t

K(t,x,X1, Y1) = 0 =

=limt0 tK(t ,Tx,X2, Y2) = limt tK(t ,Tx,X2, Y2) = 0,

that is, T maps (X1, Y1) into (X2, Y2).In the case whereTL(X1,X2) = 0 we get the result either replacing everywhere

TL(X1,X2) by > 0 and then letting 0, or else replacing Xi by Yi for i = 1, 2and by 1 (see (1.4) and (1.5)).

Taking X1 = X2 = X, Y1 = Y2 = Y, it follows that (X, Y),p and (X, Y) areinterpolation spaces. Another important consequence is the next corollary.

Corollary 1.1.7 Let(X, Y) be an interpolation couple. For0<


12/125

12 Chapter 1

Proof. Set K = R or K = C, according to the fact that X, Y are real or complexBanach spaces. Let y X Y, and define T by T() = y for each K. ThenTL(K,X) =yX,TL(K,Y) =yY, andTL(K,(X,Y),p) =y(X,Y),p . The statementfollows now taking X1 = Y1 = K and X2 = X, Y2 = Y in theorem 1.1.6, and recallingthat (K,K),p = K.

Another more direct proof is the following: for y X Y\ {0}, we have K(t, y)min{yX, tyY}, so that

tyXyY =K(t, y)tyY =tK(t, y)t1yY y1X yY,

and

tyXyY =K(t, y) yX=tK(t, y)

yYyX

yX=y1X yY.

Thereforey(X,Y), = supt>0 tK(t, y) y1X yY , and the statement follows forp = + with constant c(, ) = 1. For p < + we already know that (X, Y),p iscontinuously embedded in (X, Y)

,, and the proof is complete.

1.1.1 Examples

Let us see some basic examples. Cb(Rn) is the space of the bounded continuous functions

in Rn, endowed with the sup norm ;C1b (Rn) is the subset of the continuously differen-tiable functions with bounded derivatives, endowed with the normf +

ni=1 Dif.

For(0, 1),Cb (Rn) is the set of the bounded and uniformly Holder continuous functions,endowed with the norm

fCb =f+ [f]C =f+ supx=y|f(x) f(y)|

|x

y

|

.

For (0, 1), p[1, ), W,p(Rn) is the space of all f Lp(Rn) such that

[f]W,p =

RnRn

|f(x) f(y)|p|x y|p+n dxdy

1/p


13/125


Moreover for x= y and again for every decomposition f = a+ b, with a Cb(Rn),bC1b (Rn), we have

|f(x) f(y)| |a(x) a(y)| + |b(x) b(y)| 2a+ bC1|x y|,so that, taking the infimum over all the decompositions,

|f(x) f(y)| 2K(|x y|, f , C b(Rn

), C1b (R

n

))2|x y|

f,.Thereforef is -Holder continuous andfC =f+ [f]C3f,.

Conversely, letfCb (Rn). Let C(Rn) be a nonnegative function with supportin the unit ball, such that

Rn

(x)dx= 1. For every t >0 set

bt(x) = 1

tn

Rn

f(y)

x y

t

dy, at(x) =f(x) bt(x), x Rn. (1.18)

Then

at(x) = 1

tn

Rn

(f(x) f(x y))

y

t

dy

so that at[f]C1

tnRn

|y|(y/t)dy= t[f]CRn

|w|(w)dw.

Moreover,bt f, and

Dibt(x) = 1

tn+1

Rn

f(y)Di

x y

t

dy.

SinceRn

Di((x y)/t)dy= 0, we get

Dibt(x) = 1

tn+1

Rn

(f(x y) f(x))Di

y

t

dy, (1.19)

which implies

Dibtt1[f]CRn

|w||Di(w)|dw.

Therefore,tK(t, f)t(at+ tbtC1)CfC , 0< t1.

For t1 (see remark (c) after definition 1.1.1), we can take at = f,bt = 0 which impliestK(t, f)tf f, t1.

The embeddingCb (Rn)(Cb(Rn), C1b (Rn)), follows.

The proof of the second statement is similar. We recall that for every b

W1,p(Rn)

and h Rn \ {0}we have (see e.g. [8])Rn

|b(x + h) b(x)||h|

pdx

1/p DbLp .

For every f (Lp(Rn), W1,p(Rn)),p and h Rn let a = a(h) Lp(Rn), b = b(h)W1,p(Rn) be such that f=a + b, and

aLp+ |h| bW1,p2K(|h|, f).


14/125

14 Chapter 1

Then |f(x + h) f(x)|p|h|p+n 2

p1

|a(x + h) a(x)|p|h|p+n +

|b(x + h) b(x)|p|h|p+n

so that Rn

|f(x + h) f(x)|p|h|p+n dx

2p1Rn

|a(x + h) a(x)|p|h|p+n +

|b(x + h) b(x)|p|h|p+n

dx

22p2apLp

|h|p+n + 2p1 |h|p |Db| pLp

|h|p+n

Cp|h|pn(aLp+ |h| bW1,p)p Cp|h|pnK(|h|, f).It follows that

RnRn

|f(x + h) f(x)|p|h|p+n dxdh

CpRn

|h|pnK(|h|, f)pdh

=Cp

0

K(r, f)p

rp+1 dr

B(0,1)

dn1= Cp,nfp,p.

Therefore, (Lp(Rn), W1,p(Rn)),p is continuously embedded in W,p(Rn). To be precise, we

have estimated so far only the seminorm [f],p. But we already know that each (X, Y),pis continuously embedded in X+ Y; in our case X+ Y = X = Lp(Rn) so that we havealsofLpCf,p.

To prove the other embedding, for each fW,p defineat and bt by (1.18). Then

atpLp =

Rn

Rn|f(y) f(x)|

1

tnx yt dyp

dx

Rn

Rn

|f(y) f(x)|p 1tn

x y

t

dydx.

were for p > 1 we applied the Holder inequality to the product

|f(x) f(y)|(tn((x y)/t))1/p (tn((x y)/t))11/p.So we get

0tpatpLp

dt

t

0

RnRn

|f(y) f(x)|p 1tn

x y

t

dydx

dt

t

=RnRn

|f(y) f(x)|p 0

tp1

tnx y

t

dtt

dydx

=

RnRn

|f(y) f(x)|p

|xy|tp

1

tn

x y

t

dt

t

dydx

p + n

RnRn

|f(y) f(x)|p|y x|p+n dxdy= C[f]

pW,p

.


15/125


Using (1.19) and arguing similarly, we get also0

t(1)pDibtpLpdt

t C

p1i Di

p + n [f]p

W,p,

with Ci=

Rn

|Di(y)|dy, whilebtLp fLpL1 =fLp .Therefore,tK(t ,f ,Lp, W1,p)

t

at

Lp+ t

1

bt

W1,p

Lp(0, 1), with norm esti-

mated by CfW,p , and the second part of the statement follows.

Note that the proof of (1.16) yields also

(L(Rn),Lip(Rn)),= (BU C(Rn), B U C 1(Rn)), = C

b (R

n).

We shall see later (3.2) another method to prove (1.16) and (1.17).Example 1.1.9 Let Rn be an open set with the following property: there exists anextension operatorEsuch thatEL(Cb(), Cb(Rn)) L(C1b (), C1b (Rn)), and also, forsome(0, 1),EL(Cb (), Cb (Rn))(by extension operator we mean thatE f|= f(x),for allf Cb()). Then

(Cb(), C1b ()), = C

b ().

Proof.Theorem 1.1.6 implies that

EL((Cb(), C1b ()),, (Cb(Rn), C1b (Rn)),).We know already that (Cb(R

n), C1b (Rn)), = C

b (R

n). So, for every f(Cb(), C1b ()),the extension Ef is in Cb (R

n) andEfCb (Rn) Cf(Cb(),C1b ()), . Since f = Ef|,then fCb () andfCb ()Cf(Cb(),C1b ()),.

Conversely, iff Cb () then Ef Cb (Rn) = (Cb(Rn), C1b (Rn)),. The retractionoperator Rg= g| belongs obviously to L(Cb(R

n), Cb())

L(C1b (R

n), C1b ()). Again by

theorem 1.1.6,f=R(Ef)(Cb(), C1b ()),, with norm not exceeding CEfCb (Rn)CfCb ().

Such a good extension operator exists if is an open set with uniformly C1 boundary. is said to be uniformly C1 if there are N N and a (at most) countable set of ballsBk whose interior parts cover, such that the intersection of more thanNof these ballsis empty, and diffeomorphisms k : Bk B(0, 1) Rn such that k(Bk) ={yB(0, 1) : yn 0}, andkC1+ 1k C1 are bounded by a constant independent ofk.(In particular, each bounded with C1 boundary has uniformlyC1 boundary).

It is sufficient to construct Ewhen = Rn+. The construction ofE for any open setwith uniformly C1 boundary will follow by the usual method of local straightening the

boundary.If = Rn+ ={x= (x, xn) Rn : xn > 0}we may use the reflection method: we set

Ef(x) =

f(x), xn0,

1f(x, xn) + 2f(x, 2xn), xn < 0,

where1,2satisfy the continuity condition1+2= 1 and the differentiability condition1 22= 1, that is 1= 3, 2=2.


16/125

16 Chapter 1

Then E L(C(Rn+), Cb(Rn)) L(C(Rn+), Cb (Rn)) L(C1(Rn+), C1b (Rn)), for every(0, 1).

Let now (, ) be a -finite measure space. To define the Lorentz spaces Lp,q() weintroduce the rearrangements as follows. For every measurablef : R or f : Cset

m(, f) =

{x

:

|f(x)

|>

},

0,

andf(t) = inf{: m(, f)t}, t0.

Both m(, f) and f are nonnegative, decreasing (i.e. nonincreasing), right continuous,and f,|f|are equi-measurable, that is for each 0> 0 we have

|{t >0 :f(t)> 0}|= m(0, f) ={x :|f(x)|> 0},

and consequently|{t >0 :f(t)[1, 2]}|= {x :|f(x)| [1, 2]}, etc. Therefore,for each p1,

|f(x)|p(dx) =

0 (f

(t))p

dt; sup ess|f(x)|= f

(0) = sup essf

(t), (1.20)

and for each measurable set E,E

|f(x)|(dx) = (E)0

f(t)dt.

f is called the nonincreasing rearrangement offonto (0, ).The Lorentz spaces Lp,q() (1p , 1q ) are defined by

Lp,q() =

fL1() + L() :fLp,q =

0(t1/pf(t))q

dt

t 1/q


17/125


The proof of (1.21) is based on the equality

K(t ,f ,L1(), L()) =

t0

f(s)ds, t >0. (1.22)

Once (1.22) is established, (1.21) follows easily. Indeed, since f is decreasing thenK(t, f)

tf(t), so that for q 0

tK(t, f)Lsupt>0

t1f(t)L =fL1/(1),().

The opposite inequality follows from the Hardy-Young inequality (A.10) (i) forq 0, x we set

a(x)

=f(x) f(t) f(x)|f(x)| , if|f(x)|> f

(t),

= 0 otherwise,

b(x) =f(x) a(x).

The function a is defined in such a way that|a(x)| =|f(x)| f(t) if|f(x)| > f(t),|a(x)|= 0 if|f(x)| f(t). Then

aL1 =

E(|f(x)| f(t))(dx),

where E ={x :|f(x)| > f(t)} has measure (E) =|{s > 0 :> f(s) > f(t)}|(because |f| andf are equi-measurable) =m(f(t), f)t, andf is constant in [(E), t].Therefore,

aL1 = (E)0

(f(s) f(t))ds t0

(f(s) f(t))ds.


18/125

18 Chapter 1

Moreover,

|b(x)|

=|f(x)| if|f(x)| f(t),

=f(t) if|f(x)|> f(t),so that

|b(x)

| f(t) =

1

t t

0

f(t)ds, x

.

Therefore,

K(t ,f ,L1, L) aL1+ tbL t0

f(s)ds.

To prove the opposite inequality we use the fact that for every decomposition f=a +bwe have (see exercise 7,1.1.2)

f(s)a((1 )s) + b(s), s0, 0<


19/125


6) Following the method of example 1.1.8 show that

(Lp(Rn), W1,p(Rn)),q =B

p,q(Rn),

defined by Bp,q(Rn) ={fLp(Rn) : [f]Bp,q


20/125

20 Chapter 1

with normuV(p,,Y,X) =uLp(0,+;Y)+ vLp(0,+;X).

Moreover, forp= + define a subspace ofV(, , Y , X ), byV0(, , Y , X ) ={uV(, , Y , X ) : lim

t0tu(t)X= lim

t0tu(t)Y = 0}.

It is not difficult to see that V(p, , Y, X) is a Banach space endowed with the norm V(p,,Y,X), and thatV0(, , Y , X ) is a closed subspace ofV(, , Y , X ). Moreover anyfunction belonging toV(p, , Y, X) has aX-valued continuous extension att = 0. Indeed,for 0< s < t from the equality u(t) u(s) = ts u()d it follows, for 1< p 0 there areatX,btY such thatatX+ tbtY2K(t, x). It holdst1btY2tK(t, x), and the function ttK(t, x) is in Lp(0, +). Moreover, wealready know (see the proof of proposition 1.1.3) that limt0 bt = x in X+Y. So, thefunction t bt looks a good candidate for u. But in general it is not measurable withvalues in Y, and it is not in W1,ploc(0, ) with values in X. So we have to modify it, andwe proceed as follows.

For every n N letanX, bnYbe such that an+ bn= x, and

anX+ 1n

bnY2K(1/n,x).

For t >0 set

u(t) =

n=1

bn+1( 1n+1 ,1n ]

(t) =

n=1

(x an+1)( 1n+1 ,1n ]

(t),

whereIis the characteristic function of the interval I, and

v(t) =1

t

t0

u(s)ds.


21/125


Since (X, Y),p is contained in (X, Y),thentK(t, x) is bounded, so that limt0 K(t, x)

= 0. Therefore, limn anX= 0, so thatx bnX+Y anX 0 as n , andx= limt0 u(t) = limt0 v(t) in X+ Y. Moreover,

t1u(t)Yt1

n=1

( 1n+1 ,1n ]

(t)2(n + 1)K(1/(n + 1), x)4tK(t, x), (1.23)

so that t t1u(t) Lp(0, +; Y). By Corollary A.3.1, t t1v(t) belongs toLp(0, +; Y), and

t1vLp(0,+;Y)41x,p.On the other hand,

v(t) =x 1t

t0

n=1

( 1n+1 ,1n ]

(s)an+1ds,

so thatv is differentiable almost everywhere with values in X, and

v(t) = 1

t2

t0

g(s)ds 1t

g(t),

whereg (t) = n=1 ( 1n+1 ,1n ](t)an+1 is such thatg(t)X

n=1

( 1n+1 ,1n ]

(t)2K(1/(n + 1), x)2K(t, x).

It follows that

t1v(t) t sup0


22/125

22 Chapter 1

Example 1.2.3 Choosing X = Lp(Rn), Y = W1,p(Rn), and = 1 1/p, 1 < p


23/125


2. Let x X+Y be the trace at t = 0 of a function v V(p, 1 , Y , X ). Fix anyC0 ([0, +)) such that1 in a right neighborhood of 0, say in (0, 1]. The functiont (t)v(t) is in V(p, 1 , Y , X ), its norm does not exceed CvV(p,1,Y,X), with Cdepending only on, and its trace att = 0 is still x. Moreover, it has compact support in[0, +). This shows that in the defintion of the trace spaces one could consider just thesubset ofV(p, 1 , Y , X ) consisting of the functions with compact support, obtaining anequivalent trace space (i.e., the same space with an equivalent norm).

By means of the trace method it is easy to prove some important density properties.

Proposition 1.2.5 Let 0 < < 1. For1 p


24/125

24 Chapter 1

and

supt>0

t1z(t)Y = sup00.

In this case we writeEK(X, Y).If(0, 1) this means thatE is continuously embedded in(X, Y),.

A useful characterization ofJ(X, Y) is the following one.


25/125


Proposition 1.3.2 Let0< 0,


26/125

26 Chapter 1

so that

Dif 2fh

+1

2Diifh, h > 0.

Taking the minimum on hover (0, +) we get

Dif2(f)1/2(Diif)1/2, fC2b (Rn)

so that

fC1b (f)1/2

(f)1/2 + 2

ni=1(Diif)1/2

C(f)1/2(fC2b )

1/2.

This implies thatC1b (Rn) belongs toJ1/2(Cb(R

n), C2b (Rn)). To prove that it belongs also to

K1/2(Cb(Rn), C2b (R

n)), namely that it is continuously embedded in (Cb(Rn), C2b (R

n))1/2,,we argue as in example 1.1.8: for every f C1b (Rn) the functions at, bt defined in (1.18)are easily seen to satisfy

at

C t[f]Lip,

bt

C1b

C

f

C1b

,

Dijbt

C t1[f]Lip.

Therefore,K(t ,f ,C b(Rn), C2b (R

n)) at1/2+ tbt1/2C2b C t1/2fC1b so thatC

1b (R

n)

is in K1/2(Cb(Rn), C2b (R

n)).But C1b (R

n) is not a real interpolation space between Cb(Rn) and C2b (R

n), even forn = 1. More precisely, there does not exist any constant C such thatTL(C1b (R))C(TL(C2b (R)))

1/2(TL(Cb(R)))1/2 for all T L(C2b (R)) L(Cb(R)).Indeed, consider the family of operators

(Tf)(x) =

11

x

x2 + y2 + 2(f(y) f(0)) dy, xR.

It is easy to see thatTL(Cb(R))andTL(C2b (R)) are bounded by a constant independentof. Indeed, for every continuous and bounded f,

|(Tf)(x)| 2 11

|x|x2 + y2 + 2

fdy2f,

(Tf)(x) =

11

x2 + y2 + 2(x2 + y2 + 2)2

(f(y) f(0))dy,

and for every fC1b (R),

(Tf)(x) =

1

1

2x(x2 + 3y2 + 32)(x

2

+ y2

+ 2

)3

y

0

f(s)dsdy

=

11

2x(x2 + 3y2 + 32)(x2 + y2 + 2)3

y0

(f(s) f(0))dsdy,

so that, iffC2b (R),

|(Tf)(x)| |x| 11

x2 + 3y2 + 32

(x2 + y2 + 2)2 dy f3f.


27/125


On the contrary, choosing f(x) = (x2 + 2)1/2(x), with C0 (R), 1 in [1, 1], we

get

(Tf)(0) =

11

(y2 + 2)1/2 y2 + 2

dy=1

1/1/

(s2 + 1)1/2 1s2 + 1

ds

so that lim0(Tf)(0) = +, while the C1b norm off is bounded by a constant inde-

pendent of. ThereforeTL(C1b (R)) blows up as 0. By theorem 1.1.6, C1

b (R

) cannotbe a real interpolation space between Cb(R) and C

2b (R).

This counterexample is due to Mitjagin and Semenov, it shows also thatC1([1, 1]) isnot a real interpolation space betweenC([1, 1]) andC2([1, 1]), and it may be obviouslyadapted to show that for any dimensionn,C1b (R

n) is not a real interpolation space betweenCb(R

n) and C2b (Rn).

Remark 1.3.4 Arguing similarly one sees that Ckb (Rn) is in J1/2(C

k1b (R

n), Ck+1b (Rn))

K1/2(Ck1b (Rn), Ck+1b (Rn)), for every k N. It follows easily that for m1 < k < m2N, Ckb (R

n) belongs to J(km1)/(m2m1)(Cm1b (R

n), Cm2b (Rn)). For instance, knowing that

C

1

b (R

n

) belongs toJ1/2(Cb(R

n

), C

2

b (R

n

)) andC

2

b (R

n

) belongs toJ1/2(C

1

b (R

n

), C

3

b (R

n

)) onegets, for every fC3b (Rn),

fC1b Cf1/2f1/2C2b C

f1/2 (f1/2C1b f1/2

C3b)1/2

so thatf3/4C1b

Cf1/2f1/4C3b , which implies

fC1b Cf2/3f1/3C3b

that is, C1b (Rn) belongs to J1/3(Cb(R

n), C3b (Rn)).

Now we are able to state the Reiteration Theorem. It is one of the main tools ofgeneral interpolation theory.

Theorem 1.3.5 Let00, 1 1, 0=1. Fix(0, 1) and set = (1 )0+ 1.The following statements hold true.

(i) IfEi belong to the classKi (i= 0, 1) betweenX andY, then

(E0, E1),p(X, Y),p, p[1, ], (E0, E1)(X, Y).

(ii) IfEi belong to the classJi (i= 0, 1) betweenX andY, then

(X, Y),p(E0, E1),p, p[1, ], (X, Y)(E0, E1).

Consequently, ifEi belong to Ki(X, Y) Ji(X, Y), then

(E0, E1),p = (X, Y),p, p[1, ], (E0, E1) = (X, Y),

with equivalence of the respective norms.


28/125

28 Chapter 1

Proof. Without loss of generality (recalling that (E1, E0),p = (E0, E1)1,p and (E1, E0)=(E0, E1)1) we may assume that 0< 1.

Let us prove statement (i). Let ki be such that K(t, x)kitixEi for every xEi,i= 0, 1. For each x(E0, E1),p, letaE0,bE1 be such that x = a + b. Then

K(t,x,X,Y)K(t,a,X,Y) + K(t,b,X,Y)k0t0aE0+ k1t1bE1.

Sincea and bare arbitrary, it follows that

K(t,x,X,Y)max{k0, k1}t0K(t10, x , E 0, E1).

Consequently,

tK(t,x,X,Y)max{k0, k1}t(10)K(t10, x , E 0, E1). (1.26)

By the change of variable s = t10 we see that t tK(t,x,X,Y) is in Lp(0, +),which means that x belongs to (X, Y),p, and

x

(X,Y),p

max

{k0, k1

}(1

0)

1/p

x

(E0,E1),p , ifp 0,

belongs to V(p, 1 , E0, E1): since g(0) = x, this will imply, through proposition 1.2.2,that x(E0, E1),p.

To this aim we preliminarly estimate

v(t)

Ei, i= 0, 1. Letci be such that

yEiciy1iX yiY yY, i= 0, 1.

Thenv(s)Ei

cisi+1

s1v(s)1iX s2v(s)iY, i= 0, 1,so that from the equalities

0+ 1 = 1 (1 0), 1+ 1 = 1 + (1 )(1 0),


29/125


we get (i) s1(10)v(s)Lp(0,+;E0)c0k xT r(X,Y),p ,

(ii) s1+(1)(10)v(s)Lp(0,+;E1)c0k xT r(X,Y),p .(1.27)

From the equality v(t) =

t v(s)dsand 1.27(ii), using the Hardy-Young inequality

(A.10)(ii) ifp 0

({y :|T f(y)|> })1/q MfLp(),

for all f Lp(). This is equivalent to say that the restriction of T to Lp() is abounded operator from Lp() to Lq,(). Indeed, by the properties of the nonincreasingrearrangements,

sup>0

({y :|g(y)|> })1/q = supt>0

t1/qg(t) =gLq,. (1.28)

T is said to be of strong type (p, q) if its restriction to Lp() is a bounded operatorfromLp() to Lq().

Since Lq() = Lq,q () Lq,(), then any operator of strong type (p, q) is also ofweak type(p, q).

Theorem 1.3.12 LetT :L1() + L()L1() + L() be of weak type(p0, q0) and(p1, q1), with constantsM0, M1 respectively, and

1p0, p1 , 1< q0, q1 ,q0=q1, p0q0, p1q1.

For every(0, 1) definep andqby1

p=

1 p0

+

p1,

1

q =

1 q0

+

q1.

ThenT is of strong type(p, q), and there isC independent of such that

T fLq()C M10 M1 fLp(), fLp().


32/125

32 Chapter 1

Proof. For i = 1, 2, T is bounded from Lpi() to Lqi,(), with norm not exceed-ing CMi. By the interpolation theorem 1.1.6,T is bounded from (L

p0(), Lp1()),p to(Lq0,(), Lq1,()),p, and

T(Lp0(),Lp1()),p,(Lq0,(),Lq1,()),p)C M10 M1 .On the other hand, we know from example 1.3.10 that

(Lp0(), Lp1()),p = Lp,p() =Lp(),

andLqi,() = (L1(), L())11/qi,, i= 1, 2

(it is here that we need qi >1: L1,() is not a real interpolation space between L1()

and L()), so that by the Reiteration Theorem

(Lq0,(), Lq1,()),p = (L1(), L())(1)(11/q0)+(11/q1),p

= (L1(), L())11/q,p.

The last space is nothing but Lq,p(), again by example 1.1.10. Since p0q0 andp1q1then pq, so that Lq,p()Lq,q () =Lq(). It follows that T is bounded from Lp()to Lq(), with norm not exceeding CM10 M

1 .

Theorem 1.3.12 is slightly less general than the complete Marcinkiewicz Theorem,which holds also for q0 or q1= 1.

Since everyTof strong type (p, q) is also of weak type (p, q) we may recover a part ofthe RieszThorin Theorem from theorem 1.3.12: we get that ifTis of strong type (pi, qi),i= 0, 1, with pi, qi subject to the restrictions in theorem 1.3.12, then T is of strong type(p, q). The full RieszThorin Theorem will be proved in Chapter 2.

Letfbe a locally integrable function. For each measurable subsetA

with positive

measure(A) we define the mean value off on A by

fA = 1

(A)

A

f(x)(dx).

The spaceBM O() (BM Ostands for bounded mean oscillation) consists of those locallyintegrable functions f such that

sup0


33/125


Example 1.3.13 Let be a bounded domain inRn with Lipschitz continuous boundary.Then for0<


34/125

34 Chapter 1

(i) if the coefficients aij are in C() for some (0, 1) and the functions fj are

in BM O(), then each derivative Diu belongs to BM O(), andDiuBMO,2Cn

j=0 fjBMO,2;

(ii) if the coefficientsaij are in C1+() for some (0, 1), fj = 0 and f0BM O(),

thenuH2() satisfies the equation a.e., each second order derivativeDijubelongsto B M O(), andDij uBMO,2Cf0BMO,2.

In case (i), applying theorem 1.3.14 with r = 2 to the operators Ti, i = 1, . . . , n,defined by Ti(f0, . . . , f n) = Diu, u being the solution of the Dirichlet problem, we getthat if the fjs are in L

p(), 2< p


35/125

Chapter 2

Complex interpolation

The complex interpolation method is due to Calderon [13]. It works in complex interpola-tion couples. It may sound artificial compared to the more natural real interpolationmethod of chapter 1, see next definitions 2.1.1 and 2.1.3. It is in fact an abstraction and ageneralization of the method used in the proof of the RieszThorin interpolation theorem,

which we show below.The theorem of RieszThorin. Let (, ), (, ) be -finite measure spaces. Let1p0, p1, q0, q1 and letT :Lp0()+ Lp1()Lq0()+ Lq1()be a linear operatorsuch that

T L(Lp0(), Lq0()) L(Lp1(), Lq1()).Then

T L(Lp(), Lq()), 0<


36/125

36 Chapter 2

whereb: C is any simple function. To this aim, for every zS={z = x + iy C :0x1}, we define

f(z)(x) =

|a(x)|p1zp0

+ zp1

a(x)|a(x)| , ifx, a(x)= 0,

0, ifx

, a(x) = 0,

g(z)(x) =

|b(x)|q

1zq0

+ zq1

b(x)|b(x)| , ifx, b(x)= 0,

0, ifx, b(x) = 0.Then f() = a, g() =b and for each z S, f(z)Lp() for every p, g(z)Lq() foreveryq. In particular,f(z)Lp0()Lp1() so thatT f(z)Lq0()Lq1(), and thefunction

F :S C, F(z) =

T f(z) g(z) (dx)

is well defined, holomorphic in the interior ofS, continuous and bounded in S, andF() =(T a)(x)b(x) (dx) is the integral that we want to estimate. For every t R we have|F(it)| T f(it)Lq0()g(it)Lq0() TL(Lp0(),Lq0())a

p/p0Lp ()b

q/q0

Lq ()

,

|F(1 + it)| T f(1 + it)Lq1()g(1 + it)Lq1()

TL(Lp1(),Lq1())ap/p1Lp ()bq/q

1

Lq ()

.

By the three lines theorem (see exercise 2,2.1.3) we get

|F()| =

T a b (dx) (suptR

|F(it)|)1(suptR

|F(1 + it)|)

T1L(Lp0(),Lq0())TL(Lp1(),Lq1())aLp ()bLq ().

SinceT aLq () is the supremum of|F()|/bLq () when b 0 runs in the set of thesimple functions on , we get

T aLq () T1L(Lp0(),Lq0())TL(Lp1(),Lq1 ())aLp (),

for every simple function a : C. Since the set of such as is dense in Lp() thestatement follows.

Taking in particularpi= qi, we get that ifT L(Lp0(), Lp0())L(Lp1(), Lp1())then T L(Lr(), Lr()) for every r[p0, p1].

The crucial part of the proof is the use of the three lines theorem for the function F.The explicit expression ofFis not important; what is important is that F is holomorphicin the interior ofS, continuous and bounded inS, thatF() leads to the norm T aLq (),and that the behavior ofF iniR and in 1+iR is controlled. Banach space valued functionsof this type are precisely those used in the construction of the complex interpolation spaces.


37/125

Complex interpolation 37

2.1 Definitions and properties

Throughout the section we shall use the maximum principle for holomorphic functions withvalues in a complex Banach space X: if is a bounded open subset ofCand f : Xis holomorphic in and continuous in , thenf()X max{f(z)X : z }, forevery . This is well known ifX = C, and may be recovered for general X by thefollowing argument. For every let x

X

be such thatf()X =f(), x

andxX = 1. Applying the maximum principle to the complex functionz f(z), x weget

f()X = |f(), x| max{|f(z), x|: z}

max{f(z)X : z}.The maximum principle holds also for functions defined in strips. Dealing with complex

interpolation, we shall consider the strip

S={z= x + iy C : 0x1}.

Iff :S

X is holomorphic in the interior ofS, continuous and bounded in S, then for

each Sf()Xmax{sup

tRf(it)X, sup

tRf(1 + it)X}.

See exercise 1,2.1.3.Let (X, Y) be an interpolation couple of complex Banach spaces.

Definition 2.1.1 LetSbe the strip{z =x+iy C: 0x1}.F(X, Y) is the spaceof all functionsf :SX+ Y such that

(i) f is holomorphic in the interior of the strip and continuous and bounded up to itsboundary, with values inX+ Y;

(ii) tf(it)Cb(R; X), tf(1 + it)Cb(R; Y), and

fF(X,Y) = max{suptR

f(it)X, suptR

f(1 + it)Y}


38/125

38 Chapter 2

in Cb(R; Y), then fn converges to f inF(X, Y). Moreover, sinceF0(X, Y) is closed inF(X, Y), thenF0(X, Y) is a Banach space too.

An important technical lemma about the space F0(X, Y) is the following one. Itsintricate proof is due to Calderon [13, p. 132-133]. A very detailed proof is in the book ofKreinPetuninSemenov [28, p. 217-220].

Lemma 2.1.2 The linear hull of the functions e

z2+z

a, > 0, R

, a X Y, isdense inF0(X, Y).

The complex interpolation spaces [X, Y]are defined through the traces of the functionsinF(X, Y).

Definition 2.1.3 For every[0, 1] set[X, Y] ={f() : f F(X, Y)}, a[X,Y] = inf

fF(X,Y), f()=afF(X,Y).

[X, Y] is isomorphic to the quotient spaceF(X, Y)/N, whereN is the subset of

F(X, Y) consisting of the functions which vanish atz = . Since

Nis closed, the quotient

space is a Banach space and so is [X, Y].Some immediate consequences of the definition are listed below.

(i) For every (0, 1),[X, Y] = [Y, X]1.

(ii) We get an equivalent definition of [X, Y] replacing the spaceF(X, Y) by the spaceF0(X, Y). Indeed, for eachf F(X, Y) and >0 the function f(z) =e(z)2f(z)is inF0(X, Y), f() =f(), andfF(X,Y) max{e2, e(1)2}fF(X,Y). Let-ting 0 we obtain also

inffF(X,Y), f()=a

f

F(X,Y) = inf

fF0(X,Y), f()=a

f

F(X,Y).

(iii) IfY =Xthen [X, X] = X, with identical norms (see exercise 1,2.1.3).(iv) For every t R and for every f F(X, Y), then f( +it) [X, Y] for each

(0, 1), andf(+ it)[X,Y] =f()[X,Y] . (this is easily seen replacing thefunctionf byg (z) =f(z+ it))

Finally, from lemma 2.1.2 it follows thatX Y is dense in [X, Y] for every (0, 1).In the present chapter, this fact will be used only in example 2.1.11.

Proposition 2.1.4 Let0<


39/125


The embedding [X, Y]X+Yfollows again from the maximum principle: ifa = f()with f F(X, Y) then

aX+Ymax{suptR f(it)X+Y, suptR f(1 + it)X+Y}

max{suptR f(it)X, suptR f(1 + it)Y}=fF(X,Y)

so thataX+Y a[X,Y] .

Remark 2.1.5 LetV(X, Y) be the linear hull of the functions of the type (z)x, with F0(C,C) andxX Y. IfaX Y, its [X, Y]-norm may be obtained also as

a[X,Y] = inf fV(X,Y), f()=a

fF(X,Y). (2.2)

Proof For each > 0 let f0 F0(X, Y) be such that f0() = a andf0F(X,Y)a[X,Y]+ . Letzr(z) be a function continuous in Sand holomorphic in the interiorofSwith values in the unit disk, and such that r() = 0, r()

= 0, r(z)

= 0 for z

= .

For instance, we may take

r(z) =z z+

, zS.

Set

f1(z) =f0(z) e(z)2a

r(z) , zS.

Thenf1 F0(X, Y), so that by lemma 2.1.2 there exists a function

f2(z) =n

k=1exp(kz

2 + kz)xk,

with k >0, k R,xkX Y, such thatf1 f2F(X,Y). Set

f(z) =e(z)2

a + r(z)f2(z), zS.

Thenf V(X, Y) and

fF(X,Y) f0F(X,Y)+ f f0F(X,Y) a[X,Y]+ 2.

Remark 2.1.5 will be used in theorem 2.1.7 and in theorem 4.2.6.

We prove now that the spaces [X, Y] are interpolation spaces.

Theorem 2.1.6 Let(X1, Y1), (X2, Y2)be complex interpolation couples. If a linear oper-atorT belongs to L(X1, X2) L(Y1, Y2), then the restriction ofT to [X1, Y1] belongs toL([X1, Y1], [X2, Y2]) for every(0, 1). Moreover,

TL([X1,Y1],[X2,Y2])(TL[X1,X2])1(TL[Y1,Y2]). (2.3)


40/125

40 Chapter 2

Proof First let TL(X1,X2)= 0 and TL(Y1,Y2)= 0. Ifa[X1, Y1], letf F(X1, Y1)be such that f() =a. Set

g(z) =

TL(X1,X2)TL(Y1,Y2)

zT f(z), zS.

Theng F(X2, Y2), andg(it)X2(TL(X1,X2))1(TL(Y1,Y2))f(it)X1,

g(1 + it)Y2(TL(X1,X2))1(TL(Y1,Y2))f(1 + it)Y1,so thatgF(X2,Y2) (TL(X1,X2))1(TL(Y1,Y2))fF(X1,Y1). ThereforeT a = g()[X2, Y2], and

T a[X2,Y2] gF(X2,Y2)(TL(X1,X2))1(TL(Y1,Y2))fF(X1,Y1).

Taking the infimum over all f F(X1, Y1) we get

T a[X2,Y2] (TL[X1,X2])1(TL[Y1,Y2])a[X1,Y1] .If eitherTL(X1,X2) orTL(Y1,Y2) vanishes, replace it by >0 in the definition ofg

and then let 0 to get the statement.If bothTL(X1,X2) andTL(Y1,Y2) vanish, set g (z) =T f(z). Theorem 2.1.6 has an interesting extension to linear operators depending on zS.

Theorem 2.1.7 Let(X1, Y1), (X2, Y2) be complex interpolation couples.For every z S let Tz L(X1 Y1, X2+ Y2) be such thatz Tzx is holomorphic

in S and continuous and bounded in S for every x X1 Y1, with values in X2+ Y2.Moreover assume thattTitxC(R; L(X1, X2)), tT1+itxC(R; L(Y1, Y2))and thatTitL(X1,X2),T1+itL(Y1,Y2) are bounded by a constant independent oft.Then, setting

M0= suptR

TitL(X1,X2), M1= suptR

T1+itL(Y1,Y2)

for every(0, 1) we have

Tx[X2,Y2] M10 M1x[X1,Y1]so that T has an extension belonging to L([X1, Y1], [X2, Y2]) (which we still call T)satisfying

T

L([X1,Y1],[X2,Y2])

M10 M

1 . (2.4)

Proof The proof is just a modification of the proof of theorem 2.1.6.Assume first thatM0 andM1 are positive. For every aX1 Y1 letf V(X1, Y1) be

such that f() =a (see remark 2.1.5), and set

g(z) =

M0M1

zTzf(z), zS.


41/125


Theng F(X2, Y2) andg(it)X2M10 M1f(it)X1,

g(1 + it)Y2M10 M1f(1 + it)Y1,so thatgF(X2,Y2)M10 M1 fF(X1,Y1). Therefore Ta= g()[X2, Y2], and

Ta[X2,Y2]M10 M1 inffV(X1,Y1), f()=a

fF(X1,Y1),

but by remark 2.1.5,

a[X1,Y1] = inf fV(X1,Y1), f()=a

fF(X1,Y1),

and the statement follows.If either M0 or M1 vanishes, replace it by in the definition ofg , and then let0.

If bothM0 andM1 vanish, define g byg(z) =Tzf(z) and follow the above arguments.

Let us come back to theorem 2.1.6 and to its consequences. The same proof of corollary1.1.7 (through the equality [C,C] = Cwith the same norm) yields

Corollary 2.1.8 For every(0, 1) we havey[X,Y] y1X yY, yX Y. (2.5)

Therefore, [X, Y]J(X, Y). This means that (X, Y),1[X, Y], thanks to propo-sition 1.3.2. It is also true that [X, Y] (X, Y),; to prove it we need the followinglemma, which gives a Poisson formula for holomorphic functions in a strip with values inBanach spaces.

Lemma 2.1.9 For every boundedf :SXwhich is continuous inSand holomorphicinSwe have

f(z) =f0(z) + f1(z), z= x + iyS,where

f0(z) =

R

e(yt) sin(x) f(it)

sin2(x) + (cos(x) exp((y t)))2dt,

f1(z) =

R

e(yt) sin(x) f(1 + it)

sin2(x) + (cos(x) + exp((y t)))2dt.(2.6)

Sketch of the proof Let first X= C. Then (2.6) may be obtained using the Poissonformula for the unit circle,

f() = 12

||=1

f() 1 ||2| |2d, ||< 1,(which holds for everyf which is holomorphic in the interior and continuous up to theboundary), and the conformal mapping

(z) =eiz ieiz + i

, zS,


42/125

42 Chapter 2

which transforms S into the unit circle. IfX is a general Banach space (2.6) follows asusual, considering the complex functions z f(z), x for every x X, and applying(2.6) to each of them.

Proposition 2.1.10 For every(0, 1), [X, Y]K(X, Y), that is [X, Y] is continu-ously embedded in(X, Y),.

Proof Leta[X, Y]. For everyf F(X, Y) such thatf() =asplita= f0()+f1()according to (2.6).

Note that for z = x + iySwe have

0 0 such that for eachf Lp0() Lp1(),suptR

T(t)fLq0 ()M0fLp0 (), suptR

T(t)fLq1 ()M1fLp1 ().

Then for each(0, 1) and for eachfLp0() Lp1() we have

TfLq ()M10 M

1 fLp ().

Therefore, T may be extended to a bounded operator (which we still callT) fromLp()

to Lq(), withp andq defined in (2.1), and

TL(Lp (),Lq ())M10 M1 .

Theorem 2.1.14 has a slightly sharper version, stated below, obtained modifying thedirect proof of the RieszThorin theorem. We recall that if (, ) is a measure space, asimple function is a (finite) linear combination of characteristic functions of measurable

sets with finite measure.

Theorem 2.1.15 Let (, ), (, ) be -finite measure spaces. Assume that for everyzS, Tz is a linear operator defined in the set of the simple functions on, with valuesinto measurable functions on, such that for every couple of simple functionsa: Candb: C, the productTza b is integrable on and

z

(Tzf)(x)g(x)(dx),

is continuous and bounded inS, holomorphic in the interior ofS.Assume moreover that for somepj, qj[1, +], j= 0, 1, we have

TitaLq0()M0aLp0(), T1+itaLq1()M0aLp1(), t R,for every simple functiona. Then for each (0, 1), T may be extended to a boundedoperator (which we still callT) fromL

p() to Lq(), withp and q defined in (2.1),and

TL(Lp (),Lq ())M10 M1 .

Proof The proof is just a modification of the proof of the RieszThorin theorem. Forevery couple of simple functions a C, b : C, we apply the three lines theorem tothe function

F(z) =

Tzf(z) g(z) (dx)

wheref and g are defined as in the proof of the RieszThorin theorem, i.e.

f(z)(x) =|a(x)|p1zp0

+ zp1

a(x)

|a(x)| , ifx, a(x)= 0;

f(z)(x) = 0, ifx, a(x) = 0.

g(z)(x) =|b(x)|p1zp0

+ zp1

b(x)

|b(x)| , ifx, b(x)= 0,


47/125


g(z)(x) = 0, ifx, b(x) = 0.We get

|F()|=

(Ta)(x)b(x) (dx)

M10 M1aLp ()bLq (),so that

T a

Lq

()M1

0 M

1aL

p

(),

for every simple a defined in . Since the set of such as is dense in Lp() the statementfollows.

2.1.3 Exercises

1)The maximum principle for functions defined on a strip. Letf :SXbe holomorphicin the interior ofS, continuous and bounded in S. Prove that for each S

f() max{suptR

f(it), suptR

f(1 + it)}.

(Hint: for each(0, 1) letz0be such thatf(z0) f(1); consider the functionsf(z) = exp((z z0)2)f(z), and apply the maximum principle in the rectangle [0 , 1] [M, M] with M large).2) The three lines theorem. Let f :SXbe holomorphic the interior ofS, continuousand bounded in S. Show that

f()X(suptiR

f(it)X)1(suptiR

f(1 + it)X), 0< 0properly).

3) Show that [X, X] =X, with identical norms.

4) Using Lemma 2.1.2 prove that X Y is dense in [X, Y] for every (0, 1).


48/125

48 Chapter 2


49/125

Chapter 3

Interpolation and domains of

operators

3.1 Operators with rays of minimal growth

Let Xbe a real or complex Banach space with norm . In this section we consider alinear operator A: D(A)XX such that

(A)(0, ), M :R(, A)L(X)M, >0. (3.1)Since(A) is not empty, then A is a closed operator, so that D(A) is a Banach space withthe graph normxD(A) =x + Ax. Moreover for every m N also Am is a closedoperator (see exercise 1,3.2.1).

This section is devoted to the study of the real interpolation spaces (X, D(A)),p, and,more generally, (X, D(Am)),p. Since for every t, R the graph norm of D(Am) isequivalent to the graph norm ofD(Bm) with B = eit(A + I), the case of an operator Bsatisfying

(B) {ei : > 0}, M : R(ei, B)L(X)M, > 0for some [0, 2), 0 0, may be easily reduced to this one. The halfliner ={ei : > 0} is said to be a ray of minimal growthof the resolvent ofB. See [2, Def. 2.1].Proposition 3.1.1 LetA satisfy (3.1). Then

(X, D(A)),p ={xX : () =AR(, A)x Lp(0, +)}and the normsx,p and

x,p =x + Lp(0,+)are equivalent.

Proof. Letx(X, D(A)),p. Then ifx= a+b with aX, bD(A), for every > 0we have

AR(, A)x AR(, A)a + R(, A)Ab

(M+ 1)a + M 1Ab

(M+ 1)(a + 1bD(A)),

49


50/125

50 Chapter 3

so that

AR(, A)x (M+ 1)K(1, x).With the changement of variables 1 we see that the right hand side belongs toLp(0, ), with norm equal to (M + 1)x,p. Therefore () = AR(, A)x is inLp(0, ), and

x

,p(M+ 1)

x

,p.

Conversely, ifLp(0, ), set for every 1

x= a+ b=AR(, A)x + R(, A)x,

so thatK(1, x)(AR(, A)x + 1R(, A)xD(A))

=(2AR(, A)x + R(, A)x).The right hand side belongs to Lp(1, ), with norm estimated by

2x,p+ M 1(1 )p1/p

x, p


51/125

Interpolation and domains of operators 51

Proof. It is sufficient to remark that (I A)k is an isomorphism from D(Ak) to X, andalso from D(Ak+1) to D(A). By the interpolation theorem 1.1.6, it is an isomorphismbetween (D(Ak), D(Ak+1)),p and (X, D(A)),p, and the statement follows.

It is also important to characterize the real interpolation spaces betweenXand D(A2),or more generally, between Xand D(Am). The following proposition is useful.

Proposition 3.1.4 LetA satisfy (3.1). ThenD(A)J1/2(X, D(A2))K1/2(X, D(A2)).

Proof.Let us prove that D(A)J1/2(X, D(A2)). For everyxD(A) it holdslim

R(, A)x = lim

R(, A)Ax + x= x.

Settingf() =R(, A)xfor >0, we have

f() =R(, A)x R(, A)2x= R(, A)(I R(, A))x=R(, A)2Axand f(+) =x, so that

x R(, A)x=

R(, A)2Axd, >0,

and ifxD(A2),

Ax= AR(, A)x

R(, A)2A2xd, >0.

Therefore,

Ax (M+ 1)x + M2

A2x, >0.

Taking the infimum for (0, ) we get

Ax 2M(M+ 1)1/2x1/2A2x1/2, xD(A2), (3.2)so that

xD(A)Cx1/2x1/2D(A2), xD(A2),that is, D(A)J1/2(X, D(A2)).

Let us prove that D(A)K1/2(X, D(A2)). For every xD(A) split x asx=R(, A)Ax + R(, A)x, >0,

where

R(, A)Ax MxD(A),

R(, A)xD(A2)=R(, A)x + AR(, A)Ax

Mx + (M+ 1)Axso that settingt= 2

K(t,x,X,D(A2)) R(t1/2, A)Ax + tt1/2R(t1/2, A)xD(A2)

M t1/2xD(A)+ M tx + (M+ 1)t1/2Ax, t >0


52/125

52 Chapter 3

which implies that tK(t,x,X,D(A2)) is bounded in (0, 1] by (2M+ 1)xD(A). Sinceit is bounded byx in (1, ), then x(X, D(A2))1/2, and

x(X,D(A2))1/2,(2M+ 1)xD(A).

But in general D(A) is not an interpolation space between X and D(A2). As a coun-terexample we may take X=Cb(R), A= realization of/x in X. See example 1.3.3.

As a corollary of proposition 3.1.4 we get a useful characterization of (X, D(A2)),p.

Proposition 3.1.5 LetA satisfy (3.1). Then for= 1/2

(X, D(A2)),p = DA(2, p).

Proof. Taking into account that X belongs to J0(X, D(A2)) K0(X, D(A2)) and D(A)

belongs toJ1/2(X, D(A2))

K1/2(X, D(A

2)), and applying the Reiteration Theorem with

E0=X, E1= D(A) we get

DA(, p) = (X, D(A)),p= (X, D(A2))/2,p, 0<


53/125


We know that K(, x) Lp(0, ). With the change of variable = 2 we getthat 2K(2, x) Lp(0, ), with norm equal to 21/px,p. Therefore() =2(AR(, A))2xis in Lp(0, ), and

xe,p21/p(M+ 1)2x,p.

(The formula is true also for p=

if we set 1/

= 0).Conversely, ifLp(0, ), from the obvious identity

x= 2R(, A)2x 2AR(, A)2x + A2R(, A)2x,

whereAR(, A)2x= ( A)AR(, A)3x

=AR(, A)2R(, A)2x R(, A)A2R(, A)2xwe get

x= (I 2AR(, A))2R(, A)2x + (2R(, A) + I)A2R(, A)2x, 1,

where

(I 2AR(, A))2R(, A)2xD(A2)

=(I 2AR(, A))2R(, A)2x + (I 2AR(, A))2A2R(, A)2x

(2M+ 3)M2x + (2M+ 3)2A2R(, A)2xand

(2R(, A) + I)A2R(, A)2x (2M+ 1)A2R(, A)2x.Therefore,

2

K(2

, x , X , D(A2

))

2((2R(, A) + I)A2R(, A)2x

+2(I 2AR(, A))2R(, A)2xD(A2))

(4M+ 4)2A2R(, A)2x + (2M+ 3)M222x.The right hand side belongs to Lp(1, ), with norm estimated by

(4M+ 3)xe,p + (2M+ 2)M2

1

(2

2)p

1/p

x,

which is true also for p = with the convention (1/)1/ = 1. It follows that ttK(t,x,X,D(A2))Lp(0, 1), and hence x(X, D(A2)),p and

x(X,D(A2)),pCp(xe,p + x).

Propositions 3.1.4 and 3.1.5 may be generalized as follows.


54/125

54 Chapter 3

Proposition 3.1.7 Let A satisfy (3.1), and let r, m N, r > m. Then D(Ar) Jr/m(X, D(A

m)) Kr/m(X, D(Am)).

Proposition 3.1.8 Let A satisfy (3.1), and let m N. Then for (0, 1) such thatm / N, and for1p

(X, D(A

m

)),p = DA(m,p).

3.1.1 Two or more operators

Let us consider now two operators A: D(A)X, B :D(B)X, both satisfying (3.1).Throughout the section we shall assume that A and B commute, in the sense that

R(, A)R(, B) =R(, B)R(, A), >0.

It follows that D(AkBh) = D(BhAk) for all natural numbers h, k, and that AkBhx =

Bh

Ak

xfor everyx in D(Ak

Bh

).

Definition 3.1.9 For everymN set

Km =m

j=0

D(AjBmj), xKm =x +m

j=0

AjBmjx.

The main result of the section is the following.

Theorem 3.1.10 Letm N, p [1, ] and (0, 1) be such that m is not integer,and setk= [m], ={m}. Then we have

(X, Km),p ={xKk : AjBkjxDA(, p) DB(, p), j = 0, . . . , k},

and the norms

x x(X,Km),p,

x x +kj=0(AjBkjxDA(,p)+ AjBkj xDA(,p))are equivalent.

The theorem will be proved in several steps. The first one is the case m= 1.

Proposition 3.1.11 For everyp[1, ] and(0, 1) we have

(X, K1),p = DA(, p) DB(, p),

and the norms

x(X,K1),p, xDA(,p)+ xDB(,p)are equivalent.


55/125


Proof. The embedding (X, K1),p DA(, p) DB(, p) is obvious, since K1 = D(A) D(B) is continuously embedded both in D(A) and in D(B).

LetxDA(, p) DB(, p). We recall (see proposition 3.1.1) that the functionsAR(, A)x, BR(, B)x, >0,

belong to Lp(0,

) and their norms are less than C

x

DA(,p), C

x

DB(,p), respectively.

For every >0 set

v() =2R(, A)R(, B)x, >0, (3.3)

and split x = x v() + v(). It holdsv() x R(, A)(R(, B)x x) + R(, A)x x

MBR(, B)x + AR(, A)x,and

v()K1 =v() + Av() + Bv()

M2

x + MAR(, A)x + MBR(, B)x.Therefore, for1

K(1,x,X,K 1)2M (AR(, A)x + BR(, B)x) + M21x,so that K(1, x , X , K 1) Lp(1, ), with norm estimated by const. (x+xDA(,p)+ xDB(,p)). Then K(,x,X,K1)Lp(0, 1), with the same norm,and the statement follows.

As a second step we show that

Proposition 3.1.12 For everyp[1, ] and(0, 1) we have

(K1, K2),p ={xK1 : Ax, BxDA(, p) DB(, p)}=DA(+ 1, p) DB(+ 1, p),

and the norms

x x(K1,K2),p ,

x x + AxDA(,p)+ AxDB(,p)+ BxDA(,p)+ BxDB(,p),

x xDA(+1,p)+ xDB(+1,p)are equivalent.

Proof. Let us prove the embeddings. Since K1 D(A) and K2 D(A2) then(K1, K2),p(D(A), D(A2)),p = DA(+ 1, p). Similarly, (K1, K2),pDB(+ 1, p). Itremains to show that eachx(K1, K2),p is such thatAxDB(, p) andBxDA(, p).For every aK1, bK2 such that x = a + b we have

BR(, B)Ax BR(, B)Aa + BR(, B)Ab

(M+ 1)Aa + M1BAb (M+ 1)(aK1+ 1bK2)


56/125

56 Chapter 3

so thatBR(, B)Ax (M+ 1)K(1, x , K 1, K2), >0.

It follows thatBR(, B)Ax Lp(0, ) with norm not exceeding (M+1)x(K1,K2),p ,and the embeddingis proved.

The proof of the embedding{xK1 : Ax, BxDA(, p) DB(, p)} (K1, K2),pis similar to the corresponding proof in proposition 3.1.11, and is omitted.

Let us prove that DA(+1, p)DB(+1, p) {xK1 : Ax, BxDA(, p)DB(, p)}.We have only to show that ifx DA(+ 1, p) DB(+ 1, p) then Ax DB(, p) andBxDA(, p). Indeed, for each >0 we have

B2R(, B)2Ax

+1B2R(, B)2R(, A)Ax + B2R(, B)2AR(, A)Ax

M(M+ 1)BR(, B)Bx + (M+ 1)2AR(, A)Axso thatB2R(, B)2Ax Lp(0, ) with norm not exceeding

M(M+ 1)Bx

DB(,p)+ (M+ 1)2

Ax

DA(,p).

Thanks to proposition 3.1.6, AxDB2(/2, p), which coincides with DB(, p) thanks toproposition 3.1.5. So, Ax DB(, p) andAxDB(,p) C(xDA(+1,p)+ xDB(+1,p).Similarly, BxDA(, p) andBxDA(,p)C(xDB(+1,p)+ xDA(+1,p)).

In the last part of the proof of proposition 3.1.12 we have shown that ifxDA(+1, p) DB(+ 1, p) then AxDB(, p) and BxDA(, p), a sort of mixed regularityresult. However it is not true in general that D(A2) D(B2)D(AB).

For instance, let A be the realization of/x and let B be the realization of /yin X = C(R2). Then DA( + 1, ) consists of the functions f X such that x f(x, y)C+1(R), uniformly with respect to y R, and similarly DB(+ 1, ) consistsof the functionsf Xsuch thatyf(x, y)C

+1

(R

), uniformly with respect to x R.Proposition 3.1.12 states that iffDA( +1, )DB( +1, ) thenf/xDB(, ),that is it is Holder continuous also with respect to y, and f/y DA(, ), that is itis Holder continuous also with respect to x. On the other hand, it is known that in thisexampleD(A2) D(B2) is not embedded in D(AB).

A similar proof yields

Proposition 3.1.13 For everyk N, p[1, ] and(0, 1) we have(Kk, Kk+1),p ={xKk : AjBkjxDA(, p) DB(, p), j = 0, . . . , k}

and the norms

x(Kk,Kk+1),p , x +kj=0(AjBkjxDA(,p)+ AjBkjxDB(,p))are equivalent.

Next step consists in proving that K1 belongs toJ1/2(X, K2) and to K1/2(X, K

2).

Proposition 3.1.14K1 J1/2(X, K2) K1/2(X, K2).


57/125


Proof.We already know that D(A)J1/2(X, D(A2)) and that D(B)J1/2(X, D(B2)).Therefore there isC >0 such that

xK1 xD(A)+ xD(B)

Cx1/2(x1/2D(A2)+ x1/2D(B2))Cx1/2x

1/2K2 ,

which means that K1

J1/2(X, K2

).To prove thatK1 K1/2(X, K2), for every xK1 we split again x = x v() + v()for every >0, where v is the function defined in (3.3). Then

v() x R(, A)(R(, B)x x) + R(, A)x x

=R(, A)R(, B)Bx + R(, A)Ax

1(M(M+ 1)Bx + MAx),and

v()K2 =v() + A2v() + ABv() + B2v()

M2x + 2M(M+ 1)(Ax + Bx).Setting= t1/2 we deduce that

t1/2K(t,x,X,K2)t1/2(x v(t1/2) + tv(t1/2)K2)

C(xK1+ t1/2x),is bounded in (0, 1). We know already that tt1/2K(t,x,X,K2) is bounded in [1, ).ThereforeK1 is in the class K1/2 between Xand K

2.

Arguing similarly one shows that

Proposition 3.1.15 For everyk N {0}, Kk+1 J1/2(Kk, Kk+2) K1/2(Kk, Kk+2).More generally, Kk+1 J1/s(Kk, Kk+s) K1/s(Kk, Kk+s).

The Reiteration Theorem and proposition 3.1.14 yield now

Proposition 3.1.16 Letp[1, ], (0, 1), = 1/2. Then(X, K2),p = DA(2, p) DB(2, p),

and for >1/2 we have also

(X, K2),p ={xK1 : Ax, BxDA(2 1, p) DB(2 1, p)},with equivalence of the respective norms.

Proof.For 1/2 we apply the ReiterationTheorem with Y = K2, E0 =K1, E1 =K

2, and the statement follows from proposition3.1.12.

The above proposition is a special case of theorem 3.1.10, with m = 2. Theorem 3.1.10in its full generality may be proved by recurrence, arguing similarly. See the exercises of3.1.2.

The results and the procedures of this section are easily extended to the case of a finitenumber of operators.


58/125

58 Chapter 3

3.1.2 Exercises

1) Let A : D(A) X X satisfy (3.1). Prove that for everym N, Am is a closedoperator. (Hint: use estimate (3.2)).

2) Prove proposition 3.1.7. Hint: to show thatD(Ar)Kr/m(X, D(Am)) prove prelimi-narly that D(Ar) K1/(mr)(D(Ar1), D(Am)), using a procedure similar to the one ofproposition 3.1.4, and then argue by reiteration.3) Prove proposition 3.1.8.

4) Prove proposition 3.1.13.

5) Prove proposition 3.1.15. Hint: for the first statement, follow step by step the proof of3.1.14; for the second statement replace v() byw() =2sR(, A)sR(, B)sx.

6) Prove theorem 3.1.10 by recurrence on m, using the procedure of proposition 3.1.16and the results of propositions 3.1.13 and 3.1.15.

7) Prove that (0, +) is a ray of minimal growth for the following operators:

(a) A : D(A) = C

1

b (R

) Cb(R

) (resp. A : D(A) = W

1,p

(R

) Lp

(R

), 1 p 0. (3.5)

Proposition 3.2.1 LetA generate a semigroup T(t). Then

(X, D(A)),p ={xX : t(t) =tT(t)x x Lp(0, )}

and the normsx,p andx,p =x + Lp(0,)

are equivalent.


59/125


Proof.Recall that for everybD(A) we have

T(t)b b= t0

AT(s)b ds=

t0

T(s)Ab ds, t >0.

Letx(X, D(A)),p. Then ifx = a + bwithaX,bD(A), for everyt >0 we have

t

T(t)x x t

(T(t)a a + T(t)b b)t((M+ 1)a + tMAb)(M+ 1)tK(t, x).

Therefore (t) =tT(t)x x Lp(0, ) andx,p(M+ 1)x,p.

Conversely, ifLp(0, ) let us use (3.5) to get

AR(, A)x 0

+1t+1etT(t)x x

tdt

t,

that is, is the multiplicative convolution between the functions f(t) = t+1et and(t) = tT(t)x x. Since f L1(0, ) and Lp(0, ), then Lp(0, ) andLp(0,) fL1(0,)Lp(0,), so that

x,p(+ 1)x,p,and the statement follows.

Proposition 3.2.2 Under the assumptions of proposition 3.2.1, for every(0, 1) andp[1, ] we have

(X, D(A2)),p ={

x

X : t(t) =t2(T(t) I)2x Lp(0, )}

and the normsx,p andxee,p =x + Lp(0,)

are equivalent.

Proof.Recall that for everybD(A2) we have

(T(t) I)2b= (T(t) I) t0

T()Abd=

t0

t0

T(s + )A2b ds d, t >0,

so that

(T(t)

I)2b

t2M

A2b

.

Letx(X, D(A2)),p. Then ifx = a + b with aX, bD(A2), for every t >0 we havet2(T(t) I)2x t2((T(t) I)2a + (T(t) I)2b)

t2((M+ 1)2a + t2M2A2b)so that

t2(T(t) I)2x (M+ 1)2t2K(t2, x).


60/125

60 Chapter 3

Therefore(t) =t2(T(t) I)2x Lp(0, ) andxee,p21/p(M+ 1)2x,p.

Conversely, letxbe such that(t)Lp(0, ). Then from (3.5) it follows that(AR(, A))

2

x= 2

0

0 e(t+s)

(T(t + s) T(t) T(s) + I)xdsdt

= 220

e2udu

2u0

(T(2u) T(t) T(2u t) + I)xdt

= 220

e2u(T(2u) 2T(u) + I)x 2u0

dtdu

+22 0

e2udu

2u0

(2T(u) T(t) T(2u t))xdt.

The first integral is nothing but

42 0

ue2u(T(u) I)2xdu.

To rewrite the second one we note that 2u0

(2T(u) T(t) T(2u t))dt= 2u0

(2T(u) 2T(t))dt

=

u0

+

2uu

(2T(u) 2T(t))dt

= u0

2(T(u t) I)T(t)dt + 2uu

2T(u)(I T(t u))dt

=

u0

2(T(s) I)T(u s)ds + u0

2T(u)(I T(s))ds

= 2

u0

(T(s) I)(T(u s) T(u))ds=2 u0

(T(s) I)2T(u s)ds.

Therefore,2(AR(, A))2x 4|(f

)()| + 2|(f

1)()|,

where stands for the multiplicative convolution and

f(u) =u2+2e2u,1(u) = Mu1+2

u0

(T(s) I)2xds.

Let us remark now that the Hardy-Young inequality (A.10)(i) implies that if a functionz is such that ttz(t)Lp(0, ) the same is true for its mean v(t) =t1

t0z(s)ds,

with

ttv(t)Lp(0,) 1

( + 1)ttz(t)Lp(0,),


61/125


and this is easily seen to be true also for p = . Therefore,1 Lp(0, ) and 1Lp(0,) (2+ 1)1 Lp(0,). It follows that () =2(AR(, A))2x Lp(0, ), and

Lp(0,)=x,p4fL1(0,)(

Lp(0,)+

1Lp(0,))Cpx,p,

and the statement follows.

Remark 3.2.3 In the proof of proposition 3.2.1 we have not used the fact that T(t) isstrongly continuous or that the domain ofA is dense. The only essential assumption isthat T(t) is a semigroup such thatT(t)L(X)Mand for >0 the operators

R() =

0

etT(t)dt

are well defined and invertible. Indeed, in that case due to the semigroup property R()satisfies the resolvent identity R() R() = ( )R()R(), for , > 0. From the

general spectral theory it follows that there exists a unique closed operator A such that(A)(0, ) andR() =R(, A), for every >0. The results of propositions 3.2.1 and3.2.2 hold also for such semigroups.

The operator A may still be called generator of T(t), even if it is the infinitesimalgenerator in the usual sense if and only ifT(t) is strongly continuous.

This is the case of the translations semigroups (Ti(t)f)(x) =f(x + tei) inX=Cb(Rn),

of the Gauss-Weierstrass semigroup

P(t)f(x) = 1

(4t)n/2

Rn

e|y|2

4t f(x y)dy,

again in Cb(Rn), of the Ornstein-Uhlenbeck semigroup

T(t)f(x) = 1

(4t)n/2(det Kt)1/2

Rn

e|K

1/2t y|

2

4t f(etBx y)dy,

with Q0, B= 0 arbitrary n n matrices,

Kt=1

t

t0

esBQesB

ds,

both in Cb(Rn) and in B U C(Rn), etc. None of these semigroups is strongly continuous.

A useful embedding result in applications to PDEs is the following.

Theorem 3.2.4 Let T(t) be a semigroup in X. Assume moreover that there exists aBanach spaceEX andm N, 0< 0 such that

T(t)L(X,E) Ctm, t >0,

and that t T(t)x is measurable with values in E, for each x X. Then E J(X, D(A

m)), so that(X, D(Am)),p(X, E),p, for every(0, 1), p[1, ].


62/125

62 Chapter 3

Proof.Let xD(Am), >0 and set (I A)mx= y. Then x= (R(, A))my so that

x=(1)m1(m 1)!

dm1

dm1R(, A)y=

1

(m 1)! 0

essm1T(s)y ds,

so that for every >0

xE C(m 1)! 0 essm(1)1dsy= C(m(1 ))(m 1)! mmy

=C(m(1 ))

(m 1)! mm

mr=0

m

r

mr(1)rArx

C mr=0

mrAru.

Let us recall that D(Ar) belongs to Jm/r(X, D(Am)) so that there is C such that

xD(Ar)Cxr/mD(Am)x1r/mX . Using such inequalities and thenabC(ap +bp

) with

p= n/r,p =r/(n r) we getxEC m(muD(Am)+ u), >0,

so that taking the minimum for >0

xECu1uD(Am)and the statement holds.

3.2.1 Examples and applications. Schauder type theorems

Example 3.2.5 Let us apply propositions 3.2.1, 3.2.2 to the caseX=Lp(R), 1p


63/125


which coincides with W1,p(R) only for p= 2.Choosing X=Cb(R), A: D(A) =C

1b (R)Cb(R), Af=f, we get, recalling remark

3.2.3,(Cb(R), C

2b (R)),= C

2b (R), = 1/2,

(Cb(R), C2b (R))1/2, =

= fCb(R) : supt=0, xR |f(x + 2t) 2f(x + t) + f(x)|t 0, f X, x Rn, so that M = 1 for every i, and R(, Ai)R(, Aj) =R(, Aj)R(, Ai) for every i, j. We apply theorem 3.1.10 for those such that m isnot integer, m= k+ , k= [m], 0<


64/125

64 Chapter 3

Proof. The embeddings are easy consequences of example 3.2.6. Indeed, let X =Lp(Rn), 1 p 0, x Rn. (3.6)

P(t) may be seen as a (strongly continuous) semigroup in X=Lp

(Rn

), 1p 0.

In particular,

P(t)L(Lp,W1,p)C

1 + 1

t1/2

, t >0,

P(t)

L(Lp,W3,p)

C1 +

1

t1/2

+1

t

+ 1

t3/2, t >0,

(3.7)

and similarlyP(t)L(BU C(Rn),BUC1(Rn))C

1 +

1

t1/2

, t >0,

P(t)L(BU C(Rn),BUC3(Rn))C

1 + 1

t1/2+

1

t+

1

t3/2

, t >0.

(3.8)

ReplacingP(t) byT(t) =P(t)et (the semigroup generated by A I) we get

T(t)L(Lp,W1,p)

C

t1/2, t >0,

T(t)L(Lp,W3,p) C

t3/2, t >0,

and T(t)L(BU C(Rn),BUC1(Rn))

C

t1/2, t >0,

T(t)L(BU C(Rn),BUC3(Rn)) C

t3/2, t >0.


65/125


LetX=Lp(Rn), 1p


66/125

66 Chapter 3


67/125

Chapter 4

Powers of positive operators

The powers (with real or complex exponents) of positive operators are important tools inthe study of partial differential equations. The theory of powers of operators is very closeto interpolation theory, even if in general the domain of a power of a positive operator isnot an interpolation space.

Through the whole chapter Xis a complex Banach space.

4.1 Definitions and general properties

Definition 4.1.1 A linear operatorA: D(A)XX is said to be a positive operatorif the resolvent set ofA contains(, 0] and there isM >0 such that

R(, A)L(X) M1 + || , 0. (4.1)

Note that A is a positive operator iff (, 0) is a ray of minimal growth for theresolvent R(, A) and 0(A). So, ifA is a positive operator, thenAsatisfies (3.1) sothat all the results of3.1 are applicable.

Examples of unbounded positive operators are readily given: for instance, the realiza-tion of the first order derivative with Dirichlet boundary condition at x = 0 in C([0, 1])or in Lp(0, 1), 1p is positive. More generally, ifA is the generator of a stronglycontinuous or analytic semigroup T(t) such thatT(t) M et for some > 0, thenA is a positive operator. This can be easily seen from the already mentioned resolventformula

R(, A) =R(, A) = 0

etT(t)dt, >.

This section is devoted to the construction and to the main properties of the powers

Az

, where z is an arbitrary complex number.IfA : XXis a bounded positive operator the powers Az are readily defined by

Az = 1

2i

zR(, A)d,

where is any piecewise smooth curve surrounding (A), avoiding (, 0], with index1 with respect to every element of(A). Several properties ofAz follow easily from thedefinition: for instance, z Az is holomorphic with values in L(X); ifz = k Z then

67


68/125

68 Chapter 4

Az defined above coincides with Ak; for each z, w C we have AzAw = AwAz =Az+w;(A1)z =Az , etc.

In the case where A is unbounded the theory is much more complicated. To defineAz

we shall use an elementary but important spectral property, stated in the next lemma.

Lemma 4.1.2 LetA be a positive operator. Then the resolvent set ofA contains the set

={C : Re0,|Im|< (|Re| + 1)/M} { C : ||< 1/M},whereMis the number in formula (4.1), and for every0(0, arctan 1/M),r0(0, 1/M)there isM0> 0 such that

R(, A) M01 + ||

for allC with|| r0, and for all C withRe 0, let r, be the curve defined by r, =(1)r, (2)r, +(3)r, ,where

(1)r, ,

(3)r, are the half lines parametrized respectively by z= e

i,z= ei, r,and

(2)r, is the arc of circle parametrized by z= re

i,. See the figure.

0

r,

r

Fig. 1. The curve r, .

Now we are ready to define Az for Re z


69/125

Powers of positive operators 69

Since R(, A) is holomorphic in \ (, 0] with values in L(X), the integralis an element ofL(X) independent ofr and . Writing down the integral we get

A = 1

2i

r

(ei(+1)R(ei, A) + ei(+1)R(ei, A))d

r+1

2 ei(+1)R(rei, A)d(4.3)

for every r(0, 1/M), ( arctan 1/M,).Of course formula (4.3) may be reworked to get simpler expressions for A. For

instance, if1< Re 0 and (1, 0) we have

a

=sin()

0

+ ad, (4.5)

which agrees with (4.4) of course, and will be used later.From the definition it follows immediately that the function z Az is holomorphic

in the half plane Re z


70/125

70 Chapter 4

rk

Fig. 2. The curve k.

For every k N the function R(, A) is holomorphic in the bounded regionsurrounded by k. For every k Nwe have

1

2i

k

nR(, A)d= 1(n 1)!

dn1

dn1R(, A)

=0

= An,

and lettingk , 12i

r,

nR(, A)d= An.

(ii) Letk= 1, Re z r1 > r2 > 0, so that r1,1 is on the right hand side ofr2,2. Then

Az1Az2 = 1

(2i)2

r1,1

z1R(, A)d

r2,2

wz2R(w, A)dw

= 1

(2i)2

r1,1r2,2

z1wz2R(, A) R(w, A)

w ddw

= 1(2i)2

r1,1

z1R(, A)dr2,2

wz2

w dw

1(2i)2

r2,2

wz2R(w, A)dw

r1,1

z1

w d

= 1

2i

r1,1

z1+z2R(, A)d= Az1+z2 .


71/125


Statement (iv) of the proposition implies immediately thatAz is one to one. Indeed, ifAzx= 0 and n Nis such thatn < Rez, then Anx= AnzAzx= 0, so that x= 0.Therefore it is possible to define A if Re > 0 as the inverse ofA. But in this waythe powers Ait,t R, remain undefined. So we give a unified definition for Re 0.

Definition 4.1.5 Let0 Re < n, n N. We set

D(A) ={xX : AnxD(An)}, Ax= AnAnx.

From proposition 4.1.4 it follows that the operator A is independent of n: indeed,ifn, m > Re , then Amx = AnmAnx both for n < m (by proposition 4.1.4(iv))and for n > m (by proposition 4.1.4(ii), taking z = n and k = nm), so thatAmxD(Am) iffAnmAnxD(Am) i.e. AnxD(An).

For = 0 we get immediately A0 =I. Moreover for Re >0 we get

D(A) =A(X); A = (A)1.

Indeed, Anx D(An) iff there is y X with Anx = Any. Such a y is obviouslyunique, and Ax = y by definition. MoreoverAnAy = AAny = AAnx =Anx so that x = Ay is in the range ofA and A = (A)1.

SinceA has a bounded inverse, then it is a closed operator, so that D(A) is a Banachspace endowed with the graph norm. Again, since A has a bounded inverse, its graphnorm is equivalent to

x Ax,which is usually considered the canonical norm ofD(A).

If Re = 0, = it with t R, Ait is the inverse ofAit in the sense that for eachx

D(Ait),Aitx

D(Ait) andAitAitx= x. Indeed, ifx

D(Ait) thenAit1x

D(A),

and Aitx = A(Ait1x) by definition. Therefore A1itAitx = A1itAAit1x = AA1itAit1x= A A2xD(A), which implies that AitxD(Ait) and AitAitx= x.

But in general the operators Ait are not bounded, see next example 4.2.1. However,they are closed operators, becauseA1+it is bounded and A is closed (see next exercise 6,4.2.1). Therefore also D(Ait) is a Banach space under the graph norm.

From the definition it follows easily that for 0 Re < n N, the domain D(An)is continuously embedded in D(A): indeed for each x D(An), Anx D(An) byproposition 4.1.4(iii), andAx= AnAnx= AnAnxso thatAx An Anx.This property is generalized in the next theorem.

Theorem 4.1.6 Let , C be such thatRe < Re . Then D(A) D(A), andfor everyxD(A),

Ax= AAx.

Moreover for eachxD(A), AxD(A) and

AAx= Ax.

Conversely, ifxD(A) andAxD(A), thenxD(A) and againAAx =Ax.


72/125

72 Chapter 4

Proof. The embedding D(A) D(A) is obvious if Re < 0; it has to be proved forRe0.

IfxD(A), An+xD(An) for n > Re . ThereforeAn+x= AAn+xD(An), thanks to proposition 4.1.4(iii), so thatxD(A), andAx= AnAAn+x=AAx. Since A is a bounded operator,Ax AL(X)Ax, and D(A)is continuously embedded in D(A).

Let again xD(A

), and let n >max{Re , Re ( )}. ThenAn+Ax= An+AAx= AnAxD(An),

so thatAxD(A) and AAx= Ax.Let nowxD(A) be such that AxD(A), and fix n >max{Re , Re }.

ThenA2nx= AnAn+x= AnAnAx= AnAnAx

is in D(A2n), so that xD(A) and Ax= A2nA2nx= AAx. The condition Re < Re is essential in the above theorem when Re >0. In fact

for every > 0, t

R we have D(A) = D(A+it) if and only if Ait is bounded. See

exercise 2,4.2.1.Now we give some representation formulas for AxwhenxD(A). We consider first

the case where 0< Re < 1. Takingn = 1 in the definition, we see that xD(A) ifand only ifA1xD(A). Letting r0 and in the representation formula (4.3)for A1x(i.e., using formula (4.4) with replaced by 1) we get

A1x=sin()

0

1(I+ A)1xd. (4.6)

ThereforexD(A) if and only if the integral 0 1(I+ A)1x dis in the domainofA, and in this case

Ax= sin() A 0 1(I+ A)1x d=

1

()(1 ) A 0

1(I+ A)1xd,

(4.7)

which is the well-known Balakrishnan formula.Another important representation formula holds for1 < Re < 1. The starting

point is again formula (4.3) for A1x. We let and then we integrate by parts inthe integrals between r and, getting

A1x=sin()

r(I+ A)2x d r sin()

(rI+ A)1x

r

2

ei(reiI+ A)1x d

(with (sin())/() replaced by 1 if = 0) and letting r 0 we get (both for Re(0, 1) and for Re (1, 0])

A1x= 1

(1 )(1 + ) 0

(I+ A)2xd. (4.8)


73/125


ThereforexD(A) if and only if the integral 0 (I+ A)2x dis in the domain ofA, and in this case

Ax= 1

(1 + )(1 ) A0

(I+ A)2xd. (4.9)

The most general formula of this type may be found as usual in the book of Triebel:for n

N

{0}

,mN,

n < Re < m

nwe have

Ax= (m)

( + n)(m n )Amn

0

t+n1(tI+ A)mx dt

for every xD(A). See [36,1.5.1].We already know that the domain D(A) is continuously embedded in D(A) for Re

[0, 1). With the aid of the representation formulas (4.7) and (4.9) we are able to provemore precise embedding properties ofD(A).

Proposition 4.1.7 For0< Re 0

A1(I+ A)1x= Re1A(I+ A)1xand for every x (X, D(A))Re,1 the function ReA(A+I)1x is in L1(0, )thanks to proposition 3.1.1. Using the representation formula (4.6) for A1x, we getA1xD(A), i.e. xD(A) and by (4.7)

Ax 1|()(1 )| 0

ReA(A + I)1xd Cx(X,D(A))Re,1.

Let nowxD(A). Then x = Ay, withy = Ax, so that x = A A1y. We usethe representation formula (4.8) forA1y, that gives

x=

A

(1 )(1 + )

0 t

(A + tI)2

y dt.

On the other hand, by proposition 3.1.1 we have

x(X,D(A))Re,C()sup>0

ReA(A + I)1x,

so that

x(X,D(A))Re,C()sup>0ReA2(A + I)1(1 )(1 + )

0

t(A + tI)2y dt

.For every >0 we have

ReA2(A + I)1

0

t(A + tI)2y dtRe M

1 +

0

tRe(M+ 1)2y dt

+Re(M+ 1)

tReM(M+ 1)

1 + t ydt

Cy


74/125

74 Chapter 4

so thatx(X, D(A))Re, andx(X,D(A))Re,Cy= CAx,

which implies that D(A)(X, D(A))Re,.

Remark 4.1.8 Arguing similarly (using formula (4.9) instead of (4.7)) we see easily thatfor every(0, 1) andt R, (X, D(A)),1is contained inD(Ait). Indeed the functionitA(I+A)2x M(1+)1A(I+A)1x is inL1(0, ) for everyxin (X, D(A)),1,so that the integral

0

it(I+ A)2x dbelongs to the domain ofA. Therefore, for every(0, 1) andp[1, ],t R, (X, D(A)),p is continuously embedded inD(Ait) (becauseit is continuously embedded in (X, D(A))/2,1).

Remark 4.1.9 Let 0 < < 1. It is possible to show that in its turn A is a positiveoperator, and that

R(, A) = 1

2i

r,

R(z, A)

z dz , 0. (4.10)

(see exercise 5,4.2.1). Using the above formula for the resolvent, one shows that A

is asectorial operator. This may be surprising, sinceAis not necessarily sectorial. This alsomay help in avoiding mistakes driven by intuition. Consider for instance the case whereX=L2(0, ) and A is the realization ofd2/dx2 with Dirichlet boundary condition, i.e.A: D(A) =H2(0, ) H10 (0, ), Au =u. One could think that A1/2 is a realization ofid/dxwith some boundary condition, but this cannot be true because such operators arenot sectorial. See next example 4.3.10.

4.1.1 Powers of nonnegative operators

A part of the theory of powers of positive operators may be extended to nonnegativeoperators.

Definition 4.1.10 A linear operatorA: D(A)XXis said to be nonnegative if theresolvent set ofA contains(, 0) and there isM >0 such that

(I+ A)1 M

, >0.

In other words, Ais a nonnegative operator iff(, 0)is a ray of minimal growth for theresolvent ofA.

An important example of nonnegative operator is the realizationA of (the Laplaceoperator) in Lp(Rn), 1 p . But A is not positive because 0 (A). However ifp


75/125


for eachxD(A) R(A) (note that in the case where 0(A), Bzxcoincides with Azxsince formula (4.11) is obtained easily from (4.9)). Then one checks that Bz : D(A)R(A)His closable, and defines Az as the closure ofBz .

Another way to define A for 0< 0 one defines (I+ A)1 by

R= sin() 0

2 +

An Introduction to Interpolation Theory, 2007 [Lunardi_A.]

Documents

Transcript of An Introduction to Interpolation Theory, 2007 [Lunardi_A.]