Lecture Notes on Function Spaces

Home , Interpolation space, Schwartz space

space invaders

Karoline Götze

TU Darmstadt, WS 2010/2011 Contents

1 Basic Notions in Interpolation Theory 5

2 The K-Method 15

3 The Trace Method 19 3.1 Weighted Lp spaces ...... 19 3.2 The spaces V (p, θ, X0,X1) ...... 20 3.3 Real interpolation by the trace method and equivalence ...... 21

4 The Reiteration Theorem 26

5 Complex Interpolation 31 5.1 X-valued holomorphic functions ...... 32 5.2 The spaces [X,Y ]θ and basic properties ...... 33 5.3 The complex interpolation functor ...... 36

5.4 The space [X,Y ]θ is of class Jθ and of class Kθ...... 37 6 Examples 41 6.1 Complex interpolation of Lp -spaces ...... 41 6.2 Real interpolation of Lp-spaces ...... 43 6.2.1 Lorentz spaces ...... 43 6.2.2 Lorentz spaces and the K-functional ...... 45 6.2.3 The Marcinkiewicz Theorem ...... 48 6.3 Hölder spaces ...... 49 6.4 Slobodeckii spaces ...... 51 6.5 Functions on domains ...... 54

7 Function Spaces from Fourier Analysis 56

8 A Trace Theorem 68

9 More Spaces 76 9.1 Quasi-norms ...... 76 9.2 Semi-norms and Homogeneous Spaces ...... 76

9.3 Orlicz Spaces LA ...... 78 9.4 Hardy Spaces ...... 80 9.5 The Space BMO ...... 81

2 Motivation

Examples of function spaces:

n 1 α p k,p s,p s C(R ; R),C ([−1, 1]),C (Ω; C),L (X, ν),W (Ω),W ,Bp,q,

n where Ω domain in R , α ∈ R+, 1 ≤ p, q ≤ ∞, (X, ν) measure space, k ∈ N, s ∈ R+. Philosophy: Objectify function spaces Example: We can consider H1/2 as

• interpolation space: L2 ∩ H1 ,→ H1/2 ,→ L2 + H1, H1/2 = F(L2,H1)

• embedded space: H1 ⊂ H1/2 ⊂ L2

• space of functions f ∈ L2 with one half of a derivative: |ξ|1/2fˆ ∈ L2 trace space: 1 n 1/2 n−1 is bounded and surjective. • γ : H (R+) → H (R )

3 Books

Which topics in the lecture? Author: Title, Reference Where in the Notes? Comment Adams/Fournier Sobolev Sobolev Spaces, Orlicz Spaces Sections 6.5, 9.3 Spaces, [1] Functions on Domains Bennett/Sharpley: very nice elaborate proofs Sections 6.2, 9.4, 9.5 Interpolation of Operators, [2] Bergh/Löfström: often short proofs, Chapters 7, 8 Interpolation Spaces, [3] nice structure for us Section 9.2 Lunardi: Interpolation very good introduction, Chapters 2,3,4,5 Theory, [5] very nice proofs Lunardi: Analytic Semigroups book on a dierent topic, and Optimal Regularity of Section 6.3 for us: Hölder spaces Parabolic Problems, [4] Tartar: An Introduction to nd history, references and Sobolev Spaces and inbetween ideas Interpolation Spaces, [6] Triebel: Interpolation Theory, nd everything, dicult for Chapter 1, everywhere Function Spaces, Dierential proofs inbetween Operators, [7]

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Picture:

Let A, B be linear Hausdor spaces and A0,A1 ⊂ A, B0,B1 ⊂ B be Banach spaces. Let be a linear operator such that and are T : A → B T |A0 : A0 → B0 T |A1 : A1 → B1 bounded. Then:

•{A0,A1}, {B0,B1} are called interpolation couples.

• Question: Can we nd Banach spaces A ⊂ A, B ⊂ B, such that for all T as above, T ∈ L(A, B)?

• If the answer is yes, then A, B are said to have the interpolation property with respect to {A0,A1}, {B0,B1}.

• We will often consider the special case A = B, A0 = B0, A1 = B1. In short: A has the interpolation property with respect to {A0,A1}.

Example 1.1. Let (X, µ) be a complete measure space, where µ is σ-nite. We write p p L = L (X, µ) for the Lebesgue spaces of complex or real-valued functions f : X → C or f : X → R. It holds that

1. Lpθ has the interpolation property with respect to {Lp0 ,Lp1 } if 1 = 1−θ + θ , pθ p0 p1 0 < θ < 1.

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1 1 2. The space C ([−1, 1]) (= C ([−1, 1]; R)) does not have the interpolation property with respect to {C([−1, 1]),C2([−1, 1])}. Theorem 1.2. (Convexity Theorem of Riesz/Thorin)

Let 1 ≤ p0, p1, q0, q1 ≤ ∞, p0 6= p1, q0 6= q1 and T a linear operator such that T : Lpi (X, µ) → Lqi (Y, ν), i ∈ (0, 1) is bounded linear. Then for every 0 < θ < 1,

T : Lpθ (X, µ) → Lqθ (Y, ν) is linear and bounded, for 1 = 1−θ + θ , 1 = 1−θ + θ . Furthermore, the estimate pθ p0 p1 qθ q0 q1

1−θ θ p q kT kL(L θ ,L θ ) ≤ CkT kL(Lp0 ,Lq0 )kT kL(Lp1 ,Lq1 ) holds true.

Remark 1.3. About the above theorem:

• If the Lp are complex-valued, C = 1. If they are real-valued, C = 2.

• Example 1 immediately follows from Theorem 1.2.

• Riesz 1926, Thorin 1939/48: Interpolation result which existed before interpolation theory. The direct proof contains ideas for general constructions of interpolation spaces (As and Bs in the picture). For us, this means that the proof will be given in a later Chapter :-).

Why convexity: The theorem shows that the function given by 1 1 • f f( p , q ) = kT kL(Lp,Lq) is logarithmically convex, i.e.

1 1 1 1 1 1 1 1 f (1 − θ)( , ) + θ( , ) ≤ f( , )1−θf( , )θ p0 q0 p1 q1 p0 q0 p1 q1 or, in other words, g = log f is convex.

k Reminder: f ∈ C ([−1, 1]) ⇔ f :[−1, 1] → R is k-times continuously dierentiable and Pk 1 (k) (we can replace by ). kfkCk([−1,1]) := l=0 l! supx∈[−1,1] |f (x)| < ∞ sup max Theorem 1.4. (Mitjagin/Semenov '76)

For every ε ∈ (0, 1] let Vε : C([−1, 1]) → C([−1, 1]) be given by ¢ 1 x (1.1) (Vεf)(x) := 2 2 2 (f(y) − f(0)) dy. −1 x + y + ε Then for all ε ∈ (0, 1], it holds that

∞ 1. Vε ∈ C ([−1, 1]),

2. kVεkL(C([−1,1])) < 2π,

6 1 Basic Notions in Interpolation Theory

3. kVεkL(C2([−1,1])) < 5π + 2,

4. given p 2 2 , we get , but 0 1 . fε(y) = y + ε − ε kfεkC1([−1,1]) ≤ 2 (Vεf) (0) > 2 ln( 5ε ) Corollary 1.5. From Theorem 1.4 we get Example 2.

Proof. The proof of the corollary will be given as an exercise. Proof of Theorem 1.4: In the following, we write Ck for Ck([−1, 1]) and C0 for C([−1, 1])

1. Dierentiation in the integral in (1.1).

2. Calculate: ¢ 1 |x| |V f(x)| ≤ 2kfk 0 dy ε x2 + y2 + ε2 C −1 ¢ 1 1 ≤ 4kfkC0 |x| 2 2 dy symmetry 0 x + y 1 y 1 ≤ 4kfk 0 |x|[ arctan( )] C |x| |x| 0

≤ 2πkfkC0 .

3. Identity map: h(y) = y, then ¢ 1 xy (1.2) (Vεh)(x) = 2 2 2 dy = 0 −1 x + y + ε

for all x ∈ [−1, 1]. Taylor Theorem: If f ∈ C2, then

0 f(y) = f(0) + f (0)y + r2(f, y) (1.3)

0 for some r2(f, ·) ∈ C and

00 |f (ϑy)| 2 2 |r (f, y)| = y ≤ kfk 2 y . (1.4) 2 2 C It follows from (1.2) and (1.3) that ¢ 1 x (V f)(s) = [f(y) − f(0) − f 0(0)y] dy ε x2 + y2 + ε2 ¢−1 1 x = 2 2 2 r2(f, y) dy. −1 x + y + ε Note that

d x y2 + ε2 − x2 1 1 | ( )| = | | < < dx x2 + y2 + ε2 (x2 + y2 + ε2)2 x2 + y2 + ε2 y2 + ε2

7 1 Basic Notions in Interpolation Theory

and d2 x 2|x||x2 − 3y2 − 3ε2| 6|x| | ( )| = < . dx2 x2 + y2 + ε2 (x2 + y2 + ε2)3 (x2 + y2 + ε2)2 In conclusion, from (1.4), we get ¢ 1 0 d x |(Vεf) (x))| ≤ ( ) |r2(f, y)| dy dx x2 + y2 + ε2 −1 ¢ 1 y2 ≤ kfkC2 2 2 dy ≤ 2kfkC2 −1 y + ε and ¢ 1 00 6|x| 2 |(V f) (x))| ≤ kfk 2 y dy ε (x2 + y2 + ε2)2 C −1 ¢ 1 |x| ≤ 12kfkC2 2 2 dy ≤ 6πkfkC2 , 0 x + y

2 so kVεfkC2 ≤ (2π + 2 + 3π)kfkC2 for all f ∈ C .

4. We see: The fε approximate | · |. Note that

ε2 1 f 00(0) = | = −→ε→0 ∞ ε (y2 + ε2)3/2 y=0 ε

2 2 y2 0 (i.e. Vε : C → C is ok!). Moreover, |fε(y)| = √ < |y| ≤ 1 and |f (y)| = y2+ε2+ε ε √ |y| < 1. The interesting part is: y2+ε2 ¢ 1 y2 + ε2 − x2 p (V f )0(x) = ( y2 + ε2 − ε) dy ε ε (x2 + y2 + ε2)2 ¢−1 1 py2 + ε2 − ε x==0 dy y2 + ε2 −¢1 √ u= y 1/ε u2 + 1 − 1 =ε 2 du u2 + 1 ¢0 ¢ 1/ε 1 1/ε 1 = 2 √ du − 2 du u2 + 1 u2 + 1 ¢0 0 1/ε 1 1 ≥ 2 du − 2 arctan( ) 0 u + 1 ε 1 1 + 1 1 ≥ 2 ln(1 + ) − π > 2 ln( ε ) > 2 ln( ). ε eπ/2 5ε

8 1 Basic Notions in Interpolation Theory

Notation: We write A,→ B i id : A → B is bounded.

Lemma 1.6. Let {A0,A1} be an interpolation couple. Then

A0 + A1 = {a ∈ A : ∃a0 ∈ A0, ∃a1 ∈ A1, a = a0 + a1} with the norm

kakA0+A1 = inf (ka0kA0 + ka1kA1 ) a=a0+a1,ai∈Ai and

A0 ∩ A1 = {a ∈ A : a ∈ A0, a ∈ A1} with the norm

kakA0∩A1 = max(kakA0 , kakA1 ) are Banach spaces. It holds that

A0 ∩ A1 ,→ Ai ,→ A0 + A1.

Proof. Exercise.

Denition 1.7. (Basic denitions in category theory)

1. A category consists of

a) a class of objects A, B, C, . . . and b) a class of pairwise disjoint non-empty sets [A, B]. Each ordered pair (A, B) uniquely corresponds to a set [A, B]. The elements in [A, B] are called morphisms.

2. For every ordered triplet (A, B, C) of objects, we have the composition of morphisms via V :[B,C] × [A, B] → [A, C]. Notation: f ∈ [A, B], g ∈ [B,C], then gf = V (g, f). Moreover, we have associa- tivity f ∈ [A, B], g ∈ [B,C], h ∈ [C,D], then (hg)f = h(gf)

and for all objects A, there exists an identity idA ∈ [A, A], such that for all f ∈ [B,A], g ∈ [A, B], idAf = f and gidA = g.

3. Let C1, C2 be two categories. A map F : C1 → C2 is called a (covariant) functor, if

a) for all objects A in C2, F(A) is an object in C1,

9 1 Basic Notions in Interpolation Theory

b) for all morphisms f ∈ [A, B] in C2, F(f) ∈ [F(A), F(B)] is a morphism in C1,

c) F(idA) = idF(A) for all objects A in C2, d) F(gf) = F(g)F(f) for all morphisms f ∈ [A, B], g ∈ [B,C].

Example 1.8. The notions categories are strong in some areas of mathematics, like geometry. For more examples, check e.g. Wikipedia :). For us, the following are relevant,

• Category C1: (complex) Banach spaces A, B, C, . . . are objects, bounded linear operators T ∈ [A, B] = L(A, B) are morphisms.

• Category C2: interpolation couples {A0,A1}, {B0,B1},... are objects and

[{A0,A1}, {B0,B1}] = L({A0,A1}, {B0,B1})

are sets of morphisms, where T : A0 + A1 → B0 + B1, T ∈ L({A0,A1}, {B0,B1}) if , are bounded. T |A0 : A0 → B0 T |A1 : A1 → B1

Remark 1.9. The space L({A0,A1}, {B0,B1}) is a Banach space with the norm

kT kL({A0,A1},{B0,B1}) = max(kT |A0 kL(A0,B0), kT |A1 kL(A1,B1)) and

L({A0,A1}, {B0,B1}) ,→ L(A0 + A1,B0 + B1). (1.5)

Proof. Let

D = {(U, V ): U ∈ L(A0,B0),V ∈ L(A1,B1),U = V on A0 ∩ A1}

⊂ L(A0,B0) × L(A1,B1), where

k(U, V )kD = k(U, V )kL(A0,B0)×L(A1,B1) = max(kUkL(A0,B0), kV kL(A1,B1)). We show that

1. D is a closed subspace of L(A0,B0) × L(A1,B1), i.e. it is a Banach space,

2. D is isometrically isomorphic to L({A0,A1}, {B0,B1}).

Regarding 1.: Let (Un,Vn) → (U, V ) ∈ L({A0,A1}, {B0,B1}) for some (Un,Vn)n ⊂ D, so

max(kU − UnkL(A0,B0), kV − VnkL(A1,B1)) → 0.

We have to show that U = V on A0 ∩ A1. For every a ∈ A0 ∩ A1,

Ua − V a = (U − Un)a + (Un − Vn)a + (Vn − V )a

10 1 Basic Notions in Interpolation Theory

where Un − Vn ≡ 0, (U − Un)a → 0 in B0 and (Vn − V )a → 0 in B1. By embedding, (U − Un)a + (Vn − V )a → 0 in B0 + B1, so U = V on A0 ∩ A1. Regarding 2.: We consider , given by . j : L({A0,A1}, {A1,B1}) → D j(T ) = (T |A0 ,T |A1 ) It follows that j is linear and isometric, therefore injective. It remains to show that j is surjective. Let (U,˜ V˜ ) ∈ D. We dene T˜ : A0 + A1 → B0 + B1 by

T˜ a = T˜(a0 + a1) = Ua˜ 0 + Ua˜ 1. ˜ is well-dened: Let 0 0 , then 0 0 , so T a0 + a1 = a = a0 + a1 a0 − a0 = a1 − a1 ∈ A0 ∩ A1 ˜ ˜ ˜ ˜ 0 ˜ 0 ˜ 0 ˜ 0 ˜ 0 0 T (a0 + a1) = Ua0 + V a1 = U(a0 − a0) + Ua0 + V a1 − V (a1 − a1) = T (a0 + a1) since U˜ = V˜ on A0 ∩ A1. It is clear that T˜ ∈ L({A0,A1}, {B0,B1}) and j(T˜) = (U,˜ V˜ ). It remains to show (1.5). For all a = a0 + a1, a0 ∈ A0, a1 ∈ A1, we have

kT akB0+B1 = inf (kb0kB0 + kb1kB1 ) T a=b0+b1,bi∈Bi

≤ kT a0kB0 + kT a1kB1 ≤ max kT |A0 kL(A0,B0), kT |A1 kL(A1,B1) (ka0kA0 + ka1kA1 ).

We take the inmum over a = a0 + a1 to get that

kT akB0+B1 ≤ kT kL({A0,A1},{B0,B1})kakA0+A1 .

We are now in a position to give a formulation of interpolation of Banach spaces in the language of categories.

Denition 1.10. A functor F : C2 → C1 is called interpolation functor, if for every {A0,A1}, {B0,B1} in C2,

A0 ∩ A1 ,→ F({A0,A1}) ,→ A0 + A1 and

T ∈ L({A0,A1}, {B0,B1}) ⇒ F(T ) = T |F({A0,A1}).

If F is an interpolation functor, then the space F({A0,A1}) is called interpolation space with respect to {A0,A1}.

Note that this denition ts the picture at the beginning of the chapter, as if F({A0,A1}) is an interpolation space in the sense of the denition, it also has the interpolation property with respect to {A0,A1}. It is interesting that also something like the opposite holds true. If {A0,A1} is an interpolation couple and there is a Banach spaces A such that A0 ∩ A1 ,→ A,→ A0 + A1 and such that Im(T |A) ⊂ A for all T ∈ L({A0,A1}, {A0,A1)}), then there exists an interpolation functor F0 such that A = F({A0,A1}). The proof of this fact is an exercise, given Theorem 1.14 at the end of this chapter.

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Denition 1.11. An interpolation functor F is called of type θ, 0 < θ < 1, if there exists a constant C > 0, such that for all {A0,A1}, {B0,B1},T in C2, kT k ≤ CkT k1−θ kT kθ . L(F({A0,A1},F{B0,B1}) L(A0,B0) L(A1,B1) If we can choose C = 1, F is called exact. Note: It is always true that C ≥ 1.

Even if we do not know whether a functor is of type θ or exact, the following theorem shows that we always have some estimate of this type, uniformly in T , but depending on the interpolation space.

Theorem 1.12. Let F be an interpolation functor, {A0,A1}, {B0,B1} interpolation couples and A = F({A0,A1}), B = F({B0,B1}) interpolation spaces. Then there exists a constant C(A, B) > 0 such that kT kL(A,B) ≤ C(A, B) max kT kL(A0,B0), kT kL(A1,B1) for all T ∈ L({A0,A1}, {B0,B1}).

In order to prove this theorem, we need the following lemma, whose proof is left as an exercise.

Lemma 1.13. Let U, V, X, Y be Banach spaces such that U,→ X, V,→ Y and S : U → V such that S ∈ L(X,Y ). Then we also get S ∈ L(U, V ).

Proof. of Theorem 1.12. We use Lemma 1.13 with U = L({A0,A1}, {B0,B1}), V = L(A, B), X = L(A0 + A1,B0 + B1), Y = L(A, B0 + B1) and S : L(A0 + A1,B0 + B1) → L(A, B0 + B1) given by S(T ) = T |A. In view of (1.5), we only need to verify that V,→ Y and that S is bounded. The rst can be derived directly from the embedding B,→ B0 + B1. the boundedness of S follows from the embedding A,→ A0 + A1. More precisely, for all a ∈ A,

kT akB0+B1 ≤ kT kL(A0+A1,B0+B1)kakA0+A1

≤ C(A)kT kL(A0+A1,B0+B1)kakA, so . Clearly, the constants involved do not kST kL(A,B0+B1) ≤ C(A)kT kL(A0+A1,B0+B1) depend on T .

As a closing to this abstract chapter, we look at the following theorem, justiying the use of interpolation functors for interpolation theory. Theorem 1.14. (Aronszajn/Gagliardo '65)

Let C1, C2 be as in Example 1.8, let {A0,A1} be an interpolation couple and let A be such that

A0 ∩ A1 ,→ A,→ A0 + A1 and T (A) ⊂ A for all T ∈ L({A0,A1}). Then there exists an interpolation functor F0 : C2 → C1 such that F0({A0,A1}) = A.

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Proof. We give a proof in seven small steps.

1. Preliminary observation: By Lemma 1.13, T (A) ⊂ A implies that T ∈ L(A).

2. Construction of F0: Let {X0,X1} be an object in C2. We dene

∞ X X = F0({X0,X1}) = {x ∈ X0 + X1 : x = Tjaj abs. conv. , j=1 aj ∈ A, Tj ∈ L({A0,A1}, {X0,X1}), j ∈ N, where ∞ X kxkX = inf kTjkL({A ,A },{X ,X })kajkA < ∞}. x=P∞ T a 0 1 0 1 j=1 j j j=1

It follows that k · kX is a norm for X.

3. Show X0 ∩ X1 ,→ X: Let ϕ : A0 + A1 → C a bounded linear functional such that ∗ ∗ ∗ ϕ(a ) = 1 for some a ∈ A. Dene kϕk = c . For every x ∈ X0 ∩ X1, we dene

Tx : A0 + A1 → X0 + X1,Txa := ϕ(a)x.

It follows that for i ∈ {0, 1} and a ∈ Ai,

∗ kTxakXi = |ϕ(a)|kxkXi ≤ c kakA0+A1 kxkXi ≤ ckakAi kxkXi ,

so Tx ∈ L({A0,A1}, {X0,X1}) and

kTxkL({A ,A },{X ,X }) ≤ c max(kTxkL(A ,X )) ≤ c max kxkX = ckxkX ∩X . 0 1 0 1 i i i i i 0 1

∗ Now for every x ∈ X0 ∩ X1, x = Txa ∈ X and

∗ ∗ kxkX ≤ kTxkL({A0,A1},{X0,X1})ka kA ≤ cka kAkxkX0∩X1 .

4. Show that : Let P∞ . Then by Remark 1.9, X,→ X0 + X1 x = j=1 Tjaj ∈ X

∞ X kxkX0+X1 ≤ kTjajkX0+X1 j=1 ∞ X ≤ kTjkL(A0+A1,X0+X1)kajkA0+A1 j=1 ∞ X ≤ c kTjkL({A0,A1},{X0,X1})kajkA. j=1

Taking the , we get . inf kxkX0+X1 ≤ ckxkX

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5. is complete: If for some , P∞ then by Step 4, X (xn)n ⊂ X n=1 kxnkX < ∞ P∞ , so that the limit P∞ exists in . Moreover, n=1 kxnkX0+X1 < ∞ n=1 xn = x X0 +X1 by denition, for all , there exist n , n , such n ∈ N Tj ∈ L({A0,A1}, {X0,X1}) aj ∈ A that P∞ n n and xn = j=1 Tj aj

∞ X n n −n kTj kL({A0,A1},{X0,X1})kaj kA < kxnkX + 2 , j=1

so P∞ P∞ n n. Taking the , we get P∞ . It x = n=1 j=1 Tj aj inf kxkX < n=1 kxnkX + 1 follows that Pm m→∞ in . n=1 xn → x X

6. F0 on morphisms in C2: Let {X0,X1}, {Y0,Y1} be objects in C2, and let S ∈ L({X0,X1}, {Y0,Y1}) be a morphism. We set X = F0({X0,X1}), Y = F0({Y0,Y1}) and . We need to show that . Let P∞ , F0(S) := S|X F0(S) ∈ L(X,Y ) x = j=1 Tjaj where , . Then P∞ , where Tj ∈ L({A0,A1}, {X0,X1}) aj ∈ A Sx = j=1 STjaj STj ∈ L({A0,A1}, {Y0,Y1}). It follows that for every suitable choice of Tj, aj,

∞ X kSxkY ≤ kSTjkL({A0,A1},{Y0,Y1})kajkA j=1 ∞ X ≤ kSkL({X0,X1},{Y0,Y1}) kTjkL({A0,A1},{X0,X1})kajkA. j=1 Taking the , we get . inf kSxkY ≤ kSkL({X0,X1},{Y0,Y1})kxkX

7. We show that F0({A0,A1}) = A: It is clear that A ⊂ F0({A0,A1}). For the opposite inclusion, let P∞ be in as above. By Theorem a = j=1 Tjaj F0({A0,A1}) 1.12, there exists a constant c(A), such that for all j ∈ N,

∞ ∞ X X kakA ≤ kTjajkA ≤ c(A) kTjkL({A0,A1})kajkA. j=1 j=1 Taking the inmum, we get . kakA ≤ c(A)kakF0({A0,A1})

14 2 The K-Method

Denition 2.1. Let {A0,A1} an interpolation couple. Then (Peetre's) K-functional

K : R+ × A0 + A1 → R is dened as

K(t, a; A0,A1) = inf (ka0kA0 + tka1kA1 ). a=a0+a1,ai∈Ai

Notation: K(t, a) instead of K(t, a; A0,A1).

Lemma 2.2. Fix a ∈ A0 + A1. Then K(·, a) is positive, monotonely increasing, concave and continuous. It holds that (2.1) min(1, t)kakA0+A1 ≤ K(t, a) ≤ max(1, t)kakA0+A1 .

Proof. We see immediately: Positivity, monotonicity and (2.1). It remains to show that K is concave in t, i.e. for all 0 < λ < 1,

K ((1 − λ)t1 + λt2, a) ≥ (1 − λ)K(t1, a) + λK(t2, a). Let 0 < t < t < t < ∞ and λ = t−t1 , so that 1 − λ = t2−t and t = (1 − λ)t + λt . 1 2 t2−t1 t2−t1 1 2 For every a0 ∈ A0, a1 ∈ A1 such that a0 + a1 = a, we have

t2 − t t − t1 (ka0kA0 + t1ka1kA1 ) + (ka0kA0 + t2ka1kA1 ) = ka0kA0 + tka1kA1 . t2 − t1 t2 − t1 We rst take the inf on the left hand side to get

t2 − t t − t1 K(t1, a) + K(t2, a) ≤ ka0kA0 + tka1kA1 . t2 − t1 t2 − t1 Now taking the inf on the right hand side gives

t2 − t t − t1 K(t1, a) + K(t2, a) ≤ K(t, a). t2 − t1 t2 − t1 Since K(·, a) is concave and monotonely increasing, it is continuous.

Denition 2.3. Let {X0,X1} an interpolation couple, 0 < θ < 1 and 1 ≤ q ≤ ∞. Then we dene

(X0,X1)θ,q := {x ∈ X0 + X1 : kxk(X0,X1)θ,q < ∞}, where ( ¡ 1 ∞ −θ q dt q kxk = 0 [t K(t, x)] t , q < ∞, (X0,X1)θ,q −θ sup0

15 2 The K-Method

Theorem 2.4. Let {X0,X1} an interpolation couple, 0 < θ < 1 and 1 ≤ q ≤ ∞. Then

1. the space (X0,X1)θ,q is an interpolation space with respect to {X0,X1}, i.e. there exists an interpolation functor Kθ,q : C2 → C1 such that Kθ,q({X0,X1}) = (X0,X1)θ,q. The functor Kθ,q is exact and of type θ.

2. For all x ∈ (X0,X1)θ,q,

θ (2.2) K(t, x) ≤ cθ,qt kxk(X0,X1)θ,q .

Proof. In the following, let . We rst show (2.2). The case follows X = (X0,X1)θ,q ¡ q = ∞ immediately. Otherwise, we rst write −θq ∞ −θq dt . It follows that s = θq s t t ¢ ∞ 1 1 dt q s−θK(s, x) = (θq) q K(s, x) t−θq s t ¢ ∞ 1 1 dt q ≤ (θq) q t−θqK(t, x)q s t 1 ≤ (θq) q kxkX.

Next, we show 1. in four steps.

1. k · kX is a norm:

(2.1) • kxkX = 0 ⇒ K(t, x) ≡ 0 ⇒ x = 0,

• K(·, λx) = |λ|K(·, x) ⇒ kλxkX = |λ|kxkX , Minkowski • K(t, x0 + x1) ≤ K(t, x1) + K(t, x2) ⇒ kx0 + x1kX ≤ kx0kX + kx1kX .

2. X is complete: Let (xn)n be a Cauchy sequence in X. By (2.2) and since K(1, ·) = , has a limit in . Assume now , analogous proof k · kX0+X1 (xn)n x X0 + X1 q < ∞ works otherwise. For every , let be such that ε for every ε > 0 n0(ε) kxm − xnk < 2 m > n > n0(ε). Moreover, let L > l > 0. Then by monotonicity of K and (2.1), there is m0(ε, θ, l, L), such that for all m ≥ m0(ε, θ, l, L), m > n,

¢ 1 ¢ 1 L q L q −θ q dt ε −θ q dt [t K(t, x − xn)] ≤ + [t K(t, x − xm)] l t 2 l t ¢ 1 L q ε −θq dt ≤ + Lkx − xmkX0+X1 t 2 l t ε 1 1 ≤ + L( ) q l−θkx − x k < ε. 2 θq m X0+X1

Passing to the limit L → ∞, l → 0 yields the claim.

16 2 The K-Method

3. Show that X0 ∩ X1 ,→ X,→ X0 + X1: Let x ∈ X0 ∩ X1, then K(t, x) ≤ . W.l.o.g., let . Then min(1, t)kxkX0∩X1 q < ∞ ¢ ¢ 1 dt ∞ dt kxkq = [t−θK(t, x)]q + [t−θK(t, x)]q X t t 0 ¢ 1 ¢ 1 ∞ (1−θ)q dt −θq dt ≤ kxkX0∩X1 ( t + t ) 0 t 1 t

≤ Cθ,qkxkX0∩X1 .

(2.2) Clearly, . kxkX0+X1 = K(1, x) ≤ cθ,qkxkX

4. Kθ,q is an exact interpolation functor of type θ: Let T ∈ L({X0,X1}, {Y0,Y1}), T 6= 0, x ∈ X0 + X1, then

K(t, T a; Y0,Y1) = inf (ky0kY0 + tky1kY1 ) T x=y0+y1,yi∈Yi

≤ inf (kT x0kY0 + tkT x1kY1 ) x=x0+x1,xi∈Xi

≤ inf (kT kL(X0,Y0)kx0kX0 + tkT kL(X1,Y1)kx1kX1 ) x=x0+x1,xi∈Xi

kT kL(X1,Y1) = kT kL(X ,Y ) inf (kx0kX + t kx1kX ). 0 0 x=x +x ,x ∈X 0 1 0 1 i i kT kL(X0,Y0)

kT k We now set L(X1,Y1) to conclude that τ = t kT k L(X0,Y0)

K(t, T a; Y0,Y1) ≤ kT kL(X0,Y0)K(τ, a; X0,X1).

It follows that for all x ∈ X,

¢ 1 ∞ q −θ q dt kT xkY = [t K(t, T x; Y0,Y1)] 0 t ¢ 1 ∞ q −θ q dt ≤ kT kL(X0,Y0) [t K(τ, x; X0,X1)] 0 t ¢ 1 ∞ q kT kL(X0,Y0) −θ −θ q dτ ≤ kT kL(X0,Y0)( ) [τ K(τ, x; X0,X1)] kT kL(X1,Y1) 0 τ 1−θ θ = kxkX kT k kT k . L(X0,Y0) L(X1,Y1)

If we dene , it follows that is an interpolation Kθ,q(T ) = T |Kθ,q({X0,X1}) Kθ,q functor which is exact and of type θ.

17 2 The K-Method

Theorem 2.5. (Properties of (X0,X1)θ,q) Let {X0,X1} be an interpolation couple, 0 < θ < 1 and 1 ≤ q ≤ ∞. We obtain the following properties of (X0,X1)θ,q.

1. (X0,X1)θ,q = (X1,X0)1−θ,q, 2. for 1 ≤ q ≤ r ≤ ∞,

(X0,X1)θ,1 ,→ (X0,X1)θ,q ,→ (X0,X1)θ,r ,→ (X0,X1)θ,∞,

3. if X0 ,→ X1, then for all 0 < θ < η < 1 and 1 ≤ q, r ≤ ∞, we get (X0,X1)θ,q ,→ (X0,X1)η,r.

4. if X0 = X1, then we get (X0,X1)θ,q = X1 = X0 in the sense that the norms are equivalent,

5. there is a constant Cθ,q > 0 , such that for all x ∈ X0 ∩ X1,

kxk ≤ C kxk1−θkxkθ , (X0,X1)θ,q θ,q X0 X1

6. if we have a second interpolation couple {Y0,Y1} such that Xi ,→ Yi, then

(X0,X1)θ,q ,→ (Y0,Y1)θ,q,

7. if q < ∞ and θ ≤ 0 or θ ≥ 1 or if q = ∞ and θ < 0 or θ > 1, then (X0,X1)θ,q” = ”{0}.

Proof. Exercise.

Remark. The following three chapters 3,4 and 5 follow closely Chapters 1 and 2 in [5].

18 3 The Trace Method

Let 0 < a < b ≤ ∞, K ∈ {R, C} and let X be a Banach space. We consider in the following:

• Reminder: Bochner integral and Lp((a, b),X).

m,p • Reminder: Sobolev spaces W ((a, b); K).

• New: W 1,p((a, b); X) as the Banach space of functions f ∈ Lp((a, b); X), such that there exists a weak derivative f 0 := g ∈ Lp((a, b); X), i.e. ¢ ¢ b b 0 ∞ f(t)ϕ (t) dt = − g(t)ϕ(t) dt ∀ϕ ∈ Cc (a, b). a a

0 Norm: kfkW 1,p((a,b);X) = kfkLp((a,b);X) + kf kLp((a,b);X). In particular, we have the fundamental theorem of calculus, i.e. if f ∈ W 1,p(a, b; X) and s, t ∈ (a, b), then ¢ t f(t) = f(s) + f 0(τ) dτ. (3.1) s

3.1 Weighted Lp spaces

p Denition 3.1. Let 1 ≤ p ≤ ∞. We write L∗(0, ∞) for the space of real or complex valued functions f such that

¢ 1 ∞ p p dt kfk p = |f(t)| < ∞ if p < ∞, L∗(0,∞) 0 t kfk ∞ = ess sup |f(t)| < ∞. L∗ (0,∞) t

Lemma 3.2. (Hardy-Young inequality) Let f : (0, ∞) → R+ be measurable, α > 0, 1 ≤ p < ∞, then ¢ ¢ ¢ ∞ t p ∞ −αp ds dt 1 −αp p ds t f(s) ≤ p s f(s) . 0 0 s t α 0 s

19 3 The Trace Method

Proof. First, we substitute s to get that σ = t ¢ ¢ ¢ ¢ ∞ t ds 1/p ∞ 1 dσ 1/p t−αp−1( f(s) )p dt = t−αp−1( f(tσ) )p dt s σ 0 0 ¢ 0 0 1 −α−1/p dσ = t f(tσ) Lp(0,∞) σ ¢ 0 ¢ 1 ∞ p 1/p −αp−1 f(tσ) ≤ t p dt dσ, 0 0 σ

Next, we substitute τ = tσ, ¢ ¢ 1 ∞ 1/p = σαp−pτ −αp−1f(τ)p dτ dσ 0¢ 0 ¢ 1 ∞ dτ 1/p = σα−1 dσ τ −αpf(τ)p τ 0¢ 0 1 ∞ dτ 1/p = τ −αpf(τ)p . α 0 τ

p p Denition 3.3. For any Banach space X and 1 ≤ p ≤ ∞, we dene L∗(X) = L∗(0, ∞; X) as the space of all Bochner measurable functions f : (0, ∞) → X such that t 7→ kf(t)kX p is in L (0, ∞). The norm is given by kfk p := kt 7→ kf(t)k k p . ∗ L∗(X) X L∗(0,∞)

3.2 The spaces V (p, θ, X0,X1)

In the following, let 0 < θ < 1, 1 ≤ p ≤ ∞ and {X0,X1} an interpolation couple. Denition 3.4. We dene

1,p θ p V (p, θ, X0,X1) := {v ∈ W ((a, b); X0 + X1) ∀0 < a < b < ∞ : t 7→ t v(t) ∈ L∗(X1) and θ 0 p t 7→ t v (t) ∈ L∗(X0)} and θ θ 0 kvk := kt vk p + kt v k p . V (p,θ,X0,X1) L∗(X1) L∗(X0) Lemma 3.5. It holds that

1. V (p, θ, X0,X1) is a Banach space,

2. for all v ∈ V (p, θ, X0,X1) there exists a continuous extension of v to t = 0.

20 3 The Trace Method

Proof. The proof of the rst assertion is given as an exercise. For the second assertion, we use (3.1), the Hölder inequality and the embedding X0 ,→ X0 + X1 to get that for all , and 0 p , 0 < s < t 1 < p < ∞ p = p−1 ¢ t θ−1/p 0 1/p−θ kv(t) − v(s)kX0+X1 ≤ kτ v (τ)τ kX0+X1 dτ s ¢ ¢ 0 t 1/p t 1/p 0 ≤ kτ θ−1/pv0(τ)kp dτ τ (1/p−θ)p dτ X0+X1 s s 0 0 −1/p0 1+p0(1/p−θ) 1+p0(1/p−θ) 1/p0 ≤ Ckv k p (1 + p (1/p − θ)) (t − s ) L∗(s,t;X1) 1+p0(1/p−θ) 1+p0(1/p−θ) 1/p0 ≤ Cp,θkvkV (θ,p,X0,X1)(t − s ) .

It follows that v : R+ → X0 + X1 is continuous and if we put s = 0 and look at t → 0 in the last line, we see that it can be extended continuously to t = 0. The cases p = 1 and p = ∞ are left as an exercise.

3.3 Real interpolation by the trace method and equivalence

Theorem 3.6. Let θ, p and {X0,X1} be as above. Then

X := (X0,X1)θ,p = {x ∈ X0 + X1 : ∃v ∈ V (p, 1 − θ, X1,X0), v(0) = x} and ∼ T r kxkX = inf {kvkV (p,1−θ,X1,X0)} =: kxkX v∈V (p,1−θ,X1,X0),x=v(0) In order to prove this theorem, we want to use the Hardy-Young inequality, through the following lemma. θ p Lemma 3.7. Let u be a function such that uθ := t 7→ t u(t) ∈ L∗(0, a; X) for some Banach space X, 0 < a ≤ ∞, 0 < θ < 1 and 1 ≤ p ≤ ∞. Then also the mean value ¢ 1 t v(t) := u(s) ds, t > 0 t 0 1 has this property and kv k p ≤ ku k p . θ L∗(0,a;X) 1−θ θ L∗(0,a;X) Proof. The proof is direct if we use the Hardy-Young inequality, Lemma 3.2, as for 1 ≤ p < ∞, ¢ ¢ ¢ a a t p p dt (θ−1)p dt kvθ(t)k = t u(s) ds X t t 0 ¢0 ¢0 X a t ksu(s)k p dt ≤ t(θ−1)p X ds 0 ¢ 0 s t a 1 p (θ−1)p p ds ≤ ( ) s ksu(s)kX 1 − θ 0 s 1 p = ( ) kuθk p . 1 − θ L∗(0,a;X)

21 3 The Trace Method

The case p = ∞ also follows immediately from the denition.

We can now proceed to the proof of Theorem 3.6.

Proof. First we show that for a given x ∈ X, we can construct a function v ∈ V (p, 1 − such that is the trace of in and such that T r . This θ, X1,X0) x v t = 0 kxkX ≤ CkxkX part of the proof is devided into four steps.

1. Let x ∈ X. Then for all t > 0 there exist at ∈ X0 and bt ∈ X1 such that x = at + bt and (3.2) katkX0 + tkbtkX1 ≤ 2K(t, x). It follows that

θ kx − btkX0+X1 ≤ katkX0+X1 ≤ CkatkX0 ≤ 2CK(t, x) ≤ 2Ct kxkX

by (2.2). It therefore seems that t 7→ bt would be a candidate for v, but it is not 1−θ p necessarily dierentiable and it does not necessarily satisfy t bt ∈ L (R+; X1) or 1−θ 0 p In the next step, we use to construct a suitable candidate. t bt ∈ L (R+; X0). bt 2. Let ∞ ∞ X X u(t) := b 1 χ 1 1 (t) = (x − a 1 )χ 1 1 (t) n+1 ( n+1 ; n ] n+1 ( n+1 ; n ] n=1 n=1 and let ¢ 1 t v(t) = u(s) ds. t 0

Then it still holds that x = limt→0 u(t) = limt→0 v(t) in X0 + X1.

p 3. We show that v1−θ ∈ L∗(R+; X1). Of course, we want to use Lemma 3.7. By (3.2) and from the monotonicity of t 7→ K(t, x), we get

∞ 1−θ −θ X 1 kt u(t)kX1 ≤ t tχ( 1 , 1 ](t)2(n + 1)K , x n+1 n n + 1 n=1 ≤ 4t−θK(t, x).

It follows that ¢ ∞ 1/p −θ p dt ku k p ≤ 4 (t K(t, x)) = 4kxk , (3.3) 1−θ L∗(R+,X1) X 0 t so

kv k p ≤ Ckxk . 1−θ L∗(R+;X1) X

22 3 The Trace Method

4. We show that 0 p . We use the following remark, to ensure that we v1−θ ∈ L∗(R+; X0) nd 1,p . If on some interval we have 1 v ∈ Wloc (0, ∞,X0 + X1) I ⊂ R f, g ∈ L (I; R) and ¢ y f(x) − f(y) = g(s) ds x 1,1 for almost all x, y ∈ I, then f ∈ W (I; R) and g is its weak derivative, as for and all ∞ with , c, d ∈ I ϕ ∈ Cc (I) supp(ϕ) ∈ [c, d] ¢ ¢ ¢ ¢ ¢ ¢ d d y d d d ϕ0(y)(f(y)−f(c)) dy = ϕ0(y)g(x) dx dy = ϕ0(y) dyg(x) dx = − ϕ(x)g(x) dx. c c c c x c We use this fact on and that, by denition, kv(t)kX0+X1

¢ ∞ 1 t X v(t) = x − a 1 χ( 1 , 1 ](s) ds t n+1 n+1 n 0 n=1 and ¢ t 0 1 1 v (t) = 2 g(s) ds − g(t) t 0 t P∞ holds true almost everywhere for g(t) := a 1 χ 1 1 (t) ∈ X0. Again, from n=1 n+1 ( n+1 , n ] the monotonicity of t 7→ K(t, x), it follows that

∞ X 1 kg(t)kX0 ≤ 2K , x χ( 1 , 1 ](t) ≤ 2K(t, x) n + 1 n+1 n n=1 and therefore ¢ 1−θ t 1−θ 0 t −θ kt v (t)kX0 ≤ 2 kg(s)kX0 ds + t kg(t)kX0 t 0 ≤ 4t−θK(t, x).

As in (3.3), it follows that kv k p ≤ Ckxk . In conclusion, we see that 1−θ L∗(R+;X0) X , that in and that T r v ∈ V (p, 1 − θ, X1,X0) x = limt→0 v(t) X0 + X1 kxkX ≤ . kvkV (p,1−θ,X1,X0) ≤ CkxkX

We now look at the opposite inclusion and assume that x ∈ X0 + X1 is the trace of a function v ∈ V (p, 1 − θ, X1,X0) at t = 0. We can write ¢ t x = x − v(t) + v(t) = − v0(s) ds + v(t), t > 0, 0 cf. Lemma 3.5. Therefore, we see that ¢ t −θ 1−θ 1 0 1−θ t K(t, x) ≤ t k v (s) dskX0 + t kv(t)kX1 , t > 0. t 0

23 3 The Trace Method

By Lemma 3.7,

−θ 0 kxk = kt K(t, x)k p ≤ Ckv k p + kv k p ≤ Ckvk . X L∗(R+) 1−θ L∗(R+;X0) 1−θ L∗(R+,X1) V (p,1−θ,X1,X0)

Remark 3.8. Let x ∈ (X0,X1)θ,p be the trace of a function v ∈ V (p, 1 − θ, X1,X0).

1. We can improve the regularity of v in the following way: For¡ any smooth non- negative function with compact support and ∞ ds , we ϕ : R+ 7→ R 0 ϕ(s) s = 1 set ¢ ¢ ∞ t dτ ∞ t ds u(t) = ϕ( )v(τ) = ϕ(s)v( ) . 0 τ τ 0 s s ∞ Then we get u ∈ C (R+; X0 ∩ X1), u(0) = x and

n−θ (n) p t 7→ t u (t) ∈ L∗(R+; X0), n ∈ N, n+1−θ (n) p t 7→ t u (t) ∈ L∗(R+; X1), n ∈ N ∪ {0}, with norms estimated by . The proof is left as an exercise. c(n)kvkV (p,1−θ,X1,X0) 2. Let ∞ such that in or any right neighbourhood of . Then ψ ∈ Cc ([0, ∞)) ψ ≡ 1 (0, 1] 0 uψ : t 7→ ψ(t)u(t) ∈ V (p, 1 − θ, X1,X0) with trace x at t = 0, where u is chosen as in 1. Moreover, and it has compact kuψkV (p,1−θ,X1,X0) ≤ CψkukV (p,1−θ,X1,X0) support. This shows that we could also consider a subset of V (p, 1 − θ, X1,X0) consisting of functions with compact support in order to dene an equivalent trace space.

Corollary 3.9. Let 1 < p < ∞. Then (X0,X1)1−1/p,p is the set of the traces at t = 0 of 1,p p the functions u ∈ W (0, ∞; X1) ∩ L (0, ∞; X0).

p p Proof. Clearly, if θ = 1 − 1/p, u1−θ ∈ L∗(R+; Xi) i u ∈ L (R+; Xi). The corollary follows if we take into account Remark 3.8.

The following example gives us an important motivation to consider the trace method.

Example 3.10. Let n+1 denote the upper half-space n R+ {(t, x) ∈ R × R : t > 0}. Then for , p n 1,p n is the space of traces of functions 1 < p < ∞ (L (R ),W (R ))1−1/p,p 1,p n+1 at . (t, x) 7→ v(t, x) ∈ W (R+ ) t = 0

We close this chapter with a theoretical result on the real interpolation space X = (X0,X1)θ,p which can be derived by the trace method.

Proposition 3.11. Let 0 < θ < 1 and {X0,X1} an interpolation couple. For 1 ≤ p < ∞, X0 ∩ X1 is dense in X = (X0,X1)θ,p.

24 3 The Trace Method

∞ Proof. Let x ∈ X. By Remark 3.8, x = v(0), where v ∈ C (R+; X0 ∩ X1) ∩ V (p, 1 − 2−θ 0 p θ, X1,X0) and t 7→ t v ∈ L∗(R+,X1). We set

xε := v(ε) ∈ X0 ∩ X1, ∀ε > 0 and show that xε → x in X. We dene

zε(t) := (v(ε) − v(t))χ[0,ε](t) to get , 1,p for all and 0 0 . xε−x = zε(0) zε ∈ W (a, b; X0) 0 < a < b < ∞ zε(t) = −v (t)χ(0,ε)(t) It follows that 1−θ 0 lim kt zε(t)kLp( ;X ) = 0. ε→0 ∗ R+ 0

1−θ p We now show that t 7→ t zε(t) ∈ L∗(R+; X1) by using ¢ ∞ 0 zε(t) = χ(0,ε)(s)v (s) ds t and a modied version of the Hardy-Young inequality: For α > 0, p ≥ 1 and positive ϕ, we have that ¢ ¢ ¢ ∞ ∞ ∞ αp ds p dt 1 αp p ds (3.4) t ( ϕ(s) ) ≤ p s ϕ(s) , 0 t s t α 0 s which follows from Lemma 3.2 by substituting 1 and 1 . We get that τ = t σ = s ¢ ¢ ¢ ∞ ∞ ∞ 1−θ p dt (1−θ)p 0 ds p dt (t kzε(t)kX1 ) ≤ t ( χ(0,ε)(s)skv (s)kX1 ) 0 t 0 ¢ t s t 1 ∞ ds ≤ ( )p χ (s)s(2−θ)pkv0(s)kp , (0,ε) X1 1 − θ 0 s so that 1−θ lim kt zε(t)kLp( ;X ) = 0. ε→0 ∗ R+ 1 In conclusion, in as , so T r as . By zε → 0 V (p, 1 − θ, X1,X0) ε → 0 kxε − xkX → 0 ε → 0 Theorem 3.6, limε→0 kxε − xkX = 0.

25 4 The Reiteration Theorem

In the following, let {X0,X1} an interpolation couple. We dene two classes of intermediate spaces.

Denition 4.1. Let 0 ≤ θ ≤ 1 and let X be a Banach space such that X0 ∩ X1 ,→ X,→ X0 + X1.

1. We say that X belongs to the class Jθ between X0 and X1 if there exists a constant c > 0 such that kxk ≤ ckxk1−θkxkθ , ∀x ∈ X ∩ X . X X0 X1 0 1

We write X ∈ Jθ(X0,X1).

2. We say that X belongs to the class Kθ between X0 and X1 if there is a constant k > 0 such that

θ K(t, x, X0,X1) ≤ kt kxkX , ∀x ∈ X, t > 0.

In this case we write X ∈ Kθ(X0,X1). If θ ∈ (0, 1), this means that X,→ (X0,X1)θ,∞.

Proposition 4.2. Let θ ∈ (0, 1) and let X be an intermediate space for {X0,X1}. Then the following statements are equivalent:

1. X ∈ Jθ(X0,X1),

2. (X0,X1)θ,1 ,→ X.

Proof. The implication 2. ⇒ 1. follows directly from Theorem 2.5 (5) with q = 1. We ∞ show that 1. ⇒ 2.. For every x ∈ (X0,X1)θ,1, let u ∈ V (1, 1 − θ, X1,X0) ∩ C (R+; X1 ∩ X0) such that u(t) → 0 as t → ∞ and u(0) = x. We set ¢ 1 t v(t) = u(s) ds, t 0 so that 2−θ 0 1 and 1−θ 0 1 follows, similarly as t 7→ t v (t) ∈ L∗(R+; X1) t 7→ t v (t) ∈ L∗(R+; X0) in Remark 3.8, by Lemma 3.7. It still holds that v(0) = x and that v(t) → 0 as t → ∞, so ¢ ∞ x = − v0(t) dt. 0

26 4 The Reiteration Theorem

Let c be such that kyk ≤ ckykθ kyk1−θ for every y ∈ X ∩ X . Then X X1 X0 0 1

kv0(t)k ≤ ckv0(t)kθ kv0(t)k1−θ = ct−1kt2−θv0(t)kθ kt1−θv0(t)k1−θ. X X1 X0 X1 X0 Now by Hölder's inequality, as 1 1 , 1 = 1/θ + 1/(1−θ)

0 −θ 2−θ 0 θ −(1−θ) 1−θ 0 1−θ kxkX ≤ kv k 1 ≤ c t kt v (t)k t kt v (t)k 1 . L (R+;X) X1 X0 L (R+) 2−θ 0 θ 1−θ 0 1−θ ≤ ckt v (t)k 1 kt v (t)k 1 L∗(R+,X1) L∗(R+,X0)

≤ ckxk(X0,X1)θ,1 .

The above proposition shows that for θ ∈ (0, 1), X ∈ Jθ(X0,X1) ∩ Kθ(X0,X1) i (X0,X1)θ,1 ,→ X,→ (X0,X1)θ,∞. Example 4.3. Actually, two examples:

1. By Equation (2.2) and by Theorem 2.5, (X0,X1)θ,p ∈ Kθ(X0,X1) ∩ Jθ(X0,X1).

2. The space C1([−1, 1]) lies in

2 2 K1/2(C([−1, 1]),C ([−1, 1])) ∩ J1/2(C([−1, 1]),C ([−1, 1])), but, as we have seen, it is not an interpolation space. The proof is left as an exercise.

The following theorem shows that we can iterate the procedure of interpolating spaces, i.e. the real interpolation spaces of two suitable intermediate spaces is again a real interpolation space.

Theorem 4.4. (Reiteration Theorem)

Let 0 ≤ θ0 < θ1 ≤ 1. We x θ ∈ (0, 1) and set ω = (1 − θ)θ0 + θθ1. Then the following holds true.

1. If for an interpolation couple , there are intermediate spaces , {X0,X1} Ei ∈ Kθi (X0,X1) i ∈ {0, 1}, then

(E0,E1)θ,p ,→ (X0,X1)ω,p for all 1 ≤ p ≤ ∞.

2. If on the other hand, , then Ei ∈ Jθi (X0,X1)

(X0,X1)ω,p ,→ (E0,E1)θ,p.

27 4 The Reiteration Theorem

In conclusion, if , then Ei ∈ Jθi (X0,X1) ∩ Kθi (X0,X1)

(E0,E1)θ,p = (X0,X1)ω,p for all 1 ≤ p ≤ ∞, with equivalence of the respective norms.

Proof. We rst show 1.:

Let ki be such that

θi K(t, e, X0,X1) ≤ kit kekEi , e ∈ Ei, t > 0.

Let e0 ∈ E0 and e1 ∈ E1 be such that e = e0 + e1. It follows that

θ0 θ1 K(t, e, X0,X1) ≤ K(t, e0,X0,X1) + K(t, e1,X0,X1) ≤ k0t ke0kE0 + k1t ke1kE1 . Taking the inmum, it follows that

θ0 θ1−θ0 K(t, e, X0,X1) ≤ max{k0, k1}t (ke0kE0 + t ke1kE1 ) θ0 θ1−θ0 ≤ max{k0, k1}t K(t , e, E0,E1), so we get

−ω −θ(θ1−θ0) θ1−θ0 t K(t, e, X0,X1) ≤ max{k0, k1}t K(t , e, E0,E1).

Setting s = tθ1−θ0 , we can conclude

−θ kek ≤ max{k , k }ks K(s, e, E ,E )k p = max{k , k }kek , (X0,X1)ω,p 0 1 0 1 L∗(0,∞) 0 1 (E0,E1)θ,p if p < ∞ and

kek(X0,X1)ω,∞ ≤ max{k0, k1}kek(E0,E1)θ,p . We now show 2.:

By Theorem 3.6 and Remark 3.8, every x ∈ (X0,X1)ω,p is the trace at t = 0 of a function ∞ such that , 0 p , 0 p and v ∈ C (R+,X0 ∩ X1) v(∞) = 0 v1−ω ∈ L∗(R+,X0) v2−ω ∈ L∗(R+,X1)

0 0 T r kv k p + kv k p ≤ kkxk . (4.1) 1−ω L∗(R+,X0) 2−ω L∗(R+,X1) (X0,X1)ω,p We now consider the function g(t) = v(t1/(θ1−θ0)) and show that it belongs to V (p, 1 − θ, E1,E0), so that g(0) = v(0) = x ∈ (E0,E1)θ,p with the corresponding estimate. First, we look at 0 , , . Let such that kv (t)kEi t > 0 i ∈ {0, 1} ci

kxk ≤ c kxk1−θi kxkθi , x ∈ X ∩ X . Ei i X0 X1 0 1 We get

0 ci 1−ω 0 1−θi 2−ω 0 θi kv (t)kE ≤ kt v (t)k kt v (t)k . i t1+θi−ω X0 X1

28 4 The Reiteration Theorem

Now we calculate

1 + θ0 − ω = 1 + θ0 − (1 − θ)θ0 − θθ1 = 1 − θ(θ1 − θ0) and

1 + θ1 − ω = 1 + θ1 − (1 − θ)θ0 − θθ1 = 1 + (1 − θ)(θ1 − θ0) to obtain that 0 T r kv k p ≤ c kkxk (4.2) 1−θ(θ1−θ0) L∗(0,∞,E0) 0 (X0,X1)ω,p and that 0 T r kv k p ≤ c kkxk (4.3) 1+(1−θ)(θ1−θ0) L∗(0,∞,E1) 1 (X0,X1)ω,p by Hölder's inequality and (4.1). We now use ¢ ∞ v(t) = − v0(s) ds (4.4) t and the second Hardy-Young inequality (3.4) to get that ¢ ¢ (4.4) ∞ ∞ dsp dt1/p (1−θ)(θ1−θ0)p 0 kv k p ≤ t ksv (s)kE (1−θ)(θ1−θ0) L∗(0,∞,E1) 1 s t 0 ¢ t (3.4) 1 ∞ ds1/p ≤ s(1−θ)(θ1−θ0)p+pkv0(s)kp E1 (1 − θ)(θ1 − θ0) 0 s (4.3) 1 ≤ c kkxkT r 1 (X0,X1)ω,p. (1 − θ)(θ1 − θ0) 1/(θ −θ ) θ −θ (θ −θ )−1 With the substitution s = t 1 0 , i.e. t = s 1 0 and dt = (θ1 − θ0)s 1 0 ds, it follows that ¢ ∞ dt1/p (1−θ)p 1/(θ1−θ0) p kg1−θk p = t kv(t )k L∗(0,∞,E1) E1 t ¢0 ∞ ds1/p = s(θ1−θ0)(1−θ)pkv(s)kp (θ − θ ) E1 1 0 0 s 1/p = (θ − θ ) kv k p 1 0 (1−θ)(θ1−θ0) L∗(0,∞,E1) ≤ (1 − θ)−1(θ − θ )1/p−1c kkxkT r . 1 0 1 (X0,X1)ω,p Similarly, we look at

0 −1 (1/(θ1−θ0)−1) 0 1/(θ1−θ0) g (t) = (θ1 − θ0) t v (t ) to get that ¢ ∞ 1/p 0 1 1−θ+1/(θ −θ )−1 0 1/(θ −θ ) p dt kg k p = (t 1 0 kv (t 1 0 )k ) 1−θ L∗(0,∞,E0) E0 θ1 − θ0 0 t s=t1/(θ1−θ0) 1/p−1 0 = (θ − θ ) kv k p 1 0 1−θ(θ1−θ0) L∗(0,∞,E0) (4.2) ≤ c k(θ − θ )1/p−1kxkT r 0 1 0 (X0,X1)ω,p.

29 4 The Reiteration Theorem

In conclusion, we get that g ∈ V (p, 1 − θ, E1,E0), so that x = g(0) ∈ (E0,E1)θ,p by Theorem 3.6 and

kxk ≤ kgk ≤ max{c }k(1 − θ)−1(θ − θ )1/p−1kxkT r (E0,E1)θ,p V (p,1−θ,E1,E0) i 1 0 (X0,X1)ω,p (θ − θ )1/p = max{c }k 1 0 kxkT r . i (X0,X1)ω,p (θ1 − ω)

Corollary 4.5. From Theorem 4.4 and Example 4.3 we immediately get that for any

0 < θ0 < θ1 < 1, 0 < θ < 1 and 1 ≤ p, q0, q1 ≤ ∞,

((X0,X1)θ0,q0 , (X0,X1)θ1,q1 )θ,p = (X0,X1)(1−θ)θ0+θθ1,p.

30 5 Complex Interpolation

Idea:

•{X,Y } interpolation couple, construct interpolation space [X,Y ]θ for θ ∈ (0, 1) • only one parameter θ will do, we got two norms already...

• f : S → X + Y nice such that f(it) ∈ X and f(1 + it) ∈ Y

• [X,Y ]θ = {f(θ)} = {f(θ + it)}

• f : S → X+Y holomorphic maximum principle estimate interior by boundary

Roadmap:

1. properties of holomorphic functions f : S → X + Y

2. denition of [X,Y ]θ

3. Fθ : {X,Y } 7→ [X,Y ]θ is exact interpolation functor of type θ 4. Reiteration Theorem [X,Y ]θ ∈ Jθ ∩ Kθ

31 5 Complex Interpolation

5.1 X-valued holomorphic functions

Let X be a complex Banach space. For every set Ω ⊂ C we say that f :Ω → X is holomorphic in S ⊂ Ω, if f is dierentiable in every λ0 in a neighbourhood of S, i.e. the limit 0 f(λ) − f(λ0) f (λ0) = lim λ→λ0 λ − λ0 exists in X.

Proposition 5.1. Let Ω ⊂ C be a bounded open set and f : Ω → X a function which is holomorphic on Ω and continuous on Ω. Then

kf(ξ)kX ≤ max{kf(z)kX : z ∈ ∂Ω} ∀ξ ∈ Ω.

Proof. For every ξ ∈ Ω we can choose x0(ξ) ∈ X0 in the dual space of X such that 0 0 kf(ξ)kX = hf(ξ), x (ξ)i and kx (ξ)kX0 = 1. We apply the maximum principle for C- valued holomorphic functions to z 7→ hf(z), x0(ξ)i to get

0 0 kf(ξ)kX = |hf(ξ), x (ξ)i| ≤ max{|hf(z), x (ξ)i|, z ∈ ∂Ω}

≤ max{kf(z)kX , z ∈ ∂Ω}.

Following the idea above, we want to consider holomorphic functions on the strip

S := {x + iy ∈ C : 0 ≤ x ≤ 1}. Also for this set, the maximum principle holds.

◦ Proposition 5.2. Let f : S → X be holomorphic on S and continuous and bounded on S. Then kf(ξ)kX ≤ max{sup kf(it)kX , sup kf(1 + it)kX }. t∈R t∈R

Proof. Let ε ∈ (0, 1) and ξ0 = x0 + it0 ∈ S such that

kf(ξ0)k ≥ (1 − ε)kfkC(S,X).

2 δ(ξ−ξ0) We set fδ(ξ) := e f(ξ), ξ = x + it, so that fδ(ξ0) = f(ξ0) and

2 2 2 lim eδ(x+it−ξ0) f(ξ) = lim e−δt +2δ(it)(x−ξ0)+δ(x−ξ0) f(ξ) |t|→∞ |t|→∞ 2 = lim e−δt(t−2t0)eδRe(x−ξ0) f(ξ) |t|→∞ = 0.

32 5 Complex Interpolation

We can now apply the maximum principle, Proposition 5.1, to fδ on every domain [0, 1]× [−M,M], M > 0. By the above calculation, for large M, only the vertical boundaries will be relevant. kfδ(ξ)k ≤ max{sup kfδ(it)k, sup kfδ(1 + it)k} t∈R t∈R δRe(1−ξ )2 δRe ξ2 −δt(t−2t ) ≤ max{e 0 , e 0 } sup{e 0 } max{sup kf(it)k, sup kf(1 + it)k}. t∈R t∈R t∈R

−δt(t−2t ) δt2 It is easy to show that sup e 0 = e 0 is reached in t = t , so that in conclusion, t∈R 0 for every ε ∈ (0, 1) there exists a suciently small δ such that

(1 − ε)kfkC(S,X) ≤ kf(ξ0)k = kfδ(ξ0)k ≤ (1 + ε) max{sup kf(it)k, sup kf(1 + it)k}. t∈R t∈R

Theorem 5.3. (Three lines theorem) ◦ Let f : S → X be holomorphic on S and continuous and bounded on S. Then for all 0 < θ < 1, we have

1−θ θ kf(θ)kX ≤ (sup kf(it)kX ) (sup kf(1 + it)kX ) . t∈R t∈R

Proof. In the following, we use the abbreviations M := sup kf(it)k and M := 0 t∈R X 1 sup kf(1+it)k and we consider the function ϕ(z) = eλzf(z) where λ = log(M /M ). t∈R X 0 1 By the maximum principle on S, Proposition 5.2,

−λθ θ 1−θ kf(θ)kX = |e |kϕ(θ)k kϕ(θ)k M1 θ λit λ+λit θ λit λ+λit 1−θ ≤ ( ) max{e M0, e M1} max{e M0, e M1} M0 M1 θ θ 1−θ = ( ) M0 M0 M0 1−θ θ = M0 M1 .

5.2 The spaces [X,Y ]θ and basic properties

Denition 5.4. The space G(X,Y ) is dened as the space of all functions f : S → X+Y such that

◦ 1. f is holomorphic in S and continuous and bounded on S with values in X + Y .

33 5 Complex Interpolation

2. It holds that t 7→ f(it) ∈ Cb(R; X), t 7→ f(1 + it) ∈ Cb(R; Y ) and

kfkG(X,Y ) = max{sup kf(it)kX , sup kf(1 + it)kY } < ∞. t∈R t∈R

G0(X,Y ) is a subspace of G(X,Y ) which imposes the additional properties

lim kf(it)kX = 0, lim kf(1 + it)kY = 0. |t|→∞ |t|→∞

In the exercises, we show that both G(X,Y ) and G0(X,Y ) are Banach spaces, continuously embedded in Cb(S, X + Y ). We only cite the following technical lemma. (The reason is: the proof is dicult)

δz2+λz Lemma 5.5. The linear hull of the set of functions z 7→ e a, δ > 0, λ ∈ R, a ∈ X ∩ Y , is dense in G0(X,Y ). Denition 5.6. (Complex Interpolation Spaces) For every θ ∈ [0, 1], we set

[X,Y ]θ := {f(θ): f ∈ G(X,Y )},

kxk[X,Y ] := inf kfkG(X,Y ). θ f∈G(X,Y ),f(θ)=x

We see that [X,Y ]θ is a Banach space from the fact that it is isomorphic to the quotient space G(X,Y )/Nθ, where Nθ is the closed subspace of functions f ∈ G(X,Y ), satisfying f(θ) = 0.

Proposition 5.7. (Properties of [X,Y ]θ)

1. If θ ∈ (0, 1), it holds that [X,Y ]θ = [Y,X]1−θ.

2. The space [X,Y ]θ can be dened equivalently from the space G0(X,Y ).

3. If X = Y , then [X,X]θ = X.

4. For every t ∈ R and f ∈ G(X,Y ), f(θ + it) ∈ [X,Y ]θ for every θ ∈ (0, 1) and

kf(θ + it)k[X,Y ]θ = kf(θ)k[X,Y ]θ .

5. For every θ ∈ (0, 1), we get that [X,Y ]θ is an intermediate space of X,Y , i.e.

X ∩ Y,→ [X,Y ]θ ,→ X + Y.

6. For every θ ∈ (0, 1), X ∩ Y is dense in [X,Y ]θ.

Proof.

34 5 Complex Interpolation

1. Follows directly from the denition (reect fs)

δ(z−θ)2 2. For every f ∈ G(X,Y ) and δ > 0, we can dene fδ(z) = e f(z). We already know this function, that fδ(θ) = f(θ) and that it lies in G0(X,Y ). By denition,

δθ2 δ(1−θ)2 kfδkG(X,Y ) ≤ max{e , e }kfkG(X,Y ),

from δ → 0, we see

inf kfkG(X,Y ) = inf kfkG(X,Y ). f∈G(X,Y ),f(θ)=x f∈G0(X,Y ),f(θ)=x

3. This follows from the maximum principle. For x ∈ [X,X]θ, we have

m.p. kxkX = kf(θ)kX ≤ kfkG(X,X). Taking the gives . On the other hand, the constant function inf kxkX ≤ kxk[X,X]θ cx : z 7→ x, z ∈ S is in G(X,X), so that for every x ∈ X, kxkX = kcxkG(X,X) ≥ . kxk[X,X]θ

4. Let g(z) = f(z + it). We see directly that g ∈ G(X,Y ) and that kgkG(X,Y ) = kfkG(X,Y ). It follows that f(θ + it) = g(θ) ∈ [X,Y ]θ and that the norm doesn't change.

5. Let x ∈ X ∩ Y . Again, the constant function cx(z) = x belongs to G(X,Y ) and

kcxkG(X,Y ) ≤ max{kxkX , kxkY }, so that and . x = cx(θ) ∈ [X,Y ]θ kxk[X,Y ]θ ≤ kxkX∩Y On the other hand, if x = f(θ) ∈ [X,Y ]θ, then

kxkX+Y ≤ kf(θ)kX+Y m.p. ≤ max{sup kf(it)kX+Y , sup kf(1 + it)kX+Y } t∈R t∈R ≤ C max{sup kf(it)kX , sup kf(1 + it)kY } t∈R t∈R = CkfkG(X,Y ). Taking the inmum, we get . kxkX+Y ≤ Ckxk[X,Y ]θ

6. This follows directly from Lemma 5.5. For every x ∈ [X,Y ]θ, there is a function f ∈ G0(X,Y ) such that f(θ) = x, by 2. By Lemma 5.5, there is a sequence of functions given by mn 2 X δiz +λiz fn(z) = µie ai ∈ X ∩ Y i=1 such that in . Setting , we have fn → f G0(X,Y ) xn = fn(θ) kx − xnk[X,Y ]θ = . kf(θ) − fn(θ)k[X,Y ]θ ≤ kf − fnkG0(X,Y )

35 5 Complex Interpolation

Proposition 5.8. We let V (X,Y ) the linear hull of the functions z 7→ ϕ(z)x, S → X∩Y , where ϕ ∈ G0(C, C) and x ∈ X ∩ Y . For every x ∈ X ∩ Y , we get

kxk[X,Y ] = inf kfkG(X,Y ). θ f∈V (X,Y ),f(θ)=x

Proof. We can approximate the norm of x by choosing for every ε > 0 a function f0 ∈ such that and . We dene a function G0(X,Y ) x = f0(θ) kf0kG(X,Y ) ≤ kxk[X,Y ]θ + ε r ∈ G(C, C) by z − θ r(z) = , z ∈ S z + θ and set f (z) − e(z−θ)2 x f (z) = 0 , z ∈ S. 1 r(z)

(z−θ)2 It follows that f1 ∈ G0(X,Y ), as f0 ∈ G0(X,Y ), z 7→ e x ∈ G0(X,Y ) and |r(z)| ≤ 1, r(z) 6= 0 if z 6= 0 and r0(θ) 6= 0. By Lemma 5.5, it follows that there exists an approximating function n 2 X δiz +λiz f2(z) = µie xi i=1 with , and such that . We now set δi > 0 λi ∈ R xi ∈ X ∩ Y kf1 − f2kG(X,Y ) ≤ ε

(z−θ)2 f(z) = e x + r(z)f2(z), z ∈ S.

It follows that f ∈ V (X,Y ) and that

kfkG(X,Y ) ≤ kf0kG(X,Y ) + kf − f0kG(X,Y ) (·−θ)2 ≤ kxk[X,Y ]θ + ε + ke x + rf1 − f0kG(X,Y ) + kr(f2 − f1)kG(X,Y )

≤ kxk[X,Y ]θ + 2ε.

5.3 The complex interpolation functor

Theorem 5.9. For every θ ∈ (0, 1),

0 0 Fθ : C2 → C1, {X,Y } 7→ [X,Y ]θ is an exact interpolation functor of type , where 0 denotes the categories containing θ Ci complex Banach spaces and complex interpolation couples.

36 5 Complex Interpolation

Proof. Let {X,Y }, {X, Y } be complex interpolation couples and T ∈ L({X, X}, {Y, Y }). For every x ∈ [X,Y ]θ, let f ∈ G(X,Y ) such that f(θ) = x. We set

kT k !z−θ g(z) = L(X,X) T f(z), z ∈ S. kT kL(Y,Y )

It follows that g ∈ G(X, Y ) and that

kg(it)k ≤ kT k1−θ kT kθ kf(it)k , X L(X,X) L(Y,Y ) X kg(1 + it)k ≤ kT k1−θ kT kθ kf(1 + it)k . Y L(X,X) L(Y,Y ) Y

Therefore, kgk ≤ kT k1−θ kT kθ kfk and so T x = g(θ) ∈ [X, Y ] . We G(X,Y ) L(X,X) L(Y,Y ) G(X,Y ) θ have the estimate

kT xk ≤ kgk ≤ kT k1−θ kT kθ kfk , [X,Y ]θ G(X,Y ) L(X,X) L(Y,Y ) G(X,Y ) so that by taking the inmum over f, we get kT k ≤ kT k1−θ kT kθ . L([X,Y ]θ,[X,Y ]θ) L(X,X) L(Y,Y )

Following the ideas for Theorem 5.9, we can prove the following result.

Theorem 5.10. Let {X,Y }, {X, Y } again be interpolation couples. For every z ∈ S let Tz ∈ L(X ∩ Y, X + Y ) be such that z 7→ Tzx is in G(X, Y ) for every x ∈ X ∩ Y . Moreover, assume that Tit ∈ L(X, X) and that T1+it ∈ L(Y, Y ). Assume further that the following suprema are nite:

M0 := sup kTitkL(X,X),M1 := sup kT1+itkL(Y,Y ). t∈R t∈R In this case, for every θ ∈ (0, 1), we get

kT xk ≤ M 1−θM θkxk , θ [X,Y ]θ 0 1 [X,Y ]θ so that Tθ extends to an operator in L([X,Y ]θ, [X, Y ]θ).

Proof. The proof is an exercise, using the proof of Theorem 5.9 and Proposition 5.8.

5.4 The space [X,Y ]θ is of class Jθ and of class Kθ.

Corollary 5.11. For every θ ∈ (0, 1) we have

1−θ θ kyk[X,Y ]θ ≤ kykX kykY , y ∈ X ∩ Y, i.e. [X,Y ]θ ∈ Jθ(X,Y ).

37 5 Complex Interpolation

Proof. The idea is the same as for real interpolation spaces. We consider the operators Tyλ = λy, T ∈ L({C, C}, {X,Y }) and use Theorem 5.9, i.e. the exactness of the interpolation functor, to get

kyk = kT k ≤ kyk1−θkykθ . [X,Y ]θ y L(C,[X,Y ]θ) X Y

To prove that [X,Y ]θ ∈ Kθ(X,Y ) needs more work. The main idea is to use a Poisson integral formula for Banach-space valued holomorphic functions on the strip S.

Theorem 5.12. For θ ∈ (0, 1), the spaces [X,Y ]θ are in the class Kθ(X,Y ).

Proof. We do not give a detailed proof, but name the basic steps and ingredients.

1. Preliminary observation: for a ∈ [X,Y ]θ, in order to estimate

K(t, a, X, Y ) = inf {kxkX + tkykY }, a=x+y

we split a and therefore f ∈ G(X,Y ) with a = f(θ) into f = fX + fY , where fX : S → X, fY : S → Y , recovering estimates for fX and fY in terms of the boundary values f(it), f(1 + it).

2. The Poisson formula: The Dirichlet problem on the open unit ball D, given by

∆u = 0, in D, u(λ) = h(λ), on ∂D, (5.1) can be solved by the Poisson formula ¢ 2 1 1 − |ξ| (5.2) u(ξ) = h(λ) 2 dλ. 2π |λ|=1 |ξ − λ| If h is continuous, u is unique. Therefore if f is holomorphic on D and bounded on ∂D, it satises (5.2) with u = f and h = f|∂D. 3. The solution formula (5.2) transfers from D to S: By the Riemannian Mapping Theorem, this will work for all connected open subsets of C. If there is a conformal map ϕ :Ω → D, then the Dirichlet problem (5.1) is equivalent to the Dirichlet problem

∆v = 0, in Ω, v(ω) = (h ◦ ϕ)(ω), on ∂Ω

and v = u ◦ ϕ. In particular, eπiz − i ξ(z) = , z ∈ S, eπiz + i

38 5 Complex Interpolation

is a conformal map from S to D. The Poisson formula for S is thus given by ¢ ∞ h(it) v(z) = eπ(y−t) sin(πx) 2 π(y−t) 2 −∞ sin (πx) + (cos(πx) − e ) h(1 + it) + dt, (5.3) sin2(πx) + (cos(πx) + eπ(y−t))2

◦ where z = x + iy ∈ S and it is satised by every function f ∈ G(C, C) with v and h replaced by f.

4. It follows from the Hahn-Banach theorem, that equation (5.3) also holds in X + Y 0 0 0 for f ∈ G(X,Y ), as for every z ∈ (X + Y ) , the function fz0 : z 7→ hf(z), z i is holomorphic and satises (5.3). In particular, we can write f = fX + fY , where ¢ ∞ f(it) f (z) = eπ(y−t) sin(πx) dt, (5.4) X 2 π(y−t) 2 ¢−∞ sin (πx) + (cos(πx) − e ) ∞ f(1 + it) f (z) = eπ(y−t) sin(πx) dt. (5.5) Y 2 π(y−t) 2 −∞ sin (πx) + (cos(πx) + e )

5. The two kernels in (5.4) and (5.5) are positive and we have that ¢ ∞ 1 eπ(y−t) sin(πx) 2 π(y−t) 2 −∞ sin (πx) + (cos(πx) − e ) 1 + dt = 1, sin2(πx) + (cos(πx) + eπ(y−t))2

from considering f ≡ 1, so that ¢ ∞ 1 0 < eπ(y−t) sin(πx) dt < 1, 2 π(y−t) 2 ¢−∞ sin (πx) + (cos(πx) − e ) ∞ 1 0 < eπ(y−t) sin(πx) dt < 1. 2 π(y−t) 2 −∞ sin (πx) + (cos(πx) + e ) It follows that

kfX (z)kX ≤ sup kf(iτ)kX , z ∈ S and τ∈R kfY (z)kY ≤ sup kf(1 + iτ)kY , z ∈ S. τ∈R

6. From these estimates we get that for every t > 0, f ∈ G(X,Y ) and f(θ) = a ∈ [X,Y ]θ,

K(t, a, X, Y ) ≤ kfX (θ)kX + tkfY (θ)kY ≤ sup kf(iτ)kX + t sup kf(1 + iτ)kY . τ∈R τ∈R

39 5 Complex Interpolation

For every f, we can also apply this estimate to the function g : z 7→ tθ−zf(z), which is also in G(X,Y ) with g(θ) = a, to get that

θ θ−1 θ K(t, a) ≤ t sup kf(iτ)kX + t t sup kf(1 + iτ)kY ≤ 2t kfkG(X,Y ). τ∈R τ∈R Taking the inmum over f yields the claim.

Corollary 5.13. It follows that for every 0 < θ, θ0, θ1 < 1, 1 ≤ p ≤ ∞, we have

([X,Y ]θ0 , [X,Y ]θ1 )θ,p = (X,Y )θ0(1−θ)+θθ1,p and that

(X,Y )θ,1 ,→ [X,Y ]θ ,→ (X,Y )θ,∞. Remark 5.14. More reiteration and relations of real and complex interpolation spaces, without proof.

1. In general, it is not true that [X,Y ]θ = (X,Y )θ,p for some p. We will see later that if X and Y are Hilbert spaces,

[X,Y ]θ = (X,Y )θ,2 for 0 < θ < 1.

2. If X,→ Y or Y,→ X, OR if X and Y are reexive and X ∩ Y is dense in X and in Y , then

[[X,Y ]θ0 , [X,Y ]θ1 ]θ = [X,Y ](1−θ)θ0+θθ1

3. If X,Y are reexive, 0 < θ, θ0, θ1 < 1, and 1 < p < ∞,

[(X,Y )θ0,p, (X,Y )θ1,p]θ = (X,Y )(1−θ)θ0+θθ1,p.

40 6 Examples

6.1 Complex interpolation of Lp -spaces

Let (Ω, µ) be a measure space with a σ-nite measure µ. For 1 ≤ q ≤ ∞, we use the notation Lq = Lq(Ω, µ).

Theorem 6.1. Let θ ∈ (0, 1) and 1 ≤ p0, p1 ≤ ∞. Then we have 1 1 − θ θ p0 p1 p [L ,L ]θ = L , where = + , p p0 p1 with coinciding norms.

Proof. Note: see Theorem 1.2. Strategy: We show that

p p 1. for every 0 1 , p p p and a ∈ L ∩ L kak[L 0 ,L 1 ]θ ≤ kakL p p 2. for every 0 1 , p p p , a ∈ L ∩ L kakL ≤ kak[L 0 ,L 1 ]θ

p0 p1 p0 p1 so that the identity is an isometry between L ∩ L with the [L ,L ]θ-norm and with p p0 p1 p0 p1 p the L -norm. The claim follows from L ∩ L being dense both in [L ,L ]θ and in L p0 p1 p p0 p1 p0 p1 and from the fact that [L ,L ]θ and L embed into L + L . Let now a ∈ L ∩ L . W.l.o.g. (exercise), we say that kakLp = 1.

1. For every z ∈ S, we set

p( 1−z + z ) a(x) f(z)(x) = |a(x)| p0 p1 , for x ∈ Ω, if a(x) 6= 0, |a(x)| f(z)(x) = 0, for x ∈ Ω, if a(x) = 0. ◦ It follows that f ∈ G(Lp0 ,Lp1 ), i.e. f is holomorphic on S and continuous on S with values in Lp0 + Lp1 , t 7→ f(it) is bounded and continuous with values in Lp1 , t 7→ f(1 + it) is bounded and continuous with values in Lp1 , since we have

p/p0 p/p0 |f(it)(x)| = |a| , kf(it)kLp0 ≤ kakLp , p/p1 p/p1 |f(1 + it)(x)| = |a| , kf(1 + it)kLp1 ≤ kakLp . It follows that and so since we have kfkG(Lp0 ,Lp1 ) = 1 f(θ) = a,

p p p kak[L 0 ,L 1 ]θ ≤ 1 = kakL .

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2. We know that ¢ 0 0 p0 p1 1 = kakLp = sup | a(x)b(x) dx| : b ∈ L ∩ L ∩ (simple functions), kbkLp0 = 1 Ω

0 0 (easy: 1 1−θ θ , where 1 1 ). For all p0 p1 with , we p0 = p0 + p0 q0 = 1 − q b ∈ L ∩ L kbkLp0 = 1 again dene 0 1

0 1−z z p ( 0 + 0 ) b(x) g(z)(x) = |b(x)| p0 p1 , for x ∈ Ω, if b(x) 6= 0, |b(x)| g(z)(x) = 0, for x ∈ Ω, if b(x) = 0

and we dene for every f ∈ G(Lp0 ,Lp1 ) with f(θ) = a, ¢ F (z) := f(z)(x)g(z)(x) dx, z ∈ S. Ω It follows that F is holomorphic, by the following argument. From the denition, g p0 p0( 1−z + z ) p0 p0 0 p0 p0 is holomorphic on S with values in L 0 ∩ L 1 , as in particular, b ∈ L 0 1 ∩ p0 p0( 1−z + z ) 1 p0 p0 p0 p0 p p L 0 1 . It follows that g(z) ∈ L 0 ∩ L 1 is in the dual of L 0 + L 1 for all z ∈ S. It follows that F (z + h) − F (z) hf(z + h), g(z + h) − g(z)i lim = lim h→0 h h→0 h hf(z + h) − f(z), g(z)i + lim h→0 h = hf(z), g0(z)i + hf 0(z), g(z)i,

where h, i is the dual pairing for Lp0 +Lp1 . Moreover, F is continuous and bounded on S, so that by the maximum principle 5.2, for every z ∈ S, F (z) ≤ max{sup |F (it)|, sup |F (1 + it)|}, t∈R t∈R where

0 0 p /p0 |F (it)| ≤ kf(it)kLp0 kg(it)k p0 = kf(it)kLp0 kbk 0 = kf(it)kLp0 , L 0 Lp 0 0 p /p1 |F (1+it)| ≤ kf(1+it)kLp1 kg(1+it)k p0 = kf(1+it)kLp1 kbk 0 = kf(1+it)kLp1 , L 1 Lp so that

|F (z)| ≤ kfkG(Lp0 ,Lp1 ), z ∈ S. It follows that ¢

| a(x)b(x) dx| = |F (θ)| ≤ kfkG(Lp0 ,Lp1 ). Ω Taking the sup over b and then the inf over f gives

p p p kakL ≤ kak[L 0 ,L 1 ]θ .

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6.2 Real interpolation of Lp-spaces

Again, let (Ω, µ) be a measure space with a σ-nite measure µ. For 1 ≤ q ≤ ∞, we use the notation Lq = Lq(Ω, µ). Now, we may consider real- as well as complex-valued functions. As a roadmap for this section, we will

• dene Lorentz spaces Lp,q = Lp,q(Ω, µ)

• show that (Lp0 ,Lp1 ) = Lp,q, where 1 = 1−θ + θ θ,q p p0 P1 • prove the Marcinkiewicz Theorem

6.2.1 Lorentz spaces

Denition 6.2. (non-increasing rearrangement) 1 ∞ Let f ∈ L + L . We dene the distribution function µf : R+ → R+ ∪ {∞} of f by

µf (σ) = µ({x ∈ Ω: |f(x)| > σ}).

∗ The non-increasing rearrangement of f onto [0, ∞) is given by f : R+ ∪{0} → R+ ∪{∞},

∗ f (t) = inf{σ : µf (σ) ≤ t}, where inf Ø = ∞.

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∗ Proposition 6.3. (Properties of µf and f ) ∗ The functions µf and f are non-increasing, positive and right-continuous. Moreover, ∗ the functions |f| and f are equimeasurable, i.e. for every 0 < σ0 < σ1, ∗ |{t > 0 : f (t) ∈ [σ0, σ1]}| = µ{x ∈ Ω: |f(x)| ∈ [σ0, σ1]}. (6.1)

Proof. Clear: non-increasing, positive, equimeasurable.

Right-continuity of µf : We look at the sets E(σ) = {x ∈ Ω: |f(x)| > σ}. For every σ0 ∈ R+, we have [ [ 1 E(σ ) = E(σ) = E(σ + ). 0 0 n σ>σ0 n∈N It follows by the monotone convergence theorem that 1 1 µ (σ + ) = µ(E(σ + )) % µ(E(σ )) = µ (σ ). f 0 n 0 n 0 f 0

∗ ∗ Right-continuity of f : Note that f itself is the distribution function of µf with respect to the Lebesgue measure on , i.e. ∗ , so m R+ f (t) = sup{σ : µf (σ) > t} = mµf (σ) right-continuity follows as for µf . Proposition 6.4. The functions f and f ∗ have the same Lp-norms, i.e. for 1 ≤ p < ∞, ¢ ¢ ∞ |f|p dµ = (f ∗(t))p dt and Ω 0 ess sup f = ess sup f ∗ = f ∗(0). Ω R+∪{0}

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Proof. Exercise.

Denition 6.5. (Lorentz spaces) For 1 ≤ p, q ≤ ∞, we dene the Lorentz spaces by ¢ ( ∞ 1/q ) p,q 1 ∞ 1/p ∗ q dt L (Ω, µ) := f ∈ L + L : kfkLp,q = (t f (t)) < ∞ 0 t for q < ∞ and by p,∞ 1 ∞ 1/p ∗ L (Ω, µ) := f ∈ L + L : kfkLp,∞ = sup t f (t) < ∞ t>0 otherwise. The Lp,∞-spaces are also called Marcinkiewicz spaces. Note that Lp,p = Lp by Proposition 6.4.

6.2.2 Lorentz spaces and the K-functional

Theorem 6.6. For 0 < θ < 1 and 1 ≤ q ≤ ∞, we have that

1 ∞ 1 ,q (L ,L )θ,q = L 1−θ .

Proof. We prove the theorem in two steps:

1. We show that ¢ t K(t, f, L1,L∞) = f ∗(s) ds (6.2) 0 for all t > 0. Note that historically, the Lorentz spaces were used before this connection to real interpolation spaces was discovered. In the following, let t > 0 and

∗ Et = {x ∈ Ω: |f(x)| > f (t)}, ∗ Dt = {x ∈ Ω: |f(x)| = f (t)},

so that µ(Et) ≤ t ≤ µ(Et ∪ Dt). From (6.1) and Proposition 6.4, we get that for every y ∈ Dt - which we need if µ(Dt) 6= 0 -, ¢ ¢ t ∗ f (s) ds = |f(x)| dµ + (t − µ(Et))|f(y)|. (6.3) 0 Et Rigorously, this can be seen from the following argument. We may write

∗ (χEt f) (s) = inf{σ : µ({x ∈ Ω: |χEt (x)| · |f(x)| > σ}) ≤ s} = inf{σ : µ({x ∈ Ω: |f(x)| > σ} ∩ Et) ≤ s} = inf{σ : µ({x ∈ Ω: |f(x)| > max(σ, f ∗(t))}) ≤ s}. (6.4)

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In the case , if ∗ , we get ∗ , as s ≥ µ(Et) f (t) > 0 (χEt f) (s) = 0

∗ µ({x ∈ Ω: |f(x)| > f (t)}) = µ(Et) ≤ s, otherwise if ∗ , then ∗ ∗ . In the case , we see f (t) = 0 (χEt f) (s) = f (s) s < µ(Et) from the last line in (6.4) that ∗ ∗ . If ∗ , then (6.3) follows (χEt f) (s) = f (s) f (t) = 0 because |f(y)| = 0. If f ∗(t) 6= 0, then ¢ ¢ ¢ ¢ ¢ t µ(Et) t t f ∗ = f ∗ + f ∗ = |f| + f ∗ 0 0 µ(Et) Et µ(Et) by Proposition 6.4. Moreover, we have f ∗(s) = f ∗(t) = |f(y)| for every s ∈ ∗ ∗ ∗ [µ(Et), t] as clearly f (t) ≤ f (s) ≤ f (µ(Et)) and by denition,

∗ ∗ ∗ f (µ(Et)) = inf{σ > 0 : µ{x : |f(x)| > σ} ≤ µ{x : |f(x)| > |f (t)|}} ≤ f (t).

This gives (6.3). 1 ∞ It follows that for a decomposition f = f1 + f∞ with f1 ∈ L and f∞ ∈ L , we have ¢ ¢ t ∗ f (s) ds ≤ |f1| dµ + (t − µ(Et))|f1(y)| 0 Et + µ(Et) sup |f∞(x)| + (t − µ(Et))|f∞(y)| x∈Et

≤ kf1kL1 + tkf∞kL∞ .

For the other inequality, we may choose the following decomposition of f, ( f(x) ∗ for f(x) − |f(x)| f (t) x ∈ Et, f1(x) = 0 otherwise. and

f∞ = f − f1. First, we show that ¢ µ(Et) ∗ ∗ kf1kL1 = f (s) − f (t) ds. 0 This follows from Proposition 6.4 if f ∗(t) = 0. Otherwise, ¢ ¢ µ(Et) ∗ kf1kL1 = |χEt f1| = f1 (s) ds Ω 0

∗ i arg f by (6.4). On Et, we have f1 = (|f| − f (t))e , so that

∗ |f1| = (|f| − f (t))χEt .

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Since ∗ ∗, we get |f1| = f1 ∗ ∗ f1 = inf{σ : µ({x : χEt (|f| − f (t))}) ≤ s} ∗ = inf{σ : µ({x ∈ Et : |f| > σ + f (t)}) ≤ s} |f|>f ∗(t) only on E =t inf{σ : µ({x : |f| > σ + f ∗(t)}) ≤ s} = inf{σ > f ∗(t): µ({x : |f| > σ}) ≤ s} − f ∗(t). (6.5)

Now if σ ≤ f ∗(t), then

∗ µ({x : |f| > σ}) ≥ µ({x : |f| > f (t)}) = µ(Et) > s, so that from (6.5) we see that

∗ ∗ ∗ ∗ f1 (s) = inf{σ : µ({x : |f| > σ}) ≤ s} − f (t) = f (s) − f (t). It follows that ¢ ¢ µ(Et) t ∗ ∗ ∗ ∗ kf1kL1 = f (s) − f (t) ds ≤ f (s) ds − tf (t) 0 0 and that ∗ ∗ |f∞(x)| ≤ |f (t)χEt (x)k ≤ f (t), x ∈ Ω, so that ¢ t 1 ∞ ∗ K(t, f, L ,L ) ≤ kf1kL1 + tkf∞kL∞ = f (s) ds. 0 ¡ 2. From (6.2), the claim follows fairly directly. Note that t ∗ ∗ , as ∗ 0 f (s) ds ≥ tf (t) f is non-increasing, so that for q < ∞, we have ¢ ∞ 1/q −θq q dt kfk 1 ∞ = t K (t, f) (L ,L )θ,q t ¢0 ∞ dt1/q ≥ t(1−θ)q(f ∗(t))q 0 t

= kfk 1 ,q , L 1−θ and for q = ∞,

−θ 1−θ ∗ sup |t K(t, f)| ≥ sup |t f (t)| = kfk 1 ,∞ . t>0 t>0 L 1−θ On the other hand, by the Hardy-Young inequality, Lemma 3.2, we have ¢ ¢ ∞ t q −θ q −θq ∗ ds dt kt K(t, f)kLq = t sf (s) ∗ 0 ¢ 0 s t ∞ 1 (1−θ)q ∗ q ds ≤ q s (f (s)) θ 0 s 1 = kfkq q 1 ,q θ L 1−θ

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and also ¢ t −θ −θ ds 1−θ ∗ 1 sup |t K(t, f)| ≤ t 1−θ sup(s f (s)) = kfk 1 ,∞ . t>0 0 s s>0 θ L 1−θ

Theorem 6.7. For 1 < p0 < p1 < ∞, 0 < θ < 1 and 1 ≤ q ≤ ∞ we have 1 1 − θ θ p0 p1 p,q (L ,L )θ,q = L , where = + . p p0 p1 In particular, if p < q < p and θ = p1 p0−q , then 0 1 q p0−p1

p0 p1 q (L ,L )θ,q = L .

Moreover, for 1 < p0 < p1 < ∞, 0 < θ < 1 and 1 ≤ q0 ≤ q1 ≤ ∞, we have

p0,q0 p1,q1 p,q (L ,L )θ,q = L for p as above.

p 1 ∞ p 1 ∞ Proof. By Theorem 6.6, we know that L = (L ,L )1−1/p,p, so L ∈ J1−1/p∩K1−1/p(L ,L ). We can therefore apply the Reiteration Theorem 4.4 to get

p0 p1 1 ∞ (L ,L )θ,q = (L ,L )ω,q,

1 ∞ p,q where ω = (1 − θ)(1 − 1/p0) + θ(1 − 1/p1). By Theorem 6.6, (L ,L )ω,q = L , where 1 = 1 − ω = 1−θ + θ . The same argument yields the third statement. p p0 p1

6.2.3 The Marcinkiewicz Theorem

Denition 6.8. Let (Ω, µ) and (Λ, ν) be two σ-nite measure spaces. A linear operator T : L1(Ω, µ) + L∞(Ω, µ) → L1(Λ, ν) + L∞(Λ, ν) is called of weak type (p, q), if there is an M > 0 such that

1/q 1/q ∗ sup σ(ν{y ∈ Λ: |T f(y)| > σ}) = sup t (T f) (t) ≤ MkfkLp(Ω,µ). σ>0 t>0 for all f ∈ Lp(Ω, µ), i.e. the restriction of T to Lp(Ω, µ) is bounded from Lp(Ω, µ) to Lq,∞(Λ, ν). T is said to be of strong type (p, q) if its restriction to Lp(Ω, µ) is bounded from Lp(Ω, µ) to Lq(Λ, ν).

Clearly, every operator of strong type (p, q) is also of weak type (p, q).

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1 ∞ 1 ∞ Theorem 6.9. Let T : L (Ω, µ)+L (Ω, µ) → L (Λ, ν)+L (Λ, ν) be of weak type (p0, q0) and (p1, q1) with constants M0,M1, respectively and 1 < p0, p1 < ∞, 1 < q0, q1 < ∞, q0 6= q1 and p0 ≤ q0, p1 ≤ q1. For 0 < θ < 1 let 1 1 − θ θ 1 1 − θ θ = + , = + . p p0 p1 q q0 q1 Then T is of strong type (p, q) and there is a constant C, independent of θ, such that

1−θ θ kT fkLq(Λ,ν) ≤ CM0 M1 kfkLp(Ω,µ) for all f ∈ Lp(Ω, µ).

p q ,∞ Proof. For i = 0, 1 , T is bounded from L i (Ω, µ) to L i (Λ, ν) with norm Mi. Since real interpolation is exact of type θ, Theorem 2.4, it follows that

1−θ θ p p q ,∞ q ,∞ kT kL((L 0 ,L 1 )θ,p,(L 0 ,L 1 )θ,p) ≤ M0 M1 .

p0 p1 p,p p q0,∞ q1,∞ q,p By Theorem 6.7, (L ,L )θ,p = L = L (Ω, µ) and (L ,L )θ,p = L (Λ, ν). q,p q,q q Since pi ≤ qi, also p ≤ q, so that L ,→ L = L (Λ, ν) by Theorem 2.5. This yields the claim.

6.3 Hölder spaces

Reminder: 1 n is continuously dierentiable and f ∈ C (Ω) ⇔ f :Ω ⊂ R → R kfkC1(Ω) := Pn . kfk∞ + i=1 kDifk∞ < ∞ Denition 6.10. For θ ∈ (0, 1), we dene the Hölder space Cθ(Ω) for every open set n Ω ⊂ R , n ∈ N, as the space of bounded and uniformly θ-Hölder continuous functions f with the norm

|f(x) − f(y)| kfkCθ(Ω) = kfk∞ + [f]Cθ = kfk∞ + sup θ . x,y∈Ω,x6=y |x − y|

Theorem 6.11. For θ ∈ (0, 1), we have that

n 1 n θ n (C(R ),C (R ))θ,∞ = C (R ) with equivalence of norms.

Proof. For convenience, we may write C,C1 etc. in the following. First we show that 1 θ 1 (C,C )θ,∞ ,→ C : Clearly, for every f ∈ C , kfk∞ ≤ kfkC1 . For every decomposition 1 f = f0 + f1, f0 ∈ C, f1 ∈ C , we have kfk∞ ≤ kf0k∞ + kf1k∞, so

1 1 kfk∞ ≤ K(1, f, C, C ) ≤ kfk(C,C )θ,∞ .

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Moreover, if x 6= y, we have that

1 θ 1 |f(x) − f(y)| ≤ 2K(|x − y|, f, C, C ) ≤ 2|x − y| kfk(C,C )θ,∞ by (2.2), so that must be -Hölder continuous and θ 1 . f θ kfkC = kfk∞+[f]θ ≤ 3kfk(C,C )θ,∞ θ 1 θ Now we show that C ,→ (C,C )θ,∞: Let f ∈ C . We want to decompose f into a C- 1 ∞ and a C -part with good estimates. Let ϕ ∈ C such that ϕ ≥ 0, supp ϕ ⊂ B1 and kϕkL1 = 1. For every t > 0, we consider ¢ 1 x − y f1,t(x) = n f(y)ϕ( ) dy, t n t R f0,t(x) = f(x) − f1,t(x). (6.6) ¡ It follows that 1 z , so f0,t(x) = tn n (f(x) − f(x − z))ϕ( t ) dz R ¢ ¢ 1 θ y θ θ kf0,tk∞ ≤ [f]Cθ n |y| ϕ( ) dy = t [f]Cθ |w| ϕ(w) dw. t n t n R R

It also follows that kf1,tk∞ ≤ kfk∞. It remains to look at the derivatives, for i = 1, . . . , n, ¢ 1 x − y Dif1,t(x) = n+1 f(y)(Diϕ)( ) dy. t n t R ¡ Since has compact support, we have x−y . We get that ϕ n Diϕ( t ) dy = 0 ¢ R 1 z (6.7) Dif1,t(x) = n+1 (f(x − z) − f(x))(Diϕ)( ) dz, t n t R so that ¢ 1 |f(x − z) − f(x)| θ z kDif1,tk∞ ≤ n+1 θ |z| (Diϕ)( ) dz t n |z| t R ¢ θ−1 θ = t [f]Cθ |w| (Diϕ)(w) dw. n R For 0 < t ≤ 1, it follows that

−θ −θ t K(t, f) ≤ t (kf0,tk∞ + tkf1,tkC1 ) ≤ CkfkCθ .

For t > 1 , we set f0,t = f and f1,t = 0 to immediately get that

−θ −θ t K(t, f) ≤ t kfk∞ ≤ kfk∞.

1 It follows that and that 1 θ , with depending only f ∈ (C,C )θ,∞ kfk(C,C )θ,∞ ≤ CkfkC C on the choice of ϕ.

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Remark 6.12. The following results are mostly taken from [4], Chapter 2: 1. follows immediately, 2., 3., 4, could also be an exercise:

1. By Reiteration, Theorem 4.4:

α β 1 1 1 ω (C ,C )θ,∞ = ((C,C )α,∞, (C,C )β,∞)θ,∞ = (C,C )ω,∞ = C ,

where α, β ∈ (0, 1) and ω = (1 − θ)α + θβ. 2. It holds that 1 n ”(C,C )1,∞” = Lip(R ), n where Lip(R ) is the space of Lipschitz continuous and bounded functions, |f(x) − f(y)| kfk n = kfk + sup . Lip(R ) ∞ x6=y |x − y|

3. C1 is not dense in Cθ.

4. For any interval I ⊂ R, X a Banach space, 0 < θ < 1,

1 θ (C(I; X),C (I,X))θ,∞ = C (I; X).

5. Let m ∈ N and θ ∈ (0, 1). If θm is not an integer,

m θm (C,C )θ,∞ = C .

6. In the situation of 3., if θm = 1, then

m 1 (C,C )θ,∞ = C ,

where for every 0 < α < 2, Cα is given by

x+y α |f(x) − 2f( 2 ) + f(y)| α C = {f ∈ C :[f]C = sup α < ∞}, x6=y |x − y|

kfkCα = kfk∞ + [f]Cα .

For α 6= 1, these spaces are equivalent to the Cα's.

6.4 Slobodeckii spaces

α,p n Dilemma I: How to dene W (R ) for α∈ / N, 1 ≤ p < ∞ and n ∈ N?

α,p p 1,p 1. W = (L ,W )α,p for 0 < α < 1.

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2. W α,p is the set of functions f ∈ Lp, such that ¢ |f(x) − f(y)|p 1/p [f]α,p = αp+n dx dy < ∞, n n |x − y| R ×R

endowed with the norm kfkW α,p = kfkLp + [f]W α,p . Theorem 6.13. The two denitions above coincide.

α,p n Dilemma II: How to dene W (Ω) for open sets Ω ⊂ R ?

1. as in 1. above.

α,p α,p n 2. W (Ω) is the space of functions which are restrictions of functions in W (R ) to Ω. Theorem 6.14. The two denitions above coincide if Ω is a domain with uniformly C1-boundary.

n Denition 6.15. On an open set Ω ⊂ R , for α ∈ (0, 1), the (Sobolev-)Slobodeckii spaces α are dened by Wp (Ω) α p 1,p Wp (Ω) = (L (Ω),W (Ω))α,p.

Proof. of Theorem 6.13. It works similarly as the proof for Hölder spaces, Theorem 6.11. First we show that α α,p, dened as in 2.: Wp ,→ W 1,p n Note that for every f1 ∈ W and h ∈ R \{0}, we have that (exercise) ¢ p 1/p |f1(x + h) − f1(x)| dx ≤ k|Df1|kLp . n |h| R

p 1,p n p 1,p For all f ∈ (L ,W )α,p and h ∈ R , we set f0,h ∈ L and f1,h ∈ W such that f = f0,h + f1,h and kf0,hkLp + |h|kf1,hkW 1,p ≤ 2K(|h|, f). Moreover,

|f(x + h) − f(x)|p |f (x + h) − f (x)|p |f (x + h) − f (x)|p ≤ 2p−1 0,h 0,h + 1,h 1,h |h|αp+n |h|αp+n |h|αp+n and so ¢ ¢ p p p |f(x + h) − f(x)| p−1 |f0,h(x + h) − f0,h(x)| |f1,h(x + h) − f1,h(x)| αp+n dx ≤ 2 αp+n + αp+n dx n |h| n |h| |h| R R kf kp |h|pk|Df |kp ≤ 2p 0,h Lp + 2p−1 1,h Lp |h|αp+n |h|αp+n −αp−n p ≤ Cp|h| (kf0,hkLp + |h|kf1,hkW 1,p ) −αp−n p ≤ Cp|h| K(|h|, f) .

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Therefore, ¢ p p |f(x + h) − f(x)| [f]α,p = αp+n dx dh n n |h| R ¢×R −αp−n p ≤ Cp |h| K(|h|, f) dh n R¢ ∞ K(r, f)p = Cp,n αp+1 dr 0 r p = Cp,nkfk p 1,p . (L ,W )α,p

p 1,p p 1,p p From the embedding (L ,W )α,p ,→ L + W = L we also get that kfkLp ≤ p 1,p , so that in conclusion, Ckfk(L ,W )α,p

α,p p 1,p kfkW ≤ Ckfk(L ,W )α,p .

α,p We now check the other embedding: For every f ∈ W , we dene f0,t and f1,t as in (6.6). It follows that ¢ ¢ p p 1 x − y kf0,tkLp = |f(x) − f(y)| n ϕ( ) dy dx n n t t ¢R R p 1 x − y ≤ |f(x) − f(y)| n ϕ( ) dy dx n n t t R ×R from the Hölder inequality applied to

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¡ where Ci = n |Diϕ(y)| dy. We also see directly that kf1,tkLp ≤ kfkLp kϕk 1 = kfkLp , R L so ¢ 1 (1−α)p p dt 1 p t kf1,tkLp ≤ kfkLp . 0 t (1 − α)p It follows that

−α p 1,p −α 1−α p t K(t, f, L ,W ) ≤ t kf0,tkLp + t kf1,tkW 1,p ∈ L∗(0, 1) and that its norm is estimated by CkfkW α,p . Again, this suces to get that f ∈ p 1,p (L ,W )α,p.

6.5 Functions on domains

This section follows parts of Chapters 3,4 and 5 in [1], in particular, pages 147-152.

n m Denition 6.16. A domain Ω ⊂ R satises the uniform C -regularity condition if there exists a locally nite open cover {Uj} of ∂Ω and a corresponding sequence {Φj} of -Dieomorphisms taking onto the unit ball with inverses −1 such that m Uj B1(0) Ψj = Φj

1. for some nite R, every intersection of R + 1 of the sets Uj is empty, 2. for some δ > 0,

Ωδ = {x ∈ Ω : dist(x, ∂Ω) < δ} ⊂ ∪jΨj(B1/2(0)),

3. for each j, Φj(Uj ∩ Ω) = {y ∈ B1(0) : yn > 0},

4. there is a nite constant , such that for every m and M > 0 j, kΦjkC (Uj ) ≤ M m . kΨjkC (B1(0)) ≤ M

Proof. Of Theorem 6.14. We use that for every uniform C1-domain Ω, there is an extension operator E, such that

p p n 1,p 1,p n E ∈ L(L (Ω),L (R )),E ∈ L(W (Ω),W (R )), (6.8)

α,p α,p n α,p n α,p and E ∈ L(W (Ω),W (R )). Here, W (R ) is given as in Dilemma I.2 and W (Ω) α,p n is given as the space of functions which are restrictions of functions in W (R ) to Ω. This operator can be constructed similarly as before, rst for the half-space, then by α,p α,p n localization. However, in order to get that E ∈ L(W (Ω),W (R )), we take functions from 1,p n and extend by exact reection, not as in the proof of Theorem 6.19 for W (R+) continuously dierentiable functions. From (6.8), it follows that α α n as real interpolation is exact of E ∈ L(Wp (Ω),Wp (R )) type . For every α,p , we have that α,p n α n . Moreover, θ f ∈ W (Ω) Ef ∈ W (R ) = Wp (R ) p n p n p the restriction operator R : f ∈ L (R ) 7→ f|Ω clearly belongs to L(L (R ),L (Ω)) ∩

54 6 Examples

1,p n 1,p and thus to α n α . It follows that belongs L(W (R ),W (Ω)) L(Wp (R ),Wp (Ω)) R(E(f)) to α and we have the estimate Wp (Ω)

kfk α ≤ C kE(f)k α n ≤ C kE(f)k α,p n ≤ Ckfk α,p . Wp (Ω) E Wp (R ) E,α,p W (R ) W (Ω) On the other hand, for α , we know that α,p n and so f ∈ Wp (Ω) Ef ∈ W (R ) R(Ef) = f ∈ W α,p(Ω) and kfk α,p ≤ kE(f)k α n ≤ kfk α . W (Ω) Wp (R ) Wp (Ω)

n 1 Remark. In the same way, we get that for Ω ⊂ R with a uniformly C -boundary,

1 θ (C(Ω),C (Ω))θ,∞ = C (Ω).

55 7 Function Spaces from Fourier Analysis

Reminder I:

n 0 n n •S(R ) Schwartz space, S (R ) the dual space of S(R ) • Fourier transform: ¢ 1 (Fϕ)(ξ) =ϕ ˆ(ξ) = e−i(x·ξ)ϕ(x) dx, n/2 n (2π) R −1 n n 0 n 0 n 2 n and F, F : S(R ) → S(R ), S (R ) → S (R ), F ∈ L(L (R )) unitary, F : Lp → Lp0 if 1 ≤ p ≤ 2.

• Dα(Fϕ)(ξ) = (−i)|α|F(xαϕ(x)) and ξα(Fϕ)(ξ) = (−i)|α|F(Dαϕ(x)). Denition 7.1. (Bessel potential spaces) Let 1 < p < ∞ and s ∈ R, we dene

s,p n n 0 n −1 2 s o H ( ) := f ∈ S ( ): kfk s,p n := kF (1 + |ξ| ) 2 Ffk p n < ∞ . R R H (R ) L (R )

θm,p p m,p Remark 7.2. Hopefully, we will be able to show that H = [L ,W ]θ.

Reminder II: Sobolev spaces: 1 ≤ p ≤ ∞, k ∈ N,   k,p n  p n X α  W ( ) := f ∈ L ( ): kfk k,p n := kD fk p n < ∞ . R R W (R ) L (R )  |α|≤k 

k,p n k,p n Question: If 1 < p < ∞, k ∈ N, then H (R ) = W (R )?

p n 0,p n 0,p n 1. k = 0: L (R ) = H (R ) = W (R ) 2. p = 2: true, using Plancherel

3. p 6= 2: also true, using Mikhlin multiplier theorem

We immediately see 1., we look at 2. and 3. in the following.

n Note that for any multi-index |α| ≤ k and ξ ∈ R ,

α |α| 2 k |ξ | ≤ |ξ| ≤ (1 + |ξ| ) 2 (7.1)

56 7 Function Spaces from Fourier Analysis and that for every f ∈ Hk,2, we automatically have f ∈ L2. It follows that

X α X α −1 2 k kfkW k,2 ≤ kFD fkL2 ≤ kξ FfkL2 ≤ CkF (1 + |ξ| ) 2 FfkL2 = CkfkHk,2 . |α|≤k |α|≤k (7.2) ∞ Conversely, let ρ : R → [0, 1], ρ ∈ C (R) and ρ(s) = −ρ(−s) such that

( 1 2 k 0, t ≤ (1 + |ξ| ) 2 ρ(t) = 2 , so that → 1 . Pn k k 1, t ≥ 0 1 + i=1 ρ (ξi)ξi |ξ|→0,∞ It follows that n ! k X 2 2 k k (7.3) (1 + |ξ| ) ≤ Cn,k 1 + ρ (ξi)ξi . i=1 We get that

2 k kfkHk,2 = k(1 + |ξ| ) 2 Ff(ξ)kL2 2 k n ! (1 + |ξ| ) 2 X k k = k 1 + ρ (ξi)ξ Ff(ξ)k 2 1 + Pn ρk(ξ )ξk i L i=1 i i i=1 n ! X k k ≤ Cn,k kFfkL2 + kρ (ξi)ξi Ff(ξ)kL2 i=1 n k ! X k ∂ ≤ C kFfk 2 + k(−i) F( f)(x)k 2 n,k L ∂ xk L i=1 i ≤ CkfkW k,2 .

Now we look at the case p 6= 2 and introduce a preliminary concept, without proofs.

∞ n Denition 7.3. Let 1 ≤ p ≤ ∞ and m ∈ L (R ; C). Then m is called a Fourier n multiplier, m ∈ Mp, if there is a constant c > 0, such that for all f ∈ S(R ),

−1 kF mFfkLp ≤ ckfkLp , so that the operator n n −1 Tm : S(R ) → C0(R ), f 7→ F mFf can be extended to a bounded operator on Lp. Theorem 7.4. (Mikhlin-Hörmander) b n c+1 n Let 1 < p < ∞. Then m ∈ C 2 (R \{0}) is a Fourier multiplier in Mp if there is a constant cm > 0 such that |β| β |ξ| |D m(ξ)| ≤ cm for all n and multi-indices n . ξ ∈ R \{0} |β| ≤ 2 + 1

57 7 Function Spaces from Fourier Analysis

Proof. For example, [3], Theorem 6.1.6 for a proof and a more general version of the theorem.

s,p n Theorem 7.5. Let 1 < p < ∞, k ∈ N0, s ∈ R. Then H (R ) is a Banach space and we have k,p n k,p n W (R ) = H (R ).

Proof. The idea is as in the case p = 2, only we use Theorem 7.4, too. We know that X α −1 X −1 α kfkW k,p = kD (F Ff)kLp = kF (ξ Ff(ξ))kLp . |α|≤k |α|≤k It follows that , if we can show that kfkW k,p ≤ CkfkHk,p −1 α −1 2 k kF (ξ Ff(ξ))kLp ≤ Cn,kkF (1 + |ξ| ) 2 FfkLp for all α. We dene ξα mα,k(ξ) = k (1 + |ξ|2) 2 for |α| ≤ k and show that it is a Fourier multiplier. For |α| = 0, we have |mα,k(ξ)| ≤ 1 by (7.1). If |α| = 1, we get α −1 α α1 j αn 2 k/2 2 k/2−1 ∂ ξ αjξ ····· ξ ····· ξn (1 + |ξ| ) − ξjk(1 + |ξ| ) |ξ| = |ξ| 1 j 2 k/2 2 k ∂ξj (1 + |ξ| ) (1 + |ξ| )

|ξ| ξjk = αj − ≤ 2k. (1 + |ξ|2)k/2 (1 + |ξ|2) By more similar calculations and induction, the claim can be proved. It follows that −1 α −1 2 k/2 kF ξ FfkLp = kF mα,k(1 + |ξ| ) FfkLp −1 −1 2 k/2 = kF mα,kFF (1 + |ξ| ) FfkLp −1 2 k/2 ≤ Cn,kkF (1 + |ξ| ) FfkLp for every f ∈ Lp. For the converse estimate, we dene (1 + |ξ|2)k/2 m˜ (ξ) = ρ,k Pn k k 1 + i=1 ρ (ξi)ξi and show that it is a Fourier multiplier. For |α| = 0, we have |m˜ ρ,k(ξ)| ≤ Cn,k by (7.3). For , we use the notation Pn k k , 2 and calculate: |α| = 1 1 + i=1 ρ (ξi)ξi =:ρ ˜ (1 + |ξ| ) =: Mξ

k/2−1 k/2 k−1 0 k k k−1 ∂ ξjkMξ ρ˜ − Mξ [kρ (ξj)ρ (ξj)ξj + ρ (ξj)kξj ] |ξ| m˜ ρ,k(ξ) = |ξ| ∂ξ ρ˜2 j k/2−1 M ≤ C |ξ| ξ k M (k+1)/2 − M C + M (k−1)/2 n,k ρ˜2 ξ ξ k ξ M k/2−1 ≤ C M 1/2 ξ M k/2+1/2 ≤ C m˜ 2 (ξ) ≤ C . n,k ξ ρ˜2 ξ n,k ρ,k n,k

58 7 Function Spaces from Fourier Analysis

Again, we could calculate similarly for |α| > 1 and use induction to get the claim. We k k now use that ξ 7→ ρ (ξj)(−i) is a Fourier multiplier for every 1 ≤ j ≤ n, to get that

−1 −1 kfkHk,p ≤ kF m˜ ρ,kρ˜kFfkLp ≤ Cm˜ kF ρ˜kFfkLp n −1 X k k ≤ C(kfkLp + kF ( ρ (ξi)ξi )FfkLp ) i=1 n −1 X k k k = C(kfkLp + kF ( ρ (ξi)(−i) F(∂i f))kLp ) i=1 n X k ≤ C(kfkLp + k∂i fkLp ) ≤ CkfkW k,p . i=1

We will come back to Bessel potential spaces, but now want to dene and consider Besov spaces.

Denition 7.6. A sequence of functions ∞ n is called dyadic partition of (ϕj)j∈N ⊂ Cc (R ) unity, if and only if

1. supp ϕj ⊂ Aj, j ∈ N, where

A0 = B2(0), n j−1 j+1 Aj = {x ∈ R : 2 < |x| < 2 }, j 6= 0,

2. P∞ , j=0 ϕj(x) = 1 3. for all multi-indices n, there is a constant , such that γ = (γ1, ..., γn) ∈ N0 cγ j|γ| γ 2 |D ϕj(x)| ≤ cγ.

n Example: If ϕ ∈ S(R ), supp ϕ ⊂ B2(0) and ϕ(x) = 1 for |x| ≤ 1, then setting ϕ0 = ϕ −j −j+1 and ϕj(x) = ϕ(2 x) − ϕ(2 x) for j 6= 0 yields a dyadic partition of unity. Denition 7.7. (Besov spaces) Let , and a dyadic partion of unity. Then the Besov spaces s ∈ R 1 ≤ p, q ≤ ∞ (ϕj)j∈N s n are the spaces of all functions 0 n such that Bp,q(R ) f ∈ S (R )

 1/q  P∞ jsq −1 q j=0 2 kF ϕiFfkLp( n) , q < ∞ kfkBs ( n) := R < ∞. p,q R js −1  sup 2 kF ϕFfk p n , q = ∞ j∈N0 L (R ) Remark 7.8. We will show that p m,p θm, so that in particular, s s. (L ,W )θ,q = Bp,q Bp,p = Wp Proposition 7.9. s n is a Banach space and choosing two dierent dyadic parti- Bp,q(R ) tions yields equivalent norms.

59 7 Function Spaces from Fourier Analysis

Proof. Exercise. In order to get interpolation results for Besov spaces, we start from the following type of spaces.

Denition 7.10. Let be a Banach space, and . Then σ is the X σ ∈ R 1 ≤ p ≤ ∞ lp (X) space of sequences such that (xj)j∈N ⊂ X  1/p  P∞ jσp p j=0 2 kxjkX , p < ∞ kxk σ := < ∞. lp (X)  sup 2jσkx k , p = ∞ j∈N0 j X Remark 7.11. It is a Banach space. If and , then σ p. Moreover, if X = C σ = 0 lp (X) = l , then σ σ. 1 ≤ r ≤ p ≤ ∞ lr ,→ lp

Proof. Exercise.

Theorem 7.12. Let X be a Banach space, s0, s1 ∈ R, s0 6= s1, 1 ≤ p0, p1, p ≤ ∞ and 0 < θ < 1. Then we get s0 s1 s (lp0 (X), lp1 (X))θ,p = lp(X), where s = (1 − θ)s0 + θs1.

Proof. Use Theorem 2.5.6 and show that

s0 s1 s s0 s1 (7.4) (l∞, l∞)θ,p ,→ lp ,→ (l1 , l1 )θ,p to get that

s s0 s1 s0 s1 s0 s1 s lp ,→ (l1 , l1 )θ,p ,→ (lp0 , lp1 )θ,p ,→ (l∞, l∞)θ,p ,→ lp. We now rst show the rst embedding in (7.4). We know that

s0 s1 js0 0 js1 1 K(t, x, l∞, l∞) = inf (sup 2 kxj kX + t sup 2 kxj kX ) x=x0+x1 j j

min(s ,s ) for all s0 s1 0 1 . We look at x ∈ l∞ + l∞ = l∞ ( x 2js0 ≤ t2js1 , xˆ0 = j, and xˆ1 = x − xˆ0, j 0, otherwise. to see that

s0 s1 js0 js1 K(t, x, l∞, l∞) ≤ 2 sup min(2 , t2 )kxjkX . j On the other hand, clearly,

js0 js1 js0 0 js1 1 sup min(2 , t2 )kxjk ≤ sup(2 kxj kX + t2 kxj kX ) j j

60 7 Function Spaces from Fourier Analysis for all decompositions x = x0 + x1, so that

s0 s1 js0 js1 K(t, x, l∞, l∞) ∼ sup min(2 , t2 )kxjkX . j

Wlog, we assume that , since we could consider s1 s0 instead. We treat the s0 > s1 (lp1 , lp0 )1−θ,q case when and use the decomposition ∞ (k−1)(s0−s1) k(s0−s1) . p < ∞ (0, ∞) = ∪k=−∞[2 , 2 ) For s0 s1 , x ∈ (l∞, l∞)θ,p

¢ k(s −s ) ∞ 2 0 1 dt p X −θp s0 s1 p kxk s0 s1 = t K(t, x, l∞(X), l∞(X)) (l∞ ,l∞ )θ,p (k−1)(s −s ) t k=−∞ 2 0 1 ¢ k(s −s ) ∞ 2 0 1 dt X −θpk(s0−s1) js0p kp(s0−s1) js1p p ≥ 2 sup min(2 , 2 2 )kxjkX j (k−1)(s0−s1) t k=−∞ 2 ∞ X −θpk(s0−s1) ks0p kp(s0−s1) ks1p p ≥ C 2 min(2 , 2 2 )kxkkX (s0 − s1) ln(2) k=−∞ ∞ X kps p p = C 2 kxkk = Ckxk s X lp k=0

The case p = ∞ follows analogously. Now we show the second embedding in (7.4). Again, we may assume that s0 > s1. Analogously to the l∞-situation, we see that

∞ s0 s1 X js0 js1 K(t, x, l1 , l1 ) ∼ min(2 , t2 )kxjkX . j=0

For and s, we calculate p < ∞ x ∈ lp

p ¢ k(s −s )   ∞ 2 0 1 ∞ dt p X −θp(k−1)(s0−s1) X js0 k(s0−s1)+js1 kxk s0 s1 ≤ 2  min(2 , 2 )kxjkX  (l1 ,l1 )θ,p (k−1)(s −s ) t k=−∞ 2 0 1 j=0 p ∞  ∞  θp(s0−s1) X kps X (j−k)s0 (j−k)s1 ≤ C2 2  min(2 , 2 )kxjkX  k=−∞ j=0 p ∞  k ∞  X kps X (j−k)s0 X (j−k)s1 = C 2  2 kxjkX + 2 kxjkX  . k=−∞ j=−∞ j=k+1

61 7 Function Spaces from Fourier Analysis

For every t0, t1 such that s0 > t0 > s > t1 > s1, we have

1/p0 1/p k  k   k  0 X (j−k)s0 −ks0 X j(s0−t0)p X jt0p p 2 kxjkX ≤ 2  2   2 kxjkX  j=−∞ j=−∞ j=−∞ 1/p  k  −kt0 X jt0p p ≤ C2  2 kxjkX  , j=−∞ 1/p0 1/p ∞  ∞   ∞  0 X (j−k)s1 −ks1 X j(s1−t1)p X jt1p p 2 kxjkX ≤ 2  2   2 kxjkX  j=k+1 j=k+1 j=k+1 1/p  ∞  −kt1 X jt1p p ≤ C2  2 kxjkX  , j=k+1 so that

∞  k ∞  p X kps −kt0p X jt0p p −kt1p X jt1p p kxk s0 s1 ≤ C 2 2 2 kxjkX + 2 2 kxjkX  (l1 ,l1 )θ,p k=−∞ j=−∞ j=k+1 ∞ X jsp p p ≤ C 2 kxjk = Ckxk s . X lp j=−∞ This proves the theorem.

We now summarize several results on real and complex interpolation of Besov and Bessel potential spaces in the following theorem, and then set out to give a proof for the real part, 1.-4. and some arguments for the complex part, 5. and 6.

Theorem 7.13. Let 0 < θ < 1, s, s0 6= s1 ∈ R, 1 ≤ p, p0, p1, q, q0, q1 ≤ ∞ and 1 1 − θ θ 1 1 − θ θ sθ = (1 − θ)s0 + θs1, = + , = + . pθ p0 p1 qθ q0 q1 Then we get

1. s0 s1 sθ (Bp,q0 ,Bp,q1 )θ,q = Bp,q,

2. (Bs0 ,Bs1 ) = Bsθ , p = q ,q , q 6= 1 p0,q0 p1,q1 θ,pθ pθ,qθ θ θ 0 1

3. s0,p s1,p sθ (H ,H )θ,q = Bp,q, 1 < p < ∞,

4. s,p0 s,p1 s,pθ (H ,H )θ,pθ = H , 1 < p0, p1 < ∞,

5. [Bs0 ,Bs1 ] = Bsθ , p0,q0 p1,q1 θ pθ,qθ

62 7 Function Spaces from Fourier Analysis

s0,p0 s1,p1 s ,p 6. [H ,H ]θ = H θ θ , 1 < p0, p1 < ∞.

Let (ψi)i be a dyadic partition of unity. To show 1., by looking at the map

s s p −1 Sψ : Bp,q → lq(L ), f 7→ (F ψjFf)j, we get the idea that we can interpolate s as the we interpolate s p by Theorem Bp,q lq(L ) 7.12. The following notion is helpful.

Denition 7.14. An object X in a category is called a retract of another object Y , if there are morphisms S : X → Y and R : Y → X in the category, such that R ◦ S = idX . In this case, the map R is called retraction, and S is called coretraction. Clearly, the following holds true.

Lemma 7.15. If {X0,X1} is a retract of {Y0,Y1} in C2 with retraction R and coretraction S, then for every interpolation functor F, F({X0,X1}) is a retract of F({Y0,Y1}) with retraction F(R) = R and coretraction S. Proposition 7.16. The space s is a retract of s p with the coretraction and the Bp,q lq(L ) S retraction ∞ s p s X −1 Rϕ : lq(L ) → Bp,q, (xj)j 7→ F ϕjFxj. j=0

Proof. Clearly, s s p . On the other hand, Sψ ∈ L(Bp,q, lq(L ))

1/q  ∞ ∞  X jsq −1 X −1 q s kRϕ(xj)jkBp,q =  2 kF ψjF( F ϕkFxj)kLp  j=0 k=0 1/q  ∞  X jsq q ≤ c  2 kxjkLp  j=0

p if we use that ψjϕj−1, ψjϕj and ψjϕj+1 are Fourier multipliers on L . Clearly,

∞ ∞ X −1 X −1 RϕSψf = F ϕjψjFf = F ψjFf = f. j=0 j=0

Theorem 7.13.1 now follows from Theorem 7.12 and Proposition 7.16. Next, we want to show 2.

Proposition 7.17. For 1 < q0, q1 ≤ ∞ and an interpolation couple {X0,X1}, we have

(ls0 (X ), ls1 (X )) = lsθ ((X ,X ) ). q0 0 q1 1 θ,qθ qθ 0 1 θ,qθ

63 7 Function Spaces from Fourier Analysis

Proof. Unpopular but elementary exercise.

We get that by this Proposition and Theorem 6.7,

(Bs0 ,Bs1 ) = (ls0 (Lp0 ), ls1 (Lp1 )) = lsθ ((Lp0 ,Lp1 ) ) = lsθ (Lpθ ) = Bsθ , p0,q0 p1,q1 θ,pθ q0 ,q1 θ,pθ qθ θ,pθ qθ qθ,pθ which is 2. In order to show 3., we use the following embeddings.

Proposition 7.18. For 1 < p < ∞ and s ∈ R, we have s s,p s (7.5) Bp,1 ,→ H ,→ Bp,∞, and in particular, 0 p 0 (7.6) Bp,1 ,→ L ,→ Bp,∞.

Proof. We note that (7.5) follows from (7.6) if we use that the operator J σ = F −1(1 + 2 σ/2 is an isomorphism from s,p to s−σ,p and from s to s−θ. For the rst |ξ| ) F H H Bp,q Bp,q embedding in (7.6), let 0 . We can write f ∈ Bp,1 ∞ X −1 f = F ϕjFf, j=0 to see that ∞ X −1 kfkLp ≤ kF ϕjFfkLp = kfk 0 . Bp,1 j=0

For the second embedding, let (ϕj)j as in the example above Denition 7.7. For f ∈ S, we get ¢ ¢ −1 −n ix·ξ −iyξ F ϕjFf(x) = (2π) e ϕj(ξ) e f(y) dy dξ n n ¢R ¢ R −n izξ = (2π) f(x − z) e ϕj(ξ) dξ dz n n R ¢ R ¢ (j−1)n −n iz2j−1η j−1 = 2 (2π) f(x − z) e ϕj(2 η) dη dz n n R¢ R (j−1)n −n/2 −1 j−1 = 2 (2π) f(x − z)F ϕ1(2 z) dz. n R It follows that ¢ −1 (j−1)n −1 j−1 kF ϕjFfkLp ≤ C2 kfkLp |F ϕ1(2 z)| dz n R −1 = CkfkLp kF ϕ1kL1 . In conclusion, −1 kfk 0 = sup kF ϕ Ffk p ≤ Ckfk p , Bp,∞ j L L j∈N0 which shows the second embedding in (7.6).

64 7 Function Spaces from Fourier Analysis

By 1., the above Proposition and Theorem 2.5.6, we have

sθ s0 s1 s0.p s1,p s0 s1 sθ Bp,q = (Bp,1,Bp,1)θ,q ,→ (H ,H )θ,q ,→ (Bp,∞,Bp,∞)θ,q = Bp,q.

This proves 3. Note that in particular, θk p k,p by Theorem 7.5, so that Bp,q = (L ,W )θ,q s s,p if . Bp,p = W s∈ / Z Regarding 4., we know that p0 p1 pθ by Theorem 6.7. The claim follows if (L ,L )θ,pθ = L we again use the fact that J s : Hs,p → Lp is an isomorphism. For 5., the idea is to use Proposition 7.16 and the complex interpolation result

[ls0 (X ), ls1 (X )] = lsθ ([X ,X ] ). q0 0 q1 1 θ qθ 0 1 θ For 6., we have to work much harder. The idea is to show that Hs,p is a retract of the space p s , which involves dening the Fourier transform for Hilbert space-valued L (l2) functions and a corresponding Mikhlin multiplier theorem and then use the Riesz-Thorin- type result p0 p1 pθ [L (X0),L (X1)]θ = L ([X0,X1]θ).

Reminder: Sobolev embedding theorem: n n n W k,p ,→ Lq, k − ≥ − , k − > 0. p q p

Remark, without proof: We have

s s,p s Bp,p ,→ H ,→ Bp,2, 1 < p ≤ 2, but s s,p s (7.7) Bp,2 ,→ H ,→ Bp,p, 2 ≤ p < ∞.

For Besov and Bessel potential spaces, we get the following embedding results.

Theorem 7.19. Assume that n n s − = s1 − , p p1 then s s1 Bp,q ,→ Bp1,q1 1 ≤ p ≤ p1 ≤ ∞, 1 ≤ q ≤ q1 ≤ ∞, s, s1 ∈ R, and s,p s1,p1 H ,→ H , 1 < p ≤ p1 < ∞, s, s1 ∈ R. In particular, W s,p ,→ W s1,p1 .

Proof. We prove the rst embedding directly and then use interpolation to get the second. It suces to show the estimate

1 1 −1 nk( p − p ) −1 kF ϕkFfkLp1 ≤ C2 1 kF ϕkFfkLp (7.8)

65 7 Function Spaces from Fourier Analysis to get that

∞ !1/q1 X q1 s ks1q1 −1 kfk 1 = 2 kF ϕkFfk p1 Bp1,q1 L k=0 ∞ !1/q X ksq −1 q ≤ C 2 kF ϕkFfkLp k=0

s ≤ CkfkBp,q , as q1 ≥ q. We show (7.8). ρ p 0 1 1 1 Reminder: Young's inequality. Let k ∈ L , f ∈ L , 1 < p < ρ , = − 0 , then p1 p ρ

kk ∗ fkLp1 ≤ kkkLρ kfkLp . Proof: Riesz-Thorin.... −1 −1 ˆ p We write: F ϕkFf = F (ϕk · f) =ϕ ˇk ∗ f ∈ L and ∞ −1 X −1 F ϕkFf = F ϕkϕjFf j=0 k+1 X −1 = F ϕjϕkFf j=k−1 k+1 X = ϕˇj ∗ (ϕ ˇk ∗ f). j=k−1 By Young's inequality,

k+1 −1 X −1 kF ϕkFfkLp1 ≤ kϕˇjkLρ kF ϕkFfkLp . (7.9) j=k−1

We now estimate kϕˇjkLρ . Note that by the example above Denition 7.6, we can write −j −j+1 ϕj(x) = ϕ(2 x) − ϕ(2 x). It follows that ¢ ¢ ix·ξ −j ix·ξ −j+1 ϕˇj(ξ) = e ϕ(2 x) dx − e ϕ(2 x) dx n n R R = 2jnϕˇ(2jξ) − 2(j−1)nϕˇ(2j−1ξ). We directly get ¢ ¢ 1/ρ 1/ρ jn j ρ (j−1)n (j−1) ρ kϕˇjkLρ ≤ (2 ϕˇ(2 ξ)) dρ + (2 ϕˇ(2 ξ)) dρ n n R R jn(1−1/ρ) (j−1)n(1−1/ρ) = 2 kϕˇkLρ + 2 kϕˇkLρ jn(1/p−1/p1) ≤ 2 · 2 kϕˇkLρ .

66 7 Function Spaces from Fourier Analysis

This implies (7.8) via (7.9).

It remains to prove the second embedding. It suces to consider the case s1 = 0 and then apply the isomorphism s1 . For every , by denition, we see that s˜ r J 1 < p ≤ p1 Bp,q ,→ Bp,q if s˜ ≥ r. If we set n n s˜ = s − − p p1 and moreover use the rst embedding and (7.6), we get

s s˜ 0 p1 Bp,1 ,→ Bp1,1 ,→ Bp1,1 ,→ L . Now choose and 0 . We can then dene 00 0 and 00 via 0 < θ < 1 s < s s , p1 p1 s = (1 − θ)s0 + θs00, 0 n n s − = − 0 , p p1 00 n n s − = − 00 , p p1 to get

0 0 s p1 Bp,1 ,→ L , 00 00 s p1 Bp,1 ,→ L . By 7.13.1, Theorem 2.5.6 and Theorem 6.7, this implies

0 00 0 00 s s s p1 p1 p1,∞ Bp,∞ = (Bp,1,Bp,1)θ,∞ ,→ (L ,L )θ,∞ ,→ L . From (7.5), it follows that s,p s p1,∞ H ,→ Bp,∞ ,→ L . Now choose again and 0 . Dene 0 by n n , so that 0 0 . We θ 1 < q1 < p1 q s − 0 = − 0 1 < q ≤ q1 q q1 can then dene 00 and 00 via q q1 1 1 − θ θ = + , p q0 q00 1 1 − θ θ = 0 + 00 p1 q1 q1 n n and obtain s − 00 = − 00 . As above, we can get the inclusions q q1

s,q0 q0 ,∞ H ,→ L 1 , s,q00 q00,∞ H ,→ L 1 , and therefore, by Theorem 7.13.4, Theorem 2.5.6 and Theorem 6.7,

0 00 0 00 s,p s,q s,q q ,∞ q ,∞ p1,p H = (H ,H )θ,p ,→ (L 1 ,L 1 )θ,p = L .

p ,p p ,p p Since p ≤ p1, we have L 1 ,→ L 1 1 = L 1 , which proves the claim.

67 8 A Trace Theorem

We consider the trace operator

n n−1 T : S(R ) → S(R ), given by 0 0 0 T f(x ) = f(0, x ), x = (x2, . . . , xn). We want to show the following.

Theorem 8.1. Let 1 < p < ∞, 1 ≤ q ≤ ∞, s > 1/p. Then T can be extended to a bounded operator

s n s−1/p n−1 (8.1) T : Bp,q(R ) → Bp,q (R ), s,p n s−1/p n−1 (8.2) T : H (R ) → Bp,p (R ).

Proof. First note that n is dense in s,p n and s n as long as , S(R ) H (R ) Bp,q(R ) 1 ≤ p, q < ∞ p s,p p s ∈ R+. We know that S is dense in L . Given f ∈ H , we can approximate in L by and therefore we can approximate in s,p by −s . Density in s then follows gn H J gn ∈ S Bp,q n from Theorem 7.13.3. Thus, in the following, we consider functions in S(R ) and show the norm estimates.

The Theorem holds for all s > 1/p. We will only give a proof for 1 > s > 1/p, which is the crucial part. We split the proof into several steps. Step 2 is the crucial one. Its proof will be given further below, in a seperate theorem. Step 1 is a trace theorem we may obtain from the trace method. The remaining steps are concerned with using interpolation results suitably.

1−1/p n−1 Step 1. As an exercise, we deduce from the trace method, that Wp (R ) is the 1,p n 1,p n 1−1/p n−1 space of functions which are traces of W (R ), i.e. T : H (R ) → Bp,p (R ) is bounded and onto. Step 2. We show below that 1/p n p n−1 is bounded. T : Bp,1 (R ) → L (R ) Step 3. From 1 and 2, we can now deduce (8.1). We set such that θ . 0 < θ < 1 s = (1−θ)+ p Then by Theorem 7.13.1,

s n 1 n 1/p n T 1−1/p n−1 p n−1 (1−θ)(1−1/p) n Bp,q(R ) = (Bp,1(R ),Bp,1 (R ))θ,q → (Bp,p (R ),L (R ))θ,q = Bp,q (R ).

68 8 A Trace Theorem

It remains to calculate θ 1 1 . (1 − θ)(1 − 1/p) = s − p − (1 − θ) p = s − p Step 4. By the embedding (7.7), we immediately get (8.2) if 2 ≤ p < ∞:

s,p n s n T s−1/p n−1 H (R ) ,→ Bp,p(R ) → Bp,p (R ). Step 5. It remains to show (8.2) in the case 1 < p ≤ 2. We know that

s0,2 n s0−1/2 n−1 T : H (R ) → B2,2 (R ), s0 > 1/2,

1,p1 n s−1/p1 n−1 T : H (R ) → Bp1,p1 (R ), 1 < p1 < ∞. From the complex embeddings, Theorem 7.8.5 and 7.8.6, for 1 = 1−θ + θ and s = p 2 p1 (1 − θ)s0 + θ, we get

s,p n s0,2 n 1,p1 n T : H (R ) = [H (R ),H (R )]θ

T s0−1/2 n−1 1−1/p1 n−1 → [B2,2 (R ),Bp1,p1 (R )]θ = B(1−θ)(s0−1/2)+θ(1−1/p1)( n−1) 1 , 1 R 1−θ + θ 1−θ + θ 2 p1 2 p1 s−1/p n−1 = Bp,p (R ).

Now if s > 1/p and 1 < p ≤ 2, we can choose p1, 1 < p1 < p, so that 1−θ θ s − θ 1/p − θ + − θ s = > = 2 p1 0 1 − θ 1 − θ 1 − θ 1−θ θ + − θ p≤2 1 − θ > 2 p ≥ 2 1 − θ 1 − θ 1/2 1 = 1 − > . 1 − θ 2

To really show the theorem, it now remains to show Step 2. We will give an equivalent characterization of s n for , which helps. We dene the modulus of continuity Bp,q(R ) s > 0 by m m ωp (t, f) = sup k∆y fkLp , |y|

1 1 ∆yf(x) = f(x) − f(x + y), ωp(t, f) = sup kf(x) − f(x + y)kLp . |y|

69 8 A Trace Theorem

Theorem 8.2. Let s > 0, m, N ∈ N0, such that m + N > s and 0 ≤ N < s. Then for 1 ≤ p, q ≤ ∞,

¢ 1/q n ∞ N ! X N−s m ∂ f q dt kfk s n ∼ kfkLp + (t ω (t, )) . Bp,q(R ) p ∂xN t j=1 0 j

The proof follows the proof in [3], Theorem 6.2.5. Before we go into the details, we need to consider some more facts about Fourier multipliers, in particular, because we would like to cover the cases p = 1, ∞.

Theorem 8.3. We let 1 ≤ p ≤ ∞ and consider the space Mp of all Fourier multipliers 0 n with norm , where −1 . Then the following ρ ∈ S (R ) kρkMp = kTρkL(Lp) Tρf = F ρFf holds true.

1. We have Mp = Mp0 with equal norms.

2. ρ ∈ M∞ i −1 |F ρFf(0)| ≤ Ckfk∞ n for all f ∈ S(R ). 3. If , , then ρ ∈ Mp0 ∩ Mp1 1 ≤ p0, p1 ≤ ∞

1−θ θ kρkMp ≤ kρk kρkM . θ Mp0 p1

4. In particular, if 1 ≤ p < q ≤ 2, then

M1 ,→ Mp ,→ Mq ,→ M2.

n m 5. Let a : R → R be a surjective ane transformation. Then the map a˜ dened by (˜aρ)(x) = ρ(a(x))

m n is isometric from Mp(R ) to Mp(R ) (the space of Fourier multipliers which are n functions on R ). If m = n, then a˜ is bijective. 6. Let n be an integer and 2 n and α 2 for . Then for L > 2 ρ ∈ L (R , C) D ρ ∈ L |α| = L 1 ≤ p ≤ ∞, ρ ∈ Mp and

1−θ α θ kρkMp ≤ CkρkL2 ( sup kD ρkL2 ) , |α|=L where n . θ = 2L ∞ ∞ 7. We have M2 = L and for every ρ1, ρ2 ∈ Mp, ρ = ρ1ρ2 ∈ L and

kρkMp ≤ kρ1kMp kρ2kMp .

70 8 A Trace Theorem

p p0 Proof. for 1.: For f ∈ L , g ∈ L and ρ ∈ Mp we have ¢ ¢ |F −1ρFg(x) · f(x)| dx = ρˇ ∗ g · f(x) dx n n R ¢R = ρˇ ∗ f · g(x) dx n R −1 = kF ρFfkLp kgkLp0

p 0 ≤ kρkMp kfkL kgkLp . The second statement follows from the fact that the translation operators commute with

Tρ . The third statement is a consequence of the Riesz-Thorin theorem. It directly implies the fourth statement.

Regarding 5.: Ok to see: Mp isometrically invariant under non-singular linear coordinate transforms. After that, choose coordinates in a good way and look at the integrals (See [3], p. 134). Proof of 6.: From 4. and 2. we may deduce that it suces to show the estimate in the case p = 1, where ¢

kρkM1 = |ρˇ(x)| dx. n R We consider rst ¢ ¢ |ρˇ(x)| dx ≤ |x|−L|x|L|ρˇ(x)| dx |x|>t |x|>t −L+n/2 α ≤ Ct sup kD ρkL2 . |α|=L

Then we see ¢ n/2 |ρˇ(x)| dx ≤ Ct kρkL2 . |x|

1−θ α θ kρkMp ≤ kρkM1 ≤ CkρkL2 ( sup kD ρkL2 ) . |α|=L

∞ Proof of 7.: Clearly, if ρ ∈ L , then ρ ∈ M2 with equal norm. Conversely, if ρ ∈ M2, then −1 ˆ ˆ sup kF ρFfkL2 = sup kρfkL2 = kρkL∞ kfkL2 . kfk2=1 fˆ∈L2

We turn now to the proof of Theorem 8.2. We reduce it to the case where 0 ≤ s < 1 and let m = 1, N = 0. As an exercise, one can consider the situation when additionally, p = q.

71 8 A Trace Theorem

With this simplication, it remains to prove that

¢ 1/q ∞ ! −s q dt s p p kfkBp,q ∼ kfkL + n (t sup kf(·) − f(· + y)kL ) . 0 |y|

∞ ! X is 1 q 1/q (8.3) kfkLp + n (2 sup k∆yfkLp ) . −i i=−∞ |y|<2 The basic argument is that 1 is monotonely increasing in . We set ωp = ωp t (0, ∞) = −i −i+1 to get ∪i∈Z(2 , 2 ) ¢ ¢ ∞ ∞ 2−i+1 −sq q dt X −sq q dt t ωp(t, f) = t ωp(t, f) . t −i t 0 i=−∞ 2 The integral on the right hand side can then easily be estimated from above and from below by ¢ 2−i+1 −sq−1 isq q −i −sq−1 q sq (i−1)sq q −i+1 2 2 ωp(2 , f) ≤ t ωp(t, f) dt ≤ 2 2 ωp(2 , f). 2−i We now assume s and want to estimate in the Norm in (8.3). Note that for f ∈ Bp,q f n xed 0 6= y ∈ R , we can write ¢ ¢ f(x + y) = F −1Ff(x + y) = ei(x+y)·ηe−iξ·ηf(ξ) dηdξ n n R R = F −1eiy·Ff(x)

iξ·y and set ρy(ξ) = 1 − e to get

∞ X −1 −1 f(x) − f(x + y) = F ϕkFf(x) − F ϕkFf(x + y) k=0 ∞ X −1 = F ϕk · ρyFf(x). k=0

We must now consider the multipliers ϕk and ρy to show the estimate

−1 k −1 kF (ϕk · ρy ·Ff)kLp ≤ C min(1, |y|2 )kF (ϕk ·Ff)kLp . (8.4)

−1 Since ∞ , we have . As an exercise, we can show |F ρyFf(0)| = f(y) ≤ kfkL kρykM1 = 1 that , where 1 . From Theorem 8.3.6, we get that ry ∈ M1 ry(ξ) = y·ξ ρy(ξ) y k( , ·)ϕ(·)k < C, |y| M1

72 8 A Trace Theorem which implies

−k −k k k(y, ·)ϕ(2 ·)k n = ky · ϕ(2 ·)k < C|y|2 , M1(R ) M1(R) if we use the calculation ¢ ¢ |F −1y · ϕ(2−k·)Ff(0)| = yξϕ(2−kξ)e−iξηf(η) dηdξ R¢ R¢ x=2−kξ k = 2k y2kxϕ(x)e−i2 xηf(η) dxdη ¢R ¢R z=2kη = 2k yxϕ(x)e−izxf(2−kz) dxdz R R ¢ k ≤ 2 kfkL∞ F(y · ϕ(·))(z) dz R k ≤ C2 kfkL∞ . In conclusion, we have

−1 −1 kF ρyϕkFfkLp ≤ CkF ϕkFfkLp as well as

−1 −1 1 −1 −k −k+1 kF ρ ϕ Ffk p ≤ kF ρ (·) FF (y, ·)(ϕ(2 ·) − ϕ(2 ·))Ffk p y k L y (y, ·) L k −1 ≤ C|y|2 kF ϕkFfkLp , so that in conclusion, we get (8.4). The remaining step in the proof of the rst inclusion is to see why this estimate is helpful, in particular, why the case |y|2k < 1 is important. We get that

∞ is −i is X −1 2 ωp(2 , f) ≤ C2 sup kF ρyϕkFfkLp −i k=0 |y|<2 ∞ X (i−k)s −i+k sk −1 ≤ C 2 min(1, 2 )2 kF ϕkFfkLp . k=0

ks −k The right hand side is a convolution of the two sequences bk = 2 min(1, 2 ), and sk −1 ak = 2 kF ϕkFfkLp , k ∈ Z, where ak = 0 for k < 0. We can use that

∞ ∞ X X sk −k bk = 2 min(1, 2 ) < ∞ k=−∞ k=−∞ to get ∞ !1/q ∞ !1/q X is −i q X q s (2 ωp(2 , f)) ≤ C ak = CkfkBp,q . i=−∞ k=0

73 8 A Trace Theorem

It remains to prove the converse estimate, by using a similar technique. We set

−k ρjk(ξ) = ρ(2 ej )(ξ), where ej is the jth unit vector. The crucial part is to prove the estimate

n −1 X −1 kF ϕkFfkLp ≤ C kF ρjkFfkLp . (8.5) j=1

It then follows that

1/q  ∞ n  X ks X −1 q s p p kfkBp,q ≤ kfkL + C  (2 kF ρjkFfkL )  k=1 j=1 n ∞ !1/q X X ks q ≤ kfkLp + C (2 sup kf(·) − f(· + y)kLp ) −k j=1 k=1 |y|<2 ¢ ∞ 1/q −s q dt ≤ kfkLp + Cn (t ωp(t, f)) . 0 t

In order to prove (8.5), we use the following construction: For n ≥ 2 there exist functions n χj ∈ S(R ), such that

n X χj = 1 on supp ϕ1 = B2(0)\B1/2(0), j=1 1 supp χ ⊂ {ξ ∈ n||ξ | ≥ √ } j R j 3 n for 1 ≤ j ≤ n. Idea, why this is true: consider k ∈ S(R) satisfying supp k = {ξ ∈ R||ξ| ≥ √1 and taking positive values. Choose n−1 such that n−1 3 n } l ∈ S(R ) supp l = {ξ ∈ R ||ξ| ≤ j 3} taking positive values. Then with ξ˜ = (ξ1, . . . , ξj−1, ξj+1, . . . , ξn), the function

k(ξ )l(ξ˜ ) χ (ξ) = j j j Pn ˜ j=1 k(ξj)l(ξj) works. By Theorem 8.3.5, we see that

1 m : ξ 7→ χj(ξ)ϕ1(ξ) eiξj − 1 is a map in L2 and similarly, its derivatives, as we only need to consider the last multiplier

74 8 A Trace Theorem for √1 . It follows that , and so 0 < 3 n ≤ ξj ≤ 2 < 2π m ∈ Mp 1 ≤ p ≤ ∞

−1 −1 −k kF ϕkFfkLp ≤ CkF ϕ1(2 ·)FfkLp n X −1 −k −1 ≤ C kF m(2 )FF ρjkFfkLp j=1 n X −1 ≤ C kF ρjkFfkLp . j=1

Now that we have a new characterization of the Besov norm from Theorem 8.2, we are much closer to proving Step 2 in the Trace Theorem 8.1. We need the estimate (8.6) kT fkLp( n−1) ≤ Ckfk 1/p n R Bp,1 (R ) n 1/p ∞ for all f ∈ S(R ). From the Embedding Theorem 7.19, we know that Bpq (R) ,→ L (R) for every 1 ≤ q ≤ ∞. It follows that

0 0 |T f(x )| ≤ Ckf(·, x )k 1/p (8.7) Bp,1 (R) 0 n−1 for all x ∈ R . To deduce (8.6) from this, we use the following characterization given in [1, Theorem 7.47], which is not trivial and is shown by taking the s n norm from Bp,q(R ) real interpolation: for 0 < s < 1, 1 < p < ∞, 1 ≤ q < ∞, ¢ ¢ ∞ −s 1 q dt −s q dh k · k s ∼ k · k p + [t ω (·, t)] ∼ k · k p + [|h| k∆ fk p n ] . Bp,q L p L h L ( ) n t n R |h| 0 R n−1 We now take the p-th power of both sides of (8.7), integrate over R and take the 1/p-th root to get that by the Minkowski inequality,

kT fkLp( n−1) ¢ R ¢ ¢ ¢ 1/p ∞ p1/p 0 p 0 −1/p 1 0 dt ≤ C |f(x1, x )| dx1 dx + C t ωp(f(·, x ), t) n−1 n−1 0 t R R ¢ ¢ R p1/p −1/p 0 dh1 ≤ Ckfk p n + C |h | k∆ f(·, x )k p L (R ) 1 h1 L (R) n−1 |h1| ¢ R R ¢ ¢ M.I. 1/p −1/p 0 0 p 0 dh1 ≤ Ckfk p n + C |h | |f(x , x ) − f(x + h , x )| dx dx L (R ) 1 1 1 1 1 n−1 |h1| ¢R ¢ R ¢ R t=±h ∞ 1/p 1 −1/p 0 0 p 0 dt ≤ CkfkLp( n) + C t |f(x1, x ) − f(x1 + t, x )| dx1 dx R n−1 t ¢0 R R ∞ −1/p dt ≤ Ckfk p n + C t sup k∆ fk p n L (R ) y L (R ) 0 |y|≤t t

≤ Ckfk 1/p n , Bp,1 (R ) which proves the claim.

75 9 More Spaces

In this last chapter and in the last lecture, we want to briey introduce and consider a few function spaces which have not appeared before, but which are at the same time somehow related to our previous topics. There is no time for proofs, but they can be found in the references.

9.1 Quasi-norms

In contrast to a norm, a quasi-norm does not fulll the triangle inequality, but the following more general condition. There exists a constant C > 0, such that for every elements x, y of the quasi-normed space (X, k · k),

kx + yk ≤ C(kxk + kyk).

Typical examples of quasi-normed spaces are the Lp-spaces, when 0 < p < 1, which are dened as in the p ≥ 1-case.

We have met a quasi-norm in Section 6.2. For 1 ≤ p, q, p0, p1 ≤ ∞, and 0 < θ < p,q p,q 1 appropriately, the spaces (L , k · kLp,q ) are quasi-Banach spaces, whereas (L , k · p p are Banach spaces. We showed that they have equivalent quasi-norms. k(L 1 ,L 0 )θ,q ) Remark 9.1. For quasi-Banach spaces, the K-method still works as before, giving a quasi- Banach space as the interpolation space. One can choose 0 < q ≤ ∞ as interpolation parameter, [7, Remark 2, p. 27].

9.2 Semi-norms and Homogeneous Spaces

In contrast to a norm k · k, a semi-norm [·] does not satisfy

kxk = 0 ⇒ x = 0.

Typical examples of semi-normed function spaces are those which only care about the properties of (higher order) derivatives of a function. These spaces are called homogeneous.

76 9 More Spaces

n Example 9.2. For a domain Ω ⊂ R , m ∈ N, 1 ≤ q < ∞, we set

m,q 1 l q D (Ω) = {f ∈ Lloc(Ω) : D f ∈ L (Ω), |l| = m}. It follows that m,q m,q by Poincare Inequality. If is a local Lipschitz D (Ω) ,→ Wloc (Ω) Ω domain, then m,q m,q D (Ω) ,→ Wloc (Ω). We look at the semi-norm

 ¢ 1/q X l q [f]Dm,q =  |D f|  . |l|=m Ω

We have that [f]Dm,q = 0 if and only if f is a polynomial of degree at most m − 1, m,q f ∈ Pm−1, D modulo Pm−1 is a Banach space.

Example 9.3. In the Fourier analysis context, we can dene homogeneous Besov and n Bessel potential spaces, see [3, Section 6.3]. We choose ϕ ∈ S(R ), such that

supp ϕ = B2(0)\B 1 (0), 2 1 ϕ(ξ) > 0, < |ξ| < 2, 2 ∞ X ϕ(2−kξ) = 1, ξ 6= 0. k=−∞

−k We dene ϕ˜k(ξ) = ϕ(2 ξ) for all k ∈ Z.

0 n 1. Then we consider for all f ∈ S (R ),

∞ !1/q X skq −1 q [f] ˙ s = 2 kF ϕ˜kFfk p . Bp,q L k=−∞

0 The space of all f ∈ S , for which [f] ˙ s is nite is called the homogeneous Besov Bp,q ˙ s space B . We have that [·] ˙ s is a semi-norm and that [f] ˙ s = 0 if and only if p,q Bp,q Bp,q supp Ff = {0}, which happens if and only if f is a polynomial. We have that ˙ s p s if and that Bp,q = L ∩ Bp,q s > 0

¢ 1/q N ∞ N ! X N−s m ∂ f q dt [f] ˙ s ∼ (t ωp (t, )) Bp,q ∂xN t j=1 0 j as in Theorem 8.2.

77 9 More Spaces

2. Similarly, the homogeneous Bessel potential spaces H˙ s,p are dened. We set f ∈ H˙ s,p for f ∈ S0, if

∞ X −1 s [f]H˙ s,p = k F (|ξ| ϕ˜k)FfkLp < ∞, k=−∞ meaning that the sum on the right hand side converges to an Lp-function. Then ˙ s,p is also a semi-normed space and if and only if is a (H , [·]H˙ s,p ) [f]H˙ s,p = 0 f polynomial. It follows that H˙ s,p = Lp ∩ Hs,p if s > 0 and that n X ∂N f [f] ˙ N,p ∼ k k p , 1 < p < ∞, H ∂xN L j=1 j

if fˆ vanishes in a neighbourhood of the origin.

Interpolation and embedding results from the usual (inhomogeneous) spaces mostly carry over to homogeneous spaces, e.g. Theorem 7.13 and Theorem 7.19.

9.3 Orlicz Spaces LA

Orlicz spaces can be considered as a type of generalization of Lp-spaces. We only give a very brief introduction here following [1, Chapter 8] and also refer to [2, Section 4.8]. ¡ ¡ p p Idea: f ∈ L if |f(x)| dx < ∞, f ∈ LA if A(|f(x)| dx < ∞. n Problem: If mp = 1, p > 1, then for a good domain Ω ⊂ R ,

m,p q m,p ∞ W (Ω) ,→ L (Ω), p ≤ q < ∞, but W (Ω) * L (Ω), so there is no optimal target Lp-space for the embedding.

Denition 9.4. Let a : [0, ∞) → R be a function such that

1. a(0) = 0, a(t) > 0 if t > 0 and limt→∞ a(t) = ∞, 2. a is nondecreasing,

3. a is right continuous.

Then the function A : [0, ∞) → R, given by ¢ t A(t) = a(τ) dτ 0 is called an N−function.

78 9 More Spaces

It follows that is continuous, strictly increasing and convex, that A(t) and A limt→0 t = 0 A(t) and that A(·) is strictly increasing. limt→∞ t = ∞ · Examples: A(t) = tp and A(t) = e(tp) − 1 if 1 < p < ∞, A(t) = et − t − 1 and A(t) = (1 + t) log(1 + t) − t.

n Denition 9.5. Let Ω ⊂ R be a domain and A an N-function. The Orlicz class KA(Ω) consists of all equivalence classes [f] modulo equality a.e. on Ω of measurable functions f, such that ¢ A(|f(x)|) dx < ∞. Ω

The set KA(Ω) is convex, but it may not be a vector space.

Denition 9.6. (and Theorem) The Orlicz space LA(Ω) is the linear hull of KA(Ω). The function ¢ |f(x)| kfkLA(Ω) = inf k > 0 : A dx ≤ 1 Ω k is a norm on . It is called the Luxemburg norm. is a Banach space. LA(Ω) (LA, k · kLA )

Next, we look at some basic results.

Denition 9.7. Given a as in Denition 9.4 above, we consider

a˜(s) = sup t a(t)≤s ¡ and ˜ s . We have that also satises 1. - 3. in Denition 9.4 and A(s) = 0 a˜(σ) dσ a˜

a(t) = sup s, a˜(s)≤t so that A˜ is also an N-function. A and A˜ are said to be complementary. For A and B given N-functions, we say that B dominates A globally if there exists a constant C > 0 such that

A(t) ≤ B(Ct), for all t ≥ 0. (9.1)

We say that B dominates A near innity if there exists a constant t0 > 0 such that (9.1) is satised for all t ≥ t0.

Examples: For 1 1 , the functions and ˜ given by tp and 1 < p < ∞, p + p0 = 1 A A A(t) = p 0 ˜ sp are complementary -functions, as well as the functions and ˜ given by A(s) = p0 N B B B(t) = et − t − 1 and B˜(s) = (1 + s) log(1 + s) − s.

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Denition 9.8. Hölder Inequality: If A and A˜ are complementary N-functions, then ¢

u(x)v(x) dx ≤ 2kuk kvk . LA LA˜ Ω Embedding: We have that

LB(Ω) ,→ LA(Ω) if either B dominates A globally or B dominates A near innity and µ(Ω) < ∞.

n Theorem 9.9. Let Ω ⊂ R be a suitable bounded domain. Let mp = n and p > 1 and set p/(p−1) n/(n−m) A(t) = e(t ) − 1 = e(t ) − 1. Then the embedding m,p W (Ω) ,→ LA(Ω) holds true.

More: Orlicz-Sobolev spaces, trace theorems, interpolation. p(x) -spaces: For a function , and consider L¡ (Ω) p :Ω → R 1 ≤ p(x) < ∞ λ > 0 f(x) p(x) . Luxemburg norm p(x) . Ω | λ | dx L (Ω)

9.4 Hardy Spaces

Hardy spaces are complex Banach spaces of holomorphic functions, endowed with an Lp-type norm, usually dened on the unit disc or on the half plane. We consider the latter case, following the rst part of [2, Section 5.6].

Denition 9.10. Let C+ = {x + iy : x ∈ R, y > 0} be the upper complex half plane. p The Hardy space H (C+) is the Banach space of holomorphic functions F on C+ with nite norm

 ¡ 1/p ∞ p  supy>0 |F (x + iy)| ) dx , 1 ≤ p < ∞, kF k p = −∞ H (C+)  sup |F (z)|, p = ∞. z∈C+

1 ∞ We know: There is a unique solution u for the Dirichlet problem on C+ if f ∈ L +L (R):

∆u = 0, in C+, lim u(t, y) = f(x), on ∂C+. t+iy→x

p p It can be shown: if F ∈ H (C+), then its boundary value f satises f ∈ L (R; C). 1 Moreover, if F1 and F2 are functions in H (C+) which have the same real part fR at the boundary, then by Cauchy-Riemann they can only dier by an imaginary constant, which

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1 1 must be zero for F1 − F2 ∈ H . This implies: H as a real Banach space is isometrically 1 1 isomorphic to the space R(H ) of functions fR ∈ L which are real parts of boundary 1 functions of functions F ∈ H , where the norm is kfRkR(H1) = kF kH1 . Philosophy: R(H1) ⊂ L1, but nicer than L1.

Theorem 9.11. If 0 < θ < 1, 1 ≤ q ≤ ∞, and θ = 1 − 1/p, then

1 ∞ p,q (R(H ),L )θ,q = L .

9.5 The Space BMO

BMO means: bounded mean oscillation. On a domain n, we consider the mean value of a function 1 on a Ω ⊂ R f A f ∈ Lloc(Ω) (Lebesgue) measurable set A with 0 < µ(A) < ∞, ¢ 1 f A = f dµ. µ(A) A Denition 9.12. The space is the space of functions 1 which satisfy BMO(Ω) f ∈ Lloc(Ω) ¢ 1 [f]BMO = sup |f(x) − f A| dx < ∞, 0<µ(A)<∞ µ(A) A where [·]BMO is a semi-norm and [f]BMO = 0 implies that f is constant.

The following holds true:

• if µ(Ω) < ∞, then L∞(Ω) ,→ BMO(Ω) ,→ L1(Ω).

• if Ω is a bounded Lipschitz domain, then BMO(Ω) is contained in every Lp(Ω), 1 ≤ p < ∞.

• if Ω is a bounded Lipschitz domain, then for 0 < θ < 1, 1 ≤ q ≤ ∞, we have

1 1 ,q (L (Ω),BMO(Ω))θ,q = L 1−θ (Ω).

1 0 • BMO(R) modulo constants is isomomorphic to R(H ) , [2, Theorem 6.17].

81 Bibliography

[1] R. A. Adams and J. J. F. Fournier. Sobolev Spaces, volume 140 of Pure and Applied Mathematics (Amsterdam). Elsevier/Academic Press, Amsterdam, 2003.

[2] Colin Bennett and Robert Sharpley. Interpolation of operators, volume 129 of Pure and Applied Mathematics. Academic Press Inc., Boston, MA, 1988.

[3] J. Bergh and J. Löfström. Interpolation Spaces. Springer, 1976.

[4] Alessandra Lunardi. Analytic semigroups and optimal regularity in parabolic problems. Progress in Nonlinear Dierential Equations and their Applications, 16. Birkhäuser Verlag, Basel, 1995.

[5] Alessandra Lunardi. Interpolation theory. Appunti. Scuola Normale Superiore di Pisa (Nuova Serie). [Lecture Notes. Scuola Normale Superiore di Pisa (New Series)]. Edizioni della Normale, Pisa, second edition, 2009.

[6] Luc Tartar. An introduction to Sobolev spaces and interpolation spaces, volume 3 of Lecture Notes of the Unione Matematica Italiana. Springer, Berlin, 2007.

[7] H. Triebel. Interpolation Theory, Function Spaces, Dierential Operators. Johann Ambrosius Barth, Heidelberg, 1995.

Credits

I would like to thank Moritz Egert, David Meert, Hannes Meinlschmidt and Jonas Sauer for their critical proof-reading and kind suggestions. Moreover, I would like to thank Prof. Dr. Dorothee Haroske for allowing me to use her lecture notes on Reelle Interpolationstheorie.