Short Lecture Notes: Interpolation Theory and Function Spaces
Helmut Abels July 27, 2011
1 Introduction
In the following let K = R or K = C. De nition 1.1 Let be Banach spaces over . Then the pair X0,X1 K (X0,X1) is called admissible, compatible or an interpolation couple if there is a Hausdor topological vector space Z such that X0,X1 ,→ Z with continuous embeddings.
Lemma 1.2 Let (X0,X1) be an admissible pair of Banach spaces. Then X0 ∩ X1 normed by
kxkX0∩X1 = max(kxkX0 , kxkX1 ), x ∈ X0 ∩ X1, and X0 + X1 normed by
kxkX0+X1 = inf kx0kX0 + kx1kX1 , x ∈ X0 + X1, x=x0+x1,x0∈X0,x1∈X1 are Banach spaces.
Notation: In the following we write for simplicity T ∈ L(Xj,Yj), j = 0, 1, for the condition that T : X0 + X1 → Y0 + Y1 is a linear operator with for and . T |Xj ∈ L(Xj,Yj) j = 0 j = 1
De nition 1.3 Let (X0,X1) and (Y0,Y1) be admissible pairs of Banach spaces and let X,Y be Banach spaces.
1. X is called intermediate space with respect to (X0,X1) if X0 ∩ X1 ,→ X,→ X0 + X1 with continuous embeddings.
1 2. X and Y are called interpolation spaces with respect to (X0,X1) and (Y0,Y1) if X and Y are intermediate spaces with respect to (X0,X1) and (Y0,Y1), respectively, and if
T ∈ L(Xj,Yj), j = 0, 1, ⇒ T |X ∈ L(X,Y )
for all T : X0 + X1 → Y0 + Y1 linear.
3. X is called interpolation space with respect to (X0,X1) if the previous conditions hold with X = Y , (X0,X1) = (Y0,Y1).
4. Interpolation spaces X and Y with respect to (X0,X1) and (Y0,Y1) are called of exponent θ ∈ [0, 1], if there is some C > 0 such that
kT k ≤ CkT k1−θ kT kθ (1.1) L(X,Y ) L(X0,Y0) L(X1,Y1)
for all T ∈ L(Xj,Yj), j = 0, 1. If (1.1) holds with C = 1, then X and Y are called exact of exponent θ. Analogously, X is an (exact) interpolation space of exponent θ if the previous conditions hold for X = Y , (X0,X1) = (Y0,Y1). Examples 1.4 1. Let , and let 0 < α < 1 k ∈ N0
0 n n C (R ) = {f : R → K continuous and bounded}
kfkC0(Rn) = sup |f(x)| x∈Rn k n n -times cont. di. α 0 n C (R ) = f : R → K k : ∂x f ∈ C (R )∀|α| ≤ k α kfkCk(Rn) = sup sup |∂x f(x)| |α|≤k x∈Rn α n n C (R ) = f : R → K : kfkCα(Rn) < ∞ |f(x) − f(y)| α n kfkC (R ) = sup |f(x)| + sup α x∈Rn x,y∈Rn,x6=y |x − y|
Then Cα(Rn) is an intermediate space of C0(Rn) and C1(Rn). More precisely, there is some C > 0 such that
1−α α 1 n kfk α n ≤ Ckfk kfk 1 n for all f ∈ C ( ). C (R ) C0(Rn) C (R ) R
2 2. For 1 ≤ p ≤ ∞ let Lp(U, µ) denote the usual Lebesgue space on a measure 1 1−θ θ space (U, µ). Then for any 1 ≤ p0, p1, p ≤ ∞ such that = + for p p0 p1 some θ ∈ [0, 1] the space Lp(U, µ) is an intermediate space of Lp0 (U, µ) and Lp1 (U, µ). This follows from the generalized Hölder inequality
1−θ θ p0 p1 p for all kfkL (U,µ) ≤ kfkLp0 (U,µ)kfkLp1 (U,µ) f ∈ L (U, µ) ∩ L (U, µ), which itself follows easily from the standard Hölder inequality.
Remark 1.5 The last two inequalities are special cases of (1.1) if one chooses . The following theorem shows that p in the latter example is Yj = K L (U, µ) an exact interpolation space with respect to (Lp0 (U, µ),Lp1 (U, µ)) of expo- nent θ.
THEOREM 1.6 (Riesz-Thorin)
Let 1 ≤ p0, p1, q0, q1 ≤ ∞, θ ∈ [0, 1], let (U, µ) and (V, ν) be σ- nite measure 1 spaces and let K = C. Then
pj qj p q T ∈ L(L (U, µ),L (V, ν)), j = 0, 1 ⇒ T |Lp(U,µ) ∈ L(L (U, µ),L (V, ν)), where 1 1 − θ θ 1 1 − θ θ = + , = + . p p0 p1 q q0 q1 Moreover, for every T ∈ L(Lpj (U, µ),Lqj (V, ν)), j = 0, 1,
1−θ θ p q kT kL(L ,L ) ≤ kT kL(Lp0 ,Lq0 )kT kL(Lp1 ,Lq1 ). The theorem will follow from the results of Section 2 below.
Remark 1.7 The proof of the Riesz-Thorin interpolation theorem provided the fundamental ideas for the so-called complex interpolation method. Gener- ally an interpolation method is an functor F from the cathegory of admissible couples of Banach spaces to the cathegory of Banach spaces such that any pair of admissible couples X = (X0,X1), Y = (Y0,Y1) is mapped to an interpolation space F(X,Y ) with respect to X,Y .2 1Meaning that all functions will be complex valued in the following. 2Readers not familiar with the basic notion of cathegory do not have to understand this precisely since no cathegory theory will be needed in the following.
3 In some situations the following theorem due to Marcinkiewicz can be applied, when an application of the Riesz-Thorin theorem fails.
THEOREM 1.8 (Marcinkiewicz)
Let 1 ≤ p0, p1, q0, q1 ≤ ∞ with q0 6= q1, let θ ∈ (0, 1), and let (U, µ) and (V, ν) be measure spaces. Moreover, let
1 1 − θ θ 1 1 − θ θ = + , = + p p0 p1 q q0 q1 and assume that p ≤ q. Then
pj qj p q T ∈ L(L (U, µ),L∗ (V, ν)), j = 0, 1 ⇒ T |Lp(U,µ) ∈ L(L (U, µ),L (V, ν)),
pj qj Moreover, there is some Cθ such that for every T ∈ L(L (U, µ),L∗ (V, ν)), j = 0, 1, 1−θ θ p q q kT kL(L ,L ) ≤ CθkT k p q0 kT k p1 1 . L(L 0 ,L∗ ) L(L ,L∗ ) Here q is the weak q-space, de ned by L∗(V, ν) L C Lq(V, ν) = f : V → measurable : m(t; f) ≤ for all t > 0 and some C > 0 ∗ K tq if 1 ≤ q < ∞, where
m(t; f) = ν({x ∈ V : |f(x)| > t}), t > 0, denotes the distribution function of . Moreover, if , then ∞ f q = ∞ L∗ (V, ν) = ∞ . We note that q is quasi-normed space with quasi-norm by L (V, ν) L∗(V, ν)
1 q q kfkL∗ = sup t m(t; f) . t>0
Here a quasi-norm k.k on a vector space X satis es the same conditions as a norm except that the triangle inequality is replaced by
kx + yk ≤ C(kxk + kyk) for all x, y ∈ X for some C ≥ 1. The bounded operators and the operator norm is de ned in the same way for quasi-normed spaces as for normed spaces. Remark: One also denotes q,∞ q . L (V, ν) ≡ L∗(V, ν)
4 A proof of the Marcinkiewicz interpolation theorem in the special case , n n can be found in the appendix. The p0 = q0 = 1, p1 = q1 = r (U, µ) = (R , λ ) proof provided the basic ideas for the so-called real interpolation method. The real and the complex interpolation methods are the most important methods to construct interpolation spaces and will be one of the central topic of this lecture series. We will start with the complex method since it is easier to understand in the beginning, although the real interpolation method is in some sense more fexible and universal.
2 Complex Interpolation Method
2.1 Holomorphic Functions on the Strip
For the following let X be a complex Banach space and let S = {z ∈ C : 0 ≤ , . A mapping is called Re z ≤ 1} S0 = {z ∈ C : 0 < Re z < 1} f : S0 → X 0 holomorphic if z 7→ hf(z), x i, z ∈ S0, is holomorphic in the usual sense for all x0 ∈ X0.
THEOREM 2.1 (Maximum Principle/Phragmen-Lindelöf Theorem) Let f : S → X be continuous, bounded, and holomorphic in the interior of S. Then sup kf(z)kX ≤ max sup kf(it)kX , sup kf(1 + it)kX . (2.1) z∈S t∈R t∈R Proof: First let and assume that . Consider the X = C f(z) →| Im z|→∞ 0 mapping h: S → C with eiπz − i h(z) = , z ∈ S. (2.2) eiπz + i
Then h is a bijective mapping from S onto U = {z ∈ C : |z| ≤ 1} \ {±1}, that is holomorphic in S0 and maps ∂S onto {|z| = 1} \ {±1}. Therefore g(z) := f(h−1(z)) is bounded and continuous on U and analytic in the interior of U. Moreover, because of lim| Im z|→∞ f(z) = 0, limz→±1 g(z) = 0 and we can extend g to a continuous function on {|z| ≤ 1}. Hence |g(z)| ≤ max |g(w)| = max sup |f(it)|, sup |f(1 + it)| , |w|=1 t∈R t∈R
5 which implies the statement in this case. Next, if X = C and f is a general 2 function as in the assumptions, then we consider δ(z−z0) , fδ,z0 (z) = e f(z) 2 2 2 δ(z−z0) δ(x −y ) δ > 0, z0 ∈ S0. Since |e | ≤ e with z − z0 = x + iy, −1 ≤ x ≤ , we have . Therefore 1, y ∈ R fδ,z0 (z) →| Im z|→∞ 0
|f(z0)| = |fδ,z0 (z0)| ≤ max sup |fδ,z0 (it)|, sup |fδ,z0 (1 + it)| t∈R t∈R ≤ eδ max sup |f(it)|, sup |f(1 + it)| . t∈R t∈R Passing to the limit , we obtain (2.1) in the case since δ → 0 X = C z0 ∈ S was arbitrary.
Finally, if f : S → X is as in the assumption and z0 ∈ S0, we can nd 0 0 0 0 some x ∈ X with kx kX0 = 1 such that kf(z0)kX = |hf(z0), x i|. Now one 0 applies (2.1) to gx0 (z) := hf(z), x i and obtains kf(z0)kX = |gx0 (z0)| ≤ max sup |gx0 (it)|, sup |gx0 (1 + it)| t∈R t∈R ≤ max sup kf(it)kX , sup kf(1 + it)kX . t∈R t∈R for arbitrary z0 ∈ S0. As a corollary we obtain the following three line theorem, which is the basis for the proof of the Riesz-Thorin interpolation theorem and the complex interpolation method.
THEOREM 2.2 (Three Lines Theorem) Let f : S → X be continuous, bounded, and holomorphic in the interior of S. Then
1−θ θ sup kf(θ + it)kX ≤ sup kf(it)kX sup kf(1 + it)kX . (2.3) t∈R t∈R t∈R for every θ ∈ [0, 1]. Proof: Let , . Let M0 = supt∈R kf(it)kX M1 = supt∈R kf(1 + it)kX λ ∈ R and de ne λz Fλ(z) = e f(z). Then λ kFλ(z)kX ≤ max(M0, e M1)
6 by Theorem 2.1. Hence
−λθ λ(1−θ) kf(θ + it)kX ≤ max e M0, e M1 for all t ∈ . Choosing λ = ln M0 nishes the proof. R M1
2.2 Complex Interpolation Method and Lp-Spaces
Let (X0,X1) be an admissible pair of complex Banach spaces. In particular we will assume always K = C in the following!
De nition 2.3 Let F (X0,X1) be the set of all f : S → X0 + X1 that are continuous, bounded, and holomorphic in S0 and such that
t 7→ f(it) ∈ C(R; X0), t 7→ f(1 + it) ∈ C(R; X1).
We equip F (X0,X1) with the norm
kfkF (X0,X1) = max sup kf(it)kX0 , sup kf(1 + it)kX1 . t∈R t∈R
Finally, let F0(X0,X1) be the set of all f ∈ F (X0,X1) such that for lim kf(j + it)kXj = 0 j = 0, 1. |t|→∞
Remark 2.4 Because of Theorem 2.1, we have F (X0,X1) ,→ C(S; X0 +X1). Using this, it is not di cult to show that F (X0,X1) is a Banach space. Moreover, obviously F0(X0,X1) is a closed subspace of F (X0,X1). We will need the following technical lemma.
δz2+λz Lemma 2.5 The linear hull of the functions e a, a ∈ X0 ∩ X1, δ > 0, is dense in . λ ∈ R F0(X0,X1) We refer to [1, Lemma 4.2.3] for a proof.
De nition 2.6 Let θ ∈ (0, 1). Then the complex interpolation space (X0,X1)[θ] is de ned as
(X0,X1)[θ] = {f(θ): f ∈ F (X0,X1)}
kxk(X0,X1)[θ] = inf kfkF (X0,X1). f∈F (X0,X1):f(θ)=x
7 Remarks 2.7 1. Since Nθ = {f ∈ F (X0,X1): f(θ) = 0} is closed in ∼ F (X0,X1), we have that (X0,X1)[θ] = F (X0,X1)/Nθ is a Banach space. 2. For every θ ∈ (0, 1) we have
(X0,X1)[θ] = (X1,X0)[1−θ].
3. One can replace F (X0,X1) by F0(X0,X1) in the de nition of (X0,X1)[θ]. In order to see that this gives the same space and norm, one uses that δ(z−θ)2 fδ = e f(z) ∈ F0(X0,X1) if f ∈ F (X0,X1) and δ > 0. Moreover, and δθ2 δ(1−θ)2 . fδ(θ) = f(θ) kfδkF (X0,X1) ≤ max(e , e )kfkF (X0,X1)
4. We have X0 ∩ X1 ,→ (X0,X1)[θ] since we can choose f(z) = x for all z ∈ S and x ∈ X0 ∩ X1. Moreover, (X0,X1)[θ] ,→ X0 + X1 since F (X0,X1) ,→ C(S; X0 + X1). Summing up, we have (X0,X1)[θ] is an intermediate space with respect to (X0,X1), i.e.,
X0 ∩ X1 ,→ (X0,X1)[θ] ,→ X0 + X1
5. For every θ ∈ (0, 1) we have (X0,X0)[θ] = X0 with same norm. Because of Lemma 2.5 and statement 3 above, we obtain
Corollary 2.8 Let θ ∈ (0, 1). Then X0 ∩ X1 is dense in (X0,X1)[θ].
THEOREM 2.9 Let 1 ≤ p0, p1 ≤ ∞, θ ∈ (0, 1), and let (U, µ) be a measure space. Then p0 p1 p (L (U, µ),L (U, µ))[θ] = L (U, µ) with the same norm, where
1 1 − θ θ = + . p p0 p1 Proof: See [2, Example 2.1.11].
8 2.3 Properties of Complex Interpolation Spaces
THEOREM 2.10 Let θ ∈ (0, 1) and let (X0,X1) and (Y0,Y1) be two ad- missible pairs of Banach spaces. Then (X0,X1)[θ] and (Y0,Y1)[θ] are exact interpolation spaces of exponent θ with respect to (X0,X1) and (Y0,Y1). In particular, kT k ≤ kT k1−θ kT kθ L(Xθ,Yθ) L(X0,Y0) L(X1,Y1) for every T ∈ L(Xj,Yj), j = 0, 1, where Xθ = (X0,X1)[θ],Yθ = (Y0,Y1)[θ] Proof: See [2, Theorem 2.1.6]
Corollary 2.11 For every θ ∈ (0, 1) we have kxk ≤ kxk1−θkxkθ for all x ∈ X ∩ X . (X0,X1)[θ] X0 X1 0 1 Moreover, we note that Theorem 2.10 and Theorem 2.9 imply the Riesz- Thorin Interpolation Theorem 1.6.
2.4 Extensions and Generalizations The proof of the Theorem 2.9 can be easily modi ed to the case of weighted- Lp-spaces with respect to a xed reference measure µ. More precisely:
THEOREM 2.12 Let 1 ≤ p0, p1 < ∞, θ ∈ (0, 1), and let (U, µ) be a measure space. Moreover, let ωj : U → (0, ∞), j = 0, 1, be measurable and set p p p (1−θ) p θ ω(x) = ω0(x) 0 ω1(x) 1 . Then p0 p1 p (L (U, ω0 dµ),L (U, ω1 dµ))[θ] = L (U, ω dµ). with the same norm, where 1 1 − θ θ = + . p p0 p1 Here the measure ω dµ is de ned by Z ω dµ(A) = ω(x) dµ(x) A for every measurable A ⊆ U.
9 A useful generalization of Theorem 2.10 is:
THEOREM 2.13 Let (X0,X1), (Y0,Y1) be admissible couples of Banach spaces and let Tz ∈ L(X0 ∩ X1,Y0 + Y1), z ∈ S, be such that z 7→ Tzx is continuous on S and holomorphic in S0 for every x ∈ X0 ∩ X1. Moreover, assume that and t 7→ Tit ∈ C(R; L(X0,Y0)) t 7→ T1+it ∈ C(R; L(X1,Y1)) (with bounded operator norms). Then for every θ ∈ (0, 1)
1−θ θ for all kTθxk(Y0,Y1)[θ] ≤ M0 M1 kxk(X0,X1)[θ] x ∈ X0 ∩ X1, where
M0 = sup kTitkL(X0,Y0),M1 = sup kT1+itkL(X1,Y1). t∈R t∈R
In particular Tθ extends to a bounded operator from (X0,X1)[θ] to (Y0,Y1)[θ] with operator norm bounded by 1−θ θ. M0 M1 Proof: See [2, Theorem 2.1.7].
10 3 Real Interpolation Method
3.1 K-Method
In the following let (X0,X1) be an admissible pair of Banach spaces.
De nition 3.1 For t > 0, x ∈ X0 + X1 let
K(t, x) ≡ K(t, x; X0,X1) = inf kx0kX0 + tkx1kX1 . x=x0+x1,x0∈X0,x1∈X1
Lemma 3.2 For all x ∈ X0 + X1 the mapping t 7→ K(t, x) is non-negative, increasing and concave. Moreover,
t K(t, x) ≤ max 1, K(s, x) for all t, s > 0. (3.1) s
De nition 3.3 For θ ∈ (0, 1), 1 ≤ p ≤ ∞ we de ne the real interpolation space (X0,X1)θ,p as
(X0,X1)θ,p := {x ∈ X0 + X1 :Φθ,p(K(·, x)) < ∞} where
−θ Φθ,p(K(·, x)) = t K(t, x) p dt L ((0,∞), t ) ( p 1 R ∞ t−θK(t, x) dt p if 1 ≤ p < ∞, = 0 t −θ if supt>0 t K(t, x) p = ∞.
(X0,X1)θ,p is normed by kxkθ,p := Φθ,p(K(·, x)). Moreover, for θ ∈ (0, 1)
n −θ −θ o (X0,X1)θ := x ∈ (X0,X1)θ,∞ : lim t K(t, x) = lim t K(t, x) = 0 t→0 t→∞ denotes the continuous interpolation space of (X0,X1).
Remarks 3.4 1. It is easy to check that k·kθ,p is indeed a norm and that (X0,X1)θ,p is a linear subspace of X0 + X1. Moreover, (X0,X1)θ is a closed subspace of (X0,X1)θ,∞.
11 2. Important properties of the integral with respect to dt are: t Z ∞ dt Z ∞ dt Z ∞ dt 1 Z ∞ dt f(at) = f(t) , f(tα) = f(t) 0 t 0 t 0 t |α| 0 t for any a, α 6= 0 and integrable function f.3 The main result of this section is:
THEOREM 3.5 Let 1 ≤ p ≤ ∞, θ ∈ (0, 1) and let (X0,X1), (Y0,Y1) be admissible pairs of Banach spaces. Then (X0,X1)θ,p and (Y0,Y1)θ,p are interpolation spaces with respect to (X0,X1) and (Y0,Y1) that are exact of exponent θ. Moreover, (X0,X1)θ and (Y0,Y1)θ are interpolation spaces with respect to (X0,X1) and (Y0,Y1) that are exact of exponent θ The theorem follows from the following lemmas:
Lemma 3.6 Let 1 ≤ p1 ≤ p2 ≤ ∞, θ ∈ (0, 1). Then (3.2) X0 ∩ X1 ,→ (X0,X1)θ,p1 ,→ (X0,X1)θ,p2 ,→ X0 + X1 with continuous embeddings. Moreover, (X0,X1)θ,p ,→ (X0,X1)θ for any 1 ≤ p < ∞. In particular, (X0,X1)θ,p and (X0,X1)θ are intermediate spaces with respect to (X0,X1). Proof: See [2, Proposition 1.1.3].
Lemma 3.7 Let the assumptions of Theorem 3.5 be valid and let X = (X0,X1)θ,p, Y = (Y0,Y1)θ,p. Then
T ∈ L(Xj,Yj), j = 0, 1, ⇒ T |X ∈ L(X,Y ) and kT k ≤ kT k1−θ kT kθ L(X,Y ) L(X0,Y0) L(X1,Y1) for all T ∈ L(Xj,Yj), j = 0, 1. The same is true for X = (X0,X1)θ and Y = (Y0,Y1)θ.
3The rst property re ects the fact that f is a Haar measure on the topological group ((0, ∞), ·). I.e. it is a regular Borel measure that is invariant with respect to group action f 7→ f(a · .).
12 Proof: See [2, Theorem 1.1.6].
Note that the constants/the operator norms of the embeddings in (3.2) depend on (θ, p).
Lemma 3.8 For any θ ∈ (0, 1), 1 ≤ p ≤ ∞ the normed spaces (X0,X1)θ,p and (X0,X1)θ are complete. Proof: See [2, Proposition 1.1.3].
Some more properties are as follows:
−1 1. Since K(t, x; X0,X1) = tK(t , x; X1,X0), we have
(X0,X1)θ,p = (X1,X0)1−θ,p (X0,X1)θ = (X1,X0)1−θ
for all θ ∈ (0, 1), 1 ≤ p ≤ ∞. 2. If continuously, then and therefore X1 ,→ X0 K(t, x) ≤ kxkX0
−θ cθ,pkxkθ,p ≤ kxkX0 + kt K(t, x; X0; X1)k p dt ≤ Cθ,pkxkθ,p L ((0,1); t )
for some constants cθ,p,Cθ,p > 0 depending on θ ∈ (0, 1), 1 ≤ p ≤ ∞.
3. For all θ ∈ (0, 1), 1 ≤ p ≤ ∞ we have (X0,X0)θ,p = X0 with equivalent norms. More precisely,
cθ,pkxkθ,p ≤ kxkX0 ≤ Cθ,pkxkθ,p
for some constants Cθ,p, cθ,p > 0 depending on θ and p!!!
Proposition 3.9 Let X1 ,→ X0 continuously. Then for every 0 < θ1 < θ2 < 1 we have
(X0,X1)θ2,∞ ,→ (X0,X1)θ1,1. Recall that, if 1 ≤ p ≤ ∞ and Ω ⊆ Rn is a domain, then
1 p p Wp (Ω) = {f ∈ L (Ω) : ∇f ∈ L (Ω)} ,
1 p p kfkWp (Ω) = kfkL (Ω) + k∇fkL (Ω).
13 Here denotes the weak or distributional derivatives of ∇f = (∂x1 f, . . . , ∂xn f) p f. More precisely, gj ∈ L (Ω) for 1 ≤ j ≤ n is called weak derivative of f in direction xj, if Z Z
gj(x)ϕ(x) dx = − f(x)∂xj ϕ(x) dx Ω Ω for all smooth and compactly supported , i.e., for all ∞ . If ϕ:Ω → K ϕ ∈ C0 (Ω) the weak derivative exists, it is unique and it will be for simplicity be denoted by . Note that, if ∞ n and 1 n , then 1 n ∂xj f ψ ∈ C0 (R ) f ∈ Wp (R ) ψ ∗ f ∈ C (R ) and Z Z
∂xj ψ ∗ f(x) = − ∂yj ψ(x − y)f(y) dy = ψ(x − y)∂yj f(y) dy Rn Rn because of the de nition of the weak derivative. Choosing e.g. ψ(x) = −n x , , where ∞ n with R , one ϕε(x) = ε ϕ( ) ε > 0 ϕ ∈ C ( ) n ϕ(x) dx = 1 ε 0 R R obtains that in 1 n . Hence ∞ n 1 n is dense in ϕε ∗ f →ε→0 f Wp (R ) C (R ) ∩ Wp (R ) 1 n . Wp (R ) Moreover, if U ⊆ Rn is an open set, then Cθ(U), θ ∈ (0, 1), denotes the set of all bounded and Hölder continuous functions f : U → K of degree θ such that |f(x) − f(y)| kfkCθ(U) = kfk∞ + sup θ x6=y,x,y∈U |x − y| Example 3.10 For every 0 < θ < 1 and 1 ≤ p < ∞ we have n 1 n θ n (3.3) (C(R ),C (R ))θ,∞ = C (R ), n 1 n θ n (3.4) (C(R ),C (R ))θ = h (R ), p n 1 n θ n (3.5) (L (R ),Wp (R ))θ,p = Wp (R ), with equivalent norms, where hθ(Rn) is the little Hölder space de ned by ( ) |f(x) − f(y)| hθ( n) = f ∈ Cθ( n) : lim sup = 0 R R R→0+ θ x,y∈Rn,0<|x−y|≤R |x − y| equipped with the norm of θ n and θ n is the Sobolev-Slobodetskii C (R ) Wp (R ) space de ned by
θ n n p n o W ( ) = f ∈ L ( ):[f] θ n < ∞ , p R R Wp (R ) 1 Z |f(x) − f(y)|p p [f] θ n = d(x, y) . Wp (R ) θp+n Rn×Rn |x − y|
14 Example 3.11 Let Ω ⊂ Rn be a domain such that there is an extension oper- ator θ θ n for all and on . Then E ∈ L(C (Ω); C (R )) θ ∈ [0, 1] Ef|Ω = f Ω
1 θ (C(Ω),C (Ω))θ,∞ = C (Ω) for all 0 < θ < 1.
Remark 3.12 If n n−1 , then de ned by Ω = R+ = R × (0, ∞) E ( f(x) if x ≥ 0 Ef(x) = n 0 0 3f(x , −xn) − 2f(x , −2xn) if xn < 0
0 where x = (x1, . . . , xn−1) satis es the conditions of (3.11). More generally, every bounded domain Ω with C1-boundary satis es the conditions of (3.11). This can be shown by locally mapping the boundary of Ω to pieces of a half space n and applying the half-space extension operator together with a R+ suitable partition of unity.
Remark 3.13 On the basis on (3.5), we will be able to prove that
1 n p n−1 1 n−1 1 Trxn=0 W ( ) = (L ( ),W ( )) , p R+ R p R 1− p ,p where , n n−1 , and 1 < p < ∞ R+ = R × (0, ∞) 1 n p n−1 0 0 for some 1 n Trxn=0 Wp (R+) = a ∈ L (R ): a(x ) = f(x , 0) f ∈ Wp (R+) equipped with the quotient norm
kakTr W 1( n ) = inf kfkW 1( n ) xn=0 p R+ 1 n p R+ f∈Wp (R+):f|xn=0=a To this end we will need an equivalent characterization of the real interpo- lation spaces (X0,X1)θ,p, namely the so-called trace method. But before we need some preliminaries on vector-valued Sobolev spaces.
3.2 Excursion: Vector-Valued Lp and Sobolev Spaces De nition 3.14 Let X be a Banach space and let (M, µ) be a measure space. Then a mapping is called simple if PN for some f : M → X f = k=1 αkχAk measurable Ak ⊆ M and αk ∈ X. Moreover, f : M → X is called strongly
15 measurable if there is a sequence of simple functions , , fk : M → X fk k ∈ N such that fk(x) →k→∞ f(x) in X for almost every x ∈ M. Moreover, f is called Bochner integrable if there is a sequence , , of simple fk : M → X k ∈ N functions such that Z lim kfk(x) − f(x)kX dµ(x) = 0. (3.6) k→∞ M If PN is a simple function, then the -valued integral is de ned f = k=1 αkχAk X by N Z X f(x) dµ(x) := αkµ(Ak). M k=1
If f is Bochner integrable and fk are simple functions such that (3.6) holds, R then fk(x) dµ(x) is a Cauchy sequence in X and we de ne M k∈N Z Z f(x) dµ(x) = lim fk(x) dµ(x). M k→∞ M
As in the scalar case the de nition is independent of the sequence fk. A useful criterion is the following:
Proposition 3.15 Let f : M → X. Then f is Bochner integrable if and only if f is strongly measurable and x 7→ kf(x)kX is integrable. We refer to the book by Yosida, Functional Analysis, Theorem 1 in Chapter V, Section 5. From the de nitions it easily follows that, if f : M → X is Bochner integrable and A ∈ L(X,Y ), then Af : M → Y is Bochner integrable (w.r.t. Y ) and Z Z A f(x) dµ(x) = Af(x) dµ(x). M M In particular, if w0 ∈ X0 and f : M → X is Bochner integrable, then x 7→ hf(x), w0i is integrable and Z Z f(x) dµ(x), w0 = hf(x), w0i dµ(x). M M Here hw, w0i = w0(w) denotes the duality product of w ∈ X and w ∈ X0. Similarly to the scalar case we de ne for 1 ≤ p ≤ ∞ p p L (M; X) = {f : M → X strongly measurable : kfkX ∈ L (M)}
kfkLp(M;X) = kkf(.)kX kLp(M)
16 In the following we restrict ourselves for simplicity to an interval (a, b) to- gether with the Lebesgue measure.
De nition 3.16 Let f ∈ L1((a, b); X) ≡ L1(a, b; X). Then g ∈ L1(a, b; X) is called weak derivative if
Z b Z b g(t)ϕ(t) dt = − f(t)ϕ0(t) dt a a for all ∞ . ϕ ∈ C0 ((a, b)) Note that, if g ∈ L1(a, b; X) is a weak derivative of f ∈ L1(a, b; X), then gw(x) := hg(x), wi is a weak derivative of the scalar function fw(x) = hf(x), wi for every w ∈ X0. In particular, the weak derivative of f is unique if it exists. Therefore it will for simplicity be denoted by f 0. Finally, we set
1 p has a weak derivative 0 p Wp (a, b; X) = {f ∈ L (a, b; X): f f ∈ L (a, b; X)} 0 1 p p kfkWp (a,b;X) = kfkL (a,b;X) + kf kL (a,b;X) Lemma 3.17 Let −∞ < a < b < ∞, X be a Banach space and let 1 ≤ . Then 1 is dense in 1 and 1 p < ∞ C ([a, b]; X) Wp (a, b; X) Wp (a, b; X) ,→ 0 continuously. Moreover, for every 1 C ([a, b]; X) f ∈ Wp (a, b; X) Z y f(y) − f(x) = f 0(t) dt for all a ≤ x ≤ y ≤ b. x Here k , , is the Banach space of all -times dierentiable C ([a, b]; X) k ∈ N0 k functions f :[a, b] → X equipped with a standard norm, e.g.
(j) k kfkC ([a,b];X) = sup f (x) X x∈[a,b],j=0,...,k
Remark 3.18 If 1 < p < ∞, we even have
1 1 1− p Wp (a, b; X) ,→ C ([a, b]; X). where Cα([a, b]; X), α ∈ [0, 1], is de ned as in the scalar case just replacing |.| by k.kX in the norms, i.e.,
kf(x) − f(y)kX α 0 kfkC ([a,b];X) = kfkC ([a,b];X) + sup α x6=y,x,y∈[a,b] |x − y|
17 3.3 The Trace Method
De nition 3.19 Let (X0,X1) be an admissible pair of Banach spaces and let θ ∈ (0, 1), 1 ≤ p ≤ ∞. Then we de ne V (p, θ, X1,X0) as the set of all such that 1 p with u: (0, ∞) → X0 + X1 u ∈ Wp (a, b; X0 + X1) ∩ L (a, b; X1) 0 p for all and nite where u ∈ L (a, b; X0) 0 < a < b < ∞ kukV (p,θ,X1,X0)
( ∞ 1 R tθp (ku(t)k + ku0(t)k )p dt p if p < ∞, kuk = 0 X1 X0 t V (p,θ,X0,X1) θ 0 if ess supt>0 t (ku(t)kX1 + ku (t)kX0 ) p = ∞.
Remark 3.20 If u ∈ V (p, θ, X1,X0), then limt→0 u(t) exists in X0 + X1 because of Z t Z t 0 θ 0 1−θ dτ ku(t) − u(s)kX0 ≤ ku (τ)kX0 dτ = τ ku (τ)kX0 τ s s τ 1 1 t p t p0 Z dτ Z 0 dτ ≤ kτ θu0(τ)kp τ (1−θ)p X0 s τ s τ 1 p0 ≤ kukV (p,θ,X0,X1) (F (t) − F (s)) for all 0 < s < t < ∞ in the case 1 < p < ∞, where t 7→ F (t) := t(1−θ), t ∈ [0, ∞), is a continuous function. If p = 1 or p = ∞, one simply replaces the p dt -norm or p0 dt -norm above by the ∞ dt -norm. L (0, ∞; t ) L (0, ∞; t ) L (0, ∞; t )
THEOREM 3.21 Let (X0,X1) be an admissible pair of Banach spaces, θ ∈ (0, 1) and 1 ≤ p ≤ ∞. Then
(X0,X1)θ,p = {x ∈ X0 + X1 : x = u(0), u ∈ V (p, 1 − θ, X1,X0)} . Moreover, the norm de ned by
Tr kxkθ,p = inf kukV (p,1−θ,X1,X0) : x = u(0), u ∈ V (p, 1 − θ, X1,X0) is equivalent to the norm of (X0,X1)θ,p. For the proof we need the following Hardy-type inequalities:
Lemma 3.22 Let α > 0 and let 1 ≤ p < ∞. Then there is a constant Cα,p such that Z ∞ Z t p Z ∞ −αp ds dt −αp p dt t ϕ(s) ≤ Cα,p t ϕ(t) (3.7) 0 0 s t 0 t Z ∞ Z ∞ p Z ∞ αp ds dt αp p dt t ϕ(s) ≤ Cα,p t ϕ(t) (3.8) 0 t s t 0 t 18 for all non-negative measurable functions ϕ: (0, ∞) → R. −p Remarks: It can be shown that Cα,p = |α| is the optimal constant in the latter inequalities, cf. the book by Hardy et al. Inequalities, Theorem 330. Choosing p−1 , 0 and using R t 0 in α = p ϕ(t) = t|f (t)| |f(t)| ≤ 0 |f (τ)| dτ Lemma 3.22, one easily derives the classical Hardy inequality:
f(t) 0 ≤ Cpkf kLp(0,∞) t Lp(0,∞) for all ∞ , where . f ∈ C0 (0, ∞) 1 < p < ∞
Remark 3.23 The proof of Theorem 3.21 even shows that for x ∈ (X0,X1)θ,p there some let v ∈ V (p, 1 − θ, X1,X0) be such that v(0) = x and dt t 7→ t2−θkv0(t)k ∈ Lp 0, ∞; . X1 t
Moreover, there is some Cθ,p > 0 such that
2−θ 0 kvkV (p,1−θ,X1,X0) + t kv (t)kX1 p dt ≤ Cθ,pkxkθ,p. L (0,∞; t )
Corollary 3.24 Let θ ∈ (0, 1) and let 1 ≤ p < ∞. Then X0 ∩ X1 is dense in (X0,X1)θ,p.
Corollary 3.25 Let and let n n . Then 1 ≤ p < ∞ R+ = {x ∈ R : xn > 0} 1− 1 for every 1 n we have p n−1 . Moreover, v ∈ Wp (R+) Trxn=0 v := v|xn=0 ∈ Wp (R ) 1− 1 the mapping 1 n p n−1 is bounded, linear and onto. Trxn=0 : Wp (R+) → Wp (R ) Remark 3.26 Let θ 0 V0(∞, θ, X1,X0) = u ∈ V (∞, θ, X1,X0) : lim t (ku(t)kX1 + ku (t)kX0 ) = 0 . t→0/∞ Then
(X0,X1)θ = {x ∈ X0 + X1 : x = u(0) for some u ∈ V0(∞, 1 − θ, X1,X0)} . We refer to [2, Proposition 1.13] for a proof.4
4 The proof is correct if X1 ,→ X0 and can be modi ed for the general case.
19 3.4 Intermediate Spaces and Reiteration
De nition 3.27 Let θ ∈ [0, 1], let (X0,X1) be an admissible pair of Banach spaces and let Y be an intermediate space with respect to (X0,X1). Then
1. Y is said to belong to the class Jθ if there is a constant C > 0 such that
kxk ≤ Ckxk1−θkxkθ for all x ∈ X ∩ X . Y X0 X1 0 1
In this case we write Y ∈ Jθ(X0,X1).
2. Y is said to belong to the class Kθ if there is a constant k > 0 such that
θ K(t, x) ≤ kt kxkY for all x ∈ Y.
In this case we write Y ∈ Kθ(X0,X1).
Note that, if θ ∈ (0, 1), then Y ∈ Kθ(X0,X1) means that Y,→ (X0,X1)θ,∞ continuously.
Proposition 3.28 Let θ ∈ (0, 1) and let Y be an intermediate space with re- spect to an admissible couple of Banach spaces (X0,X1). Then Y ∈ Jθ(X0,X1) if and only if (X0,X1)θ,1 ,→ Y .
Hence we have Y ∈ Jθ(X0,X1) ∩ Kθ(X0,X1) if and only if
(X0,X1)θ,1 ,→ Y,→ (X0,X1)θ,∞.
In particular, (X0,X1)θ,p ∈ Jθ(X0,X1) ∩ Kθ(X0,X1) for every p ∈ [1, ∞].
Example 3.29 k n k−1 n k+1 n k−1 n k+1 n C (R ) ∈ J1/2(C (R ),C (R ))∩K1/2(C (R ),C (R )) for all . More generally, k n m1 n m2 n k ∈ N C (R ) ∈ J(k−m1)/(m2−m1)(C (R ),C (R )) ∩ m1 n m2 n for all with K(k−m1)/(m2−m1)(C (R ),C (R )) m1, m2, k ∈ N0 m1 < k < m2.
Remark: It can be shown that C1(Rn) is not an interpolation space with respect to (C(Rn),C2(Rn)), cf. [2, Example 1.3.3]! The main result of this section is the following fundamental theorem:
Theorem 3.30 (Reiteration Theorem)
Let (X0,X1) be an admissible pair of Banach spaces and let 0 ≤ θ0, θ1 ≤ 1 with θ0 6= θ1. Moreover, let θ ∈ (0, 1) and set ω = (1 − θ)θ0 + θθ1. Then:
20 1. If for , then Yj ∈ Kθj (X0,X1) j = 0, 1
(Y0,Y1)θ,p ,→ (X0,X1)ω,p for all 1 ≤ p ≤ ∞. 2. If for , then Yj ∈ Jθj (X0,X1) j = 0, 1
(X0,X1)ω,p ,→ (Y0,Y1)θ,p for all 1 ≤ p ≤ ∞. In particular, if for , then Yj ∈ Jθj (X0,X1) ∩ Kθj (X0,X1) j = 0, 1
(Y0,Y1)θ,p = (X0,X1)ω,p for all 1 ≤ p ≤ ∞ with equivalent norms.
Remark 3.31 Since (X0,X1)θ,q ∈ Jθ(X0,X1)∩Kθ(X0,X1) for all θ ∈ (0, 1), 1 ≤ q ≤ ∞, Theorem 3.30 yields
((X0,X1)θ0,q0 , (X0,X1)θ1,q1 )θ,p = (X0,X1)(1−θ)θ0+θθ1,p for all 0 < θ0 6= θ1 < 1 and 1 ≤ p, q0, q1 ≤ ∞. Moreover,
((X0,X1)θ0,q0 ,X1)θ,p = (X0,X1)(1−θ)θ0+θ,p
(X0, (X0,X1)θ1,q1 )θ,p = (X0,X1)(1−θ)+θθ1,p for all 0 < θ0 6= θ1 < 1 and 1 ≤ p, q0, q1 ≤ ∞ since Xj ∈ Jj(X0,X1) ∩ Kj(X0,X1) for j = 0, 1. As consequences of the previous identity we obtain
θ0 n θ1 n 0 n 1 n (C (R ),C (R ))θ,∞ = (C (R ),C (R ))(1−θ)θ0+θθ1,∞ (1−θ)θ +θθ n = C 0 1 (R ), θ0 n θ1 n p n 1 n (Wp (R ),Wp (R ))θ,p = (L (R ),Wp (R ))(1−θ)θ0+θθ1,p (1−θ)θ0+θθ1 n = Wp (R ) for all 0 ≤ θ0 6= θ1 ≤ 1, θ ∈ (0, 1), and 1 ≤ p < ∞ due to Example 3.10. Moreover, since
1 n 0 n 2 n 0 n 2 n C ( ) ∈ J 1 (C ( ),C ( )) ∩ K 1 (C ( ),C ( )), R 2 R R 2 R R cf. Example 3.29, we have
0 n 2 n 0 n 1 n 2ω n (C (R ),C (R ))ω,∞ = (C (R ),C (R ))2ω,∞ = C (R ) for all 1 . ω ∈ (0, 2 ) In order to show (X0,X1)[θ] ∈ Kθ(X0,X1) we will need the following integral representation for bounded and holomorphic function on the strip:
21 Lemma 3.32 (Poisson formula) Let X be a Banach space and let f : S → X be continuous, bounded, and holomorphic in S0. Then f(z) = f0(z) + f1(z), where Z fj(z) = Pj(z, t)f(j + it) dt for all z ∈ S0, j = 0, 1, (3.9) R and eπ(y−t) sin(πx) P (x + iy, t) = . j sin2(πx) + (cos(πx) − (−1)j exp(π(y − t)))2 Proof: (Sketch) First let X = C, recall the Poisson formula for the unit circle: Z 2 1 1 − |z| for all f(z) = f(w) 2 dw |z| < 1, 2π |w|=1 |w − z| which can be derive from the Cauchy formula, cf. e.g. Remmert: Funktio- nentheorie I. The representation on S can be derived from the latter Poisson formula with the aid of the mapping h: S → {|z| ≤ 1} as in (2.2). Finally, the case that X is a general Banach space can be easily reduced to the scalar case by considering z 7→ hf(z), x0i for an arbitrary x0 ∈ X0.
Proposition 3.33 For every θ ∈ (0, 1) we have (X0,X1)[θ] ∈ Kθ(X0,X1). As a consequence we obtain
THEOREM 3.34 Let 0 < θ0, θ1, θ < 1 with θ0 6= θ1 and let 1 ≤ q ≤ ∞. Then
((X0,X1)[θ0], (X0,X1)[θ1])θ,q = (X0,X1)(1−θ)θ0+θθ1,q with equivalent norms. For completeness we note that
((X0,X1)[θ0], (X0,X1)[θ1])[θ] = (X0,X1)[(1−θ)θ0+θθ1] if is dense in , , and , cf. [1, Theo- X0 ∩ X1 X0 X1 (X0,X1)[θ0] ∩ (X0,X1)[θ1] rem 4.6.1]. Note that, if X1 ,→ X0, the latter conditions are valid. Moreover, for all 0 < θ0 6= θ1 < 1, 0 < θ < 1, 1 ≤ p0, p1 ≤ ∞, and 1 = 1−θ + θ we have p p0 p1
((X0,X1)θ0,p0 , (X0,X1)θ1,p1 )[θ] = (X0,X1)(1−θ)θ0+θθ1,p, cf. [1, Theorem 4.7.2].
22 3.5 Summary of Further Useful Results 3.5.1 Lorentz Spaces and Real Interpolation
In the following let (U, µ) be a measure space. Recall that for given measur- able f : U → K m(t; f) = µ ({x ∈ U : |f(x)| > t}) , t ≥ 0, is the distribution function of f.
De nition 3.35 The decreasing rearrangement of f is the function f ∗ : (0, ∞) → [0, ∞] de ned by f ∗(t) = inf {σ > 0 : m(σ; f) ≤ t} . (3.10) Then f ∗ : (0, ∞) → [0, ∞) is decreasing and continuous from the right. A fundamental property of f ∗ is: Lemma 3.36
m(ρ; f ∗) = m(ρ, f) for all ρ ≥ 0. (3.11) Proof: First of all, (3.10) implies f ∗(m(ρ; f)) ≤ ρ. Since f ∗ is decreasing, this implies m(ρ; f ∗) ≤ |[0, m(ρ; f)]| = m(ρ; f). In order to prove the con- verse inequality, we note that f ∗(m(ρ; f ∗)) ≤ ρ since f ∗ is continuous from the right. Hence m(ρ; f) ≤ m(ρ, f ∗) due to (3.10).
Because of the latter lemma and [3, Theorem 8.16], we obtain Z Z ∞ |f(x)|p dµ(x) = p tp−1m(t; f) dt U 0 Z ∞ Z ∞ = p tp−1m(t; f ∗) dt = |f ∗(t)|p dt 0 0 for all 1 ≤ p < ∞ and f ∈ Lp(U, µ). Moreover,
∗ kfkL∞(U,µ) = inf{t > 0 : m(t; f) = 0} = kf kL∞(0,∞). De nition 3.37 Let 1 ≤ p, q ≤ ∞. Then the Lorentz space Lp,q(U, µ) is de ned as p,q measurable L (U, µ) = {f : U → K : kfkLp,q < ∞} 1 ∗ kfkLp,q = t p f (t) . q dt L (0,∞; t )
23 As a direct consequence of the de nition we obtain
p,p p p,∞ p L (U, µ) = L (U, µ),L (U, µ) = Lweak(U, µ) for all 1 ≤ p ≤ ∞. Moreover, since f ∗ is monotone, one obtains Lp,1(U, µ) ,→ Lp,q1 (U, µ) ,→ Lp,q2 (U, µ) ,→ Lp,∞(U, µ) (3.12) for every 1 ≤ q1 ≤ q2 ≤ ∞ similarly as in the proof of Lemma 3.6. The Lorentz spaces occur naturally as real interpolation spaces of Lp- spaces:
THEOREM 3.38 Let 1 ≤ p0, p1, q0, q1, r ≤ ∞. Then for every θ ∈ (0, 1) we have p0,q0 p1,q1 p,r (L (U, µ),L (U, µ))θ,r = L (U, µ) 1 1−θ θ with equivalent norms if p0 6= p1, where = + . If p0 = p1, the same p p0 p1 conclusion holds if additionally 1 = 1−θ + θ . q q0 q1
In the case 1 < p0 6= p1 < ∞, the latter theorem follows the Reiteration Theorem 3.30 and the following special case:
THEOREM 3.39 For every θ ∈ (0, 1)
1 ∞ 1 ,r (L (U, µ),L (U, µ))θ,r = L 1−θ (U, µ). The proof is based on the identity Z t K(t, f; L1,L∞) = f ∗(τ) dτ for all t > 0 0 the monotonicity of f ∗(t) and the Hardy inequality. We refer to [2] for the details. For the general case, a proof of Theorem 3.38 can be found in [1, Theo- rem 5.3.1].
3.5.2 Dual Spaces
THEOREM 3.40 Let (X0,X1) be an admissible couple of Banach spaces such that X0 ∩ X1 is dense in X0 and X1 and let 1 ≤ p < ∞, θ ∈ (0, 1). Then
0 0 0 ((X0,X1)θ,p) = (X0,X1)θ,p0 , 0 0 0 ((X0,X1)θ) = (X0,X1)θ,1.
24 Proof: See [2, Theorem 1.18].
We note that, if X0 ∩ X1 is dense in X0 and X1, then
0 0 0 0 0 0 (X0 ∩ X1) = X0 + X1, (X0 + X1) = X0 ∩ X1 with same norms, cf. [1, Theorem 2.7.1]. Therefore 0 and 0 0 ((X0,X1)θ,p) (X0,X1)θ,p0 are intermediate spaces with respect to 0 0 and the equalities above (X0,X1) make sense.
THEOREM 3.41 Let (X0,X1) be an admissible couple of Banach spaces such that X0 ∩ X1 is dense in X0 and X1 and let θ ∈ (0, 1). Moreover, let X0 or X1 be re exive. Then
0 0 0 (X0,X1)[θ] = (X0,X1)[θ] Proof: See [1, Corollary 4.5.2].
25 4 Bessel Potential and Besov Spaces
4.1 Mikhlin Multiplier Theorem
Recall that the Fourier transformation F and the inverse Fourier transfor- mation F −1 are de ned by Z F[f](ξ) := fˆ(ξ) := e−ix·ξf(x) dx, n Z R −1 1 ix·ξ F [f](x) := n e f(ξ) dξ, (2π) Rn where f ∈ L1(Rn). We note that the de nitions are the same for f ∈ L1(Rn; X), where X is an arbitrary Banach space. Moreover, recall that by Plancharel's theorem F : L2(Rn) → L2(Rn) is an isomorphism with inverse F −1 and that Z Z 1 ˆ (4.1) f(x)g(x) dx = n f(ξ)gˆ(ξ) dξ Rn (2π) Rn for all f, g ∈ L2(Rn). Here (4.1) holds true for f ∈ L2(Rn; X), g ∈ L2(Rn; X0) if the products of the functions are understood as duality product pointwise. The proof is the same as in the scalar case or one can proof (4.1) in the vector-valued case by approximating f ∈ L2(Rn; X), g ∈ L2(Rn; X0) by simple functions and applying the identity in the scalar case. Furthermore, if f, g ∈ L2(Rn; H) and H is a Hilbert space, then F : L2(Rn; H) → L2(Rn; H) is an isomorphism with inverse F −1. THEOREM 4.1 (Mikhlin Multiplier Theorem) Let be Hilbert spaces and let . Moreover, let n H0,H1 N = n + 2 m: R \{0} → L(H0,H1) be an N-times dierentiable function such that α −|α| (4.2) k∂ξ m(ξ)kL(H0,H1) ≤ C|ξ| for all ξ 6= 0 and |α| ≤ N and let h i −1 ˆ for all ∞ n m(Dx)f = F m(ξ)f(ξ) f ∈ C0 (R ; H0).
Then m(Dx) extends to a bounded linear operator p n p n for all (4.3) m(Dx): L (R ; H0) → L (R ; H1) 1 < p < ∞.
26 The theorem is proved in Appendix B. We just note that (4.1) implies that, if , H0 = H1 = C for all 2 n km(Dx)fkL2(Rn) ≤ kmkL∞(Rn)kfkL2(Rn) f ∈ L (R ). Hence 2 n . In order to prove Theorem 4.1, the main step is m(Dx) ∈ L(L (R )) to show that 1 n 1 n . Once this is proved, m(Dx) ∈ L(L (R ),Lweak(R )) m(Dx) ∈ L(Lp(Rn)) for all 1 < p ≤ 2 by the Marcinkiewicz interpolation theorem and the case 2 < p < ∞ will follow by duality. Using that the Plancharel Theorem holds also for 2 n , the proof can be generalized to the f ∈ L (R ; Hj) case of general Hilbert spaces H0,H1.
4.2 A Fourier-Analytic Characterization of Hölder con- tinuity First of all, we recall that:
1. If f : Rn → C is a continuously dierentiable function such that f ∈ 1 n and 1 n , then L (R ) ∂jf ∈ L (R ) ˆ (4.4) F[∂xj f] = iξjF[f] = iξjf(ξ).
2. If 1 n such that 1 n , then ˆ is continuously partial f ∈ L (R ) xjf ∈ L (R ) f(ξ) dierentiable with respect to ξj and ˆ (4.5) ∂ξj f(ξ) = F[−ixjf(x)].
The latter identities show the duality between dierentiability of f and decay of fˆ(ξ) as |ξ| → ∞ as well as decay of f for large x and dierentiability of fˆ. In the following we will use a Littlewood-Paley partition of unity ϕj(ξ), (on n). This is a partition of unity , , on n with j ∈ N0 R ϕj(ξ) j ∈ N0 R ∞ n such that ϕj ∈ C0 (R ) n j−1 j+1 for all (4.6) supp ϕj ⊆ ξ ∈ R : 2 ≤ |ξ| ≤ 2 j ≥ 1.
The partition of unity can be constructed such that supp ϕ0 ⊂ B2(0), ϕj(ξ) = −j+1 ϕ1(2 ξ) for all j ≥ 1 and (4.6) holds. Then we have α α −|α|j for all n (4.7) |∂ξ ϕj(ξ)| ≤ Ck∂ξ ϕ1kL∞(Rn)2 α ∈ N0 , j ≥ 1.
27 Moreover, we note that
−1 h ˆ i ϕj(Dx)f = F ϕj(ξ)f(ξ) =ϕ ˇj ∗ f for all ∞ n , , where −1 and f ∈ C0 (R ) j ∈ N0 ϕˇj = F [ϕj] (j−1)n j−1 for all and n ϕˇj(x) = 2 ϕˇ1(2 x) j ∈ N x ∈ R . For ∞ n we will de ne by f ∈ L (R ) ϕj(Dx)f
ϕj(Dx)f =ϕ ˇj ∗ f.
Furthermore, since ϕj−1 + ϕj + ϕj+1 ≡ 1 on supp ϕj (where ϕ−1 ≡ 0), we have
ϕj(Dx)f = (ϕj−1(Dx) + ϕj(Dx) + ϕj+1(Dx)) ϕj(Dx)f (4.8) for all n . f ∈ S(R ), j ∈ N0 Using this decomposition, we obtain the following characterization of Hölder continuous functions.
THEOREM 4.2 Let 0 < s < 1. Then f ∈ Cs(Rn) if and only if f ∈ L∞(Rn) and js s ∞ n kfkC∗ := sup 2 kϕˇj ∗ fkL (R ) < ∞. j∈N0 s n Moreover, s is an equivalent norm on . k · kC∗ C (R ) Proof: First let f ∈ Cs(Rn). Then
s sup |f(x − y) − f(x)| ≤ kfkCs |y| x∈Rn for all n. Since we have chosen such that 1−j for y ∈ R ϕj ϕj(ξ) = ϕ1(2 ξ) jn j −1 j ≥ 1, we have ϕˇj(x) = ψ2−j (x) = 2 ψ(2 x), where ψ = F [ϕ1(2·)]. This implies that
j for all (4.9) kϕˇjkL1(Rn) ≤ C, k∇ϕˇjkL1(Rn) ≤ C2 j ∈ N0. Moreover, Z Z ϕˇj(y)dy = ψ(y)dy = F[ψ](0) = ϕ1(0) = 0 Rn Rn
28 for all j ≥ 1. Hence Z Z ϕj(Dx)f = f(x − y)ψ2−j (y)dy = (f(x − y) − f(x))ψ2−j (y)dy (4.10) Rn Rn and therefore Z s kϕj(Dx)fk∞ ≤ kfkCs |y| ψ2−j (y)dy n R Z −js s −js = 2 kfkCs |z| ψ(z)dz = C2 kfkCs Rn for all . The latter inequality implies s s since also j ∈ N kfkC∗ ≤ CkfkC kϕ0(Dx)fk∞ ≤ Ckfk∞. ∞ n Conversely, let be such that s . Now, if , f ∈ L (R ) kfkC∗ < ∞ |y| ≤ 1 X X f(x − y) − f(x) = (fj(x − y) − fj(x)) + (fj(x − y) − fj(x)) , 2j ≤|y|−1 2j >|y|−1 where fj = ϕj(Dx)f. In order to estimate the rst sum, we use the mean value theorem to conclude that
|fj(x − y) − fj(x)| ≤ |y|k∇fjk∞. (4.11) Moreover, since
∂xk fj = ∂xk ϕj−1(Dx)fj + ∂xk ϕj(Dx)fj + ∂xk ϕj+1(Dx)fj, due to (4.8) and
l k∂ ϕ (D )gk ≤ k∂ ϕˇ k 1 n kgk ∞ n ≤ C2 kgk ∞ n xk l x L∞(Rn) xk l L (R ) L (R ) L (R ) for general , ∞ n , we obtain l ∈ N0 g ∈ L (R ) X X |fj(x − y) − fj(y)| ≤ C |y|k∇fjk∞ 2j ≤|y|−1 2j ≤|y|−1 X j(1−s) s s s ≤ C|y| 2 kfkC∗ ≤ C|y| kfkC∗ . 2j ≤|y|−1 The second sum is simply estimated by X X |fj(x − y) − fj(y)| ≤ 2 kfjk∞ 2j >|y|−1 2j >|y|−1 X −js s s s ≤ 2kfkC∗ 2 = C|y| kfkC∗ 2j >|y|−1
29 Altogether s s . kfkC ≤ CkfkC∗
Remark 4.3 Because of (4.9) we get
(4.12) kϕj(Dx)fkLp(Rn) ≤ kϕˇjkL1(Rn)kfkLp(Rn) ≤ kfkLp(Rn), j (4.13) k∇ϕj(Dx)fkLp(Rn) ≤ k∇ϕˇjkL1(Rn)kfkLp(Rn) ≤ C2 kfkLp(Rn) for any p n , , . f ∈ L (R ) 1 ≤ p ≤ ∞ j ∈ N0
4.3 Bessel Potential and Besov Spaces De nitions and Basic Properties
2 1 In the following let hξi := (1+|ξ| ) 2 . We note that for any s ∈ R the function hξis is a smooth function satisfying
α s s−|α| (4.14) |∂ξ hξi | ≤ Cs,α(1 + |ξ|) for all n and some . The latter estimate can be proved by α ∈ N0 Cs,α > 0 considering n+1 de ned by s. Since is f : R \{0} → R fs(ξ, t) = |(t, ξ)| fs smooth and homogeneous of degree s in (t, ξ), we have
α s−|α| |∂ξ fs(ξ, t)| ≤ Cs,α(|t| + |ξ|) uniformly in and for all n. (ξ, t) 6= 0 α ∈ N0 Using (4.14) it is easy to show, that hξisfˆ(ξ) ∈ S(Rn) for all f ∈ S(Rn). By duality hξisfˆ ∈ S0(Rn) for all f ∈ S0(Rn). Therefore we can de ne s 0 n 0 n as hDxi : S (R ) → S (R ) h i s −1 s ˆ for all 0 n hDxi f = F hξi f f ∈ S (R ).
De nition 4.4 Let s ∈ R and let 1 < p < ∞. Then the Bessel potential space s n is de ned by Hp (R )
s n 0 n s p n Hp (R ) = {f ∈ S (R ): hDxi f ∈ L (R )} s kfk s n = khD i fk . Hp(R ) x Lp(Rn)
Remarks 4.5 1. If , then we have s n if and only if s ˆ p = 2 f ∈ H2 (R ) hξi f ∈ L2(Rn) by Plancharel's theorem.
30 2. By de nition s s n p n is an isomorphism with inverse hDxi : Hp (R ) → L (R ) −s. Since n is dense in p n and −s n n , hDxi S(R ) L (R ) hDxi : S(R ) → S(R ) n is dense in s n for any , . S(R ) Hp (R ) s ∈ R 1 < p < ∞ As a consequence of the Mikhlin multiplier theorem we obtain
Theorem 4.6 Let and let . Then m n m n m ∈ N0 1 < p < ∞ Hp (R ) = Wp (R ) with equivalent norms.
Proof: We rst prove the embedding m n m n . Let n . Hp (R ) ,→ Wp (R ) f ∈ S(R ) Then h i (iξ)β ∂βf = F −1 (iξ)βfˆ(ξ) = F −1 hξimfˆ(ξ) x hξim Hence in order to obtain
β m p n p n m n (4.15) k∂x fkL (R ) ≤ CpkhDxi fkL (R ) ≡ CpkfkHp (R ) for n with we apply Theorem 4.1 to (iξ)β . Therefore β ∈ N0 |β| ≤ m mβ(ξ) = hξim we have to verify (4.2) for m = mβ. To this end, we use (4.14) and α β |β|−|α| (4.16) |∂ξ (iξ) | ≤ Cα,β|ξ| .
Moreover, (1 + |ξ|)−m−|α| ≤ |ξ|−|β|−|α| if |β| ≤ m. Therefore α −|α| (4.17) |∂ξ mβ(ξ)| ≤ Cα,β|ξ| follows from (4.16), (4.14), and the following claim: Claim: Let , and let n be -times s1, s2 ∈ R N ∈ N p1, p2 : R \{0} → C N continuously dierentiable satisfying
α sj −|α| for all |∂ξ pj(ξ)| ≤ C|ξ| |α| ≤ N, j = 1, 2. Then α 0 s1+s2−|α| (4.18) |∂ξ (p1(ξ)p2(ξ))| ≤ C |ξ| for all |α| ≤ N. Proof of Claim: The claim follows directly from the Leibniz's formula.
Because of (4.17), the conditions of the Mikhlin Multiplier Theorem 4.1 are satis ed and (4.15) follows for all , which proves m n m n |β| ≤ m Hp (R ) ,→ Wp (R ) since n is dense in m n . S(R ) Hp (R )
31 Hence it remains to prove m n m n . If , , is Wp (R ) ,→ Hp (R ) m = 2k k ∈ N0 m 2 k m even, then hξi = (1 + |ξ| ) is a polynomial of degree m. Therefore hDxi is a dierential operator of order m and
m X α khDxi fkLp(Rn;H) ≤ C k∂x fkLp(Rn;H), |α|≤m which proves the embedding in this case. If , , is odd, then m = 2k + 1 k ∈ N0
n 2 ! n 1 X ξj 1 X ξj hξim = hξim + = hξi2k + hξi2kξ , hξi2 hξi2 hξi hξi j j=1 j=1
2k 2k where hξi and hξi ξj are polynomials of degree at most 2k + 1. Hence
n m X X α khDxi fkLp(Rn;H) ≤ C kmj(Dx)∂x fkLp(Rn;H), |α|≤m j=0
−1 −1 where m0(ξ) = hξi and mj(ξ) = ξjhξi , j = 1, . . . , n. Hence it remains to verify the Mikhlin condition (4.2) for mj(ξ). If j = 0, then (4.2) for −1−|α| −|α| m(ξ) = m0(ξ) follows from (4.14) with s = −1 because of hξi ≤ |ξ| . If j = 1, . . . , n, then (4.2) follows for m(ξ) = mj(ξ) from (4.14) with s = −1, (4.16) with β = ej, and (4.18).
Theorem 4.2 gives a motivation for the following de nition of the Besov space s n . Bpq(R )
De nition 4.7 Let s ∈ R, 1 ≤ p, q ≤ ∞. Then
s n n 0 n o s n Bpq(R ) = f ∈ S (R ): kfkBpq(R ) < ∞ , where
1 ∞ ! q X jsq q kfk s n = 2 kϕ (D )fk if q < ∞, Bpq(R ) j x Lp(Rn) j=0 js s n p n if kfkBpq(R ) = sup 2 kϕj(Dx)fkL (R ) q = ∞. j∈N0
32 Remarks 4.8 1. Because of Theorem 4.2, s n s n for C (R ) = B∞∞(R ) 0 < . More generally, s n s n , , are called Hölder- s < 1 C∗ (R ) := B∞∞(R ) s > 0 Zygmund spaces.
2. Note that s n if and only if f ∈ Bpq(R ) s p n (ϕj(Dx)f)j∈N0 ∈ `q(N0; L (R )), Here s , , is the space of all -valued sequences `q(M; X) M ⊆ Z X x = such that (xj)j∈M 1 P js q q (2 kxjkX ) if q < ∞, s j∈M kxk`q(M;X) = js if supj∈M 2 kxjkX q = ∞. Moreover, we set q 0 . Of course s ` (M; X) = `q(M; X) (xj)j∈M ∈ `q(M; X) if and only if js q . (2 xj)j∈M ∈ ` (M; X) 3. Using Plancharel's theorem, it is not di cult to show that
s n s n B22(R ) = H2 (R ). The proof is left to the reader as an exercise. But the statement will also follow from Corollary 4.15 below. Some simple relations between the Besov spaces are summarized in the fol- lowing:
Lemma 4.9 Let , , and let . Then s ∈ R 1 ≤ p, q1, q2 ≤ ∞ ε > 0 s n s n if s+ε n s n Bpq1 (R ) ,→ Bpq2 (R ) q1 ≤ q2,Bp∞ (R ) ,→ Bp1(R ). Proof: The rst embedding follows from
q1 q2 s s if (4.19) ` (N0; X) ,→ ` (N0; X), `q1 (N0; X) ,→ `q2 (N0; X) q1 ≤ q2. The second embedding follows s+ε s because of `∞ (N0; X) ,→ `1(N0; X) ∞ X sj s k(aj)k`q(N0;X) = 2 kajkX j=0 ∞ ! X −εj (s+ε)j ≤ 2 sup 2 ka k ≤ C k(a )k s+ε j X ε j `∞ (N0;X) j∈ j=0 N0
33 Remark 4.10 The latter lemma shows that q measures regularity of f on a ner scale than , meaning, if 0, then s n s0 n with s s > s Bpq1 (R ) ,→ Bpq2 (R ) arbitrary 1 ≤ q1, q2 ≤ ∞. Exercise 1 Show that n p n 0 n Bp1(R ) ,→ C (R ). j n Hint: Use that , where q0 uniformly in ϕj(Dx)f =ϕ ˇj ∗ f kϕˇjkLq(Rn) ≤ C2 j ∈ Z. In order to get a sharp comparision of Besov and Bessel potential spaces we prove:
THEOREM 4.11 Let s ∈ R, 1 < p < ∞. Then there are constants c, C > 0 such that
1 ∞ ! 2 X 2js 2 ckfkHs( n) ≤ 2 |ϕj(Dx)f(x)| ≤ CkfkHs( n) p R p R j=0 Lp(Rn) for all s n . f ∈ Hp (R ) Remark 4.12 Because of the latter equivalent norm on s n , one de nes Hp (R ) more generally the Triebel-Lizorkin space s n , , as Fpq(R ) s ∈ R, 1 < p, q < ∞
s n n 0 n o s n Fpq(R ) = f ∈ S (R ): kfkFpq(R ) < ∞ , 1 ∞ ! q X qjs q kfkF s ( n) = 2 |ϕj(Dx)f(x)| . pq R j=0 Lp(Rn) Hence the latter theorem shows that s n s n . Finally, note that Hp (R ) = Fp2(R )
kfkF s ( n) = k(ϕj(Dx)f)j∈ k p n s . pq R N L (R ;`q(N0))
Proof of Theorem 4.11: First we will show that s n s n . kfkF (R ) ≤ CkfkHp(R ) To this end we de ne a mapping p2
n p n p n 2 Q: S(R ) ⊂ L (R ) → L (R ; ` (N0))
34 by
js −s 2 for all n (Qg)(x) = (2 ϕj(Dx)hDxi g(x))j∈N0 ∈ ` (N0) x ∈ R . Then −1 (Qg)(x) = Fξ7→x [m(ξ)ˆg(ξ)] where 2 is de ned by m(ξ) ∈ L(C, ` (N0)) js −s for all n m(ξ)a = (2 ϕj(ξ)hξi )j∈N0 a a ∈ C, ξ ∈ R . In order to show that Q extends to a bounded operator
p n p n 2 for all (4.20) Q: L (R ) → L (R ; ` (N0)) 1 < p < ∞, we verify the condition for the Mikhlin multiplier theorem 4.1:
∞ α 2 X 2js α −s 2 k∂ m(ξ)k 2 = 2 |∂ (ϕ (ξ)hξi )| ξ L(C,` (N0)) ξ j j=0 2js −2s−2|α| −2|α| ≤ Cα,s2 hξi χsupp ϕj (ξ) ≤ Cα,s|ξ| for all n, where we have used that j−1 j+1 on if α ∈ N0 2 ≤ |ξ| ≤ 2 supp ϕj j ≥ 1 and α −s −s−|α| |∂ξ (ϕj(ξ)hξi )| ≤ Cα,shξi uniformly in , which follows from (4.7), (4.14), and the product rule. j ∈ N0 Hence (4.20) follows and therefore
s s kfk s n = kQhD i fk p n q ≤ CkhD i fk p n ≡ Ckfk s n . Fp2(R ) x L (R ;` (N0)) x L (R ) Hp(R ) Note that we have shown that
s n p n s Qe : Hp (R ) → L (R ; `2(N0)) is bounded, where
−js s (4.21) Qfe = (2 QhDxi f)j∈N0 = (ϕj(Dx)f)j∈N0 . Conversely, we de ne a mapping
n 2 p n 2 p n R: S(R ; ` (N0)) ⊂ L (R ; ` (N0)) → L (R )
35 by
∞ X −js s for all n n 2 (Ra)(x) = 2 ϕej(Dx)hDxi aj(x) x ∈ R , a ∈ S(R ; ` (N0)). j=0 Here , , and . Note that ϕej(ξ) = ϕj−1(ξ) + ϕj(ξ) + ϕj+1(ξ) j ∈ N0 ϕ−1(ξ) = 0 since on . Then ϕ˜j(ξ)ϕj(ξ) = ϕj(ξ) ϕej(ξ) = 1 supp ϕj −1 (Ra)(x) = Fξ7→x [m(ξ)ˆaj(ξ)] where 2 is de ned by m(ξ) ∈ L(` (N0), C)
∞ X m(ξ)a = 2−jsϕ (ξ)hξisa for all (a ) ∈ `2( ). ej j j j∈N0 N0 j=0 Similarly, as before
∞ α 2 X −2js α s 2 k∂ m(ξ)k 2 ≤ 2 |∂ (ϕ ˜ (ξ)hξi )| ξ L(` (N0),C) ξ j j=0 ∞ X −2js 2s−2|α| −2|α| ≤ Cq 2 hξi χ{2j−2≤|ξ|≤2j+2} ≤ Cq|ξ| , j=0 where we have used that for each ξ ∈ Rn at most 5 terms in the sum above are non-zero and that −2js −s on j−2 j+2 . Hence, 2 ≤ Chξi supp ϕej ⊆ {2 ≤ |ξ| ≤ 2 } applying Theorem 4.1 once more, we obtain that R extends to a bounded operator
p n 2 p n for all R: L (R ; ` (N0)) → L (R ) 1 < p < ∞. Now we apply to js , . Then R aj = 2 ϕj(Dx)f j ∈ N0
∞ X −js s js s Ra = 2 ϕej(Dx)hDxi 2 ϕj(Dx)f = hDxi f j=0 since P∞ P∞ . Thus j=0 ϕej(Dx)ϕj(Dx) = j=0 ϕj(Dx) = I s s n p n p n kfkHp(R ) = khDxi fkL (R ) = kRakL (R ) js s n ≤ C (2 ϕj(Dx)f)j∈ 0 p n 2 = CkfkF ( ), N L (R ;` (N0)) p2 R 36 which proves the lemma. Finally, we note that the previous estimates imply that p n s s n Re: L (R ; `2(N0)) → Hp (R ) is bounded, where
∞ X R(a ) := hD isR(2jsa ) = ϕ (D )a (4.22) e j j∈N0 x j j∈N0 ej x j j=0 and therefore ˜ ˜ on s n . RQ = I Hp (R )
Remark 4.13 Note that the latter proof shows that s n is a retract of Hp (R ) p n s . In general: L (R ; `2(N0)) De nition 4.14 A Banach space X is called a retract of a Banach space Y if there are bounded, linear operators R: Y → X and Q: X → Y such that RQ = idX . If , are de ned by (4.22) and (4.21), then on s n . Note Re Qe ReQe = I Hp (R ) that the mappings are independent of p and s. Moreover, using the same mappings and as in the previous proof it is easy to show that s n R Q Bpq(R ) is a retract of q p n . ` (N0; L (R ))
Corollary 4.15 Let 1 < p < ∞, s ∈ R. Then s n s n s n if (4.23) Bpp(R ) ,→Hp (R ) ,→ Bp2(R ) 1 < p ≤ 2, s n s n s n if (4.24) Bp2(R ) ,→Hp (R ) ,→ Bpp(R ) 2 ≤ p < ∞. In particular, s n s n for all . H2 (R ) = B22(R ) s ∈ R Proof: The statements follows from Theorem 4.11, the embeddings
q p n p n q if (4.25) ` (N0; L (R )) ,→ L (R ; ` (N0)) 1 ≤ q ≤ p ≤ ∞ p q n q n p if (4.26) L (N0; ` (R )) ,→ ` (R ; L (N0)) 1 ≤ p ≤ q ≤ ∞
37 as well as from (4.19). Here (4.25) follows from
p n q k(fj)j∈N0 kL (R ;` (N0)) 1 p p 1 Z ∞ ! q ∞ q X q X q = |fj(x)| dx = |fj(.)| n R j=0 j=0 p L q (Rn) 1 1 ∞ ! q ∞ ! q X q X q p ≤ k|fj(.)| k q n = kfjkLp( n) , L (R ) R j=0 j=0 where we have used Minkowski's inequality. The inequality (4.26) is proved analogously.
4.4 Interpolation of Vector-Valued Lp-Spaces, Bessel Po- tential, and Besov Spaces
Recall that s n and s n are retract of p n s and s p n Hp (R ) Bpq(R ) L (R ; `2(N0)) `q(N0; L (R )) with same retraction and co-retractions, which are independent of p, s. More precisely,
s n p n s Qe : Hp (R ) → L (R ; `2(N0)), s n s p n Qe : Bpq(R ) → `q(N0; L (R )) and
p n s s n Re: L (R ; `2(N0)) → Hp (R ), s p n s n Re: `q(N0; L (R )) → Bpq(R ) are bounded linear operators satisfying ReQe = I where
s n s n Qfe = (ϕj(Dx)f)j∈N0 , f ∈ Hp (R ) ∪ Bpq(R ). and
∞ X R(a ) := ϕ (D )a , (a ) ∈ Lp( n; `s( )) ∪ `s( ; Lp( n)). e j j∈N0 ej x j j j∈N0 R 2 N0 q N0 R j=0 Generally we have:
38 Proposition 4.16 Let (X0,X1) and (Y0,Y1) be admissible Banach spaces and let Q: X0 + X1 → Y0 + Y1, R: Y0 + Y1 → X0 + X1 be linear mappings such that Q ∈ L(Xj,Yj), R ∈ L(Yj,Xj) and RQx = x for all x ∈ Xj and j = 0, 1. Then
R(Y0,Y1)θ,p = (X0,X1)θ,p R(Y0,Y1)[θ] = (X0,X1)[θ] with equivalent norms for all θ ∈ (0, 1), 1 ≤ p ≤ ∞, where R(Y0,Y1)θ,p is equipped with the quotient norm
kxkR(Y0,Y1)θ,p = inf kykθ,p. y∈(Y0,Y1)θ,p:Ry=x Proof: The proof of the proposition is similar to the proof of Example 3.11.
Because of the latter proposition, it is su cient to obtain interpolation results for p n s and s p n to characterize the real and L (R ; `2(N0)) `q(N0; L (R )) complex interpolation spaces of s n and s n . Hp (R ) Bpq(R ) We start with a result for the real interpolation method:
THEOREM 4.17 Let be a Banach spaces, let , , X s0 6= s1 ∈ R 0 < θ < 1 and let 1 ≤ q0, q1 ≤ ∞. Then for every 1 ≤ q ≤ ∞
s0 s1 s (`q0 (Z; X), `q1 (Z; X))θ,q = `q(Z; X), s0 s1 s (`q0 (N0; X), `q1 (N0; X))θ,q = `q(N0; X) with equivalent norms, where s = (1 − θ)s0 + θs1. As a consequence we obtain
THEOREM 4.18 Let , and let . s0 6= s1 ∈ R 0 < θ < 1 s = (1 − θ)s0 + θs1 Then for all 1 ≤ p, q0, q1, q ≤ ∞
s0 n s1 n s n (Bpq0 (R ),Bpq1 (R ))θ,q = Bpq(R ) with equivalent norms. Moreover, for any 1 ≤ q ≤ ∞, 1 < p < ∞ we have
s0 n s1 n s n (Hp (R ),Hp (R ))θ,q = Bpq(R ) with equivalent norms.
39 Remark 4.19 Because of Example 3.10, we know that
p n 1 n θ n (L (R ),Wp (R ))θ,p = Wp (R ) for all 1 ≤ p < ∞. On the other hand, we have by the previous theorem:
p n 1 n 0 n 1 n θ n (L (R ),Wp (R ))θ,p = (Hp (R ),Hp (R ))θ,p = Bpp(R ) if , . Hence s n s n for all , 1 < p < ∞ θ ∈ (0, 1) Bpp(R ) = Wp (R ) s ∈ (0, 1) 1 < p < ∞ and
1 Z |f(x) − f(y)|p p kfk s n = kfk p n + d(x, y) Wp (R ) L (R ) θp+n Rn×Rn |x − y| is an equivalent norm on s n . A more general statement of this kind Bpp(R ) will be given below. For the complex interpolation method we have:
THEOREM 4.20 Let 1 ≤ p0, p1, q0, q1 < ∞, 0 < θ < 1, and let (U, µ), (V, ν) be two measure spaces. Moreover, let 1 = 1−θ + θ , 1 = 1−θ + θ . Then p p0 p1 q q0 q1
p0 q0 p1 q1 p q (L (U; L (V )),L (U; L (V )))[θ] = L (U; L (V )) with equal norms.
Remark 4.21 If (X0,X1) is an admissible pair of Banach spaces and 1 ≤ p0, p1 < ∞, 0 < θ < 1, then
p0 p1 p (L (U; X0),L (U; X1))[θ] = L (U;(X0,X1)[θ]) with equal norms, where 1 = 1−θ + θ , cf. [1, Theorem 5.1.2]. p p0 p1 Corollary 4.22 Let , , and let . s0, s1 ∈ R 0 < θ < 1 1 ≤ q0, q1, p0, p1 < ∞ Moreover, let 1 = 1−θ + θ , 1 = 1−θ + θ . Then p p0 p1 q q0 q1
s0 p0 n s1 p1 n s p n (`q0 (N0; L (R )), `q1 (N0; L (R )))[θ] = `q(N0; L (R )) p0 n s0 p1 n s1 p n s (L (R ; `q0 (N0)),L (R ; `q1 (N0)))[θ] = L (R ; `q(N0)) with equal norms, where s = (1 − θ)s0 + θs1. As an application we obtain:
40 THEOREM 4.23 Let , and let . s0, s1 ∈ R 0 < θ < 1 s = (1 − θ)s0 + θs1 Then s0 n s1 n s n (Hp0 (R ),Hp1 (R ))[θ] = Hp (R ) 1 1−θ θ with equal norms for any 1 < p0, p1 < ∞ and = + . Moreover, for p p0 p1 any 1 ≤ q0, q1, p0, p1 < ∞ we have
s0 n s1 n s n (Bp0q0 (R ),Bp1q1 (R ))[θ] = Bpq(R ) with equal norms where 1 = 1−θ + θ and 1 = 1−θ + θ . p p0 p1 q q0 q1 4.5 Sobolev Embeddings and Traces We start with a Sobolev-type embedding theorem for Besov and Bessel po- tential spaces.
THEOREM 4.24 Let with and such that s, s1 ∈ R s ≤ s1 1 ≤ p1 ≤ p ≤ ∞ n n s − ≤ s1 − . p p1 Then
s1 n s n for all (4.27) Bp1q1 (R ) ,→ Bpq(R ) 1 ≤ q1 ≤ q ≤ ∞, s1 n s n if (4.28) Hp1 (R ) ,→ Hp (R ). 1 < p1 ≤ p < ∞ Proof: See [1, Theorem 6.51]. Remarks on the proof: The embedding (4.27) follows from:
ϕk(Dx)f =ϕ ˜k(Dx)ϕk(Dx)f with ϕ˜k(Dx) = ϕk−1(Dx) + ϕk(Dx) + ϕk+1(Dx), ϕ−1(Dx) := 0 as well as
ϕk(Dx)g = ψ2−k ∗ g, which implies
−k n q0 kϕk(Dx)gkLp(Rn) ≤ kψ2−k kLq(Rn)kgkLp1 (Rn) ≤ C2 kgkLp1 (Rn), where 1 1 1 1 . Then (4.28) follows from (4.27) by a clever q0 = 1 − q = p − p interpolation. 1
41 THEOREM 4.25 Let , , 1 and let 1 ≤ p < ∞ 1 ≤ q ≤ ∞ s > p Tr f = f|xn=0 for any continuous s n s n . Then can be extended to a f ∈ Hp (R ) ∪ Bpq(R ) Tr bounded linear operator
s− 1 s n p n Tr: Bpq(R ) → Bpq (R ), s− 1 s n p n Tr: Hp (R ) → Bpp (R ). Proof: See [1]. Remarks on the proof: If m = 1, then
m− 1 s− 1 m n m n p n−1 p n−1 Tr: Hp (R ) = Wp (R ) → Wp (R ) = Bpp (R )
m− 1 p n−1 follows from the trace method as before. Using a ∈ Bpp (R ) if and only 1− 1 if α p n−1 for all , the same follows for general . ∂x0 a ∈ Bpp (R ) |α| ≤ m − 1 m ∈ N Then the statement follows for s ≥ 1 by interpolation. 1 In the case 1 one uses that p 0 . Therefore p < s ≤ 1 Bp1(R) ,→ C (R) 0 0 | Tr f(x )| ≤ Ckf(x ,.)k 1 p Bp1(R) 0 ⇒ k Tr fkLp( n−1) ≤ Ckfk 1 ≤ C kfk 1 . R p n−1 p p n L (R ;Bp1(R)) Bp1(R ) An iterated interpolation then yields the statement of the theorem for 1 p < s < 1.
4.6 Equivalent Norms The following theorem is the direct generalization of Theorem 4.2 for general Besov spaces s n with . Bpq(R ) 0 < s < 1 THEOREM 4.26 Let 0 < s < 1 and let 1 ≤ p, q ≤ ∞. Then there are constants c, C (depending on s, p, q) such that
1 Z ∞ q q ωp(t; f) dt ckfk s n ≤ kfk p n + ≤ Ckfk s n (4.29) Bpq(R ) L (R ) sq Bpq(R ) 0 t t if q < ∞ and
q ωp(t; f) dt ckfk s n ≤ kfk p n + sup ≤ Ckfk s n (4.30) Bpq(R ) L (R ) sq Bpq(R ) t>0 t t
42 if q = ∞, where
ωp(t; f) = sup kf(. + h) − fkLp(Rn) |h|≤t is the Lp-modulus of continuity of f.
Remark 4.27 We refer to [1, Theorem 6.2.5] for a more general statement in the case s > 0. Proof of Theorem 4.26: We will only prove the case q < ∞ since the proof in the case q = ∞ is a simple variant of the proof of Theorem 4.2. First of all, since t 7→ ωp(t; f) is a monotone increasing function and t is proportional to 2−j on [2−j+1, 2−j],
Z 1 q ∞ ωp(t; f) dt X ≤ C 2sjqω (2−j; f)q tsq t p 0 j=0 and ∞ Z 2 q X ωp(t; f) dt 2sjqω (2−j; f)q ≤ C . p tsq t j=0 0 Moreover,
Z ∞ ω (t; f)q dt Z ∞ p ≤ 2qkfkq t−sq−1 dt = Ckfkq sq Lp(Rn) Lp(Rn) 1 t t 1 since . Hence we can replace the middle term in (4.29) ωp(t; f) ≤ 2kfkLp(Rn) by 1 ∞ ! q X sjq −j kfkLp(Rn) + 2 ωp(2 ; f) . j=0 First we prove the second inequality in (4.29). For s n we denote f ∈ Bpq(R ) fk = ϕk(Dx)f. Then
kfk(. + h) − fkkLp(Rn) ≤ |h|k∇fkkLp(Rn) due to (4.11) and therefore
ωp(t; fk) ≤ tk∇fkkLp(Rn) = tk∇ϕk(Dx)ϕek(Dx)fkkLp(Rn) k ≤ Ct2 kϕek(Dx)fkkLp(Rn)
43 because of (4.13), where , , ϕek(Dx) = ϕk−1(Dx) + ϕk(Dx) + ϕk+1(Dx) k ∈ N0 and . On the other hand, and ϕ−1(Dx) = 0 ωp(t, fk) ≤ 2kfkkLp(Rn) f = P∞ . Therefore k=0 fk
∞ ! sj −j X sj −j+k 2 ωp(2 ; f) ≤ C 2 min(1, 2 )kϕek(Dx)fkkLp(Rn) j=0 ∞ ! X s(j−k) −j+k sk ≤ C 2 min(1, 2 )2 kϕk(Dx)fkkLp(Rn) . j=0
Now, de ning sj −j , , sj if aj = C2 min(1, 2 ) j ∈ Z bj = 2 kϕk(Dx)fjkLp(Rn) j ≥ 0 sj −j and bj = 0 else, we see that 2 ωp(2 ; f) ≤ (a ∗ b)j, where X (a ∗ b)j = aj−kbk k∈Z is the convolution of two sequences. Hence
1 ∞ ! q X sjq −j q q 1 q s n 2 ωp(2 ; f) ≤ ka ∗ bk` (Z) ≤ kak` (Z)kbk` (Z) ≤ CkfkBpq(R ), j=0 where a ∈ `1(Z) since s ∈ (0, 1). Here we have used the discrete version of Young's inequality ka ∗ bk`r ≤ kak`1 kbk`r , which can be proved in the same way as for the usual convolution using Hölder's inequality. In order to prove the rst inequality in (4.29), we use that Z −j ϕj(Dx)f = (f(x − 2 z) − f(x))ψ(z) dz, Rn cf. (4.10). Therfore Z −j kϕj(Dx)fkLp(Rn) ≤ kf(. − 2 z) − fkLp(Rn)|ψ(z)| dz n ZR −j ≤ ωp(2 |z|; f)|ψ(z)| dz. Rn
44 and
1 ∞ ! q X 2sjqkϕ (D )fkq j x Lp(Rn) j=1 1 Z ∞ ! q X sjq −j q ≤ 2 ωp(2 |z|; f) |ψ(z)| dz n R j=1 1 Z Z ∞ q q ωp(t|z|; f) dt s ≤ C sq |z| |ψ(z)| dz Rn 0 (|z|t) t 1 Z ∞ q q Z ωp(t; f) dt s = C sq |z| |ψ(z)| dz, 0 t t Rn where we can estimate P∞ sjq −j q by the corresponding inte- j=0 2 ωp(2 |z|; f) gral by the same arguments as in the beginning of the proof. Finally, , which nishes the proof. kϕ0(Dx)fkLp(Rn) ≤ CkfkLp(Rn)
45 A Proof of the Marcinkiewicz Interpolation The- orem
Proof of Theorem 1.8 if , n n : p0 = q0 = 1, p1 = q1 = r (U, µ) = (R , λ ) We note that the assumptions imply that
kfkq λ(t; T f) := |{x : |T f(x)| > t}| ≤ C (A.1) q tq for q = 1 and q = r if r < ∞ and kT fk∞ ≤ C∞kfk∞ if r = ∞. First we consider the case r < ∞. Let f ∈ Lp(Rn) and consider the distribution function λ(t; T f), t > 0, de ned as above. For given t > 0 we de ne f = f1 + f2 by ( f(x) if |f(x)| > t, f1(x) = 0 else.
Then 1 n and r n since f1 ∈ L (R ) f2 ∈ L (R ) Z Z p 1−p p |f1(x)| dx = |f1(x)| |f1(x)| dx ≤ Cp(t)kfkp Rn Rn and similarly Z r p |f2(x)| dx ≤ Cr,p(t)kfkp. Rn Now, since |T f(x)| ≤ |T f1(x)| + |T f2(x)|, we have
{x : |T f(x)| > t} ⊆ {x : |T f1(x)| > t/2} ∪ {x : |T f2(x)| > t/2} . Therefore
λ(t; T f) ≤ λ(t/2; T f1) + λ(t/2; T f2) Z r Z C1 Cr r ≤ |f1(x)| dx + r |f2(x)| dx t/2 Rn (t/2) Rn Z r r Z 2C1 2 Cr r = |f(x)| dx + r |f(x)| dx t |f(x)|>t t |f(x)|≤t where we have used the weak type (1, 1) and (r, r) estimate and (A.1). Next we use that Z Z ∞ |T f(x)|p dx = p tp−1λ(t; T f) dt, Rn 0 46 cf. e.g. [3, Theorem 8.16]. We combine this with the estimate before. To this end we calculate
Z ∞ Z Z Z |f| tp−1t−1 |f| dx dt = |f| tp−2 dt dx 0 |f|>t Rn 0 1 Z = |f||f|p−1 dx p − 1 Rn since p > 1 and similarly
Z ∞ Z Z Z ∞ tp−1t−r |f|r dx dt = |f|r tp−1−r dt dx 0 |f|≤t Rn |f| 1 Z = |f|r|f|p−r dx r − p Rn since p < r. Altogether
for all p n kT fkp ≤ Cpkfkp f ∈ L (R ).
Finally, if r = ∞, we assume for simplicity that C∞ = 1. Otherwise replace by −1 . Then we use the same splitting of as before, but cut at height T C∞ T f instead of . Hence t since . Therefore t/2 t |T f2(x)| ≤ 2 kT kL(L∞(Rn)) ≤ 1
{x : |T f(x)| > t} ⊆ {x : |T f1(x)| > t/2} and λ(t, T f) ≤ λ(t/2, T f1). The rest of the proof in this case is done as before having only the rst term.
47 B Proof of the Mikhlin Multiplier Theorem
For simplicity we will rst prove the theorem in the case and H0 = H1 = C n ∼ therefore m: R \{0} → L(C) = C. How to modify the proof for the general case will be discussed below. Note that (4.1) implies that
for all 2 n km(Dx)fkL2(Rn) ≤ kmkL∞(Rn)kfkL2(Rn) f ∈ L (R ). Hence 2 n . In order to prove Theorem 4.1, the main step m(Dx) ∈ L(L (R )) is to show that 1 n 1 n , which is the content of the m(Dx) ∈ L(L (R ),Lweak(R )) next lemma. Once this is proved, p n for all by m(Dx) ∈ L(L (R )) 1 < p ≤ 2 the Marcinkiewicz interpolation theorem and the case 2 < p < ∞ will follow by duality.
Lemma B.1 Let M ≡ m(Dx) be as in Theorem 4.1. Then
Ckfk 1 n |{x ∈ n : |Mf(x)| > t}| ≤ L (R ) for all t > 0, f ∈ L1( n) (B.1) R t R for some C > 0 independent of t > 0. In order to prove Lemma B.1, an essential ingredient will be the following kernel estimate:
Proposition B.2 Let m: Rn \{0} → C be as in Theorem 4.1. Then there is a locally integrable, continuously dierentiable function k : Rn \{0} → C such that Z m(Dx)f(x) = k(x − y)f(y) dy for all x 6∈ supp f, (B.2) Rn for all f ∈ L2(Rn) with compact support which satis es |k(z)| ≤ C|z|−n |∇k(z)| ≤ C|z|−n−1 (B.3) for all z 6= 0. We postpone the proof of Proposition B.2 to the end of this section.
48 Corollary B.3 Let Q be a cube and a ∈ L1(Rn) with supp a ⊆ Q and R . Then there is a constant independent of and Q a(x) dx = 0 C > 0 a Q such that Z |m(Dx)a(x)| dx ≤ Ckak1, Rn\Qe √ √ where Qe = Q2 n denotes the cube with same center as Q and 2 n times the side-length of Q. √ 2 n Proof: Let x0 denote the center of Q. Then x 6∈ Qe = Q and y ∈ Q implies |x − x0| > 2|y − x0|. Therefore Z Z |m(Dx)a(x)| dx = |k ∗ a(x)| dx Rn\Qe Rn\Qe Z Z ≤ |k(x − y) − k(x − x0)| dx|a(y)| dy Q |x−x0|>2|y−x0| Z ≤ C |a(y)| dy, Q since R provided that Q a(x) dx = 0 Z n |k(x − y) − k(x)|dx ≤ C for all y ∈ R . (B.4) |x|>2|y| In order to prove the latter estimate, we use
Z 1 k(x − y) − k(x) = − y · ∇k(x − ty) dt. 0 If |x| > 2|y|, then
−n−1 |k(x − y) − k(x)| ≤ sup |∇xk(x − ty)||y| ≤ C|x| |y| t∈[0,1] since 1 for all . Hence |x − ty| ≥ 2 |x| t ∈ [0, 1] Z Z |k(x − y) − k(x)| dx ≤ C |x|−n−1 dx|y| ≤ C0 |x|>2|y| |x|>2|y| uniformly in y 6= 0.
49 Proposition B.4 Let f ∈ L1(Rn) be continuous and let t > 0. Then there are cubes , , with disjoint interior and parallel to the axis such Qj j ∈ N ⊆ N0 that 1. 1 Z t < |f(y)| dy ≤ 2nt for all j ∈ N. (B.5) |Qj| Qj 2. for (almost) all S . |f(x)| ≤ t x 6∈ j∈N Qj Proof: In the following , , denotes the set of all dyadic cubes with Dk k ∈ Z side length 2−k meaning the collection of all (closed) cubes Q with corners on neighboring points of the lattice 2−k n. Moreover, we set D = S D . Z k∈Z k Let 0 for given be the set of all satisfying the condition Ct t > 0 Q ∈ D 1 Z t < |f(x)| dx |Q| Q and let be the subset of all 0 that are maximal with respect to Ct Q ∈ Ct inclusion in 0. Every 0 is contained in some 0 since −1 Ct Q ∈ Ct Q ∈ Ct |Q| ≤ t kfk1 for all 0. Next, if and 0 , then by the Q ∈ Ct Q ∈ Ct ∩ Dk Q ⊂ Q ∈ Dk−1 maximality of we have 0 0, i.e., Q Q 6∈ Ct 1 Z 0 |f(x)| dx ≤ t. |Q | Q0 Moreover, since |Q0| = 2n|Q|, we get 1 Z 2n Z n for all t < |f(x)| dx ≤ 0 |f(x)| dx ≤ 2 t Q ∈ Ct. |Q| Q |Q | Q0 Hence , where are non-overlapping and Ct = {Qj : j ∈ N} Qj, j ∈ N ⊆ N0, (B.5) is satis ed for all j ∈ N. Now let n S . If , then 1 R for every F := R \ j∈N Qj x ∈ F |Q| Q f(y) dy ≤ t Q ∈ D such that x ∈ Q. Hence, choosing a sequence of cubes Qk ∈ Dk with x ∈ Qk, we obtain Z 1 |f(x)| = lim f(y) dy ≤ t for all x ∈ F. k→∞ |Qk| Qk
50 Remark B.5 The latter proposition holds for a general f ∈ L1(Rn) if one applies Lebesgue's dierentiation theorem in the last step of the proof.
Proof of Lemma B.1: First of all, since ∞ n is dense in 1 n , it C0 (R ) L (R ) is enough to consider ∞ n . Moreover, let be xed and let f ∈ C0 (R ) t > 0 , be the cubes due to Proposition B.4, S and Qj j ∈ N ⊆ N0 Ω = j∈N Qj F = Rn \ Ω. We de ne g, b ∈ L1(Rn) ∩ L2(Rn) by ( f(x) if x ∈ F g(x) = 1 R f(y) dy if x ∈ Qj |Qj | Qj and b(x) = f(x) − g(x). Note that this implies
1. |g(x)| ≤ 2nt almost every in Rn, 2. b(x) = 0 for every x ∈ F and R b(x) dx = 0 for each j ∈ N. Qj Then
|{x : |Mf(x)| > t}| ≤ |{x : |Mg(x)| > t/2}| + |{x : |Mb(x)| > t/2}| and it is su cient to estimate each term separately. In order to estimate Mg, we use that |g(x)| ≤ 2nt for almost every x ∈ Rn, for , , and that 2 n . More f(x) = g(x) x ∈ F t|Ω| ≤ kfk1 M ∈ L(L (R )) precisely, 4 Z 4 Z |{x : |Mg(x)| > t/2}| ≤ |Mg(x)|2 dx ≤ kmk |g(x)|2 dx t2 t2 ∞ Z −2 2 ≤ Ckmk∞t t|f(x)| dx + t |Ω| F −1 ≤ Ct kfk1 In order to estimate , we apply Corollary B.3 to and T b bj(x) := b(x)χQj (x) conclude Z Z |Mbj(x)| dx ≤ Ckbjk1 ≤ 2C |f(x)| dx n R \Qej Qj √ where 2 n. On the other hand, since 2 n , P and Qej = Qj b ∈ L (R ) j∈N bj therefore P converge in 2 n to and , respectively. Hence j∈N Mbj L (R ) b T b X |Mb(x)| ≤ |Mbj(x)| almost everywhere j∈N
51 and Z X Z |Mb(x)| dx ≤ 2C |f(x)| dx ≤ 2Ckfk1, n R \Ωe j∈N Qj where S . Finally, Ωe = j∈N Qej 2 Z C |{x : |Mb(x)| > t/2}| ≤ |Ωe| + |Mb(x)| dx ≤ kfk1, t Rn\Ωe t where we have used that √ Z ˜ X n X C X C |Ω| ≤ |Qej| ≤ (2 n) |Qj| ≤ |f(x)| dx ≤ kfk1 t t j∈N j∈N j∈N Qj This nishes the proof of (B.1).
Proof of Theorem 4.1: Since and because of km(Dx)kL(L2(Rn)) ≤ kmk∞ (B.1), we can apply the Marcinkiewicz interpolation theorem to conclude p n for all . The statement for follows m(Dx) ∈ L(L (R )) 1 < p ≤ 2 2 < p < ∞ by duality since Z Z 1 ˆ m(Dx)f(x)g(x) dx = n m(ξ)f(ξ)gˆ(ξ) dξ Rn (2π) Rn Z 1 ˆ = n f(ξ)m(ξ)ˆg(ξ) dξ (2π) Rn Z = f(x)m(Dx)g(x) dx Rn for all ∞ n due to (4.1), where satis es the same f, g ∈ C0 (R ) m(ξ) = m(ξ) estimates as . Therefore 0 p0 n if m(ξ) m(Dx) = m(Dx) ∈ L(L (R )) 2 < p < ∞ by the rst part.
It remains to prove Proposition B.2. To this end we will use the so-called Littlewood-Paley partition of unity, which is nowadays a standard tool in the theory of function spaces and harmonic analysis. It is frequently used to analyze mapping properties of certain operators. Usually a Littlewood-Paley or dyadic partition of unity on n is a decomposition of unity , R \{0} ϕj(ξ) j ∈ Z of Rn \{0} such that n j j for all supp ϕj(ξ) ⊆ {ξ ∈ R : c2 ≤ |ξ| ≤ C2 } j ∈ Z.
52 Here are some suitable xed numbers often chosen to be 1 c, C > 0 c = 2 ,C = 2, which we will assume in the following. Such a partition can be easily constructed by choosing some non-negative ∞ n such that ψ ∈ C0 (R ) ψ(ξ) > 0 if and only if 1 . Then de ning −j , we have 2 < |ξ| < 2 ψj(ξ) := ψ(2 ξ) j ∈ Z X Φ(ξ) = ψj(ξ) > 0 for all ξ 6= 0, j∈Z where we note that for each ξ 6= 0 the sum above contains at most two non-vanishing terms. Hence
−1 ϕj(ξ) = Φ(ξ) ψj(ξ) de nes a decomposition of unity with the desired properties. Moreover, in −j −j this case ϕj(ξ) = ϕ0(2 ξ) since Φ(2 ξ) = Φ(ξ), which implies that α α −|α|j (B.6) |∂ξ ϕj(ξ)| ≤ Ck∂ξ ϕ0kL∞(Rn)2 for all n and . α ∈ N0 j ∈ Z The idea of the proof of Proposition B.2 is to decompose X m(ξ) = mj(ξ), ξ 6= 0, j∈Z where mj(ξ) = ϕj(ξ)m(ξ). Then
X α |∂αm (ξ)| ≤ |∂α−βϕ (ξ)||∂βm(ξ)| ≤ C2−j|α| (B.7) ξ j β ξ j ξ 0≤β≤α
−j−1 −j+1 because of (4.2), (B.6), and since 2 ≤ |ξ| ≤ 2 on supp ϕj. For each part mj(ξ), we have Z −1 h ˆ i mj(Dx)f = F mj(ξ)f(ξ) = kj(x − y)f(y) dy, Rn where −1 0 n kj(x) = F [mj](x) ∈ C (R ) since 1 n . Here we have used that mj(ξ) ∈ L (R )
n 1 n F[f ∗ g](ξ) = fˆ(ξ)ˆg(ξ) for all ξ ∈ R , f, g ∈ L (R ).
53 Hence formally X Z m(Dx)f = kj(x − y)f(y) dy, n j∈Z R where it remains to show that the sum on the right-hand side converges for x 6∈ supp f and that X k(z) = kj(z), z 6= 0, j∈Z converges to a function satisfying (B.3). To this end, we need some suitable uniform estimates of kj(z). These are a consequence of the following simple lemma:
Lemma B.6 Let and let n be an -times dierentiable N ∈ N0 g : R → C N function, , with compact support. Then N ∈ N0 −1 −N β (B.8) |F [g](x)| ≤ CN | supp g||x| sup k∂ξ gkL∞(Rn) |β|=N uniformly in x 6= 0 and g, where CN depends only on n, N. Proof: Let n with . Then β ∈ N0 |β| = N β −1 −1 β (−ix) F [g] = F [∂ξ g] and therefore
β −1 β β |x ||F [g](x)| ≤ Ck∂ξ gkL1(Rn) ≤ C| supp g| sup k∂ξ gkL∞(Rn). |β|=N Since n with was arbitrary, β ∈ N0 |β| = N N −1 β |x| |F [g](x)| ≤ CN | supp g| sup k∂ξ gkL∞(Rn) |β|=N for all x ∈ Rn, which completes the proof.
Corollary B.7 Let m be as in the assumptions of Theorem 4.1 and let , , where , is the dyadic decomposition mj(ξ) = m(ξ)ϕj(ξ) j ∈ Z ϕj, j ∈ Z of unity of n as above. Then −1 satis es R \{0} kj(x) := Fξ7→x[mj] α j(n+|α|−M) −M (B.9) |∂z kj(z)| ≤ Cα,n,m2 |z| for all , , n for some constant independent z 6= 0 M = 0, . . . , n+2 α ∈ N0 Cα,n,m of j ∈ Z, z 6= 0.
54 Proof: First of all,
α −1 α ∂z kj(z) = F [(iξ) mj(ξ)] .
j−1 j+1 Using (B.7) and 2 ≤ |ξ| ≤ 2 on supp ϕj, one easily veri es
β α j(|α|−|β|) |∂ξ ((iξ) mj(ξ)) | ≤ Cα,β2
α for all |β| ≤ n + 2. Hence, applying Lemma B.6 to (iξ) mj with N = M = 0, . . . , n + 2, we obtain
α j(|α|−M) −M 0 j(n+|α|−M) −M |∂z kj(z)| ≤ Cn,m| supp ϕj|2 |z| ≤ Cn,m2 |z| , which nishes the proof.
Proof of Proposition B.2: Firstly, we will show that P ∂αk (z), |α| ≤ 1 j∈Z z j converges absolutely and uniformly on every compact subset of Rn \{0} to a function α satisfying (B.3). The main idea of the proof is to split for ∂z k(z) given z 6= 0 the sum P k (z) into the two parts j∈Z j X α and X α ∂z kj(z) ∂z kj(z) 2j ≤|z|−1 2j >|z|−1 and to show convergence and the estimate (B.3) separately. For the rst sum we use (B.9) with |α| ≤ 1 and M = 0. Then
X α X j(n+|α|) 0 −n−|α| |∂z kj(z)| ≤ C 2 ≤ C |z| j≤ld |z|−1 j≤ld |z|−1 where ld denotes the logarithm with respect to basis 2. For the second sum we apply (B.9) with |α| ≤ 1 and M = n + |α| + 1 and obtain
X α X −j −n−|α|−1 0 −n−|α| |∂z kj(z)| ≤ 2 2 |z| ≤ C |z| . ld |z|−1 Hence P ∂αk (z) converges absolutely and uniformly on every closed sub- j∈Z z j set of Rn \{0} to a function k(z) that satis es (B.3) for all |α| ≤ 1. Finally, it remains to show that k(z) satis es (B.2). First of all, ˆ X ˆ ∞ n m(ξ)f(ξ) = mj(ξ)f(ξ), f ∈ C0 (R ) j∈Z 55 since P ϕ (ξ) is locally nite. Moreover, since |m (ξ)| is uniformly bounded j∈Z j j w.r.t. ξ and fˆ(ξ) ∈ L2(Rn), the sum on the right-hand side converges in L2(Rn) to the left-hand side by Lebesgue's theorem on dominated conver- gence. Hence X X Z m(Dx)f = lim mj(Dx)f = lim kj(x − y)f(y) dy, k→∞ k→∞ n |j|≤Nk |j|≤Nk R in 2 n and almost everywhere for every ∞ n and some L (R ) f ∈ C0 (R ) Nk → ∞ as k → ∞ since Fourier transformation is bounded operator from L2(Rn) to L2(Rn). Therefore it only remains to interchange the summation and inte- gration in last term above provided that x 6∈ supp f. But this can be done since P k (z) converges absolutely and uniformly on every closed subset j∈Z j of Rn \{0} as shown above. Hence (B.2) follows. Comments on the proof of the vector-valued case: In the case that n satis es the assumptions of Theorem 4.1 for general m: R \{0} → L(H0,H1) Hilbert spaces H0,H1, we still conclude immediately that 2 n 2 n m(Dx): L (R ; H0) → L (R ; H1) is a bounded operator due Plancharel theorem.5 The rest of the proof can be modi ed in a straight-forward manner to the operator-valued case since all arguments are based on direct estimates involving only the size of f(x), m(ξ), and k(z). One just has to replace the pointwise absolute value |.| by the corresponding norms, i.e., , , and . Finally, we note k.kH0 k.kH1 k.kL(H0,H1) that the Marcinkiewicz interpolation theorem holds for general X-valued Lp- spaces for general Banach spaces X since its proof is only based on arguments involving the size of the function. 5 Actually, this is the only step in the proof, where it is needed that H0,H1 are Hilbert space and not general Banach spaces. 56 References [1] J. Bergh and J. Löfström. Interpolation Spaces. Springer, Berlin - Hei- delberg - New York, 1976. [2] Alessandra Lunardi. Interpolation Theory. Appunti. Scuola Normale Superiore, Pisa, 1999. [3] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, third edition, 1987. [4] Luc Tartar. An introduction to Sobolev spaces and interpolation spaces, volume 3 of Lecture Notes of the Unione Matematica Italiana. Springer, Berlin, 2007. [5] H. Triebel. Interpolation Theory, Function Spaces, Dierential Opera- tors. North-Holland Publishing Company, Amsterdam, New York, Ox- ford, 1978. 57