Fine properties of functions: an introduction

Petru Mironescu April 30, 2005

2

Chapter 0

Introduction

These lecture notes, intended as support to an intensive course at Şcoala Normalǎ Superioarǎ din Bucureşti, cover classical properties of function spaces. There is no unity of subjects, but it is a course that I would have loved to entitle ”Some theorems in Analysis that fascinate me”. Beauty, which is a subjective matter, was the main guide in choosing the topics. The other point was the required background, that I wanted to be the normal one for a fourth year student: a good knowledge of standard measure theory (Radon-Nikodym and Hahn decomposition theorems, Riesz representation theorem), the standard theory of distributions and basics about Sobolev spaces. Two classical textbooks were the source of the presentation (which does not pretend to the originality):

[1] Elias M. Stein, Harmonic Analysis: real variable methods, orthogonality, and oscillatory integrals, Princeton University Press, 1993

[2] William P. Ziemer, Weakly Differentiable Functions, Springer, 1989 My hope is that, after tasting the results proved in these notes, the reader will want to take a closer look to these wonderful books, which are a must in the library of an analyst.

The notes are divided into three parts: Part I is introductory: all the (simple but) non standard tools required for further developments are proved in this part; the aim was to provide the reader a self contained text (apart for the prerequisites). We talk, in this part, about: the distribution function, Lorentz spaces, Hardy’s inequality, Marcinkiewicz’s interpolation theorem, coverings of sets with balls and cubes (Vitali’s and Whitney’s lemma). The purpose was not to discuss in full generality these results: only the simplified versions required in the remaining part of the notes were presented. Part II could have been named: ”L^{1} and L^{∞} are wrong spaces”. We introduce and prove the most useful properties of two other spaces, the Hardy space and the bounded mean oscillation space. (The first one is slightly smaller than L^{1}, the second one slightly larger than L^{∞}.) We also try to briefly explain why these spaces are ”good” by considering a regularity problem in PDE’s, for which these spaces are more appropriate than L^{1} and L^{∞}. The presentation is very much influenced by [1]. Part III could be briefly described as follows: ”Young’s inequality is wrong”. Recall that Young’s inequality asserts that, if f ∈ L^{p} and g ∈ L^{q}, then f ∗ g ∈ L^{r}, where 1/p + 1/q = 1 + 1/r. The main result of this part is that we can weaken the assumption on, say, f, and yet obtain a better information concerning f ∗ g. This part follows essentially [2].

Enjoy the reading!

Part I Basic Tools

5

The distribution function

Throughout this course, we consider on ℝ^{N} the usual Lebesgue measure dx. The measure of a set A ⊂ ℝ^{N} will be simply denoted by |A|. Let f : ℝ^{N} → ℂ be a measurable function. We consider the distribution function of f,

F (t) = |{ x ∈ ℝ^{N} ; |f(x)| > t }|.  (1.1)

Clearly, F : [0, ∞) → [0, ∞] is non in creasing and thus measurable. F is related to various norms of f via

Proposition 1. For 1 ≤ p < ∞ we have

a) ∥f∥_{L^{p}}^{p} = p ∫_{0}^{∞}t^{p−1}F (t)dt;

b) (Chebyshev’s inequality) F (t) ≤ \frac{∥f∥_{L^{p}}^{p}}{t^{p}}.

Proof. a) We have

∥f∥_{L^{p}}^{p} = ∫ |f(x)|^{p}dx = ∫ |f(x)| ; ∫ ; 0pt^{p−1}dtdx = p ∫_{0}^{∞}t^{p−1} ∫ ; {x; |f(x)| > t} dxdt = p ∫_{0}^{∞}t^{p−1}F (t)dt.

(1.2)

b) Chebyshev’s inequality follows from

∥f∥_{L^{p}}^{p} ≥  ∫ ; {x; |f(x)| > t} |f(x)|^{p}dx ≥  ∫ ; {x; |f(x)| > t} t^{p}dx = t^{p}F (t).  (1.3)

By copying the proof of a) above, we obtain the following

Proposition 2. Let Φ : [0, ∞) → [0, ∞), Φ ∈ C^{1}, be a non decreasing function s. t. Φ(0) = 0. Then

∫_{ℝ^{N}} Φ(|f(x)|)dx = ∫_{0}^{∞}Φ′(t)F (t)dt.  (1.4)

1.1 Lorentz spaces

One may read the property a) in Proposition 1 as ∥f∥_{L^{p}}^{p} = p∥tF^{1/p}(t)∥L^{p}((0, ∞); dt/t)^{p}. This

suggests a more general definition: a measurable function f belongs to the Lorentz space L^{p,q} (1 ≤ p < ∞, 1 ≤ q ≤ ∞) if

∥f∥_{L^{p,q}} = ∥tF^{1/p}(t)∥L^{q}((0, ∞); dt/t) < ∞.  (1.5)

Despite this notation, ∥ · ∥_{L^{p,q}} is not a norm (but almost: it is a quasi-norm). When q = p, the corresponding Lorentz space coincides with L^{p}. When q = ∞, the corresponding space L^{p,∞} is called the weak L^{p}, also denoted by L_{w}^{p} (the Marcinkiewicz space). Clearly, a function f belongs to L_{w}^{p} if and only if its distribution function F satisfies a Chebyshev type inequality:

F (t) ≤ \frac{C}{t^{p}} for each t > 0.

It is well known that there is no inclusion relation for the L^{p} spaces. However, for fixed p, the Lorentz spaces are monotonic in q:

Proposition 3. Let 1 ≤ q < r ≤ ∞. Then L^{p,q} ⊂ L^{p,r}.

Proof: Assume first that r = ∞. If f ∈ L^{p,q}, then

∥f∥_{L^{p,q}}^{q} = ∫_{0}^{∞}t^{q−1}F^{q/p}(t)dt ≥ ∫_{0}^{s} t^{q−1}F^{q/p}(t)dt ≥ ∫_{0}^{s} t^{q−1}F^{q/p}(s)dt = \frac{1}{q}s^{q}F^{q/p}(s).  (1.6)

Thus F^{1/p}(s) ≤ \frac{C}{s}, i.e. f ∈ L^{p,∞}.

Let now r < ∞ and let f ∈ L^{p,q}. Then, using Hölder’s inequality and the case r = ∞, we find that

∥tF^{1/p}(t)∥L^{r}((0, ∞); dt/t) ≤ ∥tF^{1/p}(t)∥L^{q}((0, ∞); dt/t)^{q/r}∥tF^{1/p}(t)∥L^{∞}((0, ∞); dt/t)^{1−q/r} < ∞. (1.7)

Incidentally, we proved the stronger statement

∥f∥_{L^{p,r}} ≤ C∥f∥_{L^{p,q}}, 1 ≤ p < ∞, 1 ≤ q ≤ r ≤ ∞.  (1.8)

We complete the scale of Lorentz spaces by setting L^{∞,q} = L^{∞} for all q. The above inequality, combined with the fact that ∥ · ∥_{p,q} is a quasi-norm yields immediately the following

Corollary 1. The inclusion L^{p,q} ⊂ L^{p,r} is continuous, 1 ≤ q < r ≤ ∞.

Remark 1. One should understand the Lorentz spaces as ”microscopic” versions of the L^{p} spaces. We mean that the properties of L^{p,q} are very close to those of L^{p}. Here is an example: if Ω is a bounded set in ℝ^{N}, one may define in an obvious way the spaces L^{p,q}(Ω). It is easy to prove that, if p_{1} < p_{2}, then L^{p_{2},q_{2}}(Ω) ⊂ L^{p_{1},q_{1}}(Ω) for all the possible values of q_{1}, q_{2}. This is exactly the inclusion relation we have for the standard L^{p} spaces.

10 CHAPTER 1. THE DISTRIBUTION FUNCTION

Elementary interpolation

Theorem 1. (Marcinkiewicz’ interpolation theorem; simplified version) Let 1 < q < ∞ and let T : L^{1} ∩ L^{q}(ℝ^{N}) → D′ be linear and s. t.

∥T f∥_{L_{w}^{1}} ≤ C_{1}∥f∥_{L^{1}}, ∀ f  (2.1)

and

∥T f∥_{L_{w}^{q}} ≤ C_{q}∥f∥_{L^{q}}, ∀ f.  (2.2)

(In other words, T extends by density as a continuous operator from L^{1} into L_{w}^{1} and from L^{q} into L_{w}^{q}.) Then T is a continuous operator from L^{p} into L^{p}, for each 1 < p < q, i. e.

∥T f∥_{L^{p}} ≤ C∥f∥_{L^{p}}, ∀ f ∈ L^{1} ∩ L^{q}.  (2.3)

Before proceeding to the proof of the theorem, let us note that L^{q} embeds into L_{w}^{q}, and thus we have the following consequence, which is the form we usually make use of the above theorem

Corollary 2. Let 1 < q < ∞ and let T : L^{1} ∩ L^{q}(ℝ^{N}) → D′ be linear and s. t.

∥T f∥_{L_{w}^{1}} ≤ C_{1}∥f∥_{L^{1}}, ∀ f  (2.4)

and

∥T f∥_{L^{q}} ≤ C_{q}∥f∥_{L^{q}}, ∀ f.  (2.5)

Then T extends as a continuous operator from L^{p} into L^{p}, 1 < p < q.

Proof. Let t > 0 and let f ∈ L^{1}∩L^{q}. We are going to estimate the distribution function of T f. For

this purpose, we cut f at height t, i. e. we write f = f_{1} +f_{2}, where f_{1}(x) = \begin{cases}f(x), & if |f(x)| > t \\ 0, & otherwise \end{cases}

and f_{2} = f − f_{1}. Since T f = T f_{1} + T f_{2}, we have |T f| > t ⟹ |T f_{1}| > t/2 or |T f_{2}| > t/2, and thus

|\{|T f| > t\}| ≤ |\{|T f_{1}| > t/2\}| + |\{|T f_{2}| > t/2\}| ≤ \frac{2C_{1}}{t} ∥f_{1}∥_{L^{1}} + \frac{2^{q}C_{q}^{q}}{t^{q}} ∥f_{2}∥_{L^{q}}^{q}.  (2.6)

Therefore,

∥T f∥_{L^{p}}^{p} = p ∫ t^{p−1}|{|T f| > t}| ≤ 2pC_{1} ∫ t^{p−2}∥f_{1}∥_{L^{1}} + 2^{q}pC_{q}^{q} ∫ t^{p−q−1}∥f_{2}∥_{L^{q}}^{q}.  (2.7)

Next, if F is the distribution function of f, then the distribution function of f_{1} is \begin{cases}F (α), & if α ≥ t \\ F (t), & if α < t, \end{cases}

and the one of f_{2} is \begin{cases}0, & if α ≥ t \\ F (α) − F (t), & if α < t \end{cases}. Thus

∥f_{1}∥_{L^{1}} = tF (t) + ∫_{t}^{∞}F (α)dα and ∥f_{2}∥_{L^{q}}^{q} = q ∫_{0}^{t} α^{q−1}F (α)dα − t^{q}F (t).  (2.8)

By combining (2.6) and (2.8) and applying Fubini’s theorem (to interchange the order of integration over α and t), we find that

∥T f∥_{L^{p}}^{p} ≤ p(\frac{2C_{1}}{p − 1} + \frac{2^{q}C_{q}^{q}}{q − p})∥f∥_{L^{q}}^{q}.  (2.9)

Remark 2. We see that the estimate we obtain for the norm of T from L^{p} into L^{p} blows up as p → 1 or p → q. This is not a weakness of the proof. If this norm remains bounded as, say, p → 1, then T must continuous from L^{1} into L^{1}, which may not be the case.

There is a way to improve the estimate (2.9): instead of cutting f at height t, we cut it at height at, where a > 0 is fixed. The above computations yield this time :

∥T f∥_{L^{p}}^{p} ≤ p∥f∥_{L^{p}}^{p}(\frac{2C_{1}}{p − 1}a^{1−p} + \frac{2^{q}C_{q}^{q}}{q − p}a^{q−p}).  (2.10)

Optimizing the above r. h. s. over a > 0 (it is minimal when a = 1/2(C_{1}/C_{q})^{1/(q−1)}), we find the following

Theorem 2. With the notations and under the hypotheses of the preceding theorem, let θ ∈ (0, 1)

be the (unique) number s. t. \frac{1}{p} = \frac{θ}{1} + \frac{1 − θ}{q}. Then the norm of T from L^{p} into L^{p} satisfies

∥T ∥_{L^{p}→L^{p}} ≤ c_{p,q}∥T ∥_{L^{1}→L_{w}^{1}}^{θ}∥T ∥_{L^{p}→L_{w}^{p}}^{1−θ}.  (2.11)

This conclusion is reminiscent from the one of the Riesz-Thorin convexity theorem.

Hardy’s inequality

We present two (equivalent) forms of Hardy’s inequality. The first one generalizes the usual (and historically first) Hardy’s inequality ∫_{0}^{∞} \frac{F^{2}(x)}{x^{2}}dx ≤ 4 ∫_{0}^{∞}(F′(x))^{2}dx, F ∈ C_{0}^{∞}((0, ∞)). The second one will be needed later in the study of the Lorentz spaces. Theorem 3. (Hardy) Let 1 ≤ p < ∞ and r > 0 and let f : (0, ∞) → ℝ.

a) If ∫_{0}^{∞}|f(x)|^{p}x^{p−r−1}dx < ∞, then f ∈ L_{loc}^{1}([0, ∞)).

b) With F (x) = ∫_{0}^{x} f(t)dt, we have

∫_{0}^{∞}|F (x)|^{p}x^{−r−1}dx ≤ (\frac{p}{r})^{p} ∫_{0}^{∞}|f(x)|^{p}x^{p−r−1}dx.  (3.1)

Proof. In view of the conclusions, we may assume f ≥ 0. In this case, it suffices to prove (3.1). We want to apply Jensen’s inequality in order to estimate the integral ( ∫_{0}^{x} f(t)dt)^{p}. We consider, on (0, x), the normalized measure μ = \frac{r}{p}x^{−r/p}t^{r/p−1}dt. Then (with u ↦ u^{p} playing the role of the convex function)

( ∫_{0}^{x} f(t)dt)^{p} = (\frac{p}{r})^{p}x^{r}( ∫_{0}^{x} f(t)t^{1−r/p}dμ)^{p} ≤ (\frac{p}{r})^{p}x^{r} ∫_{0}^{x}[f(t)t^{1−r/p}]^{p}dμ,  (3.2)

14 CHAPTER 3. HARDY’S INEQUALITY

which yields

( ∫_{0}^{x} f(t)dt)^{p} ≤ (\frac{p}{r})^{p−1}x^{r(1−1/p)} ∫_{0}^{x} f^{p}(t)t^{p−r−1+r/p}dt.  (3.3)

Thus

∫_{0}^{∞}F^{p}(x)x^{−r−1}dx ≤ (\frac{p}{r})^{p−1} ∫_{0}^{∞}x^{−1−r/p} ∫_{0}^{x} f^{p}(t)t^{p−r−1+r/p}dtdx = (\frac{p}{r})^{p} ∫_{0}^{∞}f^{p}(t)t^{p−r−1}dt, (3.4)

by Fubini’s theorem. Corollary 3. (Hardy) With 1 ≤ p < ∞ and r > 0, we have

∫_{0}^{∞} | ∫_{x}^{∞}f(t)dt|^{p}x^{r−1}dx ≤ (\frac{p}{r})^{p} ∫_{0}^{∞}|f(x)|^{p}x^{p+r−1}dx.  (3.5)

Proof. We may assume f ≥ 0. We apply the preceding theorem to the map g given by g(t) = t^{−2}f(t^{−1}) and find that

∫_{0}^{∞}( ∫_{0}^{x} t^{−2}f(t)dt)^{p}x^{−r−1}dx ≤ (\frac{p}{r})^{p} ∫_{0}^{∞}f^{p}(x^{−1})x^{−p−r−1}dx = (\frac{p}{r})^{p} ∫_{0}^{∞}f^{p}(y)y^{p+r−1}dx. (3.6)

We next perform, in the integral ∫_{0}^{x} t^{−2}f(t)dt, the substitution t = s^{−1}, next we substitute, in the first integral in (3.6), y = x^{−1}, and obtain the desired result.

Coverings

4.1 The Vitali covering lemma (simplified version)

Let F be a finite family of balls in ℝ^{N}.

Lemma 1. (Vitali’s lemma). F contains a subfamily F′ of disjoint balls such that

∑_{B∈F′} |B| ≥ C| ⋃_{B∈F} B|.

Here, C depends only on the space dimension N, not on the family F.

Proof. Let B_{1} be the largest ball in F. Let B_{2} be the largest ball in F that does not intersect B_{1}, B_{3} the largest ball in F that does not intersect neither B_{1} nor B_{2}, and so on. Let F′ = { B_{1}, B_{2}, ... }. Note that, for each B ∈ F, there is some j s. t. B∩B_{j} ≠ ∅. For each ball B in F′, let B̃ be the ball having the same center as B and the radius thrice the one of B. We claim that ⋃_{B∈F} B ⊂ ⋃_{B∈F′} B̃.

Indeed, let B ∈ F and let j be the smallest integer such that B ∩ B_{j} ≠ ∅. Since B ∩ B_{j−1} = ∅, the radius of B is at most the one of B_{j}, for otherwise we would have picked B instead of B_{j} at step j in the construction of F′. Since B ∩ B_{j} ≠ ∅, we find that B ⊂ B̃_{j}. It follows that

| ⋃_{B∈F} B| ≤ | ⋃_{B∈F′} B̃| ≤ 3^{N} ∑_{B∈F′} |B|,  (4.1)

which is the desired result with C = 3^{−N}.

4.2 Whitney’s covering

Throughout this section, the norm we consider on ℝ^{N} is the ∥ · ∥_{∞} one.

Let F ⊂ ℝ^{N} be a non empty closed set and let O = ℝ^{N} \ F. If C is a (closed) cube, let l(C) be its size, i. e. the length of its edges.

Lemma 2. (Whitney’s covering lemma) There is a family F of closed cubes s. t.

a) O = ⋃_{C∈F} C;

b) distinct cubes in F have disjoint interiors;

c) c^{−1}l(C) ≤ dist(C, F) ≤ c l(C) for each C ∈ F.

Here, c depends only on N.

Proof. We may assume that 0 ∈ F. For j ∈ ℤ, let F_{j} be the grid of cubes of size 2^{j}, with sides parallel to the coordinate axes, s. t. 0 be one of the vertices. Note that each cube C ∈ F_{j} is contained in exactly one predecessor C′ ∈ F_{j+1}. In addition, each cube has an ancestor containing 0. Thus, the non increasing sequence dist(C, F), dist(C′, F), dist(C′′, F), …, becomes 0 starting with a certain range. We throw away all the cubes contained in ⋃_{j} F_{j} s. t. dist(C, F) ≤ l(C) and

call F the family of all kept cubes C s. t. their predecessors C′ were thrown away. Note that, by definition, if C ∈ F, then dist(C, F) > l(C), while there are some y ∈ C′ and z ∈ F s. t. ∥y − z∥_{∞} ≤ 2l(C). Let x ∈ C; then dist(C, F) ≤ ∥x − z∥_{∞} ≤ 3l(C), so that c) holds with c = 3. Let x ∈ O. If j is sufficiently close to −∞, then we have dist(C, F) > l(C) whenever C ∈ F_{j} and x ∈ C. Pick any such j and C and set k = sup{l ∈ ℕ ; dist(C^{(l)}, F) > l(C^{(l)})}. Then k is finite and it is clear from the definition that x ∈ C^{(k)} ∈ F. Thus a) holds. Finally, if C, D ∈ ⋃_{j} F_{j} are distinct cubes s. t. C^{∘} ∩ D^{∘}≠ ∅, then one of these cubes is contained

in the other one. Assume, e. g., that C ⊂ D. Then C′ ⊂ D. Therefore, we can not have at the same time C ∈ F and D ∈ F, for otherwise l(C′) ≥ dist(C′, F) ≥ dist(D, F) > l(D) ≥ l(C′).

For a cube C, let C_{∗} be the cube concentric with C and of size three halves the one of C.

Proposition 4. Let F, O and F be as in the proof of the above lemma. Then:

a) O = ⋃_{C∈F} C^{∘}_{∗};

b) we have d^{−1}l(C_{∗}) ≤ dist(C_{∗}, F) ≤ d l(C_{∗}) for each C ∈ F;

c) if x ∈ C_{∗}, then e^{−1} dist(x, F) ≤ dist(C_{∗}, F) ≤ e dist(x, F);

d) each point x ∈ O belongs to at most M cubes C_{∗}.

Here, d, e and M depend only on N.

Proof. On the one hand, we have dist(C_{∗}, F) ≤dist(C, F) ≤ 3l(C) = 2l(C_{∗}). On the other hand, if x ∈ C_{∗}, then there is some y ∈ C s. t. ∥x − y∥_{∞} ≤ 1/2l(C). In addition, dist(y, F) > l(C).

Thus dist(x, F) ≥ 1/2l(C) = 1/3l(C_{∗}). Thus b) holds with d = 3. Property c) is a straightforward consequence of b). Clearly, C_{∗} ⊂ O, by b), which implies a) with the help of a) of Whitney’s lemma. If x ∈ C_{∗}, then (by b) and c)) 3/2l(C) = l(C^{∗}) ≥ (de)^{−1} dist(x, F), and therefore C_{∗} ⊂ B(x, r), with r = de dist(x, F). Thus, if k is the number of cubes C_{∗} containing x, we have

r^{N} = |B(x, r)| ≥ | ⋃_{C_{∗}∩F ≠∅} C_{∗}| ≥ | ⋃_{C_{∗}∩F ≠∅} C| = ∑_{C_{∗}∩F ≠∅} |C| ≥ k(2/3(de)^{−1}dist(x, F))^{N},

whence conclusion d).

Proposition 5. With the above notations, there is, in O, a partition of the unit 1 = ∑_{C∈F} φ_{C} s. t.:

a) for each C, supp φ_{C} ⊂ C_{∗};

b) |∂^{α}φ_{C}(x)| ≤ C_{α}|C|^{−|α|/N} ≤ C′_{α}dist(x, F)^{−|α|} when x ∈ supp φ_{C}.

Here, the constants C_{α} do not depend on O, x and C.

Proof. Fix a function ζ ∈ C^{∞}(ℝ^{N}; [0, 1]) s. t. ζ = 1 in B(0, 1/2) and supp ζ ⊂ B(0, 3/4). If C ∈ F is of size 2l and center x, set ζ_{C} = ζ((· − x)/l). Note that supp ζ_{C} ⊂ C_{∗} and that ζ_{C} = 1 in C. Moreover, |∂^{α}φ_{C}| ≤ C_{α}l^{−|α|}. Set φ = ∑ ζ_{C}, which satisfies 1 ≤ φ ≤ M, by a) in Whitney’s lemma and d) in the above proposition. Finally, set φ_{C} = ζ_{C}/φ. Properties a) and b) follow immediately by combining the conclusions of Whitney’s lemma and of the above proposition.

18 CHAPTER 4. COVERINGS

The maximal function

If f is locally integrable, we define the (uncentered) maximal function of f,

Mf(x) = sup\{ \frac{1}{|B|} ∫_{B} |f(y)|dy; B ball containing x \}.  (5.1)

In this definition, one may consider cubes instead of balls. This will affect the value of Mf, but not its size. E.g., if we consider instead

M′f(x) = sup\{ \frac{1}{|Q|} ∫_{Q} |f(y)|dy; Q cube containing x \},  (5.2)

then we have C^{−1}M′f ≤ Mf ≤ CM′f, where C is the ratio of the volumes of the unit cube and of the unit ball. Thus the integrability properties of Mf remain unchanged if we change the definition. Similarly, one may consider balls centered at x; this definition yields the centered maximal function. A basic property of Mf is that it is lower semi continuous, i.e. the level sets { x; Mf(x) > t } are open.

5.1 Maximal inequalities

When f ∈ L^{∞}, we clearly have Mf ∈ L^{∞}. However, it is not obvious whether, for 1 ≤ p < ∞ and f ∈ L^{p}, the maximal function has some integrability properties or even whether it is finite a.e.

Theorem 4. (Hardy-Littlewood-Wiener) Let 1 ≤ p ≤ ∞ and f ∈ L^{p}. Then:

a) Mf is finite a.e.;

b) if 1 < p ≤ ∞, then Mf ∈ L^{p} and ∥Mf∥_{L^{p}} ≤ C∥f∥_{L^{p}};

c) if p = 1, then Mf ∈ L_{w}^{1} and ∥Mf∥_{L_{w}^{1}} ≤ C∥f∥_{L^{1}}, i.e. |\{ x; Mf(x) > t \}| ≤ \frac{C∥f∥_{L^{1}}}{t} for each

t > 0.

Here, C denotes a constant independent of f.

Proof. When p = ∞, the statement is clear and we may take C = 1. Let next p = 1. We fix some t > 0. Let O = { x; Mf(x) > t }, which is an open set. Thus |O| = sup{ |K|; K compact ⊂ O }. Let K be any compact in O. From the definition of Mf, for each x ∈ K there is some ball B

containing x such that \frac{1}{|B|} ∫_{B} |f(y)|dy > t. These balls cover K, so that we may extract a finite

covering. Using Vitali’s lemma, we may find a finite family F′ = {B_{j}} such that

B_{j} ∩ B_{k} = ∅ for j ≠ k, \frac{1}{|B_{j}|} ∫ ; B_{j} |f(x)|dx > t, ∑_{j} |B_{j}| ≥ C|K|.  (5.3)

Thus

∥f∥_{L^{1}} ≥ ∫ ; ⋃_{j} B_{j} |f(x)|dx = ∑_{j} ∫ ; B_{j} |f(x)|dx ≥ t ∑_{j} |B_{j}| ≥ Ct|K|.  (5.4)

Taking the supremum over K in the above inequality, we find that |O| ≤ \frac{C∥f∥_{L^{1}}}{t}, i.e. the property c). Letting t → ∞, we find a) for p = 1. We next prove a) for 1 < p < ∞. Let f ∈ L^{p}. We split f as f = f_{1} + f_{2}, where f_{1}(x) =

\begin{cases}f(x), & if |f(x)| ≥ 1 \\ 0, & if |f(x)| < 1 \end{cases} and f_{2} = f − f_{1}. Then f_{1} ∈ L^{1} and f_{2} ∈ L^{∞}. Since Mf ≤ Mf_{1} + Mf_{2},

we obtain a). Finally, we prove b) for 1 < p < ∞. Let t > 0. We use a splitting of f similar to the above one:

f = f_{1} + f_{2}, with f_{1}(x) = \begin{cases}f(x), & if |f(x)| ≥ t/2 \\ 0, & if |f(x)| < t/2 \end{cases} and f_{2} = f − f_{1}. Then Mf_{2} ≤ ∥f_{2}∥_{L^{∞}} ≤ \frac{t}{2}. It

follows that Mf(x) > t ⇒ Mf_{1}(x) > \frac{t}{2}. Therefore

|\{ x; Mf(x) > t \}| ≤ |\{ x; Mf_{1}(x) > \frac{t}{2} \}| ≤ \frac{2C∥f_{1}∥_{L^{1}}}{t} ,  (5.5)

using c). We find that

∥Mf∥_{L^{p}}^{p} = p ∫_{0}^{∞}t^{p−1}|{ x; Mf(x) > t }|dt ≤ C ∫_{0}^{∞}t^{p−2}∥f_{1}∥_{L^{1}}dt.  (5.6)

5.2. LEBESGUE’S DIFFERENTIATION THEOREM 21

If F is the distribution function of f, then we have

∥f_{1}∥_{L^{1}} = ∫_{0}^{∞}|\{ x; |f_{1}(x)| > s \}|ds = \frac{t}{2}F (\frac{t}{2}) + ∞ ; ∫ ; t/2 F (s)ds.  (5.7)

Thus

∫_{0}^{∞}t^{p−2}∥f_{1}∥_{L^{1}}dt = \frac{1}{2} ∫_{0}^{∞}t^{p−1}F (\frac{t}{2})dt + ∫_{0}^{∞}t^{p−2} ∞ ; ∫ ; t/2 F (s)dsdt = Cp ∫_{0}^{∞}t^{p−1}F (t)dt = C∥f∥_{L^{p}}^{p}, (5.8)

by Fubini’s theorem.

The constant in the last line of computations equals \frac{2^{p−1}}{p} + \frac{1}{p(p − 1)}. We obtain the following

Corollary 4. For 1 < p ≤ 2 we have ∥Mf∥_{L^{p}} ≤ \frac{C}{p − 1}∥f∥_{L^{p}} for some constant C depending only on N. Remark 3. The maximal function is never in L^{1} (except when f = 0). Indeed, if f ≠ 0, there is

some R > 0 s. t. ∫_{B(0,R)} |f| > 0. Then, for |x| ≥ R, we have Mf(x) ≥ \frac{1}{|B(x, 2|x|)|} ∫_{B(0,R)} |f| ≥ \frac{C}{|x|^{N}}

and thus Mf ∉ L^{1}. Remark 4. By the above remark, given f ∈ L^{1} s. t. f ≠ 0, the best we can hope is that Mf ∈ L_{loc}^{1}. However, this may not be true. Indeed, let f : ℝ → ℝ, f(x) = \frac{1}{x ln^{2} x}χ_{[0,1/2]}. Then

f ∈ L^{1}. However, for x ∈ (0, 1/2), Mf(x) ≥ \frac{1}{x} ∫_{0}^{x} f(t)dt = \frac{1}{x| ln x|}, so that Mf ∉ L_{loc}^{1}.

5.2 Lebesgue’s differentiation theorem

Theorem 5. (Lebesgue) If f ∈ Lloc^{1}, then for a.e. x ∈ ℝ^{N} we have

22 CHAPTER 5. THE MAXIMAL FUNCTION

lim_{r→0} \frac{1}{|B(x, r)|} ∫ ; B(x, r) f(y)dy = f(x).

Proof. We start by recalling the following simple measure theoretic Lemma 3. (Borel-Cantelli) Let (A_{n}) be a sequence of measurable sets such that ∑_{n} |A_{n}| < ∞. Then |\overline{lim}A_{n}| = 0, where \overline{lim}A_{n} = ⋂_{n} ⋃_{m≥n} A_{m}.

Let f(x, r) = \frac{1}{|B(x, r)|} ∫ ; B(x, r) f(y)dy. The conclusion of the theorem being local, it suffices to

prove it with f replaced by fφ for any compactly supported smooth function φ. We may thus assume that f ∈ L^{1}. Let n ≥ 1 and let f_{n} be a smooth compactly supported function such that ∥f − f_{n}∥_{L^{1}} ≤ \frac{1}{2^{n}}. Let also g_{n} = f − f_{n}. Since f_{n} is uniformly continuous, there is some δ_{n} such that |f_{n}(x, r) − f_{n}(x)| ≤ \frac{1}{n} for r ≤ δ_{n} and x ∈ ℝ^{N}. Thus, if for some r ≤ δ_{n} we have |f(x, r) − f(x)| > \frac{2}{n}, then we must have |g_{n}(x, r) − g_{n}(x)| > \frac{1}{n}, so that either |g_{n}(x)| > \frac{1}{2n} or |g_{n}(x, r)| > \frac{1}{2n}. In the latter case, we have Mg_{n}(x) > \frac{1}{2n}. Therefore

\{ x; |f(x, r) − f(x)| > \frac{2}{n} for some r ≤ δ_{n} \} ⊂ A_{n} = \{ x; |g_{n}(x)| > \frac{1}{2n} or Mg_{n}(x) > \frac{1}{2n} \}. (5.9)

By the maximal and Chebysev’s inequalities, we find that |A_{n}| ≤ \frac{Cn}{2^{n}}. If x ∉ \overline{lim}A_{n}, then clearly lim_{r→0} f(x, r) = f(x). The theorem follows from the above lemma, since ∑_{n} \frac{n}{2^{n}} < ∞.

The same argument yields the following variants of the differentiation theorem: Theorem 6. If f ∈ Lloc^{1}, then for a.e. x ∈ ℝ^{N} we have

{lim \atop x ∈ Q, |Q| → 0} \frac{1}{|Q|} ∫_{Q} f(y)dy = f(x).  (5.10)

Here, we may choose the Q′s to be balls or cubes (or, more generally, balls for some norm).

Theorem 7. If f ∈ Lloc^{1}, then for a.e. x ∈ ℝ^{N} we have

{lim \atop x ∈ Q, |Q| → 0} \frac{1}{|Q|} ∫_{Q} |f(y) − f(x)|dy = 0.  (5.11)

Yet another variant is given by the following

Theorem 8. Let φ ∈ S be s. t. ∫ φ = 1. If f ∈ L^{p} for a p ∈ [1, ∞], then f ∗ φ_{t} → f a. e. as

t → 0.

Here, φ_{t}(x) = t^{−N}φ(x/t).

Proof. Assume first that p < ∞. We consider a sequence (f_{n}) ⊂ C_{0}^{∞} s. t. ∥f_{n} − f∥_{L^{p}} < 2^{−n}. For each n, we have f_{n} ∗ φ_{t} → f_{n} in S, and thus uniformly. Consequently, there exists a t_{n} s. t. |f_{n} ∗ φ_{t} − f_{n}| < 1/n if t < t_{n}. Using the inequality |f ∗ φ_{t}| ≤ CMf (see Corollary 7 below), we find as above, that

A_{n} = {x; |f∗φ_{t}(x)−f(x)| > 3/n for a t < t_{n}} ⊂ {x; M(f−f_{n})(x) > C/n or |f_{n}(x)−f(x)| > 1/n},

(5.12)

and thus |A_{n}| ≤ Cn^{p}2^{−n}; in particular, | lim sup A_{n}| = 0. As above, if x ∉ lim sup A_{n}, then we have f ∗ φ_{t}(x) → f(x) as t → 0.

Let now p = ∞. Let A be s. t. |A| = 0 and lim_{x∈B;|B|→0} \frac{1}{|B|} ∫_{B} |f(y) − f(x)|dy = 0 for x ∉ A. We fix

any x ∉ A; we will prove that the desired conclusion holds for such an x. Since x ∉ A, we have

lim_{t→0} t^{−N} ∫_{B(x,t)} |f(y) − f(x)|dy = 0. Let R > 0 be fixed. Since φ and f are essentially bounded, we

find that

|f∗φ_{t}(x)−f(x)| = | ∫ (f(y)−f(x))φ_{t}(x−y)dy| ≤ \frac{C}{t^{N}} ∫_{B(x,Rt)} |f(y)−f(x)|dy+\frac{C}{t^{N}} ∫_{\{|x−y|>Rt\}} |φ((x−y)/t)|,

(5.13)

which implies that

lim sup_{t→0} |f ∗ φ_{t}(x) − f(x)| ≤ \frac{C}{t^{N}} ∫_{\{|x−y|>Rt\}} |φ((x − y)/t)|dy = C ∫_{\{|y|>R\}} |φ(y)|dy.  (5.14)

If we let R → ∞ in this inequality, we find that lim_{t→0} |f(x) − f ∗ φ_{t}(x)| = 0.

We end this section with two simple consequences of the maximal inequalities and of the above theorem:

Corollary 5. Let f ∈ Lloc^{1}. Then Mf ≥ |f| a.e.

Corollary 6. Let 1 < p ≤ ∞. Then ∥f∥_{L^{p}} ≤ ∥Mf∥_{L^{p}} ≤ C∥f∥_{L^{p}}.

5.3 Pointwise inequalities for convolutions

Proposition 6. Let φ be such that |φ| ≤ g for some g ∈ L^{1}, g radially symmetric and non increasing with r = |x|. Then

|f ∗ φ(x)| ≤ ∥g∥_{L^{1}}Mf(x).  (5.15)

Proof. Since |f ∗ φ| ≤ |f| ∗ g, it suffices to prove the proposition for |f| and g. We start with a special case: we assume g to be piecewise constant; the general case will follow by approximation, using, e.g., the Beppo Levi theorem. We assume thus that there is a sequence of radii r_{1} < r_{2} < ... and a sequence of non negative numbers a_{1}, a_{2}, ... such that g = ∑_{k≥j} a_{k} on B(0, r_{j}). Then

∫_{ℝ^{N}} |f(x − y)||g(y)|dy = ∑_{j} a_{j} ∫ ; B(0, r_{j}) |f(x − y)|dy ≤ ∑_{j} a_{j}|B(0, r_{j})|Mf(x) = ∥g∥_{L^{1}}Mf(x),

(5.16)

which is the desired estimate.

Let φ ∈ S(ℝ^{N}). As a consequence of the above proposition, we derive the following

Corollary 7. We have

|f ∗ φ_{t}(x)| ≤ CMf(x).  (5.17)

Here, C depends only on φ, not on t or f.

Proof. Since φ ∈ S, we have |φ(x)| ≤ g(x) = \frac{C}{1 + |x|^{N+1}} . Then clearly φ_{t} ≤ g_{t}. Since g is in L^{1}

and decreasing with r = |x|, so is g_{t}. Moreover, we have ∥g_{t}∥_{L^{1}} = ∥g∥_{L^{1}}. The corollary follows now from the above proposition.

The Calderón-Zygmund decomposition

If f ∈ L^{1}, then the set where f is large is relatively small, i.e. |\{ x; |f(x)| > t \}| ≤ \frac{∥f∥_{L^{1}}}{t}. The following result provides a nice covering of this set.

Theorem 9. (The Calderón-Zygmund decomposition) Let f ∈ L^{1}(ℝ^{N}) and t > 0. Then there is a sequence of disjoint cubes (C_{n}) such that:

a) |f(x)| ≤ t a. e. in ℝ^{N} \ (⋃_{n} C_{n});

b) for each n we have C^{−1}t ≤ \frac{1}{|C_{n}|} ∫ ; C_{n} |f(x)|dx ≤ Ct;

c) ∑_{n} |C_{n}| ≤ \frac{C∥f∥_{L^{1}}}{t}.

Here, C depends only on the space dimension N, not on f or t.

Proof. The construction looks like the Whitney decomposition. Fix some l > 0 such that l^{N} >

\frac{∥f∥_{L^{1}}}{t}. We cover ℝ^{N} with disjoint cubes of size l. We call F_{1} the family of all these cubes. We bisect the cubes in F_{1} and call F_{2} the family of cubes obtained in this way. We keep bisecting and obtain in the same way the families F_{j}, j ≥ 2. We start by throwing all the cubes in F_{1}.

For j ≥ 2, we keep a cube C in F_{j} if all its ancestors were thrown and \frac{1}{|C|} ∫_{C} |f(x)|dx > t. Let

F = (C_{n}) be the family of all kept cubes and A = ⋃_{n} C_{n}. If x ∉ A, then all the cubes containing

x were thrown. Thus |f(x)| ≤ t a.e. in ℝ^{N} \ A, by the Lebesgue differentiation theorem. Let now C ∈ F. Then C ∈ F_{j} for some j ≥ 2. The (unique) cube Q in F_{j−1} containing C was

26 CHAPTER 6. THE CALDERÓN-ZYGMUND DECOMPOSITION

thrown, so that

\frac{1}{|C|} ∫_{C} |f(x)|dx ≤ \frac{1}{|C|} ∫_{Q} |f(x)|dx = \frac{2^{N}}{|Q|} ∫_{Q} |f(x)|dx ≤ 2^{N}t.  (6.1)

Thus b) holds with C = 2^{N}. Finally, c) follows from

∥f∥_{L^{1}} ≥ ∑_{n} ∫ ; C_{n} |f(x)|dx ≥ C^{−1} ∑_{n} |C_{n}|t.  (6.2)

A variant of the above theorem is the following Theorem 10. Let f ∈ L^{1}(ℝ^{N}) and t > 0. Let (C_{n}) as above. Then f = g + ∑_{n} h_{n}, where:

a) g ∈ L^{1}, |g| ≤ Ct a. e. and g = f in ℝ^{N} \ (⋃_{n} C_{n});

b) supp h_{n} ⊂ C_{n};

c) for each n we have ∫C_{n} h_{n}(x)dx = 0;

d) for each n we have \frac{1}{|C_{n}|} ∫ ; C_{n} |h_{n}(x)|dx ≤ Ct;

e) ∥g∥_{L^{1}} + ∑_{n} ∥h_{n}∥_{L^{1}} ≤ C∥f∥_{L^{1}}.

Proof. Let g(x) = \begin{cases}f(x), & if x ∉ A \\ \frac{1}{|C_{n}|} ∫ ; C_{n} f(y)dy, & if x ∈ C_{n} \end{cases} and h_{n}(x) = f(x) − \frac{1}{|C_{n}|} ∫ ; C_{n} f(y)dy for x ∈ C_{n}. It

is easy to check that this decomposition has all the desired properties.

Part II

Hardy and bounded mean oscillations spaces Substitutes of L^{1}

7.1 The space L log L

As we have already seen, if f ∈ L^{1}, we can expect at best Mf ∈ L_{loc}^{1}, but even this could be wrong if we only assume f ∈ L^{1}. We present below a necessary and sufficient condition for having Mf ∈ L_{loc}^{1}.

A measurable function f belongs to L log L iff ∫ |f| ln(1 + |f|) < ∞. The space L log L_{loc} is

defined as the set of measurable functions s. t. f_{|K} ∈ L log L for each compact K.

Theorem 11. Let f ∈ L^{1}. Then Mf ∈ L_{loc}^{1} ⟺ f ∈ L log L_{loc}.

Remark 5. Set Φ(t) = t ln(1 + t), t ≥ 0. If F is the distribution function of f, then ∫ |f| ln(1 +

|f|) = ∫ Φ′(t)F (t)dt. It is easy to see that Φ′(λs) ≤ max{1, λ}Φ′(s) when λ, s > 0. Thus

∫ |λf| ln(1 + |λf|) = ∫ Φ′(t)F (t/|λ|)dt = |λ| ∫ Φ′(λs)F (s)ds ≤ C_{λ} ∫ |f| ln(1 + |f|) (7.1)

for each λ ∈ ℝ. On the other hand, we have |{|f + g| > t}| ≤ |{|f| > t/2}| + |{|g| > t/2}|. Therefore, if F is the distribution function of f and G the one of g, then we have, with h = f + g,

∫ |h| ln(1 + |h|) ≤ ∫ Φ′(t)(F (t/2) + G(t/2))dt ≤ 4 ∫ |f| ln(1 + |f|) + 4 ∫ |g| ln(1 + |g|). (7.2)

Thus L log L is a vector space. Similarly, L log L_{loc} is a vector space.

Proof. ”⟸” Assume that f ∈ L log L_{loc}. Fix a compact K ⊂ ℝ^{N}. We will prove that Mf ∈ L^{1}(K). Let L = {x ∈ ℝ^{N} ; dist(x, K) ≤ 1}. We split f as f = g + h, where g = fχ_{L}, h = f − g. Then Mf ≤ Mg + Mh. We note that Mh_{|K} ∈ L^{∞}. Indeed, if x ∈ K and r ≥ 1, then

30 CHAPTER 7. SUBSTITUTES OF L^{1}

\frac{1}{|B(x, r)|} ∫_{B(x,r)} |h| = 0. On the other hand, if r > 1, then \frac{1}{|B(x, r)|} ∫_{B(x,r)} |h| ≤ \frac{1}{|B(x, r)|} ∫_{ℝ^{N}} |h| ≤ C.

Thus we are bounded to prove that Mg ∈ L^{1}(K). Since g ∈ L log L, we reduced the initial problem to the case where f satisfies the stronger assumption f ∈ L log L. Let F be the distribution function of f and let G be the distribution function of Mf. Note that

∥Mf∥_{L^{1}(K)} = ∫_{0}^{2} |{x ∈ K ; Mf(x) > t}|dt + ∫_{2}^{∞}|{x ∈ K ; Mf(x) > t}|dt ≤ 2|K| + ∫_{2}^{∞}G(t)dt,

(7.3)

and thus it suffices to check that ∫_{2}^{∞}G(t)dt < ∞. By combining (5.5) and (5.7), we find that, with

some universal constant C, we have G(t) ≤ C(F (t/2)+1/t ∫_{t/2}^{∞}F (s)ds). Integrating this inequality,

we obtain

∫_{2}^{∞}G(t)dt ≤ C ∫_{1}^{∞}(ln s + 2)F (s)ds ≤ c ∫_{0}^{∞}Φ′(s)F (s)ds = c ∫ |f| ln(1 + |f|).

”⟹” Assume that Mf ∈ L_{loc}^{1} and let K, L be as above. We want to prove that ∫_{K} |f| log(1+|f|) <

∞. With g = fχ_{K}, this is the same as ∫_{K} |g| log(1 + |g|) < ∞. Since Mg ≤ Mf, we reduced

the original problem to the case where f, apart from the property Mf ∈ L_{loc}^{1}, satisfies the extra assumption supp f ⊂ K. (of course, the conclusion will be then apparently stronger :

∫ |f| log(1 + |f|) < ∞.)

Since f ∈ L^{1}, that is ∫ F (s)ds < ∞, it suffices to prove that, for some t_{0} sufficiently large, we

have ∫_{t_{0}}^{∞}(Φ′(s) − Φ′(t_{0}))F (s)ds < ∞. The key observation is that the distribution function G of

7.2. THE HARDY SPACE H^{1} 31

Mf satisfies ∫_{t_{0}}^{∞}G(t)dt < ∞ for sufficiently large t_{0}. Indeed, if x ∉ L, then Mf(x) ≤ C (this is obtained as above, by considering the average of |f| over a ball of radius r and discussing the cases r ≤ 1 or r > 1). Thus, by taking t_{0} = C, we have

∫_{C}^{∞}G(t)dt = ∫_{C}^{∞}|{x ∈ L ; Mf(x) > t}|dt ≤ ∥Mg∥_{L^{1}(L)} < ∞.  (7.4)

Let, for a fixed t > 0, O = {Mf > t}. Note that c) of the Hardy-Littlewood-Wiener theorem implies that O ≠ ℝ^{N}. Let O = ⋃_{C∈F} C be a Whitney covering of O. Recall that, if C ∈ F,

then there is an x ∈ ℝ^{N} \ O s. t. dist(x, C) ≤ 3l(C), and thus C ⊂ B(x, (3 + \sqrt{N})l(C)). Since

Mf(x) ≤ t, we find that ∫_{C} |f| ≤ ∫_{B(x,(3+\sqrt{N})l(C))} |f| ≤ c l(C)^{N}t, that is ∫_{\{Mf>t\}} |f| ≤ ctG(t). Note

that c does not depend on t. Since {|f| > t} ⊂ {Mf > t}, we find that ∫_{{|f|>t}} |f| ≤ ctG(t).

Invoking again (5.7), we obtain

∫_{t}^{∞}F (s)ds ≤ tF (t) + ∫_{t}^{∞}F (s)ds = ∫_{{|f|>t}} |f| ≤ ctG(t),  (7.5)

so that

∞ > ∫_{C}^{∞}G(t)dt ≥ c^{−1} ∫_{C}^{∞} ∫_{t}^{∞}F (s)/tdsdt = c^{−1} ∫_{C}^{∞}ln(s/C) F (s)ds ≥ d ∫_{C}^{∞}(Φ′(s) − Φ′(C))F (s)ds,

(7.6)

for some constant d.

7.2 The Hardy space H^{1}

There is a different way to come around the difficulty that Mf is never in L^{1}. Maximal functions are especially interesting because they provide pointwise estimates for convolutions. Instead of asking Mf to be in L^{1}, one could ask upper bounds for convolutions convolutions to be in L^{1}. Here

it is how it works. Fix a smooth map Φ ∈ S(ℝ^{N}) s. t. ∫ Φ ≠ 0. Set, for t > 0, Φ_{t} = t^{−N}Φ(·/t).

For u ∈ S′, let M_{Φ}u = sup_{t>0} |u ∗ Φ_{t}|. We define, for 1 ≤ p ≤ ∞,

H_{Φ}^{p} = {u ∈ S′ ; M_{Φ}u ∈ L^{p}}.  (7.7)

Note that we may assume, without loss of generality, that (H1) ∫ Φ = 1. On the other hand,

M_{Φ}u = M_{Φ_{s}}u, since f ∗ (u_{s})_{t} = f ∗ u_{st}. The condition ∫ Φ = 1 reads also Φ̂(0) = 1; replacing,

if necessary, Φ by Φ_{s} for appropriate s, we may assume that (H2) 1/2 ≤ |Φ̂(ξ)| ≤ 3/2 for |ξ| ≤ 2. We will always implicitly assume that the different test functions Φ, Ψ we will consider below are admissible, in the sense that they satisfy (H1) and (H2).

The definition (7.7) brings nothing new when 1 < p ≤ ∞.

Proposition 7. For 1 < p ≤ ∞, we have H_{Φ}^{p} = L^{p} and ∥u∥_{L^{p}} ∼ ∥M_{Φ}u∥_{L^{p}} .

Proof. Recall that, if u ∈ L^{p}, then |u ∗ Φ_{t}| ≤ CMu, and thus M_{Φ}u ∈ L^{p}. Conversely, assume that M_{Φ}u ∈ L^{p}. Then the family (u ∗ Φ_{t})_{t} is bounded in L^{p} and thus contains a sequence (u ∗ Φ_{t_{n}}) with t_{n} → 0, weakly-* convergent in L^{p}. Since, on the other hand, u ∗ Φ_{t} → u in S′ (here, we

use the assumption ∫ Φ = 1), we find that u ∈ L^{p}. Now, if u ∈ L^{p}, then u ∗ Φ_{t} → u a. e. and

thus |u| ≤ M_{Φ}u ≤ CMu, which together with the maximal theorem implies the equivalence of norms.

We next note some simple properties of H_{Φ}^{1}.

Proposition 8. a) u ↦ ∥M_{Φ}u∥_{L^{1}} is a norm on H_{Φ}^{1};

b) H_{Φ}^{1} ⊂ L^{1}, with continuous inclusion;

c) H_{Φ}^{1} is a Banach space.

Proof. The only property to be checked for a) is that ∥M_{Φ}u∥_{L^{1}} = 0 ⟹ u = 0. If ∥M_{Φ}u∥_{L^{1}} = 0, then u ∗ Φ_{t} = 0 for each t; by taking the limit in S′ as t → 0, we find that u = 0. If u ∈ H_{Φ}^{1}, then the family (u ∗ Φ_{t})_{t} is bounded in L^{1} and thus contains a sequence (u ∗ Φ_{t_{n}}) with t_{n} → 0, weakly-* convergent to some Radon measure μ. As above, this implies that u = μ, and thus u is a Radon measure. We will prove that |μ| is absolutely continuous with respect to the

Lebesgue measure. Let ε > 0. Then there is a δ > 0 s. t. ∫_{A} |M_{Φ}u| < ε whenever A is a Borel

set s. t. |A| < δ. If B is a Borel set s. t. |B| < δ, then there is an open set O containing B s. t. |O| < δ. Then

|μ|(B) ≤ |μ|(O) = sup{| ∫ φ dμ| ; φ ∈ C_{0}(O), |φ| ≤ 1} ≤ ∫_{O} M_{Φ}u < ε,  (7.8)

since

| ∫ φ dμ| = lim | ∫ u ∗ Φ_{t_{n}}φ| ≤ ∫ M_{Φ}u|φ| ≤ ∫_{O} M_{Φ}u.  (7.9)

Thus u ∈ L^{1}. Moreover, |u| ≤ M_{Φ}u, since the Lebesgue differentiability theorem implies, for a. e. x ∈ ℝ^{N}, that

|u(x)| = lim_{x∈B, |B|→0} \frac{1}{|B|} ∫_{B} |u| = lim_{x∈B, |B|→0} \frac{1}{|B|}|μ|(B) ≤ lim_{x∈B, |B|→0} \frac{1}{|B|} ∫_{B} M_{Φ}u = M_{Φ}u(x); (7.10)

here, the limit is taken over all the balls. In conclusion, ∥u∥_{L^{1}} ≤ ∥M_{Φ}u∥_{L^{1}}, which implies b).

In order to prove that H_{Φ}^{1} is a Banach space, it suffices to check that an absolutely convergent series has a sum in H_{Φ}^{1}. Assume that ∑ ∥M_{Φ}f_{n}∥_{L^{1}} < ∞. Then ∑ ∥f_{n}∥_{L^{1}} < ∞ and thus ∑ f_{n}

converges in L^{1} to some f. Clearly, M_{Φ}(f − ∑_{0}^{k} f_{n}) ≤ ∑_{k+1}^{∞} M_{Φ}f_{n} → 0 as k → ∞ and thus

∑ f_{n} = f in H_{Φ}^{1}.

There are two problems with the definition of H_{Φ}^{1}. The first one is that this space depends, in principle, on Φ. The second one is that it is not clear at all how to check that a given function belongs to H_{Φ}^{1}. A partial answer to the second question will be given in the next chapter. The answer to the first question is given by the following

Theorem 12. (Fefferman-Stein) Let Φ, Ψ be two admissible functions. Then H_{Φ}^{1} = H_{Ψ}^{1} and ∥M_{Φ}f∥_{L^{1}} ∼ ∥M_{Ψ}f∥_{L^{1}} for each f ∈ L^{1}. In addition, if f ∈ H_{Φ}^{1}, then, for every function ψ ∈ S (admissible or not!) we have sup_{t>0} |f ∗ ψ_{t}| ∈

L^{1}.

In view of this result, we may define H^{1} = H_{Φ}^{1} for some admissible Φ, and endow it with the norm f ↦ ∥M_{Φ}f∥_{L^{1}}.

The proof of the theorem is long and difficult; the remaining part of this chapter is devoted to it. However, we will first explain how the proof works; this will help understanding the technical part. Let Φ be admissible and let ψ ∈ S. We have to estimate f ∗ ψ_{t}, given the information that sup_{t>0} |f ∗ Φ_{t}| ∈ L^{1}. It would be convenient to be able to write ψ = Φ ∗ η; this is impossible in

general (pass to the Fourier transform: we have a trouble if Φ̂ vanishes at a point where ψ̂ does not vanish). However, let us forget this point, for the moment, and assume that ψ = Φ ∗ η. Then

|f ∗ ψ_{t}(x)| = |f ∗ Φ_{t} ∗ η_{t}(x)| ≤ t^{−N} ∫ |f ∗ Φ_{t}(x − y)||η(y/t)|dy.  (7.11)

Th natural way to estimate the latter integral is the following: we decompose ℝ^{N} into slices S_{n} where |y| ∼ C_{n}t. In these slices, |η| ∼ K_{n}. Then

|f ∗ ψ_{t}(x)| ≤ ∑_{n} |S_{n}|K_{n} sup_{{|x−y|∼C_{n}t}} |f ∗ Φ_{t}(x − y)|.  (7.12)

Thus, we may estimate |f ∗ ψ_{t}(x)| if we are able to estimate the function g : x ↦ sup_{{|x−y|∼Ct}} |f ∗

Φ_{t}(x − y)|. The difficulty is the hypothesis concerns the function h : x ↦ sup_{t>0} |f ∗ Φ_{t}(x)|. We may

relate h to g using Taylor’s formula:

h(x) ≤ g(x) + C sup{∑_{j=1}^{N} |y||f ∗ ∂_{j}(Φ_{t})(x − z)| ; |x − y| ∼ Ct, |x − z| ∼ Ct}.  (7.13)

We note that ∂_{j}(Φ_{t}) = t^{−1}(∂_{j}Φ)_{t}. Thus: |f ∗ ψ_{t}| can be estimated in terms h, while h can be estimated in terms of of g and |f ∗ (∂_{j}Φ)_{t}|; the latter term can be estimated in terms of h. We expect a chain of inequalities of the form:

∫ M_{ψ}f ≤ C_{1} ∫ h ≤ C_{2} ∫ g + C_{3} ∑_{j} ∫ M_{∂_{j}Φ}f ≤ C_{2} ∫ g + C_{4} ∫ h.  (7.14)

This looks like a vicious circle; the trick is to be able to adjust the constants in order to have

C_{4} < 1C_{1}, and then we find that (*) ∫ M_{ψ}F ≤ C_{5} ∫ h ≤ C_{6} ∫ g. This is essentially how the

proof works. There is one flaw in passing from (7.14) to (*): it may happen that ∫ h = ∞!

The plan of the proof is the following:

(i) we define properly h. We also define a modified h, in order to make sure that ∫ h < ∞. This

is done in Section 7.3; (ii) As already mentioned, we cannot expect to write ψ = Φ ∗ η. However, a substitute of this equality in proved in Section 7.4; (iii) Finally, we prove in Section 7.5 the right substitute of (7.14) and complete the proof of the Fefferman-Stein theorem.

7.3 More maximal functions

Let f ∈ L^{1} (we will always consider such f’s, in view of the preceding proposition) and let Φ ∈ S. We set F (x, t) = |f ∗ Φ_{t}(x)| F^{∗}(x) = M_{Φ}f(x) = sup{F (x, t) ; t > 0} F_{a}^{∗}(x) = sup{F (y, t) ; t > 0, |x − y| < at} (here, a > 0 is fixed)

F_{a,ε,M}^{∗}(x) = sup\{F (y, t)\frac{t^{M}}{(ε + t + ε|y|)^{M}} ; a^{−1}|x − y| < t < ε^{−1}\} (here, ε, M are fixed positive

constants). If we want to be more precise, we will rather write F^{Φ}, F^{∗,Φ}, and so on.

7.4. TRANSITION FROM ONE ADMISSIBLE FUNCTION TO A RAPIDLY DECREASING FUNCTION 35

The reason we introduce F_{a}^{∗} is clear, once we explained the strategy of the proof. It is less clear why we introduce F_{a,ε,M}^{∗}. These functions play the role of modified h’s: they will prove to be integrable and we be able to estimate these functions in terms of M_{Φ}f. We will next let ε → 0. We note the following elementary properties: F^{∗} ≤ F_{a}^{∗} ≤ F_{b}^{∗} if 0 < a < b lim_{ε→0} F_{a,ε,M}^{∗} = F_{a}^{∗}.

Finally, let, for α, β ∈ ℕ^{N}, p_{α,β} be the semi norm p_{α,β}(φ) = sup |x^{α}∂^{β}φ|, which is finite for φ ∈ S. F will denote a finite family of such semi norms.

The main theorem is an immediate consequence of the following

Theorem 13. If Φ is admissible, then there is a finite family F independent of ψ ∈ S s. t.

∫ F^{∗,ψ} ≤ C_{Φ} sup{p_{α,β}(ψ) ; p_{α,β} ∈ F} ∫ F^{∗,Φ}.

We end this section with a simple result we will need in the proof of Theorem 13.

Lemma 4. With a constant c depending only on N, we have

∫ F_{b,ε,M}^{∗} ≤ c(b/a)^{N} ∫ F_{a,ε,M}^{∗}, 0 < a < b.

Proof. Set, for α > 0, O_{a} = {F_{a,ε,M}^{∗} > α} and define similarly O_{b}. Then x ∈ O_{b} iff there are y, t s. t.

|y−x| < bt, t < ε^{−1} and F (y, t)\frac{t^{M}}{(ε + t + ε|y|)^{M}} > α. It follows immediately that x ∈ B(y, bt) ⊂ O_{b}

and that B(y, at) ⊂ O_{a}. Let now K ⊂ O_{b} be a fixed compact. Since the balls B(y, bt) cover O_{b}, we may find a finite collection of such balls that cover K. In addition, Vitali’s lemma implies that we may find a finite collection of such balls, say (B(y_{i}, bt_{i})), mutually disjoint and s. t. ∑ |B(y_{i}, bt_{i})| ≥ c|K|, where c depends only on N. Since a < b, the corresponding balls (B(y_{i}, at_{i})) are mutually disjoint and contained in O_{a}. Thus b^{N} ∑ t_{i}^{N} ≥ c|K|, while a^{N} ∑ t_{i}^{N} ≤ c|O_{a}|. We find that |K| ≤ c(b/a)^{N}|O_{a}|; by taking the supremum over K, we find that

|{F_{b,ε,M}^{∗} > α}| ≤ c(b/a)^{N}|{F_{a,ε,M}^{∗} > α}|.  (7.15)

The conclusion of the lemma follows by integrating the above inequality over α > 0.

7.4 Transition from one admissible function to a rapidly decreasing function

In this section, we provide the right substitute of the equality φ = Φ ∗ η.

Lemma 5. (Dyadic partition of the unit) There is a partition of the unit 1 = ∑_{k} ζ_{k} in ℝ^{N} s.

t.:

a) supp ζ_{0} ⊂ B(0, 2);

b) supp ζ_{k} ⊂ B(0, 2^{k+1}) \ B(0, 2^{k−1}) if k ≥ 1;

c) |∂^{β}ζ_{k}| ≤ C_{β} (with C_{β} independent of k).

Proof. Fix a function ζ_{0} ∈ C_{0}^{∞} s. t. supp ζ_{0} ⊂ B(0, 2) and ζ_{0} = 1 in B(0, 1). Define, for k ≥ 1, ζ_{k}(x) = ζ_{0}(2^{−k}x) − ζ_{0}(2^{−(k−1)}x). It is immediate that the ζ_{k}’s have all the desired properties.

Lemma 6. Given a semi norm p_{α,β}, there is a finite collection F of semi norms s. t., for each φ ∈ S,

p_{α,β}(φ) ≤ c sup{p_{γ,δ}(φ̂) ; p_{γ,δ} ∈ F}.  (7.16)

Proof. We have, for each α, β,

|x^{α}∂^{β}φ(x)| = (2π)^{−N}| ∫ e^{ıx·ξ}(ı∂)^{α}[(ıξ)^{β}φ̂](ξ)dξ| ≤ c sup(1+|ξ|)^{N+1}|(ı∂)^{α}[(ıξ)^{β}φ̂]| ≤ c sup_{p_{γ,δ}∈F} p_{γ,δ}(φ̂),

for some appropriate family F.

Corollary 8. For each L > 0, there is some F s. t.

|φ(x)| ≤ c(1 + |x|)^{−L} sup_{p_{γ,δ}∈F} p_{γ,δ}(φ̂), ∀ φ ∈ S.  (7.17)

Remark 6. We may, of course, reverse the roles of φ and φ̂ in the two above results.

Lemma 7. Let Φ be an admissible function. Then we may write each φ ∈ S as φ = ∑_{k=0}^{∞} Φ_{2^{−k}} ∗ η^{k}.

Here, the functions η^{k}, which depend on φ, belong to S and the series is convergent in S. In addition, given M, L > 0, there is a finite family F s. t.

|η^{k}(x)| ≤ c2^{−kM}(1 + |x|)^{−L} sup{p_{γ,δ}(φ) ; p_{γ,δ} ∈ F}.  (7.18)

Here, F and c do not depend on φ.

Proof. We start by noting that Lemma 6 and Corollary 8 imply that, in order to obtain (7.18), it suffices to establish, for each p_{α,β}, the inequality

p_{α,β}(η̂^{k}) ≤ c2^{−kM} sup{p_{γ,δ}(φ̂) ; p_{γ,δ} ∈ F},  (7.19)

for some family F which not need be the same as in the statement of the above lemma. On the other hand, we have

|∂^{β}{ \atop Φ_{2^{−k}}}(ξ)| = 2^{−k|β|}Φ̂(2^{−k}ξ)| ≤ c,  (7.20)

and therefore

∑_{k} p_{α,β}({ \atop Φ_{2^{−k}} ∗ η^{k}}) ≤ c ∑_{k ;γ≤β} p_{α,γ}(η̂^{k}).  (7.21)

Thus, if we prove (7.19), then the series ∑_{k} p_{α,β}({ \atop Φ_{2^{−k}} ∗ η^{k}}) is convergent. Consequently, the series

∑_{k} { \atop Φ_{2^{−k}} ∗ η^{k}} is convergent in S. Taking inverse Fourier transform, we find that ∑_{k} Φ_{2^{−k}} ∗ η^{k} is

convergent in S. In conclusion, it suffices to establish (7.19). Let 1 = ∑_{k} ζ_{k} be a dyadic partition of the unit. Noting that { \atop Φ_{2^{−k}}}(ξ) = Φ̂(2^{−k}ξ), we find that

φ̂(ξ) = ∑_{k=0}^{∞} φ̂(ξ)ζ_{k}(ξ) = ∑_{k=0}^{∞} { \atop Φ_{2^{−k}}}(ξ)\frac{φ̂(ξ)ζ_{k}(ξ)}{Φ̂(2^{−k}ξ)} ≡ ∑_{k=0}^{∞} { \atop Φ_{2^{−k}}}(ξ)Ψ^{k}(ξ).  (7.22)

We first note that Ψ^{k} is well-defined. Indeed, Φ being admissible, we have 1/2 ≤ |{ \atop Φ_{2^{−k}}}(ξ)| ≤ 3/2 if ξ ∈supp ζ_{k}. Moreover, ζ_{k} being compactly supported, so is Ψ^{k}. Finally, Ψ^{k} ∈ C_{0}^{∞}, and thus Ψ^{k} = η̂^{k} for some η^{k} ∈ S. It remains to establish (7.19), i. e.

p_{α,β}(Ψ^{k}) ≤ c2^{−kM} sup{p_{γ,δ}(φ̂) ; p_{γ,δ} ∈ F}  (7.23)

for some appropriate F.

Set Ψ = 1/Φ̂ ∈ C^{∞}(\overline{B}(0, 2)). Since for ξ ∈supp ζ_{k} we have 2^{−k}ξ ∈ \overline{B}(0, 2), we find, for such ξ,

|∂^{β}(1/{ \atop Φ_{2^{−k}}})(ξ)| = 2^{−|β|k}|∂^{β}Ψ(2^{−k}ξ)| ≤ c_{β}.

Since we also have |∂^{β}ζ_{k}| ≤ c_{β}, we have, for k ≥ 1 and ξ ∈supp ζ_{k},

|ξ^{α}∂^{β}Ψ^{k}(ξ)| ≤ c ∑_{γ≤β} |∂^{γ}φ̂(ξ)| ≤ c(1 + |ξ|)^{−M} sup_{p_{α,β}∈F} p_{α,β}(φ̂) ≤ c2^{−kM} sup_{p_{α,β}∈F} p_{α,β}(φ̂),

provided we choose F properly. A similar conclusion holds for k = 0, completing the proof of the lemma.

We set C_{0} = {x ; |x| ≤ 2} and, for j ∈ ℕ^{∗}, C_{j} = {x ; 2^{j−1} ≤ |x| < 2^{j}}.

Corollary 9. For each M > 0, we have

∫_{C_{j}} |η^{k}| ≤ c2^{−M(k+j)} sup{p_{α,β}(φ) ; p_{α,β} ∈ F}  (7.24)

provided F is sufficiently rich.

38 CHAPTER 7. SUBSTITUTES OF L^{1}

Proof. If we take F s. t. (7.18) holds for L = N + M, then

∫_{C_{j}} |η^{k}| ≤ ∫_{C_{j}} (1 + |x|)^{−N−M} sup_{C_{j}}(1 + |x|)^{N+M}|η^{k}(x)| ≤ c2^{−M(k+j)} sup{p_{α,β}(φ) ; p_{α,β} ∈ F}. (7.25)

7.5 Proof of Theorem 13

Proof. Step 1. F_{1,ε,M}^{∗,Φ} controls F_{2,ε,M}^{∗,φ} Lemma 8. There is a finite family F s. t. if 0 < ε < 1 and φ ∈ S, then

∫ F_{2,ε,M}^{∗,φ} ≤ c ∫ F_{1,ε,M}^{∗,Φ} sup{p_{α,β}(φ) ; p_{α,β} ∈ F}.  (7.26)

Here, c does not depend on φ or ε. Proof. Fix an ε > 0. By the transition lemma, we have

|f ∗ φ_{t}(z)| ≤ ∑_{k} ∫ |f ∗ Φ_{2^{−k}t}(z − y)||η_{t}^{k}(y)|dy = t^{−N} ∑_{k} ∫ |f ∗ Φ_{2^{−k}t}(z − y)||η^{k}(y/t)|dy. (7.27)

If 2^{−1}|z − x| < t < ε^{−1} and y ∈ tC_{j} (the sets C_{j} were defined in the preceding section), we find that |y − x| < 2^{k+j+2}2^{−k}t, while 2^{−k}t < ε^{−1}. Thus

|f ∗ Φ_{2^{−k}t}(z − y)| = F^{Φ}(z − y, 2^{−k}t) ≤ \frac{(ε + 2^{−k}t + ε|z − y|)^{M}}{(2^{−k}t)^{M}} F_{2^{k+j+2},ε,M}^{∗,Φ}(x),  (7.28)

so that

|f ∗ Φ_{2^{−k}t}(z − y)| ≤ \frac{(ε + 2^{−k}t + ε|z| + 2^{j}t)^{M}}{(2^{−k}t)^{M}} F_{2^{k+j+2},ε,M}^{∗,Φ}(x).  (7.29)

We find that

|f ∗ φ_{t}(z)| ≤ t^{−N} ∑_{k,j} \frac{(ε + 2^{−k}t + ε|z| + ε2^{j}t)^{M}}{(2^{−k}t)^{M}} F_{2^{k+j+2},ε,M}^{∗,Φ}(x) ∫_{tC_{j}} |η^{k}(y/t)|.  (7.30)

Therefore, for each fixed L > 0 we have, if F is sufficiently rich,

|f ∗φ_{t}(z)| ≤ c ∑_{k,j} \frac{(ε + 2^{−k}t + ε|z| + ε2^{j}t)^{M}}{(2^{−k}t)^{M}} F_{2^{k+j+2},ε,M}^{∗,Φ}(x)2^{−L(k+j)} sup\{p_{α,β}(φ) ; p_{α,β} ∈ F\}. (7.31)

If 0 < ε < 1 and 2^{−1}|z − x| < t < ε^{−1}, it is easy to check that

\frac{(ε + 2^{−k}t + ε|z| + ε2^{j}t)^{M}}{(2^{−k}t)^{M}} \frac{t^{M}}{(ε + 2^{−k}t + ε|z|)^{M}} ≤ 2^{M(j+k)}.  (7.32)

Combining (7.32) and the definition of F_{2,ε,M}^{∗,φ}, we obtain

F_{2,ε,M}^{∗,φ}(x) ≤ c ∑_{k,j} 2^{M(j+k)}F_{2^{k+j+2},ε,M}^{∗,Φ}(x)2^{−L(k+j)} sup{p_{α,β}(φ) ; p_{α,β} ∈ F}.  (7.33)

Recalling that L is arbitrary, we take L = M + N + 1 and find, with C = sup{p_{α,β}(φ) ; p_{α,β} ∈ F}, that

∫ F_{2,ε,M}^{∗,φ}(x) ≤ cC ∑_{k,j} 2^{−Mj−(N+1)(k+j)} ∫ F_{2^{k+j+2},ε,M}^{∗,Φ} ≤ cC ∑_{k,j} 2^{−Mj−(k+j)} ∫ F_{1,ε,M}^{∗,Φ} ≤ cC ∫ F_{1,ε,M}^{∗,Φ}.

(7.34)

Step 2. M_{Φ}f controls F_{1,ε,M}^{∗,Φ}

Lemma 9. If M > N, then F_{1,ε,M}^{∗,Φ} ∈ L^{1}.

Proof. We note that |f ∗ Φ_{t}| ≤ ∥f∥_{L^{1}}∥Φ_{t}∥_{L^{∞}} ≤ ct^{−N}. Thus

F_{1,ε,M}^{∗,Φ}(x) ≤ c sup_{|y−x|<t<ε^{−1}} \frac{t^{M−N}}{(ε + t + ε|y|)^{M}} ≤ c_{ε} sup_{|y−x|<ε^{−1}} \frac{1}{(1 + |y|)^{M}} ,  (7.35)

and the latter function belongs to L^{1} (it is bounded near the origin and behaves like |x|^{−M} at infinity).

Lemma 10. Assume that M > N and that ε < 1. Then, with some constant c that may depend on M, but not on ε or f, we have

∫ F_{1,ε,M}^{∗,Φ} ≤ c ∫ M_{Φ}f.  (7.36)

Proof. For each x, there are t and y s. t. |x − y| < t < ε^{−1} and

F_{1,ε,M}^{∗,Φ}(x) ≥ F^{Φ}(y, t)\frac{t^{M}}{(ε + t + ε|y|)^{M}} ≥ \frac{3}{4}F_{1,ε,M}^{∗,Φ}(x).

Let δ be a small constant to be fixed later. We claim that, if δ is sufficiently small and F is sufficiently rich (i. e., as in the preceding step), then

|z−y| < δt ⟹ |F^{Φ}(y, t)\frac{t^{M}}{(ε + t + ε|y|)^{M}} −F^{Φ}(z, t)\frac{t^{M}}{(ε + t + ε|z|)^{M}} | ≤ cδ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x)+ \frac{1}{4}F_{1,ε,M}^{∗,Φ}(x).

(7.37)

The above implication is an immediate consequence of the following inequalities

|\frac{t^{M}}{(ε + t + ε|y|)^{M}} − \frac{t^{M}}{(ε + t + ε|z|)^{M}} | ≤ \frac{1}{4} \frac{t^{M}}{(ε + t + ε|y|)^{M}} ,  (7.38)

respectively

|F^{Φ}(y, t) − F^{Φ}(z, t)|\frac{t^{M}}{(ε + t + ε|z|)^{M}} ≤ cδ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x).  (7.39)

Inequality (7.38) is elementary and left to the reader (it works when δ < (4/3)^{1/M} − 1). As for (7.39), we start by noting that

∂_{j}(f ∗ Φ_{t}) = \frac{1}{t}f ∗ ((∂_{j}Φ)_{t}),  (7.40)

and thus

|f ∗ Φ_{t}(y, t) − f ∗ Φ_{t}(z, t)| ≤ \frac{|y − z|}{t} sup_{1≤j≤N} sup_{|w−z|<δt} |f ∗ ((∂_{j}Φ)_{t})|(w).  (7.41)

Assuming, without loss of generality, that δ < 1, we find that

|F^{Φ}(y, t) − F^{Φ}(z, t)|\frac{t^{M}}{(ε + t + ε|z|)^{M}} ≤ δ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x) sup_{|w−z|<δt} \frac{(ε + t + ε|w|)^{M}}{(ε + t + ε|z|)^{M}} ≤ cδ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x),

(7.42)

whence (7.39). (We may take c = 2^{M}.) For each x, one of the two happens:

either (i) cδ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x) ≤ \frac{1}{4}F_{1,ε,M}^{∗,Φ}(x),

or (ii) cδ ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x) > \frac{1}{4}F_{1,ε,M}^{∗,Φ}(x).

Let A, respectively B, be the set of points s. t. (i), respectively (ii), holds. If x ∈ A, then we have

F^{Φ}(z, t) ≥ \frac{1}{4}F_{1,ε,M}^{∗,Φ}(x) whenever |z − y| < δt, and thus

\sqrt{M_{Φ}f(z)} ≥ \sqrt{F^{Φ}(z, t)} ≥ \frac{1}{2}\sqrt{F_{1,ε,M}^{∗,Φ}(x)}

7.5. PROOF OF THEOREM 13 41

for each such z. Thus

\sqrt{F_{1,ε,M}^{∗,Φ}(x)} ≤ \frac{c}{|\{|z − y| < δt\}|} ∫_{\{|z−y|<δt\}} \sqrt{M_{Φ}f(z)}.  (7.43)

Noting that {|z − y| < δt} ⊂ {|z − x| < 2t}, we find that, in case (i),

\sqrt{F_{1,ε,M}^{∗,Φ}(x)} ≤ \frac{c}{|\{|z − x| < 2t\}|} ∫_{\{|z−x|<2t\}} \sqrt{M_{Φ}f(z)} ≤ cM(\sqrt{M_{Φ}f})(x).  (7.44)

Therefore,

∫_{A} F_{1,ε,M}^{∗,Φ} ≤ c ∫_{A} (M(\sqrt{M_{Φ}f}))^{2} ≤ c ∫ M_{Φ}f,  (7.45)

by the maximal inequalities. (The above constants may depend on δ.) Concerning the set B, we have

∫_{B} F_{1,ε,M}^{∗,Φ} ≤ 4cδ ∫_{B} ∑_{j=1}^{N} F_{2,ε,M}^{∗,∂_{j}Φ}(x) ≤ c′δ ∫ F_{1,ε,M}^{∗,Φ}.  (7.46)

We finally fix δ sufficiently small in order to have (7.38), (7.39), δ < 1 and c′δ < 1/2. Then

∫_{B} F_{1,ε,M}^{∗,Φ} ≤ \frac{1}{2} ∫_{B} F_{1,ε,M}^{∗,Φ} + \frac{1}{2} ∫_{A} F_{1,ε,M}^{∗,Φ},  (7.47)

so that

∫ F_{1,ε,M}^{∗,Φ} ≤ c ∫ M_{Φ}f,  (7.48)

by combining (7.45) and (7.47).

Step 3. Conclusion

By letting ε → 0 in Lemma 10, we find that ∫ F_{1}^{∗,Φ} ≤ c ∫ M_{Φ}f. Next, letting ε → 0 in Lemma 10 yields ∫ F^{∗,φ} ≤ c ∫ M_{Φ}f sup{p_{α,β}(φ) ; p_{α,β} ∈ F}.

Let us note the following immediate consequence of Theorem 13 Corollary 10. Let F be a sufficiently rich family of semi norms. Set

F^{F}f(x) = sup{|f ∗ φ_{t}(x)|; t > 0, φ ∈ S, p_{α,β}(φ) ≤ 1 for each p_{α,β} ∈ F}.  (7.49)

Then f ↦ ∥F^{F}∥_{L^{1}} is an equivalent norm on H_{Φ}^{1}.

42 CHAPTER 7. SUBSTITUTES OF L^{1}

Proof. If Φ is admissible, then M_{Φ}f ≤ cF^{F}f for some c independent of F. On the other hand, the proof of Theorem 13 implies that ∥F^{F}∥_{L^{1}} ≤ c ∫ M_{Φ}f if F is sufficiently rich. Indeed, it suffices to note that, in inequality (7.26), we may replace F_{2,ε,M}^{∗,φ} by sup{F_{2,ε,M}^{∗,φ} ; p_{α,β}(φ) ≤ 1, p_{α,β} ∈ F} and still get the inequality

∫ sup{F_{2,ε,M}^{∗,φ} ; p_{α,β}(φ) ≤ 1, p_{α,β} ∈ F} ≤ C ∫ F_{1,ε,M}^{∗,Φ},  (7.50)

provided F is sufficiently rich. We may then follow the proof of the theorem and find that ∫ F^{F}f ≤ C ∫ M_{Φ}f.

We will need later the following simple estimate

Lemma 11. Let φ be supported in B(x, r) s. t. ∫ φ ≠ 0. Assume that |∂^{β}φ| ≤ C_{β}r^{−|β|}, with C independent of x or r. Let F be a finite family of semi norms. Then

| ∫ fφ| ≤ cr^{N}F^{F}f(x).  (7.51)

Here, c does not depend on x or r. Proof. We may assume that x = 0. Let Φ(x) = φ(−rx). Then Φ is supported in B(0, 1) and it is immediate that |x^{α}∂^{β}Φ(x)| ≤ C_{β}, which implies that we may find a constant c independent of x or r s. t. p_{α,β}(c^{−1}Φ) ≤ 1 for each p_{α,β} ∈ F. Then

| ∫ fφ| = r^{N}|f ∗ Φ_{r}(0)| = cr^{N}|f ∗ (c^{−1}Φ_{r})(0)| ≤ cr^{N}F^{F}f(0).  (7.52)

Atomic decomposition

8.1 Atoms

For the moment, we do not even know if H^{1} contains a non zero function! In this section, we will give examples of functions in H^{1}: the atoms. In the next section, we will show that this example is ”generic”. To motivate the definition of atoms, we start with the following simple

Proposition 9. If f ∈ H^{1}, then ∫ f = 0.

Proof. Argue by contradiction and assume, e. g., that ∫ f = 1. Pick some R > 0 s. t. ∫_{B(0,R)} f >

2/3 and ∫_{ℝ^{N} \B(0,R)} |f| < 1/3. Let Φ ∈ C_{0}^{∞} be s. t. Φ = 1 in B(0, 1) and 0 ≤ Φ ≤ 1. For x ∈ ℝ^{N} s.

t. |x| > R, let t = |x| + R, so that t ∼ |x|. Then

M_{Φ}f(x) ≥ f ∗ Φ_{t}(x) ≥ t^{−N} ∫_{B(0,R)} f − t^{−N} ∫_{ℝ^{N} \B(0,R)} |f| ≥ \frac{1}{3}t^{−N} ≥ c|x|^{−N},  (8.1)

and thus M_{Φ}f ∉ L^{1}.

Remark 7. H^{1} is a strict subspace of {f ∈ L^{1} ; ∫ f = 0}. To see this, it suffices to modify

the example in Remark 4 as follows: set f_{1} : ℝ → ℝ, f_{1}(x) = \frac{1}{x ln^{2} x}χ_{[0,1/2]} and let f(x) =

f_{1}(x) − f_{1}(3 − x). Then f ∈ L^{1} and ∫ f = 0. However, if we pick Φ ∈ C_{0}^{∞} s. t. 0 ≤ Φ ≤ 1, supp Φ ⊂ [0, 2] and Φ = 1 in [0, 1], then, for x ∈ [0, 1/2] we have

M_{Φ}f(x) ≥ x^{−1} ∫_{0}^{x} f_{1}(x)dx = \frac{1}{x| ln x|},  (8.2)

and thus M_{Φ}f ∉ L^{1}.

This suggests that functions in H^{1}, apart from having zero integral, can not be ”too large”.

Definition 1. An atom is a function a : ℝ^{N} → ℝ s. t.:

(i) supp a ⊂ B, where B is a ball; (ii) |a| ≤ |B|^{−1};

(iii) ∫ a = 0.

We may replace balls by cubes, in this definition, since if a is an atom with respect to a ball B, then c_{N}a is an atom with respect to any minimal cube containing B and conversely; here, c_{N} depends only on N.

Proposition 10. If a is an atom, then M_{Φ}f ∈ L^{1} and ∥M_{Φ}f∥_{L^{1}} ≤ c for some constant depending only on Φ.

Proof. Since M_{Φ}f ≤ cMf ≤ ∥f∥_{L^{∞}}, we have M_{Φ}f ≤ c|B|^{−1}. Therefore,

∫_{B^{∗}} M_{Φ}f ≤ c|B^{∗}||B|^{−1} = c2^{N};  (8.3)

here, B^{∗} is the ball concentric to B and twice larger. If x ∉ B^{∗}, we use the information (iii) and find that, with R the radius of B, we have

|f ∗ Φ_{t}(x)| = | ∫_{B} a(y)[Φ_{t}(x − y) − Φ_{t}(x)]dy| ≤ |B|^{−1} sup_{z∈B} |∇(Φ_{t})(x − z)| ∫_{B} |y|dy.  (8.4)

Taking into account the fact that |x − z| ∼ |x| and the inequality |∇Φ(x)| ≤ c|x|^{−N−1}, we obtain

|f ∗ Φ_{t}(x)| ≤ \frac{cR}{|x|^{N+1}} , and thus M_{Φ}f(x) ≤ \frac{cR}{|x|^{N+1}} . Integrating the latter inequality, we find that

∫_{ℝ^{N} \B^{∗}} M_{Φ}f ≤ c  (8.5)

and the desired inequality follows from (8.3) and (8.5).

Corollary 11. Let f = ∑ λ_{k}a_{k}, where each a_{k} is an atom and ∑ |λ_{k}| < ∞. Then f ∈ H^{1} and

∥f∥_{H^{1}} ≤ c ∑ |λ_{k}|.

More generally, we could weaken condition (ii) in the definition of an atom as follows

Definition 2. Let 1 < q ≤ ∞. A _{q}atom is a function satisfying (i), (iii) and (ii’) ∥a∥_{L^{q}} ≤ |B|^{1/q−1}.

Thus, the usual atoms are _{∞}atoms.

Proposition 11. If a is a _{q}atom, then ∥a∥_{H^{1}} ≤ c_{q}. If, in addition, q ≤ 2, then c_{q} ≤ \frac{c}{q − 1}.

Proof. We may assume that q < ∞. We repeat the reasoning in the preceding proposition. On the one hand, we have

∫_{B^{∗}} |M_{Φ}f| ≤ c ∫_{B^{∗}} |Mf| ≤ c|B^{∗}|^{1−1/q}( ∫ |Mf|^{q})^{1/q} ≤ c_{q};  (8.6)

here, we use Hölder’s inequality and the maximal theorem. In addition, we see that c_{q} ≤ \frac{c}{q − 1} if

q ≤ 2. When x ∉ B^{∗}, we find that

M_{Φ}f(x) ≤ \frac{c}{|x|^{N+1}} ∫_{B} |f(y)||y|dy ≤ \frac{c}{|x|^{N+1}} ∥f∥_{L^{q}}( ∫_{B} |y|^{q′})^{1/q′} ≤ \frac{c_{q}R}{|x|^{N+1}} ;  (8.7)

here, c_{q} remains bounded when q ≤ 2. Thus

∫_{ℝ^{N} \B^{∗}} M_{Φ}f ≤ c  (8.8)

with c independent of q ≤ 2. We conclude by combining (8.6) and (8.8).

8.2 Atomic decomposition

The following result tells that the atoms represent ”generic”H^{1} functions.

Theorem 14. (Coifman-Latter) Let f ∈ H^{1}. Then we may write f = ∑ λ_{k}a_{k}, where each a_{k}

is an atom and ∑ |λ_{k}| ∼ ∥f∥_{H^{1}}.

Proof. It suffices to write, in the sense of distributions, f = ∑ λ_{k}a_{k}, with ∑ |λ_{k}| ≤ c∥f∥_{H^{1}}.

Indeed, if we are able to do this, then on the one hand the series ∑ λ_{k}a_{k} is convergent in H^{1}, thus in D′, and therefore its sum has to be f, by uniqueness of the limit. On the other hand, we always have ∥ ∑ λ_{k}a_{k}∥_{H^{1}} ≤ c ∑ |λ_{k}|.

We fix a large family F of semi norms as in the preceding section. Let F^{F} be the corresponding maximal function, i. e.,

F^{F}(x) = F^{F}f(x) = sup{M_{Φ}f(x) ; p_{α,β}(Φ) ≤ 1, ∀ p_{α,β} ∈ F}.

Let, for j ∈ ℤ, O_{j} = {F^{F} > 2^{j}}; clearly, O_{j} is an open set and O_{j+1} ⊂ O_{j}. In addition, O_{j} ≠ ℝ^{N}, since F^{F} ∈ L^{1}. Set f_{j} = fχ_{O_{j}}.

Lemma 12. As j → ∞, f_{j} → 0 in L^{1}. As j → −∞, f_{j} − f → 0 in L^{∞}.

Proof. We have ∥f_{j}∥_{L^{1}} = ∫_{O_{j}} |f| → 0 as j → ∞, since |O_{j}| → 0 as j → ∞. On the other hand,

∥f_{j} − f∥_{L^{∞}} = sup_{ℝ^{N} \O_{j}} |f| ≤ c sup_{ℝ^{N} \O_{j}} F^{F} ≤ c2^{j} → 0  (8.9)

as j → ∞.

Corollary 12. Set g_{j} = f_{j} − f_{j+1}. Then ∑_{−∞}^{∞} g_{j} = f in the distribution sense.

Let (C_{k}^{j}) be a Whitney covering of O_{j} and let φ_{k}^{j} be the corresponding partition of the unit in O_{j}. Recall that, with 1 < a < b depending only on N, we have

(i) C_{k}^{j∗} ⊂ O_{j} (where C_{k}^{j∗} is the cube concentric with C_{k}^{j} and having a times its size); (ii) C_{k}^{j∗∗} ⊄ O_{j} (where C_{k}^{j∗∗} is the cube concentric with C_{k}^{j} and having b times its size); (iii) at most M cubes C_{k}^{j∗} meet at some point, where M depends only on N; (iv) supp φ_{k}^{j} ⊂ C_{k}^{j∗};

(v) |∂^{α}φ_{k}^{j}| ≤ c_{α} size(C_{k}^{j})^{−|α|}; (vi) φ_{k}^{j} ≥ 1/M in C_{k}^{j}. The last property implies (vii) ∫ φ_{k}^{j} ∼ size(C_{k}^{j}).

We have f_{j} = ∑ fφ_{k}^{j} (the series that appears is well-defined, at least in the sense of distributions, since on each compact there are finitely many non vanishing terms). We define the coefficient

c_{k}^{j} by the condition ∫ (f − c_{k}^{j})φ_{k}^{j} = 0. We have f_{j} = ∑(f − c_{k}^{j})φ_{k}^{j} + R_{j}, where R_{j} = ∑ c_{k}^{j}φ_{k}^{j}.

8.2. ATOMIC DECOMPOSITION 47

Lemma 13. We have ∑_{−∞}^{∞}(R_{j} − R_{j+1}) = 0 in the sense of distributions. Proof. We will see later that the coefficients c_{k}^{j} satisfy |c_{k}^{j}| ≤ c2^{j} and that ∑_{j} 2^{j}|O_{j}| < ∞. Thus, with l_{k}^{j} the size of C_{k}^{j},we have

∥R_{j}∥_{L^{1}} ≤ |c_{k}^{j}| ∑_{k} ∫ φ_{k}^{j} ≤ c2^{j} ∑_{k}(l_{k}^{j})^{N} = c2^{j} ∑_{k} |C_{k}^{j}| = c2^{j}|O_{j}| → 0 as |j| → ∞.  (8.10)

Since

∥ ∑_{j=−M}^{j=P}(R_{j} − R_{j+1})∥_{L^{1}} = ∥R_{−M} − R_{P +1}∥_{L^{1}} ≤ c(2^{−M}|O_{−M}| + 2^{P +1}|O_{P +1}|),  (8.11)

we find that ∑_{j=−M}^{j=P}(R_{j} − R_{j+1}) → 0 in L^{1} as M, P → ∞, whence the conclusion.

Corollary 13. We have f = ∑_{−∞}^{∞} [ ∑_{k}(f −c_{k}^{j})φ_{k}^{j} −∑_{l}(f −c_{l}^{j+1})φ_{l}^{j+1}] in the sense of distributions.

Using the fact that φ_{l}^{j+1} = ∑_{k} φ_{l}^{j+1}φ_{k}^{j} (since O_{j+1} ⊂ O_{j}), we may further decompose the general term of the above series as follows

∑_{k}(f−c_{k}^{j})φ_{k}^{j}−∑_{l}(f−c_{l}^{j+1})φ_{l}^{j+1} = ∑_{k}(f−c_{k}^{j})φ_{k}^{j}−∑_{k,l}[(f−c_{l}^{j+1})φ_{k}^{j}−c_{k,l}^{j}]φ_{l}^{j+1}−∑_{k,l} c_{k,l}^{j}φ_{l}^{j+1}; (8.12)

here, the coefficients c_{k,l}^{j} are chosen s. t. ∫ [(f − c_{l}^{j+1})φ_{k}^{j} − c_{k,l}^{j}]φ_{l}^{j+1} = 0.

Actually, the last sum in (8.12) vanishes. The reason is that, for fixed l, we have, with c = ∫ φ_{l}^{j+1} ≠ 0,

c ∑_{k} c_{k,l}^{j} = ∫ ∑_{k} c_{k,l}^{j}φ_{l}^{j+1} = ∫ ∑_{k}(f − c_{l}^{j+1})φ_{k}^{j}φ_{l}^{j+1} = ∫ (f − c_{l}^{j+1})φ_{l}^{j+1} = 0;  (8.13)

commuting the series with the integral in the above computations is justified by the fact that, when l is fixed, we have only finitely many non vanishing terms.

Thus f = ∑_{j=−∞}^{∞} ∑_{k} b_{k}^{j}, where

b_{k}^{j} = (f − c_{k}^{j})φ_{k}^{j} − ∑_{l}[(f − c_{l}^{j+1})φ_{k}^{j} − c_{k,l}^{j}]φ_{l}^{j+1} = fφ_{k}^{j}χ_{ℝ^{N} \O_{j+1}} − c_{k}^{j}φ_{k}^{j} + ∑_{l}[c_{l}^{j+1}φ_{k}^{j}φ_{l}^{j+1} + c_{k,l}^{j}φ_{l}^{j+1}].

(8.14)

Let C > 0 be a large constant to be specified later. We set, with l_{k}^{j} the size of C_{k}^{j}, λ_{k}^{j} = C(l_{k}^{j})^{N}2^{j} and a_{k}^{j} = (λ_{k}^{j})^{−1}b_{k}^{j}, so that

f = ∑_{j=−∞}^{∞} ∑_{k} λ_{k}^{j}a_{k}^{j};  (8.15)

this is going to be the atomic decomposition of f. Clearly, the functions a_{k}^{j} satisfy, by construction, the cancellation property (iii) required in the definition of an atom. It remains to establish three facts: a) that the support of a_{k}^{j} is contained in some ball B; b) that |a_{k}^{j}| ≤ |B|^{−1} (here, the choice of the constant C will count); c) that ∑_{j,k} |λ_{k}^{j}| ≤ C∥f∥_{H^{1}}. These information are easily obtained

by combining the conclusions of the following lemmata.

Lemma 14. There is a constant b > 0 depending only on N s. t. supp b_{k}^{j} ⊂ B_{k}^{j}, where B_{k}^{j} is the ball concentric with C_{k}^{j} and of radius b l_{k}^{j}.

Proof. If b_{k}^{j}(x) ≠ 0, then either x ∈ supp φ_{k}^{j} ⊂ C_{k}^{j∗}, or there is some l s. t. supp φ_{k}^{j} intersects supp φ_{l}^{j+1} and s. t. φ_{l}^{j+1}(x) ≠ 0. In the latter case, we have, on the one hand, x ∈ C_{l}^{j+1∗}. On the other hand, if y ∈ supp φ_{k}^{j}∩ supp φ_{l}^{j+1} ⊂ C_{k}^{j∗} ∩ C_{l}^{j+1∗}, then

l_{l}^{j+1} ≤ c_{1} dist (y, ℝ^{N} \ O_{j+1}) ≤ c_{1} dist (y, ℝ^{N} \ O_{j}) ≤ c_{2} l_{k}^{j}.  (8.16)

In both cases, we may find b s. t. the conclusion of the lemma holds.

Lemma 15. We have |c_{k}^{j}| ≤ c2^{j}, and |c_{k,l}^{j}| ≤ c2^{j}. Here, c depends only on N and F.

Proof. By definition, we have c_{k}^{j} = ∫ fφ_{k}^{j}/ ∫ φ_{k}^{j}. As already noted, we have ∫ φ_{k}^{j} ∼ (l_{k}^{j})^{N}. Let

x ∈ C_{k}^{j∗∗}\O_{j}; thus F^{F}f(x) ≤ 2^{j}, by the definition of O_{j}. We apply Lemma 11 with this x and with r = Cl_{k}^{j}, for a sufficiently large C. Using the decay properties of the functions φ_{k}^{j} in Whitney’s partition of the unit, we find that

|c_{k}^{j}| = | ∫ fφ_{k}^{j}/ ∫ φ_{k}^{j}| ≤ c(l_{k}^{j})^{N}/ ∫ φ_{k}^{j}F^{F}(x) ≤ C2^{j}.  (8.17)

The argument for c_{k,l}^{j} is similar. Since

c_{k,l}^{j} = ∫ fφ_{k}^{j}φ_{l}^{j+1}/ ∫ φ_{l}^{j+1} − c_{l}^{j+1} ∫ φ_{k}^{j}φ_{l}^{j+1}/ ∫ φ_{l}^{j+1},  (8.18)

we find that

|c_{k,l}^{j}| ≤ |c_{l}^{j+1}| + | ∫ fφ_{k}^{j}φ_{l}^{j+1}/ ∫ φ_{l}^{j+1}|;  (8.19)

the latter term appears only if φ_{k}^{j}φ_{l}^{j+1} ≢ 0. It is straightforward that |∂^{β}(φ_{k}^{j}φ_{l}^{j+1})| ≤ C_{β}(l_{l}^{j+1})^{−|β|}, with C_{β} independent of j, k, l. (It suffices to rely on the fact that l_{l}^{j+1} ≤ Cl_{k}^{j} if φ_{k}^{j}φ_{l}^{j+1} ≢ 0). As above, we find that |c_{k,l}^{j}| ≤ c2^{j}.

Lemma 16. With some constant c depending only on N and on the family F of semi norms, we have |b_{k}^{j}| ≤ c2^{j} in the support of b_{k}^{j}.

Proof. In view of the second equality in (8.14), we have

|b_{k}^{j}| ≤ |f|χ_{ℝ^{N} \O_{j+1}} + |c_{k}^{j}| + ∑_{l}(|c_{l}^{j+1}| + |c_{k,l}^{j}|) ≤ c 2^{j} + |f|χ_{ℝ^{N} \O_{j+1}}.  (8.20)

The desired conclusion is obtained by noting that |f| ≤ cF^{F}f a. e., and thus |f| ≤ c 2^{j} in ℝ^{N} \ O_{j+1}.

By combining the above results, we find immediately that the a_{k}^{j}’s are atoms, provided we chose C sufficiently large (depending only on N and F). We may now complete the proof as follows: on the one hand, we have

∑ |λ_{k}^{j}| ≤ c ∑_{j,k} 2^{j}(l_{k}^{j})^{N} = c ∑_{j,k} 2^{j}|C_{k}^{j}| = c ∑_{j} 2^{j}|O_{j}|.  (8.21)

On the other hand,

∥F^{F}f∥_{L^{1}} = ∫ |{F^{F}f > α}|dα ≥ ∑_{−∞}^{∞} ∫_{2^{j−1}}^{2^{j}} |{F^{F}f > α}|dα ≥ ∑_{−∞}^{∞} 2^{j−1}|O_{j}|.  (8.22)

If we take F sufficiently rich, we find, by combining (8.21) with (8.22), that

∥f∥_{H^{1}} ∼ ∥F^{F}f∥_{L^{1}} ≥ c ∑ |λ_{k}^{j}|.  (8.23)

Corollary 14. On H^{1}, the quantity

∥f∥ = inf{∑ |λ_{k}| ; f = ∑ λ_{k}a_{k}, the a′_{k}s are atoms}

is a norm equivalent to the usual ones.

50 CHAPTER 8. ATOMIC DECOMPOSITION

The substitute of L^{∞}: BMO

9.1 Definition of BMO

Definition 3. A function f ∈ L_{loc}^{1} belongs to BMO (=bounded mean oscillation) if

∥f∥_{BMO} = sup\{\frac{1}{|C|} ∫_{C} |f − \frac{1}{|C|} ∫_{C} f| ; C cube with sides parallel to the axes\} < ∞. (9.1)

Despite the notation, ∥·∥_{BMO} is not a norm, since ∥f∥_{BMO} = 0 when f is a constant. However, it is easy to see that, if we identify two functions in BMO when their difference is constant (a. e.), then ∥ · ∥_{BMO} is a norm on the quotient space (still denoted BMO).

It will be convenient to denote by f_{C} the average of f on C, i. e., f_{C} = \frac{1}{|C|} ∫_{C} f.

Proposition 12. a) BMO is a Banach space.

b) For each cube C and each constant m, we have

∫_{C} |f − f_{C}| ≤ 2 ∫_{C} |f − m|.  (9.2)

c) We may replace cubes by balls; the space remains the same and the norm is replaced by an equivalent one. Similarly, we may consider cubes in general position.

d) If C ⊂ Q are parallel cubes of sizes l ≤ L, then |f_{C} − f_{Q}| ≤ c(1 + ln(L/l))∥f∥_{BMO}.

e) If Ψ is a Lipschitz function of Lipschitz constant k, then ∥Ψ ∘ f∥_{BMO} ≤ 2k∥f∥_{BMO}.

Warning: In e), we do not identify two functions if there difference is constant.

Proof. a) Let ∑ f_{n} be an absolutely convergent series in BMO. Let C be a cube. The series

∑(f_{n}_{|C} − (f_{n})_{C}) is absolutely convergent (thus convergent) in L^{1}. Set f^{C} = ∑(f_{n}_{|C} − (f_{n})_{C}).

52 CHAPTER 9. THE SUBSTITUTE OF L^{∞}: BMO

Then ∫_{C} f^{C} = 0 and

\frac{1}{|C|} ∫_{C} |f^{C}| ≤ ∑ \frac{1}{|C|} ∫_{C} |f_{n} − (f_{n})_{C})| ≤ ∑ ∥f_{n}∥_{BMO}.  (9.3)

We now cover ℝ^{N} with an increasing sequence of cubes (C_{k}). We set f(x) = f^{C_{k}}(x) − (f^{C_{k}})_{C_{0}} if x ∈ C_{k}. We claim that the definition is correct (in the sense that it does not depend on the choice of C_{k}). This follows immediately from the equality

(f^{C})_{C_{0}} = ∑[f_{n}_{|C} − (f_{n})_{C_{0}}],  (9.4)

valid whenever C_{0} ⊂ C.

b) We have

|C||m − f_{C}| = | ∫_{C} (m − f)| ≤ ∫_{C} |m − f|,  (9.5)

and thus

∫_{C} |f − f_{C}| ≤ ∫_{C} |f − m| + ∫_{C} |m − f_{C}| ≤ 2 ∫_{C} |m − f|.  (9.6)

c) We prove the assertion concerning balls. The proof of the other statement is analog. Let B be a ball and let C, Q be cubes s. t. C is inscribed in B and B is inscribed in Q. Then

\frac{1}{|B|} ∫_{B} |f − f_{B}| ≤ \frac{2}{|B|} ∫_{B} |f − f_{Q}| ≤ \frac{2}{|B|} ∫_{Q} |f − f_{Q}| ≤ \frac{c}{|Q|} ∫_{Q} |f − f_{Q}|  (9.7)

and similarly

\frac{1}{|C|} ∫_{C} |f − f_{C}| ≤ \frac{2}{|C|} ∫_{C} |f − f_{B}| ≤ \frac{2}{|C|} ∫_{B} |f − f_{B}| ≤ \frac{c}{|B|} ∫_{B} |f − f_{B}|,  (9.8)

so that the supremum over the balls and the supremum over the cubes are equivalent quantities.

d) We have

|f_{C} − f_{Q}| = \frac{1}{|C|}| ∫_{C} (f − f_{Q})| ≤ \frac{1}{|C|} ∫_{C} |f − f_{Q}| ≤ \frac{1}{|C|} ∫_{Q} |f − f_{Q}| ≤ (L/l)^{N}∥f∥_{BMO},  (9.9)

which implies the desired estimate when L/l ≤ 2. If L/l > 2, let j ∈ ℕ^{∗} be s. t. L ∈ [2^{j}l, 2^{j+1}l) and consider a sequence C_{0}, …, C_{j+1} of cubes s. t. C_{0} = C, C_{j+1} = Q and the size of each cube

9.2. H^{1} AND BMO 53

is at most the double of the size of its predecessor. Then

|f_{C} − f_{Q}| ≤ ∑_{l=0}^{l=j} |f_{C_{l}} − f_{C_{l+1}}| ≤ c j∥f∥_{BMO} ≤ c′ ln(L/l)∥f∥_{BMO}.  (9.10)

e) We have

\frac{1}{|C|} ∫_{C} |Ψ ∘ f − (Ψ ∘ f)_{C}| ≤ \frac{2}{|C|} ∫_{C} |Ψ ∘ f − Ψ(f_{C})| ≤ \frac{2k}{|C|} ∫_{C} |f − f_{C}| ≤ 2k∥f∥_{BMO}. (9.11)

Remark 8. The space BMO is not trivial: L^{∞} functions are in BMO and ∥f∥_{BMO} ≤ 2∥f∥_{L^{∞}}. However, BMO is not reduced to L^{∞} functions. Here is an example: let f : ℝ^{N} → ℝ, f(x) = ln |x|. Then f ∈ BMO. Indeed, let B be a ball of radius R and center x. If |x| ≤ 2R, then there is a ball B^{∗} of radius ρ ∼ R, containing B and centered at the origin. Then

\frac{1}{|B|} ∫_{B} |f − f_{B}| ≤ \frac{2}{|B|} ∫_{B} |f − ln ρ| ≤ \frac{2}{|B|} ∫_{B^{∗}} |f − ln ρ| ≤ \frac{c}{|B^{∗}|} ∫_{B^{∗}} |f − ln ρ|.  (9.12)

Now it is easy to see that the last integral is finite and independent of ρ.

Assume now that |x| > 2R. Then | ln |y| − ln |z|| ≤ c\frac{R}{|x|} whenever y, z ∈ B, and therefore

\frac{1}{|B|} ∫_{B} |f − f_{B}| = \frac{1}{|B|^{2}} ∫_{B} | ∫_{B} (f(y) − f(z))dy|dz ≤ \frac{cR}{|x|} ≤ c.  (9.13)

We emphasize the following consequence of our above computation

lim_{|x|→∞} \frac{1}{|B(x, R)|} ∫_{B(x,R)} | ln |y| − ln |y|_{B(x,R)}|dy = 0, ∀ R > 0.  (9.14)

9.2 H^{1} and BMO

Theorem 15. (Fefferman) BMO is the dual of H^{1} in the following sense:

a) if f ∈ BMO, then the functional T (g) = ∫ fg, initially defined on the set of finite combinations

of atoms, satisfies |T (g)| ≤ c∥f∥_{BMO}∥g∥_{H^{1}} and thus gives raise (by density) to a unique element of (H^{1})^{∗} of norm ≤ c∥f∥_{BMO}.

b) Conversely, let T ∈ (H^{1})^{∗}. Then there is some f ∈ BMO s. t. T (g) = ∫ fg whenever g is a

finite combination of atoms. In addition, ∥f∥_{BMO} ≤ c∥T ∥_{(H^{1})^{∗}}.

Remark 9. Since atoms are bounded and compactly supported, ∫ fg makes sense when f ∈ L_{loc}^{1}

and g is an atom. Moreover, the definition is correct when f ∈ BMO, in the sense that if we replace f by f + const, then the value of the integral does not change, since atoms have zero integral.

Proof. a) Assume first that f is bounded. Then T (g) = ∫ fg is well-defined and continuous in

H^{1}, since the inclusion H^{1} ⊂ L^{1} is continuous. If g = ∑ λ_{k}a_{k} is an atomic decomposition of g

s. t. ∑ |λ_{k}| ≤ c∥g∥_{H^{1}} and each a_{k} is supported in some B_{k}, then

|T (g)| ≤ ∑ |λ_{k}|| ∫ fa_{k}| = ∑ |λ_{k}|| ∫ (f − f_{B_{k}})a_{k}| ≤ ∑ |λ_{k}|\frac{1}{|B_{k}|} ∫_{B_{k}} |f − f_{B_{k}}| ≤ c∥f∥_{BMO}∥g∥_{H^{1}}

(9.15)

and a) follows.

When f is arbitrary, we apply (9.15) to the truncated function f_{n}(x) = \begin{cases}n, & if f(x) ≥ n \\ f(x), & if |f(x) < n \\ −n, & if f(x) ≤ −n \end{cases}.

Noting that f_{n} = Ψ_{n} ∘ f, where Ψ_{n} is Lipschitz of Lipschitz constant 1, we find that

| ∫ f_{n}g| ≤ c∥f∥_{BMO}∥g∥_{H^{1}}.  (9.16)

When g is a finite combination of atoms, we have |f_{n}g| ≤ |fg| ∈ L^{1} and f_{n}g → fg a. e. Thus

| ∫ fg| = lim | ∫ f_{n}g| ≤ c∥f∥_{BMO}∥g∥_{H^{1}},  (9.17)

by dominated convergence. This implies a) in full generality.

b) Conversely, let T ∈ (H^{1})^{∗}. Let B be a ball and let X_{B} be the space of L^{2} functions supported in

B having zero integral. If g ∈ X_{B}, then \frac{1}{∥g∥_{L^{2}}|B|^{1/2}} g is a _{2}atom. Thus ∥g∥_{H^{1}} ≤ c∥g∥_{L^{2}}|B|^{1/2}. It

follows that T restricted to X_{B} defines a linear continuous functional of norm ≤ c∥T ∥|B|^{1/2}. Thus,

there is some f^{B} ∈ X_{B} s. t. ∥f^{B}∥_{L^{2}} ≤ c∥T ∥|B|^{1/2} and T (g) = ∫ f^{B}g when g ∈ X_{B}. We now

cover ℝ^{N} with an increasing sequence of balls B_{n} and set f(x) = f^{B_{n}}(x) − (f^{B_{n}})_{B_{0}} if x ∈ B_{n}. This definition is correct. Indeed, if j > k, then f^{B_{k}} and f_{|B_{k}}^{B_{j}} − (f^{B_{j}})_{B_{k}} yield the same functional T_{|X_{B}} and thus must coincide. Therefore, f^{B_{j}} and f^{B_{k}} differ only by a constant in B_{j} (and thus in B_{0}), which implies that the definition of f is correct. Another obvious consequence of our argument is that, on each ball B, f_{|B} and f^{B} differ with a constant. In other words, f^{B} = f − f_{B}.

9.3. BMO FUNCTIONS ARE ALMOST BOUNDED 55

We claim that, when g is a finite combination of atoms, we have T (g) = ∫ fg. Indeed, there is

some n s. t. supp g ⊂ B_{n}. Since g ∈ X_{B_{n}} for such n, we find that T (g) = ∫ f^{B_{n}}g = ∫ fg.

It remains to prove that f ∈ BMO and that ∥f∥_{BMO} ≤ c∥T ∥_{(H^{1})^{∗}}. This follows from the fact that, if B is any ball, then we have

\frac{1}{|B|}|f − f_{B}| = \frac{1}{|B|}|f^{B}| ≤ \frac{1}{\sqrt{|B|}}∥f^{B}∥_{L^{2}} ≤ c∥T ∥.  (9.18)

9.3 BMO functions are almost bounded

Strictly speaking, the assertion in the title is not even nearly true, as shows the example x ↦ ln |x|. However, we will see that, on compacts, BMO functions are in each L^{p}, p < ∞, and even better.

Proposition 13. Let f ∈ BMO. Then, for each ball B, f ∈ L^{p}(B) and ∥f − f_{B}∥_{L^{p}(B)} ≤ c_{p}|B|^{1/p}∥f∥_{BMO}. In addition, we have c_{p} ≤ c p when p ≥ 2.

Proof. We copy the proof of b) in the preceding theorem. When p = 1, the conclusion is trivial, so that we may assume that 1 < p < ∞. We may also assume that f_{B} = 0. Let q be the conjugate exponent of p. It is straightforward that, if g ∈ L^{q}(B), then ∥g − g_{B}∥_{L^{q}(B)} ≤ 2∥g∥_{L^{q}(B)}. On

the other hand, if g ∈ L^{q}(B) and ∫_{B} g = 0, then \frac{1}{|B|^{1−1/q}∥g∥_{L^{q}}} g is a _{q}atom and thus ∥g∥_{H^{1}} ≤

C_{q}|B|^{1−1/q}∥g∥_{L^{q}}. Here, C_{q} satisfies C_{q} ≤ \frac{c}{q − 1} ≤ c p when q ≤ 2 (and thus p ≥ 2). Thus

∥f∥_{L^{p}(B)} = sup{∫ f(g − g_{B}) ; ∥g∥_{L^{q}(B)} ≤ 1} ≤ 2c C_{q}∥f∥_{BMO}|B|^{1−1/q} = c_{p}∥f∥_{BMO}|B|^{1/p}. (9.19)

Theorem 16. (John-Nirenberg) There are constants c_{1}, c_{2} > 0 s. t.

|{x ∈ B |f − f_{B}| > α}| ≤ c_{1}|B| exp(−c_{2}α/∥f∥_{BMO}).  (9.20)

Proof. It is immediate that, if the conclusion holds for f, it also holds for a multiple of f. We may therefore assume that ∥f∥_{BMO} = 1. It is also clear that, if the conclusion holds for α ≥ 2c e, where c is the constant in the preceding proposition, then we may adjust the constants s. t. (9.20) holds for each α. We may therefore assume that α ≥ 2c e. We assume also that f_{B} = 0. Let

p = \frac{α}{c e} ≥ 2. Then

|\{x ∈ B ; |f| > α\}| ≤ \frac{∥f∥_{L^{p}(B)}^{p}}{α^{p}} ≤ \frac{(cp)^{p}|B|}{α^{p}} = |B| exp(−c α/e).  (9.21)

Theorem 17. (John-Nirenberg) There are constants C, k > 0 s. t. if f ∈ BMO and ∥f∥_{BMO} ≤ 1, then

\frac{1}{|B|} ∫_{B} exp(C|f − f_{B}|) ≤ k.  (9.22)

Remark 10. The normalization condition ∥f∥_{BMO} ≤ 1 is necessary. Indeed, if exp f ∈ L^{1}, there is no reason to have exp(2f) ∈ L^{1}. On the other hand, the constant C cannot be arbitrary large, as shown by the example x ↦ ln |x|.

Proof. We may assume that f_{B} = 0. Then

\frac{1}{|B|} ∫_{B} (exp(C|f|) − 1) = 1 + \frac{1}{|B|} ∑_{p=1}^{∞} C^{p}∥f∥_{L^{p}(B)}^{p} ≤ 1 + ∑_{p=1}^{∞} \frac{(cCp)^{p}}{p!} ≤ k < ∞  (9.23)

provided C < (ce)^{−1}, as it is easily seen using Stirling’s formula.

L^{p} regularity for the Laplace operator

10.1 Preliminaries

Let E be the fundamental solution of the Laplace operator in ℝ^{N}, N ≥ 2,

E(x) = \begin{cases}\frac{1}{2π} ln |x|, & if N = 2 \\ −\frac{1}{(N − 2)|S^{N−1}||x|^{N−2}} , & if N ≥ 3 \end{cases}. If f ∈ C_{0}^{∞}, then u = E ∗ f is a (classical) solution

of the equation (*) Δu = f. If f ∈ L_{0}^{p} for some 1 ≤ p ≤ ∞, we may still define u = E ∗ f and we then have u ∈ L_{loc}^{p}. Indeed, if K is a compact and L = supp f, let Φ ∈ C_{0}^{∞} be s. t. Φ = 1 in the compact K − L. Then, in K, we have E ∗ f = (ΦE) ∗ f, and thus

∥E ∗ f∥_{L^{p}(K)} ≤ ∥(ΦE) ∗ f∥_{L^{p}} ≤ ∥ΦE∥_{L^{1}}∥f∥_{L^{p}} ≤ C_{K,L}∥f∥_{L^{p}},  (10.1)

using Young’s inequality and the fact that E ∈ L_{loc}^{1}. In addition, u still satisfies (*), this time in the distribution sense. The reason is that we may approximate f with a sequence (f_{n}) ⊂ C_{0}^{∞} s. t. f_{n} → f in L^{1} and supp f_{n} ⊂ L′, with L′ a compact independent of n. Then (10.1) with p = 1 and L replaced by L′ implies that E ∗ f_{n} → E ∗ f in L_{loc}^{1}, and thus in D′. Since we also have Δ(E ∗ f_{n}) = f_{n} → f in D′, we find that Δ(E ∗ f) = f. Let now 1 ≤ j, k ≤ N and consider the operator T = T_{p} : L_{0}^{p} → D′, T f = ∂_{j}∂_{k}(E ∗ f). Note that the definition does not depend on p, in the sense that, if f ∈ L_{0}^{p} ∩ L_{0}^{q}, then T_{p}f = T_{q}f. We start by noting some simple properties of T that will be needed in the next section.

Lemma 17. If f ∈ L_{0}^{2}, then

T f = F^{−1}(\frac{ξ_{j}ξ_{k}}{|ξ|^{2}} f̂).  (10.2)

Consequently, T has a continuous extension to L^{2}, given by the r. h. s. of (10.2).

58 CHAPTER 10. L^{P} REGULARITY FOR THE LAPLACE OPERATOR

In addition, T is self-adjoint in L^{2}, i. e.

∫ T f \overline{g} = ∫ f \overline{T g}, ∀ f, g ∈ L^{2}.  (10.3)

Proof. The r. h. s. of (10.2) is continuous from L^{2} into L^{2} (and thus into D′), by Plancherel’s theorem. On the other hand, if L is a fixed compact, the l. h. s. is continuous from L_{L}^{2} (the space of L^{2} functions supported in L) into D′. Therefore, it suffices to prove the equality when f ∈ C_{0}^{∞}. Since (1 + |x|^{2})^{−N}E ∈ L^{1}, we have E ∈ S′, and thus ∂_{j}∂_{k}E ∈ S′. Therefore, T̂ f = { \atop ∂_{j}∂_{k}E}f̂, and

it suffices to prove that { \atop ∂_{j}∂_{k}E} = \frac{ξ_{j}ξ_{k}}{|ξ|^{2}} . We write E = E_{1} + E_{2}, where E_{1} = ΦE, E_{2} = (1 − Φ)E,

Φ ∈ C_{0}^{∞} and Φ = 1 near the origin. Then { \atop ∂_{j}∂_{k}E} = { \atop ∂_{j}∂_{k}E_{1}} + { \atop ∂_{j}∂_{k}E_{2}} ∈ C^{∞} + L^{2}, since ∂_{j}∂_{k}E ∈ E′, while ∂_{j}∂_{k}E_{2} ∈ L^{2}. On the other hand, Δ∂_{j}∂_{k}E = ∂_{j}∂_{k}δ, and thus |ξ|^{2}{ \atop ∂_{j}∂_{k}E} = ξ_{j}ξ_{k}. Thus

{ \atop ∂_{j}∂_{k}E} = \frac{ξ_{j}ξ_{k}}{|ξ|^{2}} + ∑_{|α|≤2} c_{α}∂^{α}δ. The coefficients c_{α} must be zero, since { \atop ∂_{j}∂_{k}E} ∈ C^{∞} + L^{2}, whence the

first conclusion of the lemma. As for (10.3), it follows from Plancherel’s theorem:

∫ T f \overline{g} = (2π)^{−N} ∫ T̂ f \overline{ĝ} = (2π)^{−N} ∫ \frac{ξ_{j}ξ_{k}}{|ξ|^{2}} f̂ \overline{ĝ} = (2π)^{−N} ∫ f̂\overline{\frac{ξ_{j}ξ_{k}}{|ξ|^{2}} ĝ} = (2π)^{−N} ∫ f̂ \overline{T̂ g} = ∫ f \overline{T g}.

(10.4)

Lemma 18. Assume that f ∈ L_{0}^{p} and let x ∉ supp f. Then, with K(x) = \frac{1}{|S^{N−1}|}(\frac{δ_{j,k}}{|x|^{N}} −\frac{Nx_{j}x_{k}}{|x|^{N+2}} ),

we have

T f(x) = ∫ K(x − y)f(y)dy.  (10.5)

In addition, K satisfies

|K(x − y) − K(x)| ≤ \frac{C|y|}{|x|^{N+1}} , if |y| < 1/2|x|.  (10.6)

Proof. If L = supp f and O is a relatively compact open set s. t. \overline{O} ∩ L = ∅, then the (pointwise) derivatives of E(x − y)f(y) with respect to x satisfy

|∂_{x}^{α}(E(x − ·)f(·)| ≤ c_{α}|f(·)| ∈ L^{1}, x ∈ O,  (10.7)

and thus E ∗ f ∈ C^{∞}(O). Moreover, we may differentiate twice under the integral sign in the formula of E ∗ f to obtain, in O, both the pointwise and the distributional derivative ∂_{j}∂_{k}(E ∗ f)

through the formula ∂_{j}∂_{k}(E ∗ f) = ∫ ∂_{j}∂_{k}E(x − y)f(y)dy. Here, ∂_{j}∂_{k}E stands for the pointwise derivative. Finally, we have ∂_{j}∂_{k}E = K, whence the first conclusion. To prove the inequality (10.6), we note that |DK(z)| ≤ C|z|^{−N−1} and thus, for x, y s. t. |y| < 1/2|x|, we have

|K(x − y) − K(x)| ≤ |y| sup_{z∈[x−y,x]} |DK(z)| ≤ C|y| sup_{z∈[x−y,x]} |z|^{−N−1} ≤ \frac{C|y|}{|x|^{N+1}} .  (10.8)

Lemma 19. Let Φ ∈ C_{0}^{∞}. Then

a) (T Φ)_{t} = T Φ_{t}, for each t > 0.

b) T Φ ∈ L^{∞}.

c) D(T Φ) ∈ L^{∞}.

d) If |y| < 1/2|x|, then |T Φ(x − y) − T Φ(x)| ≤ \frac{C|y|}{|x|^{N+1}} .

Proof. a) Actually, this holds under the sole assumption that Φ ∈ L^{2}. It suffices to check that { \atop (T Φ)_{t}} = T̂ Φ_{t}. This equality follows from

{ \atop (T Φ)_{t}}(ξ) = { \atop (T Φ)}(tξ) = \frac{ξ_{j}ξ_{k}}{|ξ|^{2}} Φ̂(tξ) = \frac{ξ_{j}ξ_{k}}{|ξ|^{2}} Φ̂_{t}(ξ) = T̂ Φ_{t}(ξ).  (10.9)

b) Since |T̂ Φ| ≤ |Φ̂|, we find that T̂ Φ ∈ L^{1} and thus T Φ ∈ L^{∞}.

c) Similarly, { \atop D(T Φ)} = ıξT̂ Φ, and thus { \atop D(T Φ)} ∈ L^{1}, which implies that D(T Φ) ∈ L^{∞}.

d) Let R > 0 be s. t. Φ = 0 outside B(0, R). If |x| ≤ 3R, then the conclusion follows from b). Assume |x| > 3R. Then both x and x − y are outside the support of Φ, which implies that

T Φ(x − y) − T Φ(x) = ∫_{B(0,R)} (K(x − y − z) − K(x − z))Φ(z)dz. Therefore,

|T Φ(x − y) − T Φ(x)| ≤ C sup_{|z|≤R} |K(x − y − z) − K(x − z)| ≤ \frac{C|y|}{|x|^{N+1}} ;  (10.10)

here, we rely on the inequality (10.6) and we take into account the fact that |x−y −z| ∼ |x−z| ∼ |x|.

In the next section, we will prove the following

Theorem 18. a) (Calderón-Zygmund) For p = 1, the operator T, initially defined on L_{loc}^{1}, has a continuous extension from L^{1} into L_{w}^{1}.

b) (Fefferman-Stein) When restricted to H^{1}, the extension of T to L^{1} maps continuously H^{1} into H^{1}. c) (Calderón-Zygmund) For 1 < p < ∞, the operator T, initially defined on L_{loc}^{p}, has a continuous extension from L^{p} into L^{p}.

d) (Spanne-Peetre-Stein) T maps BMO_{0} continuously into BMO and thus L_{0}^{∞} continuously into BMO.

The most widely used form of the above result sais that a solution u of Δu = f ”gains two derivatives with respect to f”:

Corollary 15. Assume that Δu = f in the distribution sense.

a) If f ∈ L_{loc}^{p} for some 1 < p < ∞, then u ∈ W_{loc}^{2,p}.

b) If f ∈ H^{1}, then u ∈ W_{loc}^{2,1}.

Proof. Let K be a compact in ℝ^{N} and let Φ ∈ C_{0}^{∞} be s. t. Φ = 1 in an open neighborhood O of K. Set g = Φf ∈ L_{0}^{p} and let v = E ∗ g, which satisfies Δv = f in O. Then Δ(u − v) = 0 in O, and thus u − v ∈ C^{∞}(O), by Weyl’s lemma. Now v ∈ L_{loc}^{p}, since g ∈ L_{0}^{p}, and the second order derivatives of v are in L^{p} if f ∈ L_{loc}^{p} and 1 < p < ∞, respectively in L^{1} if f ∈ H^{1}. In addition, it is easy to see that the distributional first order derivatives of E ∗ f are computed according to

the formula ∂_{j}(E ∗ g)(x) = ∫ (∂_{x_{j}}E)(x − y)f(y)dy, where ∂_{x_{j}}E stands for the pointwise derivative

(this is obtained using an integration by parts when f ∈ C_{0}^{∞}; the general case is obtained by approximation, with the help of Young’s inequality). Since ∂_{x_{j}}E ∈ L_{loc}^{1} and (in all the cases) g ∈ L_{0}^{p}, we find that ∂_{j}v ∈ L_{loc}^{p}. Therefore, u ∈ W_{loc}^{2,p}.

The above results are optimal, in the following sense:

Proposition 14. T does not map L_{0}^{1} into L^{1} and does not map L_{0}^{∞} into L^{∞}.

Proof. We fix a compact L in ℝ^{N}. We already noted that T maps continuously L_{L}^{p} into D′. We claim that, if T : L_{L}^{p} → L^{p}, then T has to be continuous. Indeed, let f_{n} → f in L_{L}^{p} be s. t. T f_{n} → g in L^{p}. Since T f_{n} → T f in D′, we find that T f = g, and thus T has closed graph. Therefore, T is continuous. Let now p = 1. We argue by contradiction. Let L be a ball containing the origin. We consider a sequence (f_{n}) ⊂ C_{0}^{∞} s. t. ∥f_{n}∥_{L^{1}} ≤ C, supp f_{n} ⊂ L and f_{n} → δ in D′ and set u_{n} = E ∗ f_{n}. Then u_{n} → E in D′ and ∥D^{2}u_{n}∥_{L^{1}} ≤ C. On the other hand, Du_{n} = (DE) ∗ f_{n} (where DE ∈ L_{loc}^{1} is the pointwise derivative of E), and thus ∥Du_{n}∥_{L^{1}(L)} ≤ C. Consequently, the sequence (Du_{n}) is bounded in W^{1,1}(L). The Sobolev embeddings imply that (Du_{n}) is bounded also in L^{N/(N−1)}(L). Since Du_{n} → DE in D′, we find that DE ∈ L^{N/(N−1)}(L); thus |DE|^{N/(N−1)} is integrable near the origin. However, if we compute the (pointwise or distributional) gradient DE, we see that |DE(x)| ∼ |x|^{−(N−1)}, a contradiction. We next consider the case p = ∞. Argue again by contradiction. Recall that there is a function u : ℝ^{N} → ℝ, u ∉ C^{2}, s. t. f = Δu (computed in the distributional sense) be continuous (example due to Weierstrass). We may assume, e. g., that u ∉ C^{2}(B(0, 1)). Let g = Φf, where Φ ∈ C_{0}^{∞}, Φ = 1 in B(0, 1), supp Φ ⊂ B(0, 2). Then g ∈ L_{0}^{∞}, and thus T f ∈ L^{∞} (for all j, k). Let (g_{n}) ⊂ C_{0}^{∞} be s. t. g_{n} → g uniformly, supp g_{n} ⊂ B(0, 2). Then T g_{n} → T g uniformly. Since T g_{n} ∈ C^{∞}, this implies that T g is continuous. Thus E ∗ g ∈ C^{2}. Since Δ(E ∗ g) = Δu in B(0, 1), Weyl’s lemma implies that u ∈ C^{2}(B(0, 1)), a contradiction.

10.2 Proof of Theorem 18

Proof. The plan of the proof is the following: a) we prove that T maps L^{1} into L_{w}^{1}; this will rely on the Calderón-Zygmund decomposition. b) Marcinkiewicz’ interpolation theorem, combined with the continuity of T from L^{2} into L^{2}, will imply the result when 1 < p < 2. c) For H^{1}, the result is obtained via the atomic decomposition. d) The remaining cases, i. e. 2 < p ≤ ∞ or BMO_{0}, will be obtained by duality; we will exploit the fact that T is a symmetric operator.

Step 1. Continuity from L^{1} into L_{w}^{1} It suffices to prove the following estimate

|\{|T f| > t\}| ≤ \frac{C}{t} ∥f∥_{L^{1}}, ∀ t > 0, ∀ f ∈ L^{1} ∩ L^{2}.  (10.11)

Indeed, assume (10.11) proved, for the moment. Let f ∈ L^{1} and consider a sequence (f_{n}) ⊂ L^{1}∩L^{2} s. t. f_{n} → f in L^{1}. Then (10.11) applied to f_{n} −f_{m} implies that |{|T f_{n} −T f_{m}| > t}| → 0 when t is fixed and m, n → ∞.Thus (T f_{n}) is a Cauchy sequence in measure, and thus it converges in measure to some g. In particular, this implies that g does not depend on the sequence (f_{n}), that g = T f if f happens to be in L^{1} ∩ L^{2} and that f ↦ g is linear. Possibly after passing to a subsequence

(f_{n_{k}}), we have T f_{n_{k}} → g a. e., and thus |\{|g| > t\}| ≤ lim inf_{k} |\{|T f_{n_{k}}| > t\}| ≤ \frac{C}{t} ∥f∥_{L^{1}}. Thus,

f ↦ g is the desired extension of T. (We needed this argument since L_{w}^{1} is not a normed space.) We now return to the proof of (10.11). Let t > 0. We write, as in Theorem 10 (with α replaced by t), a function f ∈ L^{1} ∩ L^{2} as f = g + ∑ h_{n}. We first note that g ∈ L^{2}, since g ∈ L^{1} and |g| ≤ Ct.

We claim that the series ∑ h_{n} is convergent in L^{2}. Noting that the functions h_{n} are mutually

orthogonal in L^{2}, it suffices to prove that ∑ ∥h_{n}∥_{L^{2}}^{2} < ∞. Since ∥h_{n}∥_{L^{2}} = ∥f − f_{C_{n}}∥_{L^{2}(C_{n})} ≤

∥f∥_{L^{2}(C_{n})}, we find that ∑ ∥h_{n}∥_{L^{2}}^{2} ≤ ∑ ∥f∥_{L^{2}(C_{n})}^{2} ≤ ∥f∥_{L^{2}}^{2}, whence the claim.

This allows us to write T f = T (g + ∑ h_{n}) = T g + ∑ T h_{n}. On the one hand, we have

|{|T f| > t}| ≤ |{|T g| > t/2}| + |{|T ∑ h_{n}| > t/2}|.  (10.12)

Since ∫ |g|^{2} ≤ Ct ∫ |g| ≤ Ct∥f∥_{L^{1}} (by the properties of the Calderón-Zygmund decomposition), we obtain

|\{|T g| > t/2\}| ≤ \frac{C}{(t/2)^{2}} ∥g∥_{L^{2}}^{2} ≤ \frac{C}{t} ∥f∥_{L^{1}}.  (10.13)

On the other hand, let, for each n, C_{n}^{∗} be the cube concentric with C_{n} and twice bigger than it. Then, with A = ℝ^{N} \ ⋃ C_{n}^{∗}, we have

|\{|T ∑ h_{n}| > t/2\}| ≤ | ⋃ C_{n}^{∗}| + |\{x ∈ A ; ∑ |T h_{n}| > t/2\}| ≤ C ∑ |C_{n}| + \frac{C}{t} ∑ ∥T h_{n}∥_{L^{1}(A)}.

(10.14)

We denote by \overline{x}_{n} the center of C_{n} and by l_{n} its size. For x ∈ A, we have

T h_{n}(x) = ∫ K(x − y)h_{n}(y)dy = ∫ [K(x − y) − K(x − \overline{x}_{n})]h_{n}(y)dy,  (10.15)

and thus

|T h_{n}(x)| ≤ \frac{C}{|x − \overline{x}_{n}|^{N+1}} ∫ |y − \overline{x}_{n}||h_{n}|dy ≤ \frac{Cl_{n}}{|x − \overline{x}_{n}|^{N+1}} ∥h_{n}∥_{L^{1}}.  (10.16)

Integrating the above inequality and summing over n, we find that

∑ ∥T h_{n}∥_{L^{1}(A)} ≤ C ∑ ∥h_{n}∥_{L^{1}} ≤ C∥f∥_{L^{1}},  (10.17)

by the properties of the Calderón-Zygmund decomposition. We conclude the first step by combining (10.12), (10.13), (10.14), (10.17) and the fact that

∑ |C_{n}| ≤ \frac{C}{t} ∥f∥_{L^{1}}.

Step 2. Continuity in L^{p}, 1 < p < ∞ We know that T, when defined in L^{1} ∩ L^{2}, is continuous from L^{1} into L_{w}^{1} and from L^{2} into L^{2}. Marcinkiewicz’ interpolation theorem implies that T has a unique extension continuous from L^{p} into L^{p} when 1 < p < 2. Let now 2 < p < ∞. Part c) of the theorem follows if we prove that ∥T f∥_{L^{p}} ≤ C∥f∥_{L^{p}} whenever f ∈ L^{p} ∩ L^{2}. For such an f, we have, with q < 2 the conjugate exponent of p,

∥T f∥_{L^{p}} = sup_{g∈L^{q} ; ∥g∥_{L^{q}} ≤1} ∫ T f \overline{g} = sup_{g∈L^{q}∩L^{2} ; ∥g∥_{L^{q}} ≤1} ∫ T f \overline{g} = sup_{g∈L^{q}∩L^{2} ; ∥g∥_{L^{q}} ≤1} ∫ f\overline{T g} ≤ C∥f∥_{L^{p}};

(10.18)

here, we use the continuity of T in L^{q}.

Step 3. Continuity in H^{1} In view of the properties of the atomic decomposition, it suffices to prove, with a constant C independent of a, the estimate

∥T a∥_{H^{1}} ≤ C, ∀ atom a.  (10.19)

Let a be an atom supported in B = B(\overline{x}, R). Let Φ ∈ C_{0}^{∞} be s. t. ∫ Φ = 1 and supp Φ ⊂ B(0, 1).

For each x, we have M_{Φ}a(x) ≤ CMa(x), and thus

∫_{B(\overline{x},2R)} M_{Φ}a ≤ C ∫_{B(\overline{x},2R)} Ma ≤ ∥Ma∥_{L^{2}}|B(\overline{x}, 2R)|^{1/2} ≤ C∥a∥_{L^{2}}|B|^{1/2} ≤ C.  (10.20)

We consider now an x outside B(\overline{x}, 2R) and estimate Φ_{t} ∗ (T a)(x). We have (we take a, Φ real, here)

Φ_{t}∗(T a)(x) = ∫ Φ_{t}(x−y)T a(y)dy = ∫ (T Φ_{t})(x−y)a(y)dy = ∫_{B} [(T Φ)_{t}(x−y)−(T Φ)_{t}(x−\overline{x})]a(y)dy.

(10.21)

We next note that, when y ∈ B, we have |x − y| < 1/2|x − \overline{x}|. We intend to make use of the decay properties of T Φ. To this purpose, we distinguish two possibilities concerning the size of t:

(i) t > |x − \overline{x}| and (ii) t ≤ |x − \overline{x}|. In case (i), we use the fact that T Φ is Lipschitz, and find that

|(T Φ)_{t}(x − y) − (T Φ)_{t}(x − \overline{x})| ≤ Ct^{−N−1}|y − \overline{x}|,  (10.22)

and thus

|Φ_{t} ∗ (T a)(x)| ≤ \frac{C}{t^{N+1}} ∫_{B} |y − \overline{x}||a(y)|dy ≤ \frac{Cl}{t^{N+1}} ≤ \frac{Cl}{|x − \overline{x}|^{N+1}} .  (10.23)

In case (ii), we make use of Lemma 19 d), and obtain

|(T Φ)_{t}(x − y) − (T Φ)_{t}(x − \overline{x})| ≤ \frac{C}{|x − \overline{x}|^{N+1}} |y − \overline{x}|,  (10.24)

which gives

|Φ_{t} ∗ (T a)(x)| ≤ \frac{C}{|x − \overline{x}|^{N+1}} ∫_{B} |y − \overline{x}||a(y)|dy ≤ \frac{Cl}{|x − \overline{x}|^{N+1}} .  (10.25)

(10.23) combined with (10.25) yields

M_{Φ}a(x) ≤ \frac{Cl}{|x − \overline{x}|^{N+1}} when x ∉ B(\overline{x}, 2R).  (10.26)

Integration of (10.26) over ℝ^{N} \ B(\overline{x}, 2R) combined with (10.20) gives the needed conclusion ∥M_{Φ}a∥_{L^{1}} ≤ C.

Step 4. Continuity of T in BMO_{0} We note that BMO_{0} ⊂ L_{0}^{2}, by the John-Nirenberg inequalities. We also note that the vector space V spanned by the atoms is contained in L^{2}. Thus, for f ∈ BMO_{0}, we have

∥T f∥_{BMO} ∼ sup_{g∈V,∥g∥_{H^{1}}≤1} ∫ T f \overline{g} = sup_{g∈V,∥g∥_{H^{1}}≤1} ∫ f \overline{T g} ≤ C∥f∥_{BMO};  (10.27)

here, we used the duality between H^{1} and BMO, the density of V in H^{1} and the continuity of T from H^{1} into H^{1}.

10.3 An equation involving the jacobian

We consider, in ℝ^{2}, the following equation that appears in Geometry

Δu = det(Df, Dg), f, g ∈ H^{1}(ℝ^{2}).  (10.28)

Theorem 19. a) (Wente) Equation (10.28) has one and only one distribution solution u ∈ C(ℝ^{2}) vanishing at infinity, i. e., s. t. lim_{|x|→∞} u(x) = 0. In addition, Du ∈ L^{2} and

∥Du∥_{L^{2}} ≤ C∥Df∥_{L^{2}}∥Dg∥_{L^{2}}.  (10.29)

b) (Coifman-Lions-Meyer-Semmes) In addition, we have D^{2}u ∈ H^{1}.

Proof. The main argument in the proof is that

det(Df, Dg) ∈ H^{1} and ∥ det(Df, Dg)∥_{H^{1}} ≤ C∥Df∥_{L^{2}}∥Dg∥_{L^{2}}.  (10.30)

Assuming (10.30) proved for the moment, we reason as follows: let h = det(Df, Dg). Consider sequences (f_{n}), (g_{n}) ⊂ C_{0}^{∞} s. t. f_{n} → f, g_{n} → g in H^{1}. Then h_{n} = det(Df_{n}, Dg_{n}) → h in H^{1}, by (10.30). Let u_{n} = E ∗ h_{n}, which is a solution of Δu_{n} = h_{n}. We claim that u_{n} ∈ C_{∞} (the space of continuous functions vanishing at infinity) and that (u_{n}) is a Cauchy sequence for the sup norm. Indeed, let, for fixed n, R = R_{n} > 0 be s. t. h_{n}(y) = 0 if |y| > R. Then

|u_{n}(x)| = \frac{1}{2π} | ∫ ln |x − y|h_{n}(y)dy| = \frac{1}{2π} | ∫ [ln |x − y| − (ln | · |)_{B(x,R)}]h_{n}(y)dy|,  (10.31)

and thus

|u_{n}(x)| ≤ C_{n} ∫_{B(0,R)} | ln |x − y| − (ln | · |)_{B(x,R)}|dy = C_{n} ∫_{B(x,R)} | ln y − (ln | · |)_{B(x,R)}|dy → 0 as |x| → ∞.

(10.32)

On the other hand, we have ln ∈ BMO and thus, using the H^{1}-BMO duality,

|u_{n}(x)−u_{m}(x)| = \frac{1}{2π} | ∫ ln |y|(h_{n}(x−y)−h_{m}(x−y))dy| ≤ C∥(h_{n}−h_{m})(x−·)∥_{H^{1}} = C∥h_{n}−h_{m}∥_{H^{1}},

(10.33) since the H^{1} norm is translation invariant. (Similarly, we have |u_{n}| ≤ C∥h_{n}∥_{H^{1}}.) To summarize, the sequence (u_{n}) is Cauchy in C_{∞}, and thus converges to some u ∈ C_{∞}. This u is a distribution solution of (10.28). It is also the only solution of (10.28) in C_{∞}, for if v is another solution, their difference w is, by Weyl’s lemma, a harmonic function vanishing at infinity, thus constant, by the maximum principle. We now turn to the proof of (10.29) and b) (assuming, again, (10.30) already proved). Note that b), at least when f, g ∈ C_{0}^{∞}, follows by combining (10.30) and the Fefferman-Stein regularity result concerning the equation Δu = f with f ∈ H^{1}. The general case is obtained by approximation, as above. Similarly, it suffices to establish (10.29) when f, g ∈ C_{0}^{∞}. Formally, estimate (10.29) is clear, as shown by the following (wrong, in principle) computation:

∫ |Du|^{2} = − ∫ uΔu = − ∫ uh ≤ ∥u∥_{L^{∞}}∥h∥_{L^{1}} ≤ C∥u∥_{L^{∞}}∥h∥_{H^{1}} ≤ C∥Df∥_{L^{2}}^{2}∥Dg∥_{L^{2}}^{2}. (10.34)

The point is that this computation can be transformed into a rigorous one as follows: set F (r) =

\frac{1}{2πr} ∫_{|x|=r} |u|^{2}dl. Then lim_{r→∞} F (r) = 0 and F′(r) = \frac{1}{πr} ∫_{|x|=r} u · u_{r}dl. Thus, along a subsequence

r_{n} → ∞, we must have r_{n}F′(r_{n}) → 0 (argue by contradiction; otherwise, we have F (r) ≥ C ln r for large r). Then, for large n, we have Δu = 0 outside B(0, r_{n}) and thus

∫ |Du|^{2} = lim_{n} ∫_{B(0,r_{n})} |Du|^{2} = lim_{n}{ ∫_{|x|=r} u · u_{r} − ∫ uΔu} = − ∫ uΔu ≤ C∥Df∥_{L^{2}}^{2}∥Dg∥_{L^{2}}^{2}. (10.35)

The only part of the proof left open is

Proof of (10.30) In view of the conclusion we want to obtain, we may assume that f, g ∈ C_{0}^{∞}. Let Φ ∈ C_{0}^{∞} be

supported in B(0, 1) and s. t. ∫ Φ = 1. Then, with h = det(Df, Dg), we have

Φ_{t} ∗ h(x) = ∫ Φ_{t}(y)h(x − y)dy = t^{−1} ∫ f(x − y) det((DΦ)_{t})(y), Dg(x − y))dy,  (10.36)

as shown by an integration by parts. Next, if k, l ∈ C_{0}^{∞}, then ∫ det(Dk, Dl) = 0 (again, this

follows by an integration by parts), and thus

Φ_{t} ∗ h(x) = t^{−1} ∫ [f(x − y) − f_{B(x,t)}] det((DΦ)_{t})(y), Dg(x − y))dy.  (10.37)

Using the Hölder inequality and the inequality |(DΦ)_{t}| ≤ Ct^{−2} together with the fact that Φ_{t} vanishes outside B(0, t), we find that

|Φ_{t} ∗ h(x)| ≤ t^{−3}( ∫_{B(x,t)} |f − f_{B(x,t)}|^{4})^{1/4}( ∫_{B(x,t)} |Dg|^{4/3})^{3/4}.  (10.38)

Applying Lemma 20 below to the function given by v(y) = f(x − ty), we find that

∥f − f_{B(x,t)}∥_{L^{4}(B(x,t)} ≤ C∥Df∥_{L^{4/3}(B(x,t)},  (10.39)

and thus (10.38) yields

|Φ_{t} ∗ h(x)| ≤ C(\frac{1}{|B(x, t)|} ∫_{B(x,t)} |Df|^{4/3})^{3/4}(\frac{1}{|B(x, t)|} ∫_{B(x,t)} |Dg|^{4/3})^{3/4}.  (10.40)

Recalling the definition of the maximal function, we obtain

M_{Φ}h(x) ≤ C(M|Df|^{4/3}(x))^{3/4}(M|Dg|^{4/3}(x))^{3/4},  (10.41)

and the Cauchy-Schwarz inequality implies that

∥M_{Φ}h∥_{L^{1}} ≤ C( ∫ (M|Df|^{4/3})^{3/2})^{1/2}( ∫ (M|Dg|^{4/3})^{3/2})^{1/2},  (10.42)

which may be rewritten as

∥M_{Φ}h∥_{L^{1}} ≤ C∥M|Df|^{4/3}∥_{L^{3/2}}^{3/4}∥M|Dg|^{4/3}∥_{L^{3/2}}^{3/4} ≤ C∥|Df|^{4/3}∥_{L^{3/2}}^{3/4}∥|Df|^{4/3}∥_{L^{3/2}}^{3/4} = C∥Df∥_{L^{2}}∥Dg∥_{L^{2}};

(10.43)

that is, ∥h∥_{H^{1}} ≤ C∥Df∥_{L^{2}}∥Dg∥_{L^{2}}, as claimed at the beginning of the proof.

We next recall the following Sobolev embedding and the corresponding Poincaré inequality

Lemma 20. W^{1,4/3}(ℝ^{2}) is embedded into L^{4} and, with B = B(0, 1) and v ∈ W^{1,4/3}(B), we have

∥v − v_{B}∥_{L^{4}(B)} ≤ C∥Dv∥_{L^{4/3}(B)}.  (10.44)

Proof. The above Sobolev embedding will be proved, in a slightly better form, in the next chapter. We present a proof of (10.44), which is less standard. The starting point is the usual Poincaré inequality

∥v − v_{B}∥_{L^{4/3}(B)} ≤ C∥Dv∥_{L^{4/3}(B)}.  (10.45)

We may assume, with no loss of generality, that v_{B} = 0. We extend v by reflections in a

neighborhood of B by setting ṽ(x) = \begin{cases}v(x), & if x ∈ B \\ v(x/|x|^{2}), & if 1 < |x| < 3/2 \end{cases}. The new function ṽ is in W^{1,4/3}(B(0, 3/2) and satisfies ∥ṽ∥_{L^{4/3}} ≤ C∥v∥_{L^{4/3}} and ∥Dṽ∥_{L^{4/3}} ≤ C∥Dv∥_{L^{4/3}}. Next let Φ ∈ C_{0}^{∞} be s. t. Φ = 1 in B and supp Φ ⊂ B(0, 3/2). Set w = Φṽ ∈ W^{1,4/3}(ℝ^{2}). Then

∥v∥_{L^{4}(B)} ≤ ∥w∥_{L^{4}} ≤ C∥Dw∥_{L^{4/3}} ≤ C(∥ṽ∥_{L^{4/3}} + ∥Dṽ∥_{L^{4/3}}) ≤ C(∥v∥_{L^{4/3}} + ∥Dv∥_{L^{4/3}}) ≤ C∥Dv∥_{L^{4/3}}. (10.46)

68 CHAPTER 10. L^{P} REGULARITY FOR THE LAPLACE OPERATOR

Part III

Functions in Sobolev spaces Improved Sobolev embeddings

The usual form of the Sobolev embeddings states that W^{1,p}(ℝ^{N}) ⊂ L^{Np/(N−p)}, provided 1 ≤ p < N. In this chapter, we will improve the conclusion to W^{1,p}(ℝ^{N}) ⊂ L^{Np/(N−p),p}; this is slightly better,

since \frac{NP}{N − p} > p, and thus L^{Np/(N−p),p} ⊂ L^{Np/(N−p)}.

11.1 An equivalent norm in Lorentz spaces

Let f : ℝ^{N} → ℂ be a measurable function and let F : (0, ∞) → [0, ∞] be its distribution function. Intuitively, we may think of F as a bijection of (0, ∞) into itself. Then, if p, q < ∞ and if f^{∗} = F^{−1} (which is decreasing), we may (formally) compute the L^{p,q} quasi-norm as follows:

∥f∥_{L^{p,q}}^{q} = ∫ t^{q−1}F^{q/p}(t)dt = − ∫ s^{q/p}(f^{∗})^{q−1}(s)f^{∗}′(s)ds = p^{−1} ∫ s^{q/p−1}(f^{∗})^{q}(s)ds;  (11.1)

here, the - sign at the beginning of the computation comes from the fact that F is decreasing. The second equality is obtained through the change of variable F (t) = s, the third one arises after an integration by parts.

The above equality maybe rewritten as

∥f∥_{L^{p,q}} = ∥t^{1/p}F ∥_{L^{q}((0,∞);dt/t)} ∼ ∥t^{1/p}f^{∗}∥_{L^{q}((0,∞);dt/t)}.  (11.2)

In this section, we will see that this formula is right!...provided we interpret it accurately.

Definition 4. The non increasing rearrangement f^{∗} : [0, ∞) → [0, ∞] of f is defined through the formula

f^{∗}(t) = sup{s > 0 ; F (s) ≤ t}.  (11.3)

We note that, when F is a bijection, we have f^{∗} = F^{−1}. Before going further, we warn the reader that all the functions f we will rearrange in this chapter satisfy the following property

(H) lim_{t→∞} F(t) = 0.

This is not too restrictive, since we will deal with functions in Lorentz spaces. These functions satisfy the inequality F (t) ≤ Ct^{−p}, thus they do satisfy (H). Note that, in particular, this implies that f^{∗}(t) < ∞ for each t. Hypothesis (H), though needed sometimes in the proofs, will never be mentioned explicitly as hypothesis. The elementary results we gather below explain, in particular, why f^{∗} is called the non increasing rearrangement of f.

Proposition 15. a) F is continuous from the right.

b) F (f^{∗}(t)) ≤ t everywhere (in other words, inf = min in the definition of f^{∗}).

c) f^{∗} is non increasing and continuous from the right.

d) f and f^{∗} are equally distributed, i. e., |{x ∈ ℝ^{N} ; |f(x)| > t}| = |{s ∈ (0, ∞) ; f^{∗}(s) > t}| for each t > 0.

e) f^{∗} depends continuously on f in the following sense: if (f_{n}) is a sequence of functions s. t. |f_{n}(x)| ↗ |f(x)| for a. e. x ∈ ℝ^{N}, then f_{n}^{∗}(t) ↗ f^{∗}(t) for each t > 0.

f) We have (f + g)^{∗}(2t) ≤ f^{∗}(t) + g^{∗}(t). More generally, ( ∑ f_{j})^{∗}(∑ t_{j}) ≤ ∑(f_{j})^{∗}(t_{j}).

Proof. a) follows from the equality {|f| > t} = ⋃{|f| > t + 1/n}, which implies that F (t) = lim F (t + 1/n).

b) Let s = f^{∗}(t). Then F (s + ε) ≤ t for ε > 0, and thus F (s) ≤ t.

c) The fact that f^{∗} is non increasing is clear from the definition. Concerning the second assertion, it suffices to prove that f^{∗}(t + 0) ≥ f^{∗}(t). Let t_{n} ↘ t. Then F (f^{∗}(t_{n})) ≤ t_{n}, which implies that F (f^{∗}(t + 0)) ≤ t_{n} and thus F (f^{∗}(t + 0)) ≤ t, that is f^{∗}(t + 0) ≥ f^{∗}(t).

d) Since f^{∗} is non increasing, we have |{s ∈ (0, ∞) ; f^{∗}(s) > t}| = τ, where τ is uniquely defined by f^{∗}(s) > t if s < τ and f^{∗}(s) ≤ t if s > τ. In view of the conclusion we want, it suffices to check that f^{∗}(s) > t if s < F (t) and that f^{∗}(s) ≤ t if s > F (t). If s < F (t), then F (f^{∗}(s)) ≤ s < F (t) and thus f^{∗}(s) > t. On the other hand, if s > F (t), then t ≥ f^{∗}(s), by definition of f^{∗}(s).

e) We note that |f| ≤ |g| ⟹ f^{∗} ≤ g^{∗}; therefore, the sequence (f_{n}^{∗}) is non decreasing and h(t) := lim f_{n}^{∗}(t) ≤ f^{∗}(t) for each t. Hence, it suffices to prove that h(t) ≥ f^{∗}(t), i. e., that F (h(t)) ≤ t. We note that, for each s, we have F_{n}(s) → F (s), since the set {|f| > s} is the union of the non decreasing sequence ({|f_{n}| > s}). Thus F_{n}(f_{n}^{∗}(t)) ≤ t ⟹ F_{n}(h(t)) ≤ t ⟹ F (h(t)) ≤ t, as needed.

f) Let s = f^{∗}(t) and τ = g^{∗}(t). Then |{|f| > s}| ≤ t and |{|g| > τ}| ≤ t. Since {|f + g| > s + τ} ⊂ {|f| > s} ∪ {|g| > τ}, we find that |{|f + g| > s + τ}| ≤ 2t, i. e., (f + g)^{∗}(2t) ≤ s + τ = f^{∗}(t) + g^{∗}(t).

We next justify the equality (11.1).

Proposition 16. For 1 ≤ p < ∞ and 1 ≤ q ≤ ∞, we have ∥f∥_{L^{p,q}} ∼ ∥t^{1/p}f^{∗}∥_{L^{q}((0,∞);dt/t)}. For p = ∞, we have ∥f∥_{L^{∞}} = ∥f^{∗}∥_{L^{∞}}.

Proof. We start with the case p = ∞; we will prove the equality of the quasi-norms. Indeed,

∥f∥_{L^{∞,q}} = ∥f∥_{L^{∞}} = inf{s ; F (s) = 0} = inf{s ; F (s) ≤ 0} = f^{∗}(0) = ∥f^{∗}∥_{L^{∞}},  (11.4)

since f^{∗} is non increasing and continuous from the right. Let now p < ∞ and q = ∞; once again, we will prove the equality of the quasi-norms. ” ≤ ” Let C = ∥f∥_{L^{p,∞}} = sup tF^{1/p}(t). Let t > 0. With s = f^{∗}(t), we want to prove that t^{1/p}s ≤ C. If s = 0, there is nothing to prove. If s = ∞, then F (τ) > t for each τ, and then C = ∞. If s ∈ (0, ∞), then F (s − ε) > t for small ε > 0, and thus

t^{1/p}s < f^{1/p}(s − ε)s ≤ \frac{Cs}{s − ε},  (11.5)

and the desired conclusion follows by letting ε → 0. ” ≥ ” With C = sup t^{1/p}f^{∗}(t), we will prove that tF^{1/p}(t) ≤ C for each t > 0. If F (t) = 0, there is nothing to prove. If F (t) = ∞, then f^{∗}(s) ≥ t for each s, and thus C = ∞. Finally, if u = F (t) ∈ (0, ∞), let, for small ε > 0, u_{ε} = u − ε > 0. Then F (t) > u_{ε} and thus f^{∗}(u_{ε}) > t. We find that

tF^{1/p}(t) ≤ f^{∗}(u − ε)u^{1/p},  (11.6)

and we conclude by letting ε → 0. Finally, we consider the case 1 ≤ p, q < ∞. In view of the preceding proposition, it suffices to prove the equality p∥f∥_{L^{p,q}}^{q} = ∥t^{1/p}f^{∗}∥_{L^{q}((0,∞);dt/t)}^{q} when f is a step function; the general case will follow by monotone convergence, by approximating an arbitrary function f with a sequence (f_{n}) s. t. each f_{n} is a step function and |f_{n}| ↗ |f|. In addition, since the quantities we consider do not distinguish between f and |f|, we may assume that f ≥ 0. Let then f = ∑ a_{n}χ_{A_{n}}, where a_{1} > a_{2} > … > a_{k} > 0 and the sets A_{n} are measurable and mutually disjoint. Set b_{n} = |A_{n}|, c_{l} = b_{1} + … + b_{l}, c_{0} = 0 and c_{k+1} = ∞. Then, with a_{0} = ∞ and a_{k+1} = 0, we have F (t) = c_{l} if t ∈ [a_{l+1}, a_{l}). On the other hand, f^{∗}(t) = a_{l+1} if t ∈ [c_{l}, c_{l+1}). Then

p∥f∥_{L^{p,q}}^{q} = p ∑_{l=0}^{k} ∫_{[a_{l+1},a_{l})} t^{q−1}(c_{l})^{q/p}dt = \frac{p}{q} ∑_{l=1}^{k}(c_{l})^{q/p}[(a_{l})^{q} − (a_{l+1})^{q}]  (11.7)

and

∥t^{1/p}f^{∗}∥_{L^{q}((0,∞);dt/t)}^{q} = ∑_{l=0}^{k} ∫_{[c_{l},c_{l+1})} t^{q/p−1}(a_{l+1})^{q} = \frac{p}{q} ∑_{l=0}^{k−1}(a_{l+1})^{q}[(c_{l+1})^{q/p} − (c_{l})^{q/p}],  (11.8)

so that the two quantities are equal (since c_{0} = 0 and a_{k+1} = 0).

74 CHAPTER 11. IMPROVED SOBOLEV EMBEDDINGS

11.2 Properties of f^{∗}

Lemma 21. For each t > 0 we have (with F the distribution function of f)

∫_{0}^{F (f^{∗}(t))}f^{∗}(s)ds = F (f^{∗}(t))f^{∗}(t) + ∫_{f^{∗}(t)}^{∞} F (s)ds.  (11.9)

In particular, ∫_{f^{∗}(t)}^{∞} F (s)ds ≤ ∫_{0}^{t} f^{∗}(s)ds.

Proof. Let g(x) = \begin{cases}|f(x)|, & if |f(x)| > f^{∗}(t) \\ 0, & if |f(x) ≤ f^{∗}(t) \end{cases}, whose distribution function G is given by

\begin{cases}F (s), & if s ≥ f^{∗}(t) \\ F (f^{∗}(t)), & if s < f^{∗}(t) \end{cases}. Let τ ≥ F (f^{∗}(t)). Then G(s) ≤ τ for each s and thus g^{∗}(τ) = 0. On

the other hand, if τ < F (f^{∗}(t)), then clearly g^{∗}(τ) = f^{∗}(τ). The equality ∥g∥_{L^{1}} = ∥g^{∗}∥_{L^{1}} reads then ∫ G(s)ds = ∫ g^{∗}(s)ds, which is precisely the desired equality.

Although it is actually part of the preceding proof, we emphasize for later use the following Corollary 16. Let, for α > 0, f_{α}(x) = \begin{cases}f(x), & if |f(x)| > α \\ 0, & otherwise \end{cases}. Then

∥f_{f^{∗}(t)}∥_{L^{1}} ≤ ∫_{0}^{t} f^{∗}(s)ds.  (11.10)

Lemma 22. Let F be the distribution function of f. Then

∫_{0}^{f^{∗}(t)} ∫_{0}^{F (s)}g^{∗}(u)du ds = f^{∗}(t) ∫_{0}^{t} g^{∗}(s)ds + ∫_{t}^{∞}f^{∗}(u)g^{∗}(u)du.  (11.11)

Proof. Let I be the l. h. s. of (11.11). Fubini’s theorem implies that

I = ∫_{0}^{∞}g^{∗}(u)|{s ; s < f^{∗}(t) and u < F (s)}|du.  (11.12)

Note that u < F (s) ⟺ f^{∗}(u) > s, and therefore

\{s ; s < f^{∗}(t) and u < F (s)\} = (0, min(f^{∗}(u), f^{∗}(t))) = \begin{cases}(0, f^{∗}(t)), & if u ≤ t \\ (0, f^{∗}(u)), & if u > t \end{cases} .  (11.13)

Thus

I = ∫_{0}^{t} g^{∗}(u)f^{∗}(t)du + ∫_{t}^{∞}f^{∗}(u)g^{∗}(u)du,  (11.14)

whence the result.

Lemma 23. Let A ⊂ ℝ^{N} be a measurable set of measure t. Then

∫_{A} |f| ≤ ∫_{0}^{t} f^{∗}(s)ds.  (11.15)

Proof. We may replace f by fχ_{A} (assuming thus f supported in A), since in this way the l. h. s. of (11.15) remains unchanged, while the r. h. s. is not increased. In this case, we have F (s) ≤ t for each s, and thus g^{∗}(s) = 0 if s ≥ t. Therefore,

∫_{A} |f| = ∥f∥_{L^{1}} = ∥f^{∗}∥_{L^{1}} = ∫_{0}^{∞}f^{∗}(s)ds = ∫_{0}^{t} f^{∗}(s)ds.  (11.16)

11.3 Rearrangement and convolutions

The reason we considered f^{∗} is that it is related convolution products. We start with some elementary, though tricky, results linking these objects.

Lemma 24. Let f be s. t. |f| ≤ α and f = 0 outside a set E s. t. |E| = t. Then

|f ∗ g| ≤ α ∫_{0}^{t} g^{∗}(s)ds.  (11.17)

Proof. We have

|f ∗ g(x)| ≤ ∫_{E} |f(y)||g(x − y)|dy ≤ α ∫_{x−E} |g| ≤ α ∫_{0}^{t} g^{∗}(s)ds,  (11.18)

since |x − E| = t.

Lemma 25. Let f ∈ L^{∞} and set α = ∥f∥_{L^{∞}}. Then

|f ∗ g| ≤ ∫_{0}^{α} ∫_{0}^{F (t)}g^{∗}(u)du dt.  (11.19)

Proof. Using a monotone convergence argument, we may assume that f is a step function. Each

step function may be written as f = ∑_{j=1}^{k} a_{j}χ_{A_{j}}, where a_{j} > 0, A_{k} ⊂ A_{k−1} ⊂ … ⊂ A_{1} and

0 < |A_{k}| < … < |A_{1}| < ∞. With b_{j} = a_{1} + ... + a_{j}, we have f = b_{j} in A_{j} \ A_{j+1}. We set b_{0} = 0 and A_{0} = ℝ^{N}. Note that α = b_{k}. Since |f ∗ g| ≤ ∑_{j} a_{j}χ_{A_{j}} ∗ |g|, the preceding lemma implies that

|f ∗ g| ≤ ∑ a_{j} ∫_{0}^{|A_{j}|}g^{∗}(t)dt.  (11.20)

On the other hand, we have F (t) = |A_{j}| if t ∈ [b_{j−1}, b_{j}) and thus

∫_{0}^{α} ∫_{0}^{F (t)}g^{∗}(u)du dt = ∑ ∫_{b_{j−1}}^{b_{j}} ∫_{0}^{|A_{j}|}g^{∗}(u)du dt = ∑(b_{j} −b_{j−1}) ∫_{0}^{|A_{j}|}g^{∗}(u)du = ∑ a_{j} ∫_{0}^{|A_{j}|}g^{∗}(t)dt. (11.21)

Lemma 26. (O’Neil) Let h = f ∗ g. Then

h^{∗}(3t) ≤ \frac{3}{t} ∫_{0}^{t} f^{∗}(s)ds ∫_{0}^{t} g^{∗}(s)ds + ∫_{t}^{∞}f^{∗}(s)g^{∗}(s)ds.  (11.22)

Proof. We may assume that f, g ≥ 0. We split f = f_{1} + f_{2}, g = g_{1} + g_{2} and h = h_{1} + h_{2} + h_{3}. Here,

(i) f is cut at height f^{∗}(t), i. e., we set f_{1}(x) = \begin{cases}f(x), & if f(x) > f^{∗}(t) \\ 0, & if f(x) ≤ f^{∗}(t) \end{cases} and f_{2} = f − f_{1};

(ii) similarly, g is cut at height g^{∗}(t); (iii) h_{1} = f_{2} ∗ g, h_{2} = f_{1} ∗ g_{2} and h_{3} = f_{1} ∗ g_{1}. We start by noting that f_{2} ≤ f, and thus the distribution function of f_{2} is dominated by the one of f. Lemma 25 implies that

h_{1} ≤ ∫_{0}^{f^{∗}(t)} ∫_{0}^{F (s)}g^{∗}(u)du ds ≤ f^{∗}(t) ∫_{0}^{t} g^{∗}(s)ds + ∫_{t}^{∞}f^{∗}(u)g^{∗}(u)du,  (11.23)

by Lemma 11.11. Concerning h_{2}, the inequality ∥h_{2}∥_{L^{∞}} ≤ ∥f_{1}∥_{L^{1}}∥g_{2}∥_{L^{∞}} combined with Corollary 16 yields

h_{2} ≤ g^{∗}(t) ∫_{0}^{t} f^{∗}(s)ds.  (11.24)

We next note that h_{3} satisfies

∥h_{3}∥_{L^{1}} ≤ ∥f_{1}∥_{L^{1}}∥g_{1}∥_{L^{1}} ≤ ∫_{0}^{t} f^{∗}(s)ds ∫_{0}^{t} g^{∗}(s)ds.  (11.25)

To conclude, we start with the inequality h^{∗}(3t) ≤ (h_{1})^{∗}(t) + (h_{2})^{∗}(t) + (h_{3})^{∗}(t). For h_{1} and h_{2}, we use the fact that k^{∗}(t) ≤ ∥k^{∗}∥_{L^{∞}} = ∥k∥_{L^{∞}}. For h_{3}, we rely on the inequality

k^{∗}(t) ≤ \frac{1}{t} ∫_{0}^{t} k^{∗}(s)ds ≤ \frac{1}{t} ∫_{0}^{∞}k^{∗}(s)ds = \frac{1}{t}∥k^{∗}∥_{L^{1}} = \frac{1}{t}∥k∥_{L^{1}}.  (11.26)

We find that

h^{∗}(3t) ≤ \frac{1}{t} ∫_{0}^{t} f^{∗}(s)ds ∫_{0}^{t} g^{∗}(s)ds + f^{∗}(t) ∫_{0}^{t} g^{∗}(s)ds + g^{∗}(t) ∫_{0}^{t} f^{∗}(s)ds + ∫_{t}^{∞}f^{∗}(s)g^{∗}(s)ds; (11.27)

we complete the proof noting that f^{∗}(t) ≤ \frac{1}{t} ∫_{0}^{t} f^{∗}(s)ds and a similar inequality holds for g.

Theorem 20. (O’Neil; simplified version) Let 1 < p, q, r < ∞ be s. t. \frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}. If f ∈ L^{p}

and g ∈ L_{w}^{q}, then f ∗ g ∈ L^{r,p}.

Remark 11. This statement is to be compared with the usual Young inequality, which asserts that f ∗ g ∈ L^{r} if f ∈ L^{p} and g ∈ L^{q}. Our hypothesis is weaker, since L^{q} ⊂ L_{w}^{q}, while the conclusion is stronger, since L^{r,p} ⊂ L^{r} (because p < r).

Proof. Let h = f ∗g. We have to prove that ∥t^{1/r}h^{∗}(t)∥_{L^{p}((0,∞);dt/t)} < ∞. Clearly, this is equivalent to proving that ∥t^{1/r}h^{∗}(3t)∥_{L^{p}((0,∞);dt/t)} < ∞. In view of the preceding lemma, this amounts to proving the following:

(i) ∥t^{1/r−1} ∫_{0}^{t} f^{∗}(s)ds ∫_{0}^{t} g^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)} < ∞;

(ii) ∥t^{1/r} ∫_{t}^{∞}f^{∗}(s)g^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)} < ∞.

The fact that g ∈ L_{w}^{q} is equivalent to the boundedness of the map t ↦ t^{1/q}g^{∗}(t), and thus

g^{∗}(t) ≤ Ct^{−1/q}. It follows that ∫_{0}^{t} g^{∗}(s)ds ≤ C′t^{1−1/q}, and therefore (i) and (ii) reduce to

(i’) ∥t^{1/r−1/q} ∫_{0}^{t} f^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)} < ∞;

(ii) ∥t^{1/r} ∫_{t}^{∞}s^{−1/q}f^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)} < ∞.

To deal with (i’), we apply to f^{∗} the first Hardy’s inequality (Theorem 3) with r replaced by p − 1 > 0 and find that

∥t^{1/r−1/q} ∫_{0}^{t} f^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)}^{p} = ∫_{0}^{∞}t^{−p}( ∫_{0}^{t} f^{∗}(s)ds)^{p}dt ≤ C ∫_{0}^{∞}(f^{∗}(s))^{p}ds = C∥f^{∗}∥_{L^{p}}^{p} < ∞,

(11.28)

since ∥f^{∗}∥_{L^{p}} = ∥f∥_{L^{p}}. Concerning (ii’), the second Hardy’s inequality (Corollary 3) with r replaced by p/r and f replaced by s ↦ s^{−1/q}f^{∗}(s) yields

∥t^{1/r} ∫_{t}^{∞}s^{−1/q}f^{∗}(s)ds∥_{L^{p}((0,∞);dt/t)}^{p} = ∫_{0}^{∞}t^{p/r−1}( ∫_{t}^{∞}s^{−1/q}f^{∗}(s)ds)^{p}dt ≤ C ∫_{0}^{∞}(f^{∗}(s))^{p}ds < ∞.

(11.29)

Corollary 17. Set, with p, q, r as in O’Neil’s theorem, a = N/q. If f ∈ L^{p}, then f ∗ |x|^{−a} ∈ L^{r,p}.

Proof. It suffices to prove that |x|^{−a} ∈ L_{w}^{q}. This follows from |{|x|^{−a} > t}| = Ct^{−N/a} = Ct^{−q}.

11.4. IMPROVED SOBOLEV EMBEDDINGS 79

11.4 Improved Sobolev embeddings

In the remaining part of this chapter, we assume that N ≥ 2. We start with a simple Lemma 27. Let u ∈ C_{0}^{∞}(ℝ^{N}). Then

|u(x)| ≤ \frac{1}{|S^{N−1}|} ∫ \frac{|Du(y)|}{|x − y|^{N−1}} dy.  (11.30)

Proof. Let v ∈ S^{N−1}. Then

u(x) = −[u(x + tv)]_{t=0}^{t=∞} = − ∫_{0}^{∞} \frac{d}{dt}(u(x + tv))dt = − ∫_{0}^{∞}(Du)(x + tv) · vdt,  (11.31)

and therefore

|u(x)| ≤ ∫_{0}^{∞}|Du(x + tv)|dt.  (11.32)

Integrating this inequality over v ∈ S^{N−1} we find that

|S^{N−1}||u(x)| ≤ ∫_{S^{N−1}} ∫_{0}^{∞}|Du(x + tv)|dtds_{v}.  (11.33)

We conclude by noting that the change of variables y = x + tv, t > 0, v ∈ S^{N−1}, yields

∫ \frac{|Du(y)|}{|x − y|^{N−1}} dy = ∫_{S^{N−1}} ∫_{0}^{∞}|Du(x + tv)|dtds_{v}.  (11.34)

We next recall the following well-known result Theorem 21. (converse to the dominated convergence) Let 1 ≤ p ≤ ∞. If f_{n} → f in L^{p}, then there are a subsequence (f_{n_{k}}) and a function g ∈ L^{p} s. t. f_{n_{k}} → f a. e. and |f_{n_{k}}| ≤ g. Proof. After passing, if necessary, to a subsequence, we may assume that f_{n} → f a. e. Consider

a subsequence (f_{n_{k}}) s. t. ∥f_{n_{k}} − f_{n_{k+1}}∥_{L^{p}} ≤ 2^{−k} and set g = |f_{n_{0}}| + ∑_{k≥0} |f_{n_{k}} − f_{n_{k+1}}|. Then

∥g∥_{L^{p}} ≤ ∥f_{n_{0}}∥_{L^{p}} + ∑_{k≥0} ∥f_{n_{k}} − f_{n_{k+1}}∥_{L^{p}} < ∞  (11.35)

and, clearly, |f_{n_{k}}| ≤ g for each k.

Theorem 22. (O’Neil) Let 1 < p < N and set p^{∗} = \frac{Np}{N − p}. If u ∈ W^{1,p}(ℝ^{N}), then u ∈

L^{p^{∗},p}(ℝ^{N}).

Proof. The strategy consists in proving the following generalization of (11.30)

|u(x)| ≤ \frac{1}{|S^{N−1}|} ∫ \frac{|Du(y)|}{|x − y|^{N−1}} dy, ∀ u ∈ W^{1,p}(ℝ^{N}).  (11.36)

Assume (11.36) proved, for the moment. Corollary 17 with a = N−1 implies that |Du|∗|x|^{−(N−1)} ∈ L^{p^{∗},p}. Since |u| ≤ C|Du| ∗ |x|^{−(N−1)} a. e., we obtain that u ∈ L^{p^{∗},p}. It remains to prove (11.36). This is done by approximation. Consider a sequence (u_{n}) ⊂ C_{0}^{∞} s. t. u_{n} → u in W^{1,p}. Possibly after passing to a subsequence, we may assume that u_{n} → u and Du_{n} → Du outside a null set A and that |Du_{n}| ≤ g ∈ L^{p}. Since g ∗ |x|^{−(N−1)} ∈ L^{p^{∗},p} ⊂ L^{p}, we find

that ∫ \frac{g(y)}{|x − y|^{N−1}} dy < ∞ for x outside a null set B. When x ∉ A ∪ B, we find, by dominated

convergence, that

|u(x)| = lim |u_{n}(x)| ≤ lim inf \frac{1}{|S^{N−1}|} ∫ \frac{|Du_{n}(y)|}{|x − y|^{N−1}} dy = \frac{1}{|S^{N−1}|} ∫ \frac{|Du(y)|}{|x − y|^{N−1}} dy.  (11.37)

This completes the proof of the theorem.

11.5 The limiting case p = 1

When p = 1, Theorem 20 is no longer true. To see this, we choose f = χ_{B} ∈ L^{1} (here, B is the

unit ball) and g(x) = |x|^{−α}, which belongs to L_{w}^{q} if αq = N. We have f ∗ g(x) = ∫_{B} \frac{1}{|x − y|^{α}} dy. If

|x| ≥ 2, we have |x − y| ∼ |x| when |y| ≤ 1, and thus f ∗ g(x) ∼ |x|^{−α} when |x| ≥ 2. Therefore,

∥f ∗ g∥_{L^{q,1}} ≥ ∥f ∗ g∥_{L^{q}} ≥ C ∫_{\{|x|≥2\}} \frac{1}{|x|^{N}} = ∞.  (11.38)

A remarkable fact is that the conclusion of Theorem 22 still holds; the proof requires an argument that does not involves convolution products. We start with one essential ingredient which is the isoperimetric inequality. We will not need the sharp (i. e., with the best constant) version, so that we will simply prove the following

Theorem 23. (weak form of the isoperimetric inequality) Let O be a smooth bounded domain in ℝ^{N} and let Σ be its boundary. Then

|O| ≤ C|Σ|^{N/(N−1)}.  (11.39)

11.5. THE LIMITING CASE P = 1 81

Proof. Let ρ ∈ C_{0}^{∞} be s. t. ρ ≥ 0, ∫ ρ = 1 and supp ρ ⊂ B(0, 1). We apply the Sobolev inequality

∥u∥_{L^{N/(N−1)}} ≤ C∥Du∥_{L^{1}} to the function u = χ_{O} ∗ ρ_{ε} and find that

∥χ_{O} ∗ ρ_{ε}∥_{L^{N/(N−1)}} ≤ C ∑_{j} ∫_{ℝ^{N}} | ∫_{O} ∂_{j}ρ_{ε}(x − y)dy|dx = C ∑_{j} ∫_{ℝ^{N}} | ∫_{Σ} n_{j}ρ_{ε}(x − y)ds_{y}|dx; (11.40)

here, n_{j} = n_{j}(y) is the j^{th} component of the outer normal n to O at y. Thus

∥χ_{O} ∗ ρ_{ε}∥_{L^{N/(N−1)}} ≤ C ∫_{Σ} ∫_{ℝ^{N}} ρ_{ε}(x − y)dxds_{y} = C|Σ|.  (11.41)

On the other hand, we have χ_{O} ∈ L^{N/(N−1)} and thus ∥χ_{O} ∗ ρ_{ε}∥_{L^{N/(N−1)}} → ∥χ_{O}∥_{L^{N/(N−1)}} as ε → 0. This leads us to

|O|^{(N−1)/N} = ∥χ_{O}∥_{L^{N/(N−1)}} ≤ C|Σ|,  (11.42)

which ends the proof.

Theorem 24. We have

∥u∥_{L^{N/(N−1),1}} ≤ C∥Du∥_{L^{1}}, ∀ u ∈ W^{1,1}(ℝ^{N}).  (11.43)

Proof. The strategy of the proof is the following: we first prove the inequality (11.43) when u ∈ C_{0}^{∞}; the general case will be obtained from this one by passing to the limits. Let u ∈ C_{0}^{∞}; Sard’s theorem insures that fact that, for a. e. t > 0, all the points x s. t. |u(x)| = t satisfy ∇u(x) ≠ 0; in other words, the set Σ_{t} = {|u| = t} is a smooth hyper surface. For any such t, set O_{t} = {|u| > t}, which is a bounded open set. We claim that (*) O_{t} is a smooth domain with boundary Σ_{t}. Indeed, it is obvious that ∂O_{t} ⊂ Σ_{t}. On the other hand, if x ∈ Σ_{t} and we set v = ∇u(x), then Taylor’s formula implies that u(x + sv) − t has the sign of s when s is close to 0. Thus on the one hand x ∈ ∂O_{t}, on the other hand O_{t} is locally on one side of Σ_{t}, which is the same as (*). With F the distribution function of u, we have

F (t) = |O_{t}| ≤ C|Σ_{t}|^{N/(N−1)},  (11.44)

by the weak isoperimetric inequality. Thus, with H_{t} = {u = t}, we have

∥u∥_{L^{N/(N−1),1}} = ∫ F^{(N−1)/N}(t)dt ≤ C ∫_{0}^{∞}|Σ_{t}|dt = C ∫_{ℝ} |H_{t}|dt = C ∫ |Du|;  (11.45)

the last equality follows from the coarea formula we will prove later. We next turn to a general u ∈ W^{1,1}. Consider a sequence (u_{n}) ⊂ C_{0}^{∞} s. t. u_{n} → u in W^{1,1} and pointwise outside an exceptional zero measure set B. We claim that the corresponding distribution functions, F and F_{n}, satisfy F (t) ≤ lim inf F_{n}(t) for each t. Indeed, let A = {|u| > t}, A_{n} = {|u_{n}| > t}. If x ∈ A \ B, then x ∈ A_{n} for sufficiently large n. Put it otherwise, A \ B ⊂ lim inf(A_{n} \ B), and thus

F (t) = |A| = |A \ B| ≤ lim inf |A_{n} \ B| = lim inf |A_{n}| = lim inf F_{n}(t).  (11.46)

Fatou’s lemma implies than that

∥u∥_{L^{N/(N−1),1}} = ∫ F^{(N−1)/N}(t)dt ≤ lim inf ∫ F_{n}^{(N−1)/N}(t)dt ≤ C ∫ |Du|.  (11.47)

We have thus obtained (11.43) in full generality.

11.6 The limiting case p = N

The conclusion of Theorem 20 is wrong when 1/p + 1/q = 1 (and p ≠ 1). Indeed, let f(x) = χ_{B(0,1/2)}|x|^{−N/p}| ln |x||^{−β}, where β > 1/p. Then f ∈ L^{p}. Let also g(x) = |x|^{−N/q} ∈ L_{w}^{q}. We claim that f ∗ g ∉ L^{∞} if β is well-chosen. We start by noting that Fatou’s lemma implies that f ∗ g(0) ≤

lim inf_{x→0} f ∗ g(x). Therefore, f ∗ g ∉ L^{∞} if f ∗ g(0) = ∞. Since f ∗ g(0) = ∫_{{|x|≤1/2}} |x|^{−N}| ln |x||^{−β}dx,

we find that f ∗ g(0) = ∞ if β ≤ 1. Consequently, we may not use Theorem 20 in the proof of Theorem 22 when p = N. Actually, when p = N, the expected conclusion of Theorem 22, namely W^{1,N} ⊂ L^{∞}, is wrong: it is easy to see that the function given by f(x) = χ_{B(0,1/2)}| ln |x||^{α}, where 0 < α < 1 − 1/N, belongs to W^{1,N}, but not to L^{∞}. However, we will see that each function in W^{1,N} is ”almost” bounded. We start with a simple (and non optimal) result.

Proposition 17. There are constants c, C > 0 s. t. \frac{1}{|B|} ∫_{B} exp(c|u − u_{B}|) ≤ C for each u ∈

W^{1,N}(ℝ^{N}) s. t. ∥Du∥_{L^{N}} ≤ 1.

Proof. The above estimate follows immediately from Theorem 17 and the following result.

Proposition 18. We have, for some C depending only on N,

\frac{1}{|B|} ∫_{B} |u − u_{B}| ≤ C∥Du∥_{L^{N}} , ∀ u ∈ W^{1,N}.  (11.48)

11.6. THE LIMITING CASE P = N 83

Proof. It suffices to prove (11.48) when u ∈ C_{0}^{∞}; the general case is obtained by passing to the limits in (11.48) when B is kept fixed. If B = B(x, R), then

\frac{1}{|B|} ∫_{B} |u − u_{B}| = \frac{1}{|B|^{2}} ∫_{B} | ∫_{B(0,R)} (u(y + z) − u(y))dz|dy ≤ \frac{1}{|B|^{2}} ∫_{B} ∫_{B(0,R)} |u(y + z) − u(y)|dy dz.

(11.49)

Applying Taylor’s formula in integral form, we find that

\frac{1}{|B|} ∫_{B} |u−u_{B}| ≤ \frac{1}{|B|^{2}} ∫_{B} ∫_{B(0,R)} ∫_{0}^{1} |(Du)(y+tz)||z|dt dz dy ≤ \frac{C}{R^{2N−1}} ∫_{B} ∫_{B(0,R)} ∫_{0}^{1} |(Du)(y+tz)|dt dz dy.

(11.50)

For each t and z, Hölder’s inequality implies that

∫_{B} |(Du)(y + tz)|dy ≤ ∥Du∥_{L^{N}} |B(x, 2R)|^{(N−1)/N} ≤ \frac{C}{R^{N−1}} ∥Du∥_{L^{N}} ,  (11.51)

so that

\frac{1}{|B|} ∫_{B} |u − u_{B}| ≤ \frac{C∥Du∥_{L^{N}}}{R^{N}} ∫_{B(0,R)} ∫_{0}^{1} dt dz ≤ C∥Du∥_{L^{N}} .  (11.52)

We may actually replace |u| by |u|^{N/(N−1)} in the preceding exponential integrability result. The statement we give below includes the assumption that supp u ⊂ B. This is not a crucial assumption; if we want to remove it, it suffices to apply the theorem when B is replaced by B^{∗} (the ball concentric with B and twice larger) and u is replaced by φu, where φ is a cutoff function supported in B^{∗} and that equals 1 in B. However, the resulting inequality is less elegant.

Theorem 25. (Trudinger) Let u be a W^{1,N} function supported in B. If ∥Du∥_{L^{N}} ≤ 1, then

\frac{1}{|B|} ∫_{B} exp (c|u|^{N/(N−1)}) ≤ C,  (11.53)

where c, C > 0 depend only on N.

Proof. We may assume that B = B(0, R). We start by noting that (11.30) is valid for a compactly supported function u ∈ W^{1,N}. Indeed, u being compactly supported, it belongs to W^{1,2N/3} ; we may therefore rely on (11.36). Let now f = |Du|, which belongs to L^{N} and is supported in B. Set g = f ∗ |x|^{−(N−1)}. In view of (11.30), it suffices to prove that \frac{1}{|B|} ∫_{B} exp (c g^{N/(N−1)}) ≤ C provided that ∥f∥_{L^{N}} ≤ 1. The key

result in proving this estimate is the following inequality

g(x) ≤ C(δMf(x) + (ln(2R/δ))^{(N−1)/N}), ∀ x ∈ B, ∀ δ ∈ (0, R].  (11.54)

Assume (11.54) proved, for the moment. We consider, for x ∈ B, the two following possibilities:

(i) if x is s. t. Mf(x) ≤ R^{−1}, we choose δ = R and find that g(x) ≤ C; (ii) if Mf(x) > R^{−1}, we choose δ = 1/Mf(x) and find that g(x) ≤ C(1 + ln(RMf(x))^{(N−1)/N}). Thus, in any event, we have g(x)^{N/(N−1)} ≤ C(1 + (ln(RMf(x))_{+}), so that exp (c g^{N/(N−1)(x)}) ≤ C_{1}(1 + (RMf(x))^{cC_{2}}). Choosing c s. t. cC_{2} = N, we find that

\frac{1}{|B|} ∫_{B} exp (c g^{N/(N−1)}) ≤ \frac{C}{|B|} ∫_{B} (1 + R^{N}Mf^{N}) ≤ C(1 + ∫ Mf^{N}) ≤ C(1 + ∥f∥_{L^{N}}^{N}) ≤ C, (11.55)

by the maximal inequalities. It thus remains to establish (11.54). We have

g(x) = ∫_{B(0,δ)} \frac{f(x − y)}{|y|^{N−1}}dy +  ∫_{B(x,R)\B(0,δ)} \frac{f(x − y)}{|y|^{N−1}}dy = I_{1} + I_{2}.  (11.56)

To estimate I_{1}, we note that I_{1} = f ∗ h(x), where h(y) = χ_{B(0,δ)}|y|^{−(N−1)}. Since h is integrable, radial and non increasing, we have I_{1} ≤ Mf(x)∥h∥_{L^{1}} = CδMf(x). We complete the proof of (11.54) by noting that Hölder’s inequality combined with the fact that ∥f∥_{L^{N}} = 1 yields

I_{2} ≤ ( ∫_{B(x,R)\B(0,δ)} |y|^{−N})^{(N−1)/N} ≤ (  ∫_{B(0,2R)\B(0,δ)} |y|^{−N})^{(N−1)/N} = C(ln(2R/δ))^{(N−1)/N}. (11.57)

11.7 The coarea formula

We start by recalling some simple facts from linear algebra. If A is a N × m matrix, with N ≥ m,

we set |A| = \sqrt{∑(det A_{j})^{2}}; the sum is computed over the C_{N}^{m} m × m minors obtained from A. The reason we are interested in such quantities is that, if Σ is a m-dimensional manifold parametrized as O ∋ y ↦ Φ(y) ∈ Σ, with O open set in ℝ^{m}, then

∫_{Σ} f(x)ds_{x} = ∫_{O} f(Φ(y))|DΦ(y)|dy.  (11.58)

Lemma 28. Let B be an invertible N × N matrix and let B be the N × (N − 1) matrix obtained by deleting the first column in A. If v is the first line of A^{−1}, then |B| = |v||A|.

Proof. Let A = (a_{ij}) and let Γ_{ij} be the cofactor of a_{ij}, i. e., the determinant of the (N −1)×(N −1)

matrix obtained by deleting the i^{th} line and the j^{th} column of A. Then |B| = \sqrt{∑_{i=1}^{N} (det Γ_{i1})^{2}}.

On the other hand, we have v = |A|^{−1}((−1)^{i+1}Γ_{i1})_{i=1}^{i=N}, and the conclusion is obvious.

Theorem 26. (light version of the coarea formula of Federer) Let u ∈ C^{∞}(ℝ^{N}; ℝ) and let Σ_{t} = {u = t} (which is a smooth hyper surface for a. e. t). If f ∈ L^{1}(ℝ^{N}; |Du|dx), then:

a) for a. e. t, f_{|Σ_{t}} is integrable with respect to the surface measure on Σ_{t};

b) the map t ↦ ∫_{Σ_{t}} f(x)ds_{x} is measurable;

c) ∫_{ℝ} ∫_{Σ_{t}} f(x)ds_{x}dt = ∫_{ℝ^{N}} f|Du|.

Proof. We may assume f ≥ 0. Let C be the set of the critical values of u, let Z be the set of the critical points of u and set U = ℝ^{N} \ Z and A = u^{−1}(C). Sard’s lemma implies that |C| = 0.

Since u_{|U} is an epimorphism, A \ Z = (u_{|U})^{−1}(C) is a null set. Thus ∫_{A\Z} f|Du| = 0. Therefore,

∫_{A} f|Du| = 0. On the other hand, ∫_{ℝ} ∫_{Σ_{t}} f(x)ds_{x}dt = ∫_{ℝ\C} ∫_{Σ_{t}} f(x)ds_{x}dt, provided the first integral

makes sense. Consequently, we may replace ℝ^{N} by O = ℝ^{N} \ A, and thus assume that u has no critical points in the open set where it is defined. Since u is of constant rank 1 in O, we may locally flatten the coordinates in order to have u = x_{1}. More specifically, there is a covering O = ⋃ O_{i} of O with open sets, s. t. for each i there is a diffeomorphism Φ_{i} : (0, 1)^{N} → O_{i} and there is some j = j_{i} ∈ {1, …, N} s. t. u ∘ Φ_{i}(y) = y_{j} + C_{i}, y ∈ (0, 1)^{N}. Using a partition of the unit subordinate to the covering O = ⋃ O_{i}, we may assume that: (i) u is defined in an open set O, (ii) there is a diffeomorphism Φ : (0, ε)^{N} → O s. t. u ∘ Φ(y) = y_{1} + C, y ∈ (0, ε)^{N}. We may also assume, without loss of generality, that C = 0. Then Σ_{t} is non empty iff t ∈ (0, ε), and if this is the case, then Σ_{t} is parametrized by (0, ε)^{N−1} ∋ y′ ↦ Φ(t, y′). Let B(t, y′) be the N × (N − 1) matrix obtained from the Jacobian matrix DΦ(t, y′) by deleting the first column. Then

∫_{Σ_{t}} f(x)ds_{x} = ∫_{(0,ε)^{N−1}} f(t, y′)|B(t, y′)|dy′,  (11.59)

from which it is clear that a) and b) of the theorem hold. Concerning c), it reduces to

∫_{(0,ε)^{N}} f(Φ(y))|B(y)|dy = ∫_{O} f(x)|Du(x)|dx.  (11.60)

If we perform, in the second integral, the change of variables x = Φ(y), we are led to proving the equality

∫_{(0,ε)^{N}} f(Φ(y))|B(y)|dy = ∫_{(0,ε)^{N}} f(Φ(y))|Du(Φ(y))||DΦ(y)|dy.  (11.61)

Since u ∘ Φ(y) ≡ y_{1}, we find by differentiation that ^{t}DΦ(y)Du(Φ(y)) =^{t} (1, 0, …, 0). In other words, Du(Φ(y)) is the first column of the matrix (^{t}DΦ(y))^{−1}, i. e., the first line of the matrix (DΦ(y))^{−1}. Lemma 28 implies that |Du(Φ(y))||DΦ(y)| = |B(y)|. Equality (11.61) is established.

Traces

12.1 Definition of the trace

We discuss here the properties of the ”restrictions” of Sobolev maps to hyper surfaces, e. g., to the boundary of a smooth domain. There is a standard reduction procedure which allows to replace a ”smooth” (at least Lipschitz) hyper surface with a hyperplane; this is done by flattening locally the coordinates. Since this part works without any problem and we want to insist on the analytic part, we will simply consider in this chapter maps defined in the whole ℝ^{N} and consider properties of their trace on the hyperplane H = {x = (x′, x_{N}) ∈ ℝ^{N} ; x_{N} = 0}, which we identify with ℝ^{N−1}. We start by recalling the following

Proposition 19. The map u ↦ u_{|H}, initially defined from C_{0}^{∞}(ℝ^{N}) into C_{0}^{∞}(ℝ^{N−1}), extends uniquely by density to a linear map (called trace map) u ↦ tr u from W^{1,p}(ℝ^{N}) into L^{p}(ℝ^{N}), for 1 ≤ p < ∞.

We will elude here the case of W^{1,∞}. Maps in W^{1,∞} are Lipschitz, thus continuous, and in this case the trace is simply the restriction.

Proof. Fix a function φ ∈ C_{0}^{∞}(ℝ) s. t. φ(0) = 1 and supp φ ⊂ (−1, 1). If u ∈ C_{0}^{∞}, then v = uφ(x_{N}) ∈ C_{0}^{∞}(ℝ^{N−1}×(−1, 1)) and u_{|H} = v_{|H}. In addition, it is clear that ∥v∥_{W^{1,p}} ≤ C∥u∥_{W^{1,p}}. It therefore suffices to prove that ∥v_{|H}∥_{L^{p}} ≤ C∥v∥_{W^{1,p}}. This follows from

∫_{H} |v(x′, 0)|^{p}dx′ = ∫_{H} | ∫_{0}^{1} ∂_{N}v(x′, t)dt|^{p}dx′ ≤ ∫_{H×(0,1)} |Dv|^{p} ≤ ∥Dv∥_{L^{p}}^{p}.  (12.1)

When 1 < p < ∞, the above result is not sharp, in the following sense: if f is an arbitrary map in L^{p}(ℝ^{N−1}), we can not always find a map u ∈ W^{1,p} s. t. tr u = f. In other words, the trace map is not surjective between the spaces we consider. In this chapter, we will determine the image of the trace map.

Definition 5. For 0 < s < 1 and 1 ≤ p < ∞, we define

W^{s,p} = W^{s,p}(ℝ^{N}) = \{f ∈ L^{p}(ℝ^{N}) ; ∫_{ℝ^{N}} ∫_{ℝ^{N}} \frac{|f(x) − f(y)|^{p}}{|x − y|^{N+sp}} dx dy < ∞\},  (12.2)

equipped with the norm

∥f∥_{W^{s,p}} = ∥f∥_{L^{p}} + ( ∫_{ℝ^{N}} ∫_{ℝ^{N}} \frac{|f(x) − f(y)|^{p}}{|x − y|^{N+sp}} dx dy)^{1/p}.  (12.3)

We let the reader check that W^{s,p} is a Banach space. The main result of this chapter states that tr W^{1,p}(ℝ^{N}) = W^{1−1/p,p}(ℝ^{N−1}). We start with some preliminary results.

Lemma 29. C^{∞}(ℝ^{N}) ∩ W^{s,p}(ℝ^{N}) is dense into W^{s,p}(ℝ^{N}) for 0 < s < 1 and 1 ≤ p < ∞.

Proof. Let ρ be a standard mollifier (i. e., ρ ∈ C_{0}^{∞}, ρ ≥ 0, ∫ ρ = 1, supp ρ ⊂ B(0, 1)). We will

prove that, if f ∈ W^{s,p}, then f_{ε} = f ∗ ρ_{ε} → f in W^{s,p}. Clearly, f_{ε} → f in L^{p}. It remains to prove that, with g_{ε} = f_{ε} − f, we have

I_{ε} = ∫_{ℝ^{N}} ∫_{ℝ^{N}} \frac{|g_{ε}(x) − g_{ε}(y)|^{p}}{|x − y|^{N+sp}} dx dy = ∫_{ℝ^{N}} ∫_{ℝ^{N}} \frac{|g_{ε}(x + h) − g_{ε}(x)|^{p}}{|h|^{N+sp}} dx dh → 0 as ε → 0. (12.4)

In order to estimate I_{ε}, we start by noting that

g_{ε}(x + h) − g_{ε}(x) = ∫_{B(0,ε)} (f(x + h − y) − f(x + h) − f(x − y) + f(x))ρ_{ε}(y)dy.  (12.5)

Using in addition the fact that ρ_{ε} ≤ Cε^{−N}, we find that

|g_{ε}(x + h) − g_{ε}(x)| ≤ \frac{C}{ε^{N}} ∫_{B(0,ε)} |f(x + h − y) − f(x + h) − f(x − y) + f(x)|dy.  (12.6)

We next consider the two following cases:

(i) if |h| < ε, we have

|g_{ε}(x + h) − g_{ε}(x)| ≤ \frac{C}{ε^{N}} ∫_{B(0,ε)} (|f(x + h − y) − f(x − y)| + |f(x) − f(x + h)|)dy;  (12.7)

12.1. DEFINITION OF THE TRACE 89

(ii) if |h| ≥ ε, we use the inequality

|g_{ε}(x + h) − g_{ε}(x)| ≤ \frac{C}{ε^{N}} ∫_{B(0,ε)} (|f(x + h − y) − f(x + h)| + |f(x) − f(x − y)|)dy.  (12.8)

Thus I_{ε} ≤ \frac{C}{ε^{Np}} (J_{1} + J_{2} + J_{3} + J_{4}), where

J_{1} = ∫_{ℝ^{N}} ∫_{{|h|<ε}} ( ∫_{B(0,ε)} |f(x + h − y) − f(x − y)|dy)^{p}|h|^{−(N+sp)}dh dx;  (12.9)
J_{2} = ∫_{ℝ^{N}} ∫_{{|h|<ε}} ( ∫_{B(0,ε)} |f(x + h) − f(x)|dy)^{p}|h|^{−(N+sp)}dh dx;  (12.10)
J_{3} = ∫_{ℝ^{N}} ∫_{{|h|≥ε}} ( ∫_{B(0,ε)} |f(x + h − y) − f(x + h)|dy)^{p}|h|^{−(N+sp)}dh dx;  (12.11)
J_{4} = ∫_{ℝ^{N}} ∫_{{|h|≥ε}} ( ∫_{B(0,ε)} |f(x − y) − f(x)|dy)^{p}|h|^{−(N+sp)}dh dx.  (12.12)

We will prove that ε^{Np}J_{j} → 0, j = 1, …, 4. The only ingredient we use in the proof is

lim_{ε→0} ∫_{ℝ^{N}} ∫_{B(0,ε)} \frac{|f(x + y) − f(x)|^{p}}{|y|^{N+sp}} dy = 0;  (12.13)

this follows easily by dominated convergence. We start with J_{2}. Noting that ( ∫_{B(0,ε)} |f(x + h) − f(x)|dy)^{p} = Cε^{Np}|f(x + h) − f(x)|^{p}, we find that

ε^{−Np}J_{2} = C ∫_{ℝ^{N}} ∫_{B(0,ε)} \frac{|f(x + h) − f(x)|^{p}}{|h|^{N+sp}} dh → 0.  (12.14)

For J_{1}, Hölder’s inequality with exponents p and p′ implies that

( ∫_{B(0,ε)} |f(x + h − y) − f(x − y)|dy)^{p} ≤ |B(0, ε)|^{p−1} ∫_{B(0,ε)} |f(x + h − y) − f(x − y)|^{p}dy, (12.15)

and thus

ε^{−Np}J_{1} ≤ Cε^{−N} ∫_{ℝ^{N}} ∫_{{|h|<ε}} ∫_{B(0,ε)} |f(x + h − y) − f(x − y)|^{p}dy |h|^{−(N+sp)}dh dx.  (12.16)

For fixed y and h, the change of variables x − y = z leads to

ε^{−Np}J_{1} ≤ C ∫_{ℝ^{N}} ∫_{{|h|<ε}} |f(z + h) − f(z)|^{p}|h|^{−(N+sp)}dh dz → 0.  (12.17)

We next estimate J_{3}; the computation for J_{4} is similar and will be omitted. Inequality (12.15) implies that

ε^{−Np}J_{3} ≤ Cε^{−N} ∫_{ℝ^{N}} ∫_{{|h|≥ε}} ∫_{B(0,ε)} |f(x + h − y) − f(x + h)|^{p}|h|^{−(N+sp)}dy dh dx.  (12.18)

In this integral, we fix y and h and make the change of variables x + h = z. Next we integrate in h and find that

ε^{−Np}J_{3} ≤ C ∫_{ℝ^{N}} ∫_{B(0,ε)} \frac{|f(z − y) − f(z)|^{p}}{ε^{N+sp}} dy dz ≤ C ∫_{ℝ^{N}} ∫_{B(0,ε)} \frac{|f(z − y) − f(z)|^{p}}{|y|^{N+sp}} dy dz → 0. (12.19)

Lemma 30. If u ∈ C(ℝ^{N}) ∩ W^{1,p}, then tr u = u_{|H}. Proof. Let ρ be a standard mollifier s. t. ρ(0) = 1 and set u_{ε} = ρ(ε·)(u ∗ ρ_{ε}). Clearly, u_{ε} ∈ C_{0}^{∞} and u_{ε} → u in W^{1,p}. Thus u_{ε}_{|H} = tr u_{ε} → tr u in L^{p} (and thus in D′). On the other hand, u_{ε}_{|H} converges to u_{|H} uniformly on compacts (and thus in D′), whence the conclusion.

The same argument leads to the following variant Lemma 31. Assume that u ∈ W^{1,p} is continuous in a neighborhood of H. Then tr u = u_{|H}. Lemma 32. Let u ∈ C(ℝ^{N})∩C^{1}(ℝ^{N} \H). Assume that the pointwise differential Du of u satisfies Du ∈ L^{p}(ℝ^{N}). Then Du is also the distributional differential of u.

Proof. We have to prove that ∫ D_{j}uφ = − ∫ u∂_{j}φ, φ ∈ C_{0}^{∞}, j = 1, …, N. When j ≤ N − 1,

this follows simply by Fubini’s theorem. Assume j = N. We integrate by parts ∫ D_{N}uφ in the

set {x ∈ ℝ^{N} ; |x_{N}| > ε} and next let ε → 0. We find that

∫ D_{N}uφ = lim_{ε→0} ( ∫_{{x_{N} =−ε}} uφds_{x′} − ∫_{{x_{N} =ε}} uφds_{x′} − ∫_{{|x_{N} |>ε}} u∂_{N}φdx) = − ∫ u∂_{N}φdx. (12.20)

12.2. TRACE OF W^{1,P}, 1 < P < ∞ 91

12.2 Trace of W^{1,p}, 1 < p < ∞

Theorem 27. (Gagliardo) Let p ∈ (1, ∞).

a) If u ∈ W^{1,p}(ℝ^{N}), then tr u ∈ W^{1−1/p,p}(ℝ^{N−1}) and ∥tr u∥_{W^{1−1/p,p}} ≤ C∥u∥_{W^{1,p}}.

b) Conversely, let f ∈ W^{1−1/p,p}(ℝ^{N−1}). Then there is some u ∈ W^{1,p}(ℝ^{N}) s. t. tr u = f. In addition, we may pick u s. t. ∥u∥_{W^{1,p}} ≤ C∥tr u∥_{W^{1−1/p,p}}.

Remark 12. Let T : W^{1,p}(ℝ^{N}) → W^{1−1/p,p}(ℝ^{N−1}), T u = tr u. T is linear, and the above theorem implies that T is continuous and surjective. Then the last conclusion in b) follows from the open map theorem (each surjective linear continuous map between two Banach spaces has a bounded right inverse). However, we will see during the proof a stronger conclusion: we will construct in b) a linear right inverse, i. e. , the map f ↦ u in b) will be linear.

Proof. a) By density, it suffices to prove that

∥u_{|H}∥_{W^{1−1/p,p}} ≤ C∥u∥_{W^{1,p}} ∀ u ∈ C_{0}^{∞}.  (12.21)

We start by noting that we already know that ∥u_{|H}∥_{L^{p}} ≤ C∥u∥_{W^{1,p}}; thus it suffices to establish, with f(x′) = u(x′, 0), the inequality

I = ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + h′) − f(x′)|^{p}}{|h′|^{N+p−2}} dh′ dx′ ≤ C ∫_{ℝ^{N}} |Du(x)|^{p}dx.  (12.22)

The starting point is the inequality

|f(x′ + h′) − f(x′)| ≤ |f(x′ + h′) − u(x′ + h′/2, |h′|/2)| + |f(x′) − u(x′ + h′, |h′|/2)|, (12.23)

which implies that I ≤ C(I_{1} + I_{2}), where

I_{1} = ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + h′) − u(x′ + h′/2, |h′|/2)|^{p}}{|h′|^{N+p−2}} , I_{2} = ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′) − u(x′ + h′/2, |h′|/2)|^{p}}{|h′|^{N+p−2}} .

(12.24)

If we perform, in I_{1}, the change of variables x′ + h′ = y′, next we change h′ into −h′, we see that I_{1} = I_{2}, and thus

I ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′) − u(x′ + h′/2, |h′|/2)|^{p}}{|h′|^{N+p−2}} dh′ dx′.  (12.25)

Changing h′ into 2k′ and applying the Leibniz-Newton formula, we find that

I ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} ( ∫_{0}^{|k′|}|Du(x′ + t(k′/|k′|), t)|)^{p}|k′|^{−(N+p−2)}dk′ dx′.  (12.26)

Expressing k′ in polar coordinates, we find that

I ≤ C ∫_{ℝ^{N−1}} ∫_{S^{N−2}} ∫_{0}^{∞}( ∫_{0}^{s} |Du(x′ + tω, t)|dt)^{p}s^{−p}ds ds_{ω} dx′.  (12.27)

Applying, for fixed x′ and ω, Hardy’s inequality in to the double integral in s and t, we find that

I ≤ C ∫_{ℝ^{N−1}} ∫_{S^{N−2}} ∫_{0}^{∞}|Du(x′ + tω, t)|^{p}dt ds_{ω} dx′.  (12.28)

Integrating, in the above inequality, first in x′, next in ω, we find that

I ≤ C ∫_{ℝ^{N−1}} ∫_{0}^{∞}|Du(x′, t)|^{p}dt dx′ = C ∫_{ℝ^{N}^{+}} |Du(x)|^{p}dx ≤ C∥Du∥_{L^{p}}^{p}.  (12.29)

b) It suffices to construct a linear map f ↦ u, f ∈ C^{∞}(ℝ^{N−1}) ∩ W^{1−1/p,p}, u ∈ W^{1,p}(ℝ^{N}), s. t. tr u = f and ∥u∥_{W^{1,p}} ≤ C∥f∥_{W^{1−1/p,p}}. We fix a standard mollifier ρ in ℝ^{N−1} and an even function φ ∈ C^{∞}(ℝ) s. t. φ(0) = 1, 0 ≤ φ ≤ 1 and supp φ ⊂ (−1/2, 1/2). We define, for t ≠ 0, v(x′, t) = f ∗ ρ_{|t|}(x′) and u(x′, t) = v(x′, t)φ(t). We extend u to ℝ^{N} by setting u(x′, 0) = f(x′). Clearly, the map f ↦ u is linear and u ∈ C^{∞}(ℝ^{N} \ H). In addition, u ∈ C(ℝ^{N}) when f is continuous. We also note that, for a fixed t ≠ 0, Young’s inequality implies that ∥f ∗ ρ_{|t|}∥_{L^{p}} ≤ ∥f∥_{L^{p}}, and thus ∥u∥_{L^{p}} ≤ ∥f∥_{L^{p}}. Since u is even with respect to x_{N}, it suffices to prove, in view of Lemmata 30 and 32, that the usual differential Du of u satisfies

∫_{ℝ^{N−1}} ∫_{0}^{∞}|Du(x′, t)|^{p}dt dx′ ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + y′) − f(x′)|^{p}}{|y|^{N+p−2}} dy′ dx′ + C∥f∥_{L^{p}}^{p}.  (12.30)

For 1 ≤ j ≤ N −1, we have |∂_{j}u| ≤ |∂_{j}v|. On the other hand, |∂_{N}u| ≤ C|v|χ_{ℝ^{N−1}×(−1/2,1/2)}+|∂_{N}v|. Since ∥|v|χ_{ℝ^{N−1}×(−1/2,1/2)}∥_{L^{p}} ≤ ∥u∥_{L^{p}}, it suffices to prove the estimate

∫_{ℝ^{N−1}} ∫_{0}^{∞}|Dv(x′, t)|^{p}dt dx′ ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + y′) − f(x′)|^{p}}{|y|^{N+p−2}} dy′ dx′.  (12.31)

Let 1 ≤ j ≤ N − 1. Since ∫ ∂_{j}ρ = 0, we have

∂_{j}v(x′, t) = t^{−N} ∫ f(y′)(∂_{j}ρ)((x′ − y′)/t)dy′ = t^{−N} ∫ [f(y′) − f(x′)](∂_{j}ρ)((x′ − y′)/t)dy′, (12.32)

12.3. TRACE OF W^{1,1} 93

so that

|∂_{j}v(x′, t)| ≤ \frac{C}{t^{N}} ∫_{B(0,t)} |f(x′ + y′) − f(x′)|dy′.  (12.33)

We next claim that ∫ \frac{d}{dt}[ρ_{t}(x′)]dx′ = 0. This follows from the fact that ∫ ρ_{t} ≡ 1. Thus

|∂_{N}v(x′, t)| = | ∫ [f(y′) − f(x′)]\frac{d}{dt}[ρ_{t}(x′ − y′)]dy′| ≤ \frac{C}{t^{N}} ∫_{B(0,t)} |f(x′ + y′) − f(x′)|dy′, (12.34)

since |\frac{d}{dt}ρ_{t}| ≤ Ct^{−N}. We find that |Dv(x′, t)| ≤ \frac{C}{t^{N}} ∫_{B(0,t)} |f(x′ + y′) − f(x′)|dy′, and therefore it

suffices to establish the estimate

I = ∫_{ℝ^{N−1}} ∫_{0}^{∞}( ∫_{B(0,t)} |f(x′ + y′) − f(x′)|dy′)^{p}t^{−Np}dt dx′ ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + y′) − f(x′)|^{p}}{|y|^{N+p−2}} dy′ dx′.

(12.35)

This is done as in the proof of lemma 29: Hölder’s inequality applied to the integral over B(0, t) implies that

I ≤ C ∫_{ℝ^{N−1}} ∫_{0}^{∞} ∫_{B(0,t)} |f(x′ + y′) − f(x′)|^{p}dy′ t^{−N−p+1}dt dx′.  (12.36)

Fubini’s theorem yields

I ≤ C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} |f(x′ + y′) − f(x′)|^{p} ∫_{|y′|}^{∞}t^{−N−p+1}dt dx′ dy′ = C ∫_{ℝ^{N−1}} ∫_{ℝ^{N−1}} \frac{|f(x′ + y′) − f(x′)|^{p}}{|y|^{N+p−2}} dy′ dx′.

(12.37)

Corollary 18. Let f ∈ W^{1−1/p,p}(ℝ^{N}) and set , for t ≠ 0, u(x′, t) = f ∗ρ_{|t|}(x′)φ(t). Then u ∈ W^{1,p} and tr u = f.

12.3 Trace of W^{1,1}

We start with some auxiliary results needed in the proof of the fact that the trace of W^{1,1} is L^{1}.

Lemma 33. Let u ∈ W^{1,p} ∩ W^{1,q}. Then the two traces of u (one in W^{1,p}, the other one in W^{1,q}), coincide.

Proof. If ρ is a standard mollifier s. t. ρ(0) = 1, then u_{ε} = ρ(ε·)u ∗ ρ_{ε} converges (as ε → 0) to u both in W^{1,p} and in W^{1,q}. Since, for u_{ε} ∈ C_{0}^{∞}, both traces coincide, we obtain the result by passing to the limits.

The same argument leads to the following result.

Lemma 34. Let u ∈ W^{1,p}. For λ ≠ 0 and x′ ∈ ℝ^{N−1}, we have tr u(λ · −x′) = (tr u)(λ · −x′).

Lemma 35. Let f be the characteristic function of a cube in ℝ^{N−1}. Then f ∈ W^{1−1/p,p} for 1 < p < 2.

Proof. We may assume that C = (−l, l)^{N}. If we consider in ℝ^{N−1} the ∥ · ∥_{∞} norm, then

∥f∥_{W^{1−1/p,p}}^{p} ∼ ∫_{|x′|<l} ∫_{|y′|>l} \frac{dx′ dy′}{|x′ − y′|^{N+p−2}} .  (12.38)

If |x′| < l and |y′| > l, then y′ ∈ ℝ^{N−1} \ B(x′, l − |x′|), and therefore

∫_{|y′|>l} \frac{dy′}{|x′ − y′|^{N+p−2}} ≤ ∫_{|z′|>l−|x′|} \frac{dz′}{|z′|^{N+p−2}} = C ∫_{l−|x′|}^{∞} r^{−p} = C(l − |x′|)^{1−p}.  (12.39)

Since p < 2, we find that

∥f∥_{W^{1−1/p,p}}^{p} ≤ C ∫_{|x′|<l} (l − |x′|)^{1−p} ≤ C  ∫_{\{|x_{j}|≤x_{1}<l, j=1,...,N−2\}} \frac{dx′}{(l − x_{1})^{p−1}} < ∞.  (12.40)

Lemma 36. Let C be a cube of size l in ℝ^{N−1} and set a = \frac{1}{|C|}χ_{C}. Then there is a map u ∈ W^{1,1}

s. t. tr u = a and

∥u∥_{L^{1}} ≤ c l and ∥Du∥_{L^{1}} ≤ c.  (12.41)

Proof. We start with the case where C is the unit cube (or any other cube of size 1). We fix a p ∈ (1, 2). Since a ∈ W^{1−1/p,p}, we have a = tr u_{0} for some u ∈ W^{1,p}. In addition, Corollary 18 implies that we may assume u_{0} compactly supported. Thus u ∈ W^{1,1} and tr u_{0} = a (computed in W^{1,1}). Let now C be an arbitrary cube, which we may assume with sides parallel to the unit cube Q. Let C = x′ + (0, l)^{N−1}. Set u = l^{−(N−1)}u_{0}(l^{−1}(· − x′)). Then u ∈ W^{1,1} and tr u = a. Inequality (12.41) follows from the identities ∥u∥_{L^{1}} = l∥u_{0}∥_{L^{1}} and ∥Du∥_{L^{1}} = ∥Du_{0}∥_{L^{1}}.

Theorem 28. (Gagliardo) Let f ∈ L^{1}(ℝ^{N−1}). Then there is some u ∈ W^{1,1}(ℝ^{N}) s. t. tr u = f and ∥u∥_{W^{1,1}} ≤ C∥f∥_{L^{1}}.

Remark 13. This time, the map f ↦ u we construct is not linear.

Proof. The main ingredient is the following: if f ∈ L^{1}, then we may write, in L^{1}, f = ∑ λ_{n}a_{n}, where:

(i) each a_{n} is of the form a_{n} = \frac{1}{|C_{n}|}χ_{C_{n}};

(ii) each C_{n} is of size at most 1; (iii) ∑ |λ_{n}| ≤ C∥f∥_{L^{1}}. Assuming that this can be achieved, here is the end of the proof: the preceding lemma implies that each a_{n} is the trace of some u_{n} ∈ W^{1,1} s. t. ∥u_{n}∥_{W^{1,1}} ≤ C. The linearity of the trace and property (iii) imply that the map u = ∑ λ_{n}u_{n} ∈ W^{1,1} satisfies tr u = f and ∥u∥_{W^{1,1}} ≤ C∥f∥_{L^{1}}.

It remains to perform the decomposition f = ∑ λ_{n}a_{n}. For each j ∈ ℕ, let F_{j} be the grid of cubes of size 2^{−j}, with sides parallel to the coordinate axes and having the origin among the edges. We define the linear map T_{j} : L^{1} → L^{1}, T_{j}f(x) = f_{C} if x ∈ C ∈ F_{j}. Clearly, T_{j} is of norm 1. We claim that, for each f ∈ L^{1}, we have T_{j}f → f in L^{1} as j → ∞. This is clear when f ∈ C_{0}^{∞}; the case of a general f follows by approximation using the fact that ∥T_{j}∥ = 1. We may thus find an increasing sequence of indices, (j_{k}), s. t. ∥f_{j_{0}}∥_{L^{1}} + ∑ ∥f_{j_{k+1}} − f_{j_{k}}∥_{L^{1}} ≤ C∥f∥_{L^{1}}. We

claim that f_{j_{k+1}} − f_{j_{k}} = ∑ λ_{n}^{k}a_{n}^{k}, where each a_{n}^{k} is of the form \frac{1}{|C|}χ_{C} for some cube of size at

most 1 and ∑ |λ_{n}^{k}| = ∥f_{j_{k+1}} − f_{j_{k}}∥_{L^{1}}. Indeed, f_{j_{k+1}} − f_{j_{k}} is constant on each cube C ∈ F_{j_{k+1}},

and thus f_{j_{k+1}} − f_{j_{k}} = ∑_{C∈F_{j_{k+1}}}(f_{j_{k+1}} − f_{j_{k}})_{|C}χ_{C}, so that f_{j_{k+1}} − f_{j_{k}} = ∑_{C∈F_{j_{k+1}}} λ_{C} \frac{1}{|C|}χ_{C}, with

λ_{C} = (f_{j_{k+1}} − f_{j_{k}})_{|C}|C|. We find that

∥f_{j_{k+1}} − f_{j_{k}}∥_{L^{1}} = ∑_{C∈F_{j_{k+1}}} ∫_{C} |f_{j_{k+1}} − f_{j_{k}}| = ∑_{C∈F_{j_{k+1}}} |C||(f_{j_{k+1}} − f_{j_{k}})_{|C}| = ∑_{C∈F_{j_{k+1}}} |λ_{C}|. (12.42)

Similarly, we may write f_{j_{0}} = ∑ λ_{n}^{0}a_{n}^{0}, where each a_{n}^{0} is of the form \frac{1}{|C|}χ_{C} for some cube of size

at most 1 and ∑ |λ_{n}^{0}| = ∥f_{j_{0}}∥_{L^{1}}.

Finally, we write f = ∑_{k} ∑_{n} λ_{n}^{k}a_{n}^{k}, and this decomposition has the properties (i)-(iii).