Besov spaces and self-similar solutions for the wave-map equation

Besov spaces and self-similar solutions for the wave-map
equation
P. Germain
June 27, 2007
Abstract
Self-similar solutions play a crucial role in the blow-up theory for the wave-map
equation; in space-dimension d, they correspond to data in the infinite energy space
˙B d/2
2,∞ × ˙ B d/2−1
2,∞ at the time of the blow-up. However, solutions to this equation are
generally considered for data in the standard finite energy spaces ˙ Hd/2 × ˙ Hd/2−1 .
We show in this article that it is possible to build up solutions of the covariant
wave-map equation for data which are small and of infinite energy, or large and selfsimilar.
This provides us with a general framework which includes in particular the
blowing up solutions of Shatah [15] and Bizoń [3]. As an application, we describe more
precisely the regularity breakdown phenomenon.
1 Introduction
1.1 Presentation of the equation
We start with the wave map equation for a map u from the Minkowski space R 1+d to a
general manifold N. Greek indices correspond to the Minkowski space, and they are raised
according to its metric. Latin indices correspond to the target manifold, whose Christoffel
symbols are denoted by Γ. The Cauchy problem takes then the form
(1)
⎧
⎨
⎩
∂ α ∂αu k = −Γ k i,j ∂α u i ∂αu j
u(0) = u0
ut(0) = u1 .
The problem for small data has been studied in a series of articles, see in particular
[19] [18] [8] [17] [20] ; without going into the details, it is globally well-posed for (u0, u1)
small in ˙ H d/2 × ˙ H d/2−1 ; notice that these spaces are at the scaling of the equation.
1.2 Reduction of the problem
We will focus on the case where the target manifold N is spherically symmetric,
more precisely it is a rotationnally symmetric warped product manifold (we follow Shatah
and Tahvilar-Zadeh [16]) defined as
N = [0, φ ∗ ) ×G S d ,
where φ ∗ ∈ R + ∪ {∞} and G : R → R is smooth and odd G(0) = 0, G ′ (0) = 1. On N we
have the polar coordinates
(φ, χ) ∈ [0, φ ⋆ ) × S d−1 ;
1

in these coordinates the metric reads
dφ 2 + G 2 (φ)dχ 2 ,
where dχ 2 corresponds to the standard metric of S d−1 .
We further assume that the data (and hence the solution) is covariant; in other
words, if we adopt polar coordinates on R d and on N, u is of the form
u :R 1+d −→ N
(t, r, ω) −→ (φ(t, r) , ω) ,
for some function φ. The equation 1) becomes
d − 1
(2) φtt − φrr −
r φr + 1
f(φ) = 0
r2 where f(φ) = (d − 1)G(φ)G ′ (φ).
We perform a final transformation by setting
v = φ
r
so that the Cauchy problem we will consider reads
(3)
⎧
⎨
⎩
vtt − vrr − n−1
r vr = v 3 Z(rv)
v(0) = v0
vt(0) = v1 .
where n = d + 2 and Z(x) = 1
((d − 1)x − f(x)) is a smooth, even function. It will be
x3 very helpful in the following to consider the above equation as a wave equation for a radial
function, set in R1+n .
Notice that the above problem is invariant by the transformation
(4)
(v0(x) , v1(x)) ↦→ (λv0(λx) , λ 2 v1(λx))
v(t, x) ↦→ λv(λt, λx) ,
for any λ > 0, so that scale invariant spaces for the data and solution are typically
(v0, v1) ∈ ˙ H n
2 −1 × ˙ H n
2 −2
1.3 Infinite energy and self-similar solutions
Self-similar solutions
, v ∈ L ∞ (R, ˙ H n
2 −1 ) .
In order to explain ideas better, we discuss in this section the case of the sphere N = S d .
Let us first consider the case of finite energy data
φ0 ∈ ˙ H d
2 , φ1 ∈ ˙ H d
2 −1
(notice that these spaces are scale invariant). For small data (see the references in Section
1.1), the equation is well-posed globally, and scattering occurs. For large data, we
have to distinguish between d = 2 (the critical case) and d ≥ 3 (the super-critical case).
2

• In dimension 2, it has been proved by Shatah and Tahvildar-Zadeh [16] that selfsimilarblow
up cannot occur, ie blow up cannot occur along a profile of the type
f , it has to be faster. Indeed, instances of blow-up of the form f
x
T −t
x
(T −t)| log(T −t)| −1/2
+ radiation were built up recently by Rodnianski and Sterbenz [13]; and Krieger,
Schlag and Tataru [9] proved that another possible blow-up regime is of the form
f
x
(T −t) s
+ radiation, where s > 3
2 .
• In dimension 3 (or higher) the picture is very different, since examples of exact self-
similar blow up (i.e. solutions of the form f
x
T −t
) are known, see Shatah [15],
Shatah and Tahvildar-Zadeh [16] and Bizon [3]. Furthermore, numerical experiments
[4] seem to indicate that self-similar blow-up is a generic situation.
So consider in the case of dimension 3 a blowing-up self-similar solution φ , with
φ ∈ ˙ H d/2 . It leaves the space
L ∞ t
˙H d
2
x
x
T −t
as it approaches the blow-up time T . Furthermore, at time T , the trace of the solution
looks like
φ(T ) = c0 and ∂tφ(T ) = c1
|x| ,
so that
(φ(T ) , ∂tφ(T )) does not belong to ˙ H d
2 × ˙ H d
2 −1 .
The idea of the present article is the following observation: if one replaces ˙ H d
2 and L∞ d
H˙ 2
by spaces that can accomodate self-similar data, respectively self-similar solutions, there
is in some sense no blow-up any more. Such spaces will be provided by
˙B d
2
2,∞ × ˙ B d
2 −1
2,∞ and L∞ d
B˙ 2
(see the next section for a definition of these Besov spaces). If one works with this new
functional framework, one has
x
φ ∈ L
T − t
∞
R, ˙ B d
2 −1
2,∞
and
2,∞
for any time t, (φ(t) , ∂tφ(t)) ∈ ˙ B d
2
2,∞ .
The above discussion shows that this new functional framework could be very interesting
for the study of self-similar solutions.
Infinite energy solutions
Also notice that it is interesting and natural to study infinite energy solutions for the
following reasons
• First, harmonic maps (that is, stationary wave-maps) of S 3 are of infinite energy.
• Second, data or solutions of globally finite energy necessarily correspond to functions
which at spatial infinity focus to a point; by allowing the energy to be infinite, we
can study the global behaviour of much more general data.
3

• Finally (and this is somehow the dual of the previous point), infinite energy data
enable us to consider (covariant) wave-maps such that φ(0) is not necessarily the
north or the south pole of the sphere S d , so that a singularity occurs at 0.
Plan of the article
In Section 2, we will prove existence, uniqueness and scattering of solutions for data that
are of infinite energy but small. We shall use a classical fixed-point argument.
In Section 3, we will study the case of large data. However, since we cannot resort to
perturbation techniques anymore, we will have to restrict to the case of self-similar data.
Matters reduce then to studying a (singular) ODE.
In Section 4, we shall see that we can apply the results of the last two sections in order
to describe a little more precisely the blow-up phenomenon for the wave-map equation on
the sphere S 3 .
1.4 Littlewood-Paley decomposition and associated spaces
We shall present in this section the main harmonic analysis tools that will be used in the
sequel.
A Littlewood-Paley decomposition is defined in the following way: consider Ψ a function
supported in the annulus centered in 0, of small radius 3/4, and big radius 8/3, such that
ξ
Ψ
2j
= 1 for ξ = 0 .
j∈Z
Then the Fourier multipliers ∆j and Sj are defined by
D
∆j = Ψ
2j
SN = 1 −
D
Ψ
2j
j≥N+1
Thus any distribution can be decomposed (modulo polynomials) into a sum of elementary
elements that are localized in frequency:
f =
∆jf or f = SNf +
∆jf .
j∈Z
We can now define the (homogeneous) Besov spaces ˙ B s q,r
f ˙ B s q,r
def
=
2 js ∆jf q
L
x ℓr j
.
j≥N+1
which are given by the norm
Notice that the homogeneous L 2 -based Sobolev spaces identify with Besov spaces
˙H s = ˙ B s 2,2 .
A variant of Besov spaces which are particularly well suited for the study of PDEs is given
by the spaces L p (R, ˙ B s q,r ) - we will later omit the R in order to make the notations lighter.
4
.

These spaces were introduced by Chemin and Lerner [5] in the context of PDEs (notice
that very similar spaces were considered earlier by Triebel [14] from a harmonic analysis
point of view). The norm of these spaces is defined as follows
fe Lp (R, B˙ s )
q,r
def
= 2 js ∆jf p
L
t Lq
x ℓr j
In Section 2, we will build up a solution to (3) in Chemin-Lerner spaces.
Let us record the two following results, which extend classical theorems on Besov spaces.
Theorem 1.1 (Sobolev embedding) Suppose that
we have then
q ≥ Q and s − n
q
L p ˙ B S Q,r ↩→ L p ˙ B s q,r .
The proof simply relies on Bernstein inequality.
= S − n
Q ,
Proposition 1.1 Suppose s > 0 and
f =
fk with Supp fk ⊂ B(0, C2 k ) .
One has then
k
f L p (R, ˙ B s q,r) ∼
2 Small data in Besov spaces
2 js fjf L p
t Lq
x ℓr j
2.1 Main result: existence, uniqueness, and scattering
Well-posedness for the wave-map system is only known for data in Sobolev spaces. We
want to extend this to Besov spaces, in the same spirit as Planchon [12] did for the wave
equation with a power non-linearity.
Theorem 2.1 Let n ≥ 5 (i.e. d ≥ 3), and consider data (v0, v1) ∈ ˙ B n/2−1
2,∞ × ˙ B n/2−2
2,∞ radial
and small enough. Then there exists a global solution v of (3), which is unique in a ball
of sufficiently small radius of
(5) X def
∞
= L R, ˙B n
2 −1
∩ L 2
∩ L 3
.
2,∞
R, ˙B n
2 −2
( 1
2
− 3
2n) −1 ,∞
.
.
R, ˙B n
2 −2
( 1
2
− 4
3n) −1 ,∞
Furthermore, a weak scattering takes place, in the sense that there exists (v + 0 , v+ 1 ) and
(v − 0 , v− 1 ) in ˙ B n/2−1
2,∞ × ˙ B n/2−2
2,∞ such that
∀j ,
+
∆j W (t)(v 0 , v + 1 ) − v + 2
+
∆j∂t W (t)(v 0 , v + 1 ) − v
−→ 0 2
∆j
−
W (t)(v0 , v − 1 ) − v + 2
∆j∂t
−
W (t)(v0 , v − 1 ) − v −→ 0 2
as t → +∞ ,
where we denote by W (t)(f, g) the free solution of the wave equation associated to the
initial data (f, g).
In general, no classical scattering result for the norm of ˙ B n/2−1
2,∞
5
× ˙ B n/2−2
2,∞
holds.

The proof of the above theorem will proceed in three steps: first, dealing with existence
and uniqueness in the case Z = Id (Section 2.2), then in the general case (Section 2.3),
and finally proving the scattering result (Section 2.4).
Let us make a few comments concerning this theorem.
• Actually, we will prove the above theorem for any smooth function Z in (3) ; when
the equation (3) is derived from the wave-map system however, we get a function Z
which is even
• The reason why we have to take such unpleasant exponents in the definition of X is
that we need positive regularity indices so that Proposition 2.2 holds true.
• How does the regularity of v0 and v1 translate for φ0 and φ1? It is easy to see that
(v0, v1) ∈ ˙ B n/2−1
2,∞ × ˙ B n/2−2
2,∞ does not imply (φ0, φ1) ∈ ˙ B d/2
2,∞ × ˙ B n/2−1
2,∞ . On the other
hand, it is proved in [16] that (v0, v1) ∈ ˙ Hn/2−1 × ˙ Hn/2−2 is equivalent to (φ0, φ1) ∈
˙H d/2−1 × ˙H d/2−2 . As a conclusion, the data of the above theorem correspond to
(φ0, φ1) in a space lying (strictly) between ˙ Hd/2−1 × ˙ Hd/2−2 and ˙ B d/2
2,∞ × ˙ B n/2−1
2,∞ .
2.2 Proof of existence and uniqueness in the case Z = Id
The equation under consideration is for now
(6)
⎧
⎨
⎩
v = rv 4
v(0) = v0
vt(0) = v1 .
where we denote for the wave operator = ∂ 2 t − ∆. The initial data (v0, v1) are radial
and belong to ˙ B n/2−1
2,∞
× ˙ B n/2−2
2,∞ .
We will use a fixed point argument in the space
X =
∞
L R, ˙ B n
2 −1
2,∞
∩ L 2
R, ˙ B n
2 −2
Strichartz estimates will be a crucial tool.
( 1
2
− 3
2n) −1 ,∞
Lemma 2.1 (Strichartz estimate) The solution u of
(7)
satisfies the estimate
uX ≤ C
⎧
⎨
⎩
u0 ˙B −1+ n 2
2,∞
u = F
u(0) = u0
ut(0) = u1 .
+ u1 ˙B −2+ n 2
2,∞
∩ L 3
R, ˙ B n
2 −2
( 1
2
+ F eL 1 ˙ B −2+ n 2
2,∞
− 4
3n) −1 ,∞
Proof of Lemma 2.1: We notice first that we can localize in frequency the equation (7).
The classical energy estimate for the wave equation gives then
∆juL∞H˙ 1 ≤ C
∆ju0H˙ 1 + ∆ju12 + ∆jF L1L2 .
6
.
.

Using the frequency localization, we can rewrite this as
2 j ∆juL∞L2 ≤ C 2 j
∆ju02 + ∆ju12 + ∆jF L1L2 ,
n
j( and it suffices to multiply the above by 2 2 −2) and take the supremum over j to obtain
u −1+
eL ∞ ( B˙ n
≤ C u0 2
−1+
2,∞ ) ˙B n 2
2,∞
+ u1 ˙B −2+ n 2
2,∞
+ F eL 1 ˙ B −2+ n 2
2,∞
Next, we would like to obtain the same estimate for the L2 n
B˙ 2 −2
( 1 3
− 2 2n) −1 ,∞
apply Strichartz’ inequality [7] to each one of the dyadic components
∆ju
L2L ( 1 2
−
3
2n) −1 ≤ C 2 j
∆ju02 + ∆ju12 + ∆jF L1L2 ,
.
norm. We just
n
j( and multiplying both sides by 2 2 −2 ), then taking the supremum over j, one gets
u
eL 2 ˙B n 2 −2
( 1 2
−
3
2n) −1 ,∞
≤ C
u0 ˙B −1+ n 2
2,∞
+ u1 ˙B −2+ n 2
2,∞
It remains to show this inequality for the L3 n
B˙ 2 −2
( 1 4
− 2 3n) −1 ,∞
similar to the above one, and we therefore omit it.
+ F eL 1 ˙B −2+ n 2
2,∞
norm; but the proof is very
The above lemma shows that in order to run a fixed point argument and solve (6) in the
space X for small data, it suffices to control rv 4 in L 1 ˙ B −2+ n
2
2,∞ .
Proposition 2.1 There holds
rv 4 eL 1 ˙ B −2+ n 2
2,∞
≤ Cv 4 X
Proof of Proposition 2.1: First of all we observe that by Sobolev embedding (Theorem
1.1),
X ↩→ L ∞ ˙ B −1+ n
2
2,∞
∩ L 2 ˙ B − 1
2
∞,∞ ∩ L 3 −
B˙ 2
3
∞,∞ ,
and we will be using in the proof of Proposition 2.1 only the norms appearing in the right
hand side of the above embedding.
We now split v 4 using the paraproduct algorithm introduced by Bony [1], that is we write
v 4 =
j∈Z
(Sj+1v) 4 − (Sjv) 4 .
Next we observe that the summands in the above equation are of the form (∆jv)ABC,
where A, B and C can be either Sjv or ∆jv. The most difficult case is the one where
A = B = C = Sjv, so we will try to estimate in the following
r
∆jv(Sjv) 3 ,
j
7
.

that we rewrite as
r
∆jv(Sjv) 3 =
j
Estimate for I eL 1 ˙B −2+ n 2
2,∞
j
Sj+100
= I + II .
We will use the following estimates:
• Simply by definition of X,
• By Lemma 5.1, we have
r∆jv(Sjv) 3 +
(1 − Sj+100) r∆jv(Sjv) 3
n
j(1−
∆jvL∞ L2 ≤ CvX2 2 ) .
rSjvL ∞ L ∞ ≤ Cv L ∞ ˙ B n 2 −1
2,∞
j
≤ CvX .
• Due to the negative regularity of L2 −
B˙ 1
2
∞,∞, we can bound Sjv as follows
SjvL2 L∞ ≤
∆kvL2 L∞ ≤ CvX
k

• By definition of X,
• With the same justification as above
Combining these two estimates, we get
n
1−
∆jvL∞ L2 ≤ C2 2 vX .
2
j
SjvL3 L∞ ≤ C2 3 vX .
n
j(3−
(8) fjL1 L2 ≤ 2 2 ) v 4 X .
Now to estimate ∆k(rfj) we use Plancherel theorem (in space) which yields
∆k(rfj)2 =
ψ
ξ
2k
1
|ξ − η| n+1
fj(η) dη
.
Notice that, due to the Fourier localization of the Littlewood-Paley operators, in the above
expression, η ≤ C2j and ξ ∼ 2k . As a result, ξ − η is of order 2k and we don’t have to
worry about the Fourier transform at 0 of |x|. We can bound the above by
∆k(rfj)2 ≤
ψ
ξ
2k
2
|η|≤C2 j
2
1
|ξ − η| n+1
fj(η) dη
.
∞
We now use Hölder’s inequality and the fact that ξ − η ∼ 2 k to estimate the integral in
the above expression and get
n
k
∆k(rfj)2 ≤ C2 2 2 −k(n+1) n
n
j k(−1−
2 2 fj2 = C2 2 ) n
j
2 2 fj2 .
Integrating in time and using (8),
It suffices now to sum over k to get
∆k(rfj)
and Proposition 1.1 yields
n
k(−1−
∆k(rfj)L1L2 ≤ C2 2 ) 2 3j v 4 X .
k>j+100
L 1 L 2
II eL 1 ˙B −2+ n 2
2,∞
n
j(2−
≤ C2 2 ) v 4 X ,
≤ Cv 4 X .
This concludes the proof of Theorem 2.1 in the case where Z = Id.
2.3 Proof of existence and uniqueness in the general case
We want now to consider the non-linearity Z(rv)v 3 instead of rv 4 .
We observe first that we can assume that Z vanishes at 0. If it is not the case, we split
the non-linearity into two parts as follows
Z(0)v 3 + (Z(rv) − Z(0)) v 3 .
9

The cubic term is easy to handle, so we will focus on the other one, (Z(rv) − Z(0)) v 3 ,
which is the same as assuming that Z vanishes at 0. We further notice that, in order to
apply the results of Section 2.2, it would suffice to show that Z(rv)
has the same regularity
as v. This is the content of the next Proposition, which will therefore complete the proof
of Theorem 2.1.
Proposition 2.2 Suppose that v is a radial function. We have then
Z(rv)
≤ CvX .
r
X
Proof of Proposition 2.2: The proof is classical, it uses the first linearization method
of Meyer [11]. We write, using Taylor’s formula
Z(rv)
r
Z (rSj+1v) − Z (rSjv)
=
r
j∈Z
=
∆jv
j
1
0
Z ′ (rSjv + sr∆jv) ds .
We now observe that rv is bounded in L ∞ by Lemma 5.1, and that the regularity index
of each of the spaces whose intersection is X has a positive regularity index. It therefore
suffices to follow the lines of [2] in order to prove the proposition.
2.4 Proof of the scattering result
We consider a solution v as built up in the preceding sections, and we only prove the
scattering as t → ∞, the other case being identical. Denoting F = Z(rv)v 3 , we have seen
that F ∈ L1B˙ d/2−1
2,∞ . Applying the frequency localization operator to the equation satisfied
by v, we get
∆jv = ∆jF ,
2
with ∆jF 2 ≤ C2
j( n −2) . So by the classical energy scattering result, there exists
(v + 0,j , v+ 1,j ) such that
and
It suffices to set
∆jv − W (t)(v + 0,j , v+ 1,j )
+ ∂t∆jv − ∂tW (t)(v
2 + 0,j , v+ 1,j )
−→ 0 ,
2
v + 0,j2 n
j(1−
≤ C2 2 ) , v + 1,j2 n
j(2−
≤ C2 2 ) .
v + 0
to get the weak scattering result.
=
j∈Z
v + 0,j , v + 1
=
j∈Z
v + 1,j
Let us now prove that scattering for the norm of B˙ n/2−1
2,∞
general. To do so, we consider self-similar data
v0(x) = c
|x|
, v1(x) = c
,
|x| 2
10
r
× ˙ B n/2−2
2,∞
does not occur in

where c is small enough so that the existence and uniqueness result applies. By uniqueness,
the solution v itself is self-similar, ie is left unchanged by the transformation (4). Let us
argue by contradiction and suppose that there exists (f0 , f1) ∈ ˙ B n
2 −1
2,∞ × ˙ B n
2 −2
2,∞ such that
W (t)(f0 , f1) − v ˙B n 2 −1
2,∞
n
= sup 2
j( 2
j
−1) (∆jW (t)(f0 , f1) − v)2 −→ 0 as t → ∞ .
It is easy to see that (f0 , f1) has to be scaling invariant as well, and a small computation
yields then
∆j (v − W (f0 , f1)) (t, x) = 2 j ∆0 (v − W (f0 , f1)) (2 j t , 2 j x) .
This implies that
d
sup 2
j( 2
j
−1)
∆j (W (t)(f0 , f1) − v) (t)2 = sup ∆0 (W (f0 , f1) − v) (2
j
j t) , 2
and we see that this last quantity goes to zero as t goes to infinity only if u = W (f0 , f1),
which implies u = 0.
3 Large self-similar data
3.1 Main result
For the sake of simplicity, we will restrict our investigations in this section to a classical
case: in the following, we take d = 3 (hence n = 5) and N = S 3 ; already a great
wealth of behaviours can be observed. In this case, G(φ) = sin(φ), and equation (2)
becomes
(9) φtt − φrr − 2
r φr + 1
sin(2φ) = 0 .
r2 We want to consider self-similar solutions; since we always deal with corotational solutions,
the data must have the following form
φ0 = c0 , φ1 = c1
|x| .
If for instance uniqueness holds, it is easy to see that the solution φ is itself self-similar,
i.e. given by a profile f,
x
φ(x, t) = f .
t
We will look for a solution under that form, so that the problem reduces to finding f.
The interest of self-similar solutions is that they provide, if the profile f is regular, an
example of blow up of (9) at t = 0. The first example of such a blowup was given by
Shatah [15]: for c0 = 0, c1 = 2, the solution corresponding to the profile
(10) f0(ρ) = 2 tan −1 (ρ)
solves (9). Numerical simulations of Bizoń, Chmaj and Tabor [4] suggest that f0 is a
generic blowup profile.
Bizoń [3] showed that f0 was actually the “lowest-energy” member of a family of regular
profiles (fn)n∈N. The profiles fn for n > 1 should be less stable than f0.
11

Finally, it was proved in [16] that for given data the profile may not be unique. Their
example was the following
¯f0(ρ) =
π
2
for ρ < 1
f0(ρ) for rho > 1
It is easily seen that the solutions corresponding to these data share the same initial data.
The initial data considered by Bizoń and Shatah form a countable set, namely only those
data that yield regular profiles; in the following theorem, we consider any initial data.
We now switch to the v unknown function and observe that the Cauchy problem becomes
(11)
⎧
⎪⎨
⎪⎩
vtt − vrr − n−1
r vr = v 3 Z(rv)
v(0) = c0
|x|
vt(0) = c1
|x| 2 .
What regularity should be expected for the self-similar solutions of (11)? For c0 and c1
small, we know that the solution v = φ
r
in (5) and which reads for n = 5
built up in Theorem 2.1 belongs to X defined
X =
∞
L R, ˙ B 3/2
2,∞ ∩
2
L R, ˙ B 1/2
5,∞ ∩
3
L R, ˙ B 1/2
30/7,∞ .
How does it translate for the profile f? By an argument given for instance in [12], v
belongs to X if and only if, when considered as a function over R 5 .
with
f(ρ)
ρ
∈ Y
Y def
= ˙ B 3/2
2,∞ (R5 ) ∩ ˙ B 1/2
5,2 (R5 ) ∩ ˙ B 1/2
30/7,3 .
For c0 and c1 large, the theorem which follows tells us that we can always build up self-
similar solutions such that v ∈ X or equivalently f
ρ
∈ Y .
Theorem 3.1 (i) For any real numbers c0 and c1, there exists a solution v of (11) given
by a profile f.
(ii) If one denotes v = φ
r , one has
v ∈ X and
(as explained above, both functions are here considered as functions on R 1+5 ).
(iii) v or f are unique outside of the ball of radius one, but uniqueness for v ∈ X or f ∈ Y
does not hold in general.
(iv) However, if v ∈ X or f ∈ Y , one has f(0) = k π
2 , with k an integer.
So in contrast with the result of Theorem 2.1, which dealt with small data, uniqueness
does not hold anymore for large data.
The proof of the above theorem will be given in the following subsections 3.2, 3.3, 3.4,
and 3.5.
12
f
ρ
∈ Y
.

3.2 The equation satisfied by the profile f
Due to the self-similarity assumption, the equation satisfied by f reduces to an ordinary
differential equation. It is given by the following proposition.
Proposition 3.1 Suppose that φ is given by φ(x, t) = f( |x|
t ) (we consider here the radial
function f as a function of one real variable). Then
(i) Suppose f ′ is locally integrable, and f, f ′ and ρf ′′ (ρ) are continuous at 0. Then v
solves (11) if and only if f solves
⎧
⎨ ρ
(12)
⎩
2 (1 − ρ2 )f ′′ (ρ) + 2ρ(1 − ρ2 )f ′ (ρ) − sin(2f(ρ)) = 0 in S ′ (0, ∞)
limρ→∞ f(ρ) = c0
limρ→∞ ρ 2 f ′ (ρ) = −c1 .
(ii) The first equation of (12) is satisfied if and only if it holds true in the classical sense
on (0, 1)∪(1, ∞), and f(1−) = f(1+) (these limit values make sense because if the equation
is satisfied on (0, 1) ∪ (1, ∞), f ′ is locally integrable on (0, 1) ∪ (1, ∞)).
Proof of Proposition 3.1: We begin by proving (i). It is first very easy to check that
the initial conditions of (11) and the boundary conditions of (12) are equivalent.
Next, we will work with the φ equation, which is easier to handle in this case. It it then
easy to see that v solves (11). By definition, φ is a solution of (9) if and only if for any
test function ψ ∈ D([0, ∞) × [0, ∞)),
(13)
∞ ∞
−ψtφt + ψrφr + 1
sin(2φ)
r2 0
0
ɛ
0
r 2 ∞
dr dt =
0
ψ(r, 0) c1
r r2 dr
and the initial conditions are verified. It is very easy to check that the initial conditions
of (9) and the boundary conditions of (12) are equivalent. So we will focus on the equation
above, and prove that it is equivalent to the first equation of (12). We begin by looking at
∞ ∞
− ψtφtr
0 0
2 ∞ ∞ r
dr dt =
0 0 t2 f ′ r
ψtr
t
2 dr dt
∞ ∞ r
=
t2 f ′ r
ψtr
t
2 dr dt + O(ɛ) ,
where the 0(ɛ) comes from the finite limit of ρ2f ′ (ρ) at ∞. Integrating by parts in t, we
find
∞ ∞
− ψtφtr
0 0
2 dr dt
∞ r
= O(ɛ) −
ɛ2 f ′ r
ψr
ɛ
2 ∞ ∞ r
dr dt + 2
t3 f ′ r
ψr
t
2 ∞ ∞ r
dr dt +
2
t4 f ′′ r
ψr
t
2 dr dt .
0
0
ɛ
0
Integrating by parts in r the last term of the right-hand side, we get, using the fact that
f ′ has a finite limit at 0,
∞ ∞
− ψtφtr
0 0
2 dr dt
∞ r
= O(ɛ) −
ɛ2 f ′ r
ψr
ɛ
2 ∞ ∞ r
dr dt − 2
t3 f ′ r
ψr
t
2 ∞ ∞ r
dr dt −
2
t3 f ′ r
ψrr
t
2 dr dt .
ɛ
0
13
ɛ
ɛ
0
0

We now perform the change of variable ρ = r
above equality, the left-hand side of (13) becomes
∞ ∞
−ψtφt + ψrφr +
0 0
1
sin(2φ) r
r2 2 dr dt
∞ r
= O(ɛ) −
ɛ2 f ′ r
ψr
ɛ
2 ∞ ∞
dr dt +
0
ɛ
0
t , so that in particular ψr = 1
t ψρ. Using the
−2ρ 3 f ′ ψ + ρ 2 (1 − ρ 2 )f ′ ψρ + ψ sin(2f) dρdt .
Obviously, the limit as ɛ goes to 0 of the two first terms of the right hand side is
∞
0
ψ(r, 0) c1
r r2 dr. So (13) holds true if and only if for any t
∞
This is equivalent to
0
−2ρ 3 f ′ ψ + ρ 2 (1 − ρ 2 )f ′ ψρ + ψ sin(2f) dρ = 0 .
ρ 2 (1 − ρ 2 )f ′′ (ρ) + 2ρ(1 − ρ 2 )f ′ (ρ) − sin(2f(ρ)) = 0 in S ′ (0, ∞)
Let us now prove (ii). It is convenient to switch to the function
1
g(z) = f ,
z
and one can check that the first equation of (12) is equivalent to
(z 2 − 1)g ′′ (z) = sin(2g(z)) .
Let us assume that this equation is satisfied classically on (0, 1)∪(1, ∞), and that g(1−) =
g(1+). We will show that it is then satisfied in S ′ . Of course, the only problematic point
is 1, so we will try to describe more precisely the behaviour of g near 1. First of all, we
observe that g ′ is integrable on (0, 1)∪(1, ∞), so that g is continuous. We have furthermore
the following bound
(14) |g ′ (z)| ≤ C| log |z − 1||
for z = 1; g ′ could a priori have a δ singularity at 1, but this is excluded by the continuity
of g.
Now we would like to show that
∞ 2 ′′
(z − 1)g (z) − sin(2g(z)) ψ(z) dz = 0 ,
0
if ψ ∈ D((0, ∞)). We can cut the integral into three pieces,
∞
=
0
1−ɛ
0
+
1+ɛ
1−ɛ
∞
+ .
The first and third pieces are 0 because the equation is satisfied in the classical sense, and
for the second we have, integrating by parts,
1+ɛ
1−ɛ
(z 2 − 1)g ′′ (z) − sin(2g(z)) ψ(z) dz
= O(ɛ) + g ′ (z)(z 2 − 1)ψ(z) 1+ɛ
1−ɛ −
1+ɛ
g
1−ɛ
′ (z) (z 2 − 1)ψ(z) ′
dz .
Due to (14), we see that the right-hand side goes to zero as ɛ goes to 0.
14
1+ɛ

3.3 Solving the equation for f
We will here show the existence of solutions for the system (12).
Proposition 3.2 For any real numbers c0 and c1, the system (12) has a solution f ∈ L ∞ .
It is unique on [1, ∞), but not in general on [0, 1].
Remark 3.1 As stated in the above proposition, solutions to the first equation of (12)
ρ 2 (1 − ρ 2 )f ′′ (ρ) + 2ρ(1 − ρ 2 )f ′ (ρ) − sin(2f(ρ)) = 0 ,
(with prescribed data at infinity) are not unique in S ′ . Nevertheless, one can prove that
solutions to the equation
f ′′ (ρ) + 2
ρ f ′ (ρ) − sin(2f(ρ))
ρ2 (1 − ρ2 = 0 ,
)
considered in ([16]) and ([3]) are unique in S ′ . Notice that formally, the second equation
is nothing but the first divided by ρ 2 (1 − ρ 2 ).
Proof of proposition 3.2: The problem (12) becomes, for the new unknown function
g(z) = f
1
z
⎧
⎨
⎩
(z 2 − 1)g ′′ (z) = sin(2g(z))
g(0) = c0
g ′ (0) = −c1
It is clear that this problem admits a unique solution on [0, 1], which translates into a
unique solution for (12) on [1, ∞). Notice that g ′ is locally integrable, so that g(1−) =
f(1+) is well defined.
By proposition 3.1, it suffices, in order to get a global solution of
(15) ρ 2 (1 − ρ 2 )f ′′ (ρ) + 2ρ(1 − ρ 2 )f ′ (ρ) − sin(2f(ρ)) = 0 ,
(with the given prescribed data at infinity) to solve this equation on [0, 1], in such a way
that f(1−) = f(1+), and furthermore f, f ′ , ρf ′′ have finite limits at 0.
We begin with the following lemma
Lemma 3.1 For any α ∈ R, the Cauchy problem
⎧
⎪⎨
⎪⎩
f ′′ + 2
ρf ′ − sin(2f)
ρ2 (1−ρ2 = 0 )
f(0) = 0
f ′ (0) = α
admits a solution on [0, 1). Furthermore, it depends continuously on α.
Proof: We only prove the existence part of the lemma; the continuous dependence on
the data can be handled in a similar fashion. We set f = H + αx, so that the problem
becomes ⎧ ⎪⎨
⎪⎩
H ′′ + 2
ρH′ + 2α sin(2(H+αρ))
ρ − ρ2 (1−ρ2 = 0 )
H(0) = 0
H ′ (0) = 0
15

Setting h = H ′ , the problem becomes
h = 1
ρ2 ρ
0
H = ρ
0 h
We use the following iteration scheme
hn+1 = 1
ρ2 ρ
0
Hn+1 = ρ
0 hn ,
sin(2αρ+2H)
1−ρ 2
sin(2αρ+2H n )
1−ρ 2
− 2αρ dρ
− 2αρ dρ
and we set h 0 = H 0 = 0. It is easy to see that, on a small enough interval,
|h 1 (ρ)| ≤ Cρ 2
(and H 1 = 0) .
The difference of two consecutive iterates is given by
hn+1 − hn = 1
ρ2 ρ
0
Hn+1 − Hn = ρ
0 hn − hn−1 .
sin(2αρ+2H n )−sin(2αρ+2H n−1 )
1−ρ 2
On a small enough (but fixed) interval, this implies that
|Hn − Hn−1 | ≤ Cρ3 |hn − hn−1 | ≤ Cρ2
|Hn+1 − Hn | ≤ δCρ3 =⇒
|hn+1 − hn | ≤ δCρ2 where δ < 1. This implies that the iteration scheme converges on a small interval [0, ɛ],
with ɛ > 0. It is then straightforward to extend the solution to [0, 1].
With the help of this lemma, it is easy to complete the proof of the proposition. It suffices
to show that, by choosing α (as in the statement of the lemma) in an appropriate way,
one can reach any given value f(1−). But we know two particular solutions, the solution
corresponding to α = 0 for which f(1) = 0, and f0 (defined in (10)) for which α = 2,
f(1) = π . The continuous dependence on α of the solutions built up in the above lemma
2
tells us that any real number between 0 and π
2
dρ
can be reached by f(1−).
Then the symmetries of the equation (for instance, if f is a solution, so is π−f and 2π+f)
enable to reach any desired f(1) with f(0) = kπ and f ′ (0) judiciously chosen.
3.4 Regularity questions
In this section, we complete the proof of Theorem 3.1 by examining the regularity of the
solutions.
Proposition 3.3 Any of the solutions v built up in Proposition 3.2 satisfies
v ∈ X which is equivalent to
f
ρ
∈ Y .
(here f is considered as a function on R 5 ; for the definitions of X and Y , see section 3.1).
Proof: We will prove that f
ρ belongs to Y . The space Y is the intersection of three Besov
spaces; the hardest to deal with is the first one, B˙ 3/2
, so we will focus on it, the proof for
the two other ones being similar.
The function f
ρ (seen as a function on R5 ) may be singular at three points: 0, 1 and ∞,
it is C ∞ elsewhere, so we will examine separately these three points.
16
2,∞
,

• Let us begin with the behaviour at infinity, that is we look at χf with χ a smooth
function equal to 0 on B(0, 2) and 1 outside of B(0, 3).
It is first of all clear that χ c0
ρ ∈ ˙ B 3/2
f−c0
2,∞ ; so we are left with χ ρ . Using the equation
satisfied by f and the data at ∞, we see that
′′
f − c0
χ ≤
ρ
C
.
〈ρ〉 4
In order to show that χ f−c0
, it suffices to see that the second
derivative belongs to ˙ B −1/2
2,∞ . Using the above bound along with Young’s inequality,
we get
∆j ′′
f − c0 2
χ =
ρ 25j ′′
ψ(2 j f − c0 2
·) ⋆ χ
ρ
≤ 2 5j
ψ(2 j
·)
χ
ρ belongs to ˙ B 3/2
2,∞
10/9
′′
f − c0
5/3
≤ C2
ρ
j/2 .
• We will now examine the function in a neighborhood of the point 0.
Consider first the profile identically equal to kπ
2 . Then it is straightforward to see
that f
ρ
= kπ
2ρ belongs to ˙ B 3/2
2,∞ (R5 ).
We suppose now f(0) = 0 and look at one of the solutions built up in Lemma 3.1.
An examination of the equation satisfied by f implies that, close to 0,
f(ρ) ∼ αρ , f ′ (ρ) ∼ αρ and f ′′ (ρ) ∼ − α
ρ .
We localize the function around 0 by considering φ f
ρ , with φ a smooth function equal
to 1 on B(0, 1/2) and 0 outside of B(0, 3/4). As in the last point, it suffices to see
that
′′
f
φ ∈ L
ρ
5/3 ,
and it is easy to see that this holds, since, by the estimates above on the behaviour
of f close to 0, we have f
ρ ∈ L5/3 .
• We finally investigate the situation at 1. Since f is radial, it belongs around ρ = 1
(as a function on R5 ) to ˙ B 3/2
2,∞ if and only if it does so as a function on R. So we have
reduced the question to a one-dimensional problem. To simplify matters a little bit
more, we will work with the function g(z) = f
1
z ; its regularity at 1 is obviously
the same as the regularity of f, and it satisfies the simple equation
(16) (z 2 − 1)g ′′ (z) = sin(2g(z)) .
This implies at once that g ′ behaves like sin(2g(1)| log |z − 1|| around 1, and integrating
one more time we see that
Coming back to (16), we see that
g(z) = g(1) + O(|z − 1| log |z − 1|) .
g ′′ (z) = sin(2g(1)
z 2 − 1
17
+ O(log |z − 1|) .

So that, integrating one last time we get the desired description of the singularity
at 1
g ′ (1 + δ) = sin(2g(1)) log |δ| + Csign δ + φ ,
where C± are constants, and φ is a continuous function around 1, that we will ignore
in the following. So (shifting the singularity to 0, we are now in one dimension) the
problem reduces to showing that log |z| and the Heaviside function H(z) belong to
˙B 1/2
1
2,∞ (R). Differentiating, this is equivalent to the fact that z and the Dirac function
δ belong to ˙ B −1/2
2,∞ (R).
But it is well known that the Fourier transform of these two functions are, respec-
tively, sign(ξ) and 1. It follows that
1
∆j
z 2
=
ψ
ξ
2j
sign(ξ) ∼ 2 j/2 and
∆jδ2 =
ψ 2
and this completes the study of the singularity of f at |ρ| = 1.
3.5 Uniqueness issues
ξ
2j
2
∼ 2 j/2 ,
That the solution is unique outside the ball of radius one is obvious. The first known
example of non-uniqueness consisted of the function ¯ f0 of Shatah, defined in Section 3.1.
As proved in the previous section, it belongs to X.
Due to the second point of Proposition 3.1, it very easy to build up new examples of
non-unique solutions. The profile is imposed for ρ ∈ (1, ∞), this also imposes f(1+).
Then we can complete it by any of the solutions on (0, 1) of Lemma 3.1, provided that
f(1−) = f(1+). That the resulting solution belongs to X has been proved in the previous
section.
As a last result, we would like to prove that, though no uniqueness holds on (0, 1), not
any value can be taken at 0 by the profile.
If v = f
r
t belongs to X, then, by Lemma 5.1, f ∈ L
r
∞ . We have then the following
lemma.
Lemma 3.2 Suppose f solves (15) on (0, 1)) and is bounded in L∞ . Then f is continuous
at 0, and f(0) = k π
2 , with k an integer.
If k is odd, then f is constant on (0, 1).
Proof: Take f as in the lemma, and let
A small computation shows that
M(ρ) = 1
2 (ρ2 − ρ 4 )f ′2 + cos 2 f .
M ′ = −f ′2 ρ ,
so that M is a decreasing quantity. If M is larger than 1 for some ɛ > 0, then it must be
larger than 1 on (0, ɛ), and this implies that f blows up at 0. So M is less than one on
(0, 1/2), let us denote M0 for its limit at 0. We will show that f also has a limit at 0.
We observe that the integral of M ′ is finite, this implies that
(17) ρf ′2 ∈ L 1
0, 1
;
2
18

also the boundedness of M implies that
(18) |f ′ (ρ)| ≤ C
ρ .
Fix ɛ > 0, we can suppose that ρ is taken close enough to 0 so that
|M(ρ) − M0| ≤ ɛ .
Also, we must have cos 2 f(ρ n ) − M(ρ n ) ≤ ɛ
for a sequence ρ n going to 0, due to (17). We argue by contradiction and assume (possibly
considering a subsequence of (ρ n )) that for a sequence τ n going to zero, with ρ n+1 ≤ τ n ≤
ρ n , we have
cos 2 f(τ n ) − M(τ n ) ≥ 2ɛ and hence
n
In
n
1
2 (ρ2 − ρ 4 )f ′2 ≥ ɛ .
Due to (18), f ′ remains larger than ɛ on an interval In centered at τn and of size comparable
to τ n . But then
1/2
ρf
0
′2 dρ ≥
ρf ′2 dρ ≥
1 ɛ
dρ = ∞ ,
2 In ρ
contradicting (17). So f has a finite limit at 0, that we denote f(0). Equation (15) can
be rewritten as
d
2 ′
ρ f
dρ
= sin(2f)
,
1 − ρ2 and so f cannot remain bounded unless sin(2f(0)) = 0 i.e. f(0) = k π
2 , with k an integer.
If k is odd, we have M0 = 0, hence M remains equal to 0 on (0, 1), that is, f is constant.
4 An application to the development of singularities
4.1 A refined version of Theorem 2.1
Let us begin by noticing that the proof of Theorem 2.1 can be adapted (as for instance
in [12] or [6]) to yield the following theorem.
Theorem 4.1 (i) Take (v0, v1) ∈ ˙ B 3/2
2,r × ˙ B 1/2
2,r with 1 < r < ∞. Then there exists T > 0
and a solution of (3) in
(19) Xr([0, T ]) def
=
∞
L [0, T ], ˙ B 3/2
∩
2
L [0, T ], ˙ B 1/2
which is unique in a ball of sufficiently small radius of the above space.
(ii) Consider now data of the form
with
2,r
5,r
v0 = w0 + w0 , v1 = w1 + w1 ,
∩
3
L [0, T ], ˙ B 1/2
30/7,r
( w0, w1) ∈ ˙ B 3/2
2,∞ × ˙ B 1/2
2,∞ and (w0, w1) ∈ ˙ H 3/2 × ˙ H 1/2 ,
19
,

Assume that all the norms of the above functions are small enough, then there exists by
Theorem 2.1 a unique solution v ∈ X = X∞ of (3).
Denote by w ∈ X the unique solution of (3) with data ( w0, w1); then v reads
v = w + w with v ∈ X2 ↩→ L ∞ (R, ˙ H 3/2 ) .
If r = 2, the assertion (i) corresponds simply to the well posedness result for data in
˙H 3/2 × ˙ H 1/2 proved in [16].
4.2 Regularity breakdown
As we discussed at the beginning of Section 3, we expect for the covariant wave-map
equation with N = S 3 the development of singularities along self-similar blow-up profiles.
Consider a solution v which blows up at 0, in the sense that v ∈ L ∞ loc ((−ɛ, 0) ∪ (0, ɛ), ˙ H 3/2 .
We expect the value at 0 of the function to be
with
v(0) = c0
|x| + w0(x) , ∂tv(0) = c1
+ w1(x)
|x| 2
(w0 , w1) ∈ ˙ H 3/2 × ˙ H 1/2 .
We would like to understand what are the possible values of c0 and c1. The idea is to
look at the problem not as a regularity breakdown problem, but as an “instantaneous
smoothing” problem. More precisely, we consider the Cauchy problem
(20)
⎧
⎪⎨
⎪⎩
vtt − vrr − n−1
v(0) = c0
|x| + w0
vt(0) = c1
|x| 2 + w1(x)
r vr = v 3 Z(rv)
and we ask for which (c0, c1) ∈ R 2 , (w0, w1) ∈ ˙ H 3/2 × ˙ H 1/2 the solution v belongs to
L ∞ loc ((−ɛ, 0) ∪ (0, ɛ), ˙ H 3/2 ) for some ɛ > 0.
Two obvious possibilities are (c0, c1) = 0, and the Shatah-Bizoń finite energy self similar
solutions fn (see section 3.1), in which case (w0, w1) = 0.
The following theorem states that, for small data, these are the only possibilities.
Theorem 4.2 If (c0, c1) is small enough, and (w0 , w1) ∈ ˙ H 3/2 × ˙ H 1/2 , Theorem 2.1 gives
a local in time solution v of (20).
• If (c0, c1) = (0, 0), there exists ɛ > 0 such that v ∈ L ∞ ((−ɛ, ɛ), ˙ H 3/2 ).
• If (c0, c1) = (0, 0), v does not belong to L ∞ loc ((−ɛ, 0) ∪ (0, ɛ), ˙ H 3/2 ) for any ɛ > 0.
Remark 4.1 If we do not assume (c0, c1) to be small, we do not know if a solution exists,
and even if it did, there are good chances that it would not be unique, so it is not clear
how to generalize the above theorem to large data.
Proof: The first point of the theorem is contained in Theorem 4.1; so we focus on the
second one, and assume that (c0, c1) = (0, 0).
By finite speed of propagation, we can localize matters around 0, and hence assume that
the norm of (w0 , w1) is small in ˙ H 3/2 × ˙ H 1/2 . By point (ii) of Theorem 4.1, the solution
v reads
w + w ,
20

where w ∈ L ∞ (R, ˙ H 3/2 ), and w is the unique solution of (3) with data
c0
|x| ,
c1
|x| 2
. We
will prove that w is given by a profile, and that this profile does not belong to ˙ H 3/2 .
We follow the construction of Section 3 in order to build up a profile f for the data we
are interested in. First solving from infinity to 1, we have that f(1) has to be close to 0
since the data are small. Then in order to solve on (0, 1) we apply Lemma 3.1, where α is
taken small, since g(1) is small. The solution corresponding to this profile has a small X
norm for small enough data; by the uniqueness in Theorem 2.1, it is equal to w.
All we have to do now is proving that f (and hence w) does not belong to ˙ H 3/2 . If f(1)
is not zero, this is obvious, since f ′ behaves like | log |ρ − 1|| close to 1, see Section 3.4.
If f(1) = 0, we observe first that f is 0 on [0, 1]. In order to examine the situation on
[1, ∞), we will resort to the following lemma. Its proof is very similar to that of Lemma 3.1,
we therefore omit it.
Lemma 4.1 For any β ∈ R, the Cauchy problem
⎧
⎪⎨
⎪⎩
f ′′ + 2
ρf ′ − sin(2f)
ρ2 (1−ρ2 = 0 )
f(1) = 0
f ′ (1+) = β
admits a unique solution on (1, ∞).
Now we switch to g(z) = f
1
z , which verifies
g ′′ = (z 2 − 1) sin(2g(z)) .
Integrating two times, we see that g is bounded by
|g(z)| ≤ |z − 1|| log |z − 1|| .
But this implies that g ′′ is integrable, hence g ′ has a limit at 1−, hence f is given on (1, ∞)
by a solution of the above lemma, for some β. This β cannot be 0, otherwise f would be
identically 0. Hence
f ′ (1−) = f ′ (1+) ;
this is the desired singularity, f does not belong to ˙ H 3/2 , see Section 3.4.
5 Appendix : generalized Hardy inequality and Besov spaces
It is a consequence of Lemma 1.2. of [16] that if n ≥ 4 and f is radial, one has
|f(x)| ≤ C f ˙ H n 2 −1
|x|
The next lemma extends this result to the Besov space ˙B n
2 −1
2,∞ .
Lemma 5.1 Suppose n ≥ 5. If f is a radial function,
We even have
f n
˙B 2
−1
2,∞
|f(x)| ≤ C
|x|
f n
˙B 2
−1
2,∞
|Sjf(x)| ≤ C
|x|
21
.
.
for any j .

Proof : The first inequality can be proved by real interpolation. Indeed, we know (see
Lemma 1.2 of [16]) that, for a radial function v and α small enough,
rvL∞ (r1+αdr) =
1+α
r v∞ ≤ Cv
˙H d 2
−1−α .
The conclusion follows from the two following real interpolation facts
d
˙H 2 −1−α d
, H˙ 2 −1+α
1
2 ,∞ = ˙ B d
2 −1
2,∞
∞ 1−α ∞ 1+α
L (r dr) , L (r dr) 1
2 ,∞ = L∞ (r dr) .
To prove the second inequality of Lemma 5.1, we use the fact that, denoting the standard
maximal function M,
|Sjf(x)| ≤ CMf(x) ,
and that
M
1
≤
|x|
C
|x| .
Acknowledgements: The author is indebted to Jalal Shatah for very interesting and
useful discussions during the writing of this paper.
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23

Pierre GERMAIN
Courant Institute of Mathematical Sciences
New York University
New York, NY 10012-1185
USA
pgermain@math.nyu.edu
24