Functions whose Fourier transform vanishes on a surface

0.21 MB·

by  Dmitriy M. Stolyarov

Functions whose Fourier transform vanishes on a set
Dmitriy M. Stolyarov∗
January 26, 2016
6
1
0 Abstract
2
We study the subspaces of Lp(Rd) that consist of functions whose Fourier transforms vanish on
n
a smooth surface of codimension 1. We show that a subspace defined in such a manner coincides
a with the whole L space for p > 2d . We also prove density of smooth functions in such spaces
J p d−1
when p < 2d for specific cases of surfaces and give an equivalent definition in terms of differential
4 d−1
operators.
2

]
A 1 Definitions
C
h. Definition 1. Let Σ be a closed subset of Rd, let p∈[1,∞]. Define the space ΣLp by the rule
t
ma ΣLp =closLp f ∈S(Rd) ∀ξ ∈Σ fˆ(ξ)=0 .
(cid:16)(cid:8) (cid:12) (cid:9)(cid:17)
[ Here and in what follows, we denote the class of Sc(cid:12)hwartz functions by S and the Fourier transform

of f by fˆor by F[f]. There are many specific spaces that can be described as L for various choices
2 Σ p
v of Σ, for example, the Paley–Wiener spaces (Σ is the complement of a compact set, p = 2). In the
4 presentarticle,werestrictourattentiontothecasewhereΣisaC∞-smoothcompactsubmanifoldofRd.
0 Moreover,we assume that dimΣ=d−1. Our main example is the unit sphere Sd−1.
6
4 Definition 2. Let Σ be a closed subset of Rd. We call the operator R : S(Rd) → C(Σ) acting by the
Σ
0
rule
.
1 R [f]=fˆ|
Σ Σ
0
6 the restriction operator associated with Σ. We say that the statement R(Σ,p,q) holds true if RΣ admits
1 a continuous extension as an Lp(Rd) → Lq(Σ) operator. If R(Σ,p,1) holds true, we say that Σ admits
: the L restriction.
v p
i
X The definition needs two comments. First, to define L (Σ), we need to equip Σ with a measure. We
q
r alwaysequipitwiththenaturalLebesguemeasure,whichwedenotebyσ. Second,thenumberoneinthe
a
definition of the L restriction property is not random: R(Σ,p,q ) leads to R(Σ,p,q ) provided q >q .
p 1 2 1 2
We ask for the weakest natural restriction estimate.
It is an extremly difficult problem to decide whether R(Σ,p,q) holds true or not. The most well-
studied casewhere Σ is convexandits curvature does notvanishis still farfrombeing completely solved
(of course, we omit the trivial case where Σ is a hyperplane and there is no restrictions except p = 1).
We refer the readerto the books [4], [9] for the survey and the basics of the subject. We will not use the
1
results of the restriction theory , we will usually assume that Σ admits the restriction property.
∗Supported byRFBRgrant N14-01-00198.
1However, someideasofthistheory maybefoundinSection5.
1
One can also define the space of functions whose Fourier transform vanishes on Σ by the formula
L˜ = f ∈L R [f]=0 . (1)
Σ p p Σ
Surely, such a definition needs Σ to admit th(cid:8)e Lp res(cid:12)(cid:12)triction. (cid:9)
There are two motivations for investigating such type spaces. The first one comes from the fact that
theequationRSd−1[f]=0maybeconsideredasanaturalFredholmconditionfortheHelmholtzequation,
see[2]. Second,suchspacesariseifoneconsidersinequalitiesinthe styleofSobolevembeddingtheorems
with differential operators. Let us consider an example from [7]. The author asked for which collections
of parameters does the inequality
kfk .k(∂k−σ∂l)fk (2)
Wqα,β 1 2 Lp
hold true for allfunctions f ∈S(R2) uniformly. The Sobolev potential normonthe left is defined by the
formula
kfk = F−1 F[f](ξ,η)|ξ|α|η|β .
Wqα,β Lq
(cid:13) (cid:13)
The homogeneity considerations say that (cid:13) (cid:2) (cid:3)(cid:13)
(cid:13) (cid:13)
α β 1 1 1 1
+ =1− − + . (3)
k l p q k l
(cid:16) (cid:17)(cid:16) (cid:17)
Since f is smooth, the function (∂k−σ∂l)f lies in L , where
1 2 Σ p
Σ= (ξ,η)∈R2 (2πiξ)k =σ(2πiη)l .
n (cid:12) o
The problem is interesting only when the operato(cid:12)r (∂k−σ∂l) is not elliptic and k 6=l, thus Σ is a curve
(cid:12) 1 2
in the plane. It does not satisfy the assumptions we impose on our surfaces in the present article (it is
not compact and may not be a smooth submanifold).
However,afteranapplicationoftheLittlewood–Paleytheoryandthehomogeneityconsiderations,the
problemmaybelocalizedinspectrum,i.e. wemayassumethatfˆissupportedinasmallballcenteredat
somefixedpointonΣ(hereweomitthedetails). Allinall,toproveinequality (2),onehastoinvestigate
the action of the Fourier multiplier with the symbol v.p. 1 as an L → L operator. Note
(2πiξ)k−(2πiη)l Σ p q
thattheoperatorcanbeconsideredasaBochner–Riesztypeoperatoroforder−1. Thecontinuityofsuch
type operators on Lebesgue spaces is well studied in R2, see [1]. In our case, it says that inequality (2)
holds true if (3) is true, 1 − 1 > 2, p< 4, and q >4. The classical Knapp example construction shows
p q 3 3
that the condition 1 − 1 > 2 is necessary for (2) to hold. Substitution of functions f that converge to
p q 3
the surface measure on Σ leads us to the necessity of the condition p< 4. It was conjectured in [7] that
3
the condition q >4 may be omitted if one considers inequality (2).
Theorem 1 (Easy corollary of Theorem 5 in [2]). Suppose that 1<p,q <∞. Inequality (2) holds true
if and only if (3) holds true, 1 − 1 > 2, and p< 4.
p q 3 3
The paper [2] deals with the Bochner–Riesz operators of negative order, i.e. Σ = Sd−1. It is proved
there that the continuity properties of these operators as L → L operators are better than their
Σ p q
continuity properties as L →L operators(i.e. there aremany pairs (p,q) suchthat the Bochner–Riesz
p q
operator is not an L → L operator, but it is an L → L operator). Using the same reasoning,
p q Σ p q
one can prove the same results for general convex Σ with non-vanishing curvature, and then, using the
Littlewood–Paley theory, derive Theorem 1. We note that the phenomenon has been noticed long time
ago, see [3].
The paper [2] deals with L˜ spaces2 instead of L . It is natural to conjecture that L˜ = L
Σ p Σ p Σ p Σ p
provided Σ admits the L restriction and L = L provided it does not. In other words, the Schwartz
p Σ p p
functions are dense in L˜ .
Σ p
2Moreover, ΣL˜p isdefinedwiththehelpofR(Σ,p,2). Aswehavesaid,thedefinitionwithR(Σ,p,1)ismoregeneral.
2
Theorem 2. Let p∈(1,∞). Suppose that Sd−1 admits the L restriction. Then L˜ = L .
p Sd−1 p Sd−1 p
It is well known and easy to show that R(Σ,p,q) cannot be true if p > 2d (see, e.g. [9]). Thus, Σ
d+1
cannot admit the L restriction for p> 2d .
p d+1
Theorem 3. For any p∈( 2d ,∞), we have L =L .
d+1 Σ p p
Thetwotheoremsdescribethesituationforthe cased=2,Σ=S1 forallp∈(1,∞)exceptforp= 4.
3
Using some specific geometric properties of the circumference, we can deal with the endpoint.
Theorem 4. Let Σ be a convex closed curve in the plane, suppose its curvature does not vanish.
Then L =L .
Σ 34 34
The proofs of all three theorems are based on duality, so we start with the description of the annihi-
lators of our spaces in L (q is the adjoint exponent, 1 + 1 = 1) in Section 2. After that we prove the
q p q
main theorems in Sections 3, 4, and 5. Finally, we give another definition of L in Section 6 and prove
Σ p
its equivalence with the initial one. The last section contains severalconjectures.
I am grateful to Michael Goldberg, Sergey Kislyakov, and Andreas Seeger for discussions of the
material of this article.
2 Description of annihilator
Definition 3. Define the operator Π : S(Rd)→C∞(Σ) by the rule
Σ
Π [f]=f| .
Σ Σ
Lemma 1. The operator Π admits right inverse.
Σ
Proof. It suffices to extend any function ϕ∈C∞(Σ) to the whole space,and do this linearly. We denote
the operator we are going to construct by Ext, Ext: C∞(Σ)→S(Rd). Its main property is the identity
Π [Ext[ϕ]]=ϕ, ϕ∈C∞(Σ).
Σ
Let{ζ } beafinitepartitionofunityonΣthatissubordinatedtoanatlas. Itsufficestodefinetheaction
n n
ofExtonthefunctionsofthetypeϕζ sinceϕ= ϕζ . Fromnowon,fixn,letU bethecorresponding
n n n n
chart neighborhood; we may assume that ϕ is supported in U . Without loss of generality, we may also
n
assume that Σ∩Un = {(x,f(x))| x ∈ V}, whePre V ⊂ Rd−1 = {x ∈ Rd | xd = 0} is a neighborhood
of zero and f: V → Rd is a smooth function. Let ψ be a C∞(R) function such that ψ(0) = 1 and its
0
support is sufficiently small. Define the extension of ϕ by the formula
Ext[ϕ](x,x )=ϕ(x,f(x))ψ(x −f(x)), x=(x ,x ,...,x )∈V.
d d 1 2 d−1
First, this definition is smooth and linear in ϕ, second, Ext[ϕ] is supported in U provided the support
n
of ψ is sufficiently small (and thus Ext[ϕ]| =ϕ indeed).
Σ
We will use the adjont operator Π∗ : (C∞(Σ))′ →S′ more often.
Σ
Theorem 5. Let p∈[1,∞). The annihilator of L in L , 1 + 1 =1, can be described as
Σ p q p q
AnnLq(Rd) ΣLp = g ∈Lq(Rd) ∃ζ ∈(C∞(Σ))′ such that gˆ=Π∗Σ[ζ] .
Itiseasytoseethatth(cid:0)eFo(cid:1)urierntransformof(cid:12)(cid:12)afunction g intheannihilatorof L isosupportedonΣ.
(cid:12) Σ p
Theorem 5 says that the distribution gˆ is not only supported on Σ, but can be treated as a distribution
on Σ, e.g. does not contain differentiations with respect to directions transversalto Σ.
3
Proof. Itsufficestoshowthateveryg ∈Lq(Rd)thatannihilatesΣLp canberepresentedasF−1Π∗Σ[ζ],ζ ∈
(C∞)′(Σ). In other words, we need to construct ζ satisfying the identity
hgˆ,fi=hζ,Π [f]i, for all f ∈S(Rd). (4)
Σ
Consider the following diagram:
S(Rd) ΠΣ ✲ C∞(Σ)
❅ (cid:0) .
gˆ❅❅❘ (cid:0)✠(cid:0)?
C
Then, ζ corresponds to the arrowmarked with a question. We will construct ζ as a map that makes the
diagram commutative. First, since g annihilates L , we have the inclusion kerΠ ⊂kergˆ. Construct ζ
Σ p Σ
by the formula
ζ =gˆ◦Ext,
where Ext is the right inverse of Π provided by Lemma 1. Let us prove (4), which is rewritten as
Σ
hgˆ,fi=hgˆ,Ext Π [f] i.
Σ
(cid:2) (cid:3)
This is the same as to show that f −Ext Π [f] ∈kergˆ. Since kerΠ ⊂kergˆ, it suffices to show that
Σ Σ
(cid:2) (cid:3)
Π f −Ext Π [f] =0,
Σ Σ
h (cid:2) (cid:3)i
which is trivial since Π ◦Ext is the identity operator.
Σ
3 Density of smooth functions with restriction operator
This section contains the proof of Theorem 2. We start with a similar statement for the annihilator.
Lemma 2. Let p∈[1,∞). The set
g ∈Lq(Rd) ∃ζ ∈C∞(Sd−1) such that gˆ=Π∗Sd−1[ζ] (5)
n (cid:12) o
(cid:12)
is dense in AnnLq Sd−1Lp . (cid:12)
Proof. Consider th(cid:0)e group(cid:1) O of rotations of Sd−1. Let µ be the Haar measure on O . Let {ϕ }
d d n n
be a smooth approximation of identity in O . By this we mean that {ϕ } is a collection of C∞-
d n n
smooth functions on O , each function ϕ is supported in a 1-neighborhood of the identity element,
d n n
and ϕ dµ=1 for all n. It is easy to construct such a system of functions on the Euclidean space of
Od n
appropriatedimension;thenonecantransferthe constructedfunctionsto aneighborhoodofthe identity
R
element in O using a chart mapping (and multiplying each function by a suitable constant to fulfill the
d
condition ϕ dµ=1).
Od n
Our aim is to approximate a function g ∈ Ann L by functions belonging to the set (5). By
R Lq Sd−1 p
Theorem5, there existsa distributionζ ∈ C∞(Sd−1) ′ suchthatgˆ=Π∗ [ζ]. The idea is to define the
(cid:0) (cid:1) Sd−1
functions g and ζ by the formulas
n n
(cid:0) (cid:1)
ζ = ζ ϕ (T)dµ(T), g =F−1 Π∗ [ζ ] . (6)
n T n n Sd−1 n
Z
Od (cid:2) (cid:3)
4
where ζ is the rotationof ζ by T. Note thatthe integral(6) shouldbe understoodin a special way(the
T
integrandis distributional-valued). We define the averagingoverO withthe weightϕ inthe same way
d n
one defines the convolutionof distributions with some fixed test function: the operatorof averagingacts
on C∞(Sd−1) and is formally self-adjoint, thus, one can define the averaging (6) as a conjugate operator
to the averaging of test functions.
We note that the ζ are smooth functions (it is instructive to consider the case d = 2, where the
n
averaging (6) turns into the classical convolution on the circumference; the proof for the general case is
verbatim the reasoning for this classical case). Therefore, it remains to prove that gn →g in Lq(Rd).
Since rotations commute with the Fourier transform,
gn(x)= g(Tx)ϕn(T)dµ(T) for all x∈Rd.
Z
Od
We see that gn is an Lq-function. Using polar coordinates Rd ∋x=(r,σ)∈(0,∞)×Sd−1, we can write
∞
kg−g kq = |g(r,σ)−g (r,σ)|qdσ rd−1dr. (7)
n Lq n
Z0 (cid:18)SdZ−1 (cid:19)
(There is an ambiguity in this formula: we denote by g and g both the initial functions and the ones
n
they turn into after the change of variables; we also disregard the constant factor.) We see that g (r,·)
n
is the averagingof g(r,·) with respectto ϕ dµ for everyfixed r. By the properties of the approximation
n
of identity, g (r,·) → g(r,·) in L for every r. Thus, we have a pointwise convergence of the integrands
n q
in the outer integral in (7). Moreover,
|g(r,σ)−g (r,σ)|qdσ 62 |g(r,σ)|q,
n
SdZ−1 SdZ−1
so, the integrands also posses a summable majorant. Therefore, by the Lebesgue convergence theo-
rem, kg −gk →0.
n Lq
Proof of Theorem 1. Assume that Sd−1L˜p *Sd−1Lp. By the Hahn–Banach theorem, there exist
functions f and g such that
f ∈ L˜ , g ∈Ann L , hf,gi6=0.
Sd−1 p Lq Sd−1 p
Lemma 2 says that we may make g belong to the set(cid:0)(5). Let(cid:1){f } be a sequence of Schwartzfunctions
n n
approximating f in Lp(Rd). On the one hand,
|hfn,gi|=|hfˆn,Π∗Sd−1[ζ]i|=|hΠSd−1[fˆn],ζi|6kRSd−1[fn]kL1(Sd−1)kζkL∞ .kf −fnkLpkζkL∞ →0
since we have assumed that Sd−1 admits the Lp restriction and RSd−1[f] = 0. On the other hand,
hf ,gi→hf,gi6=0. A contradiction.
n
Remark 1. It was noticed by Michael Goldberg that one can actually omit the usage of Theorem 5 in the
reasoning above, i.e. can avoid passing to duals and directly construct the approximation of a function
in L˜ using the same averaging and properties of the extention operator built in Lemma 1.
Sd−1 p
5
4 Density of smooth functions without restriction operator
This section contains the proof of Theorem 3.
Definition 4. Let µ be a complex measure of locally bounded variation on Rd. The number
dimµd=ef inf α ∃F — Borel set, µ(F)6=0,dimF 6α
n (cid:12) o
is called the lower Hausdorff dimension(cid:12)of µ.
(cid:12)
We will use only the Hausdorff dimension for sets and measures, so we do not emphasize this. We
will also call the lower dimension of measure simply the dimension. The proof of Theorem 3 relies on
Lemma 2 in [8]. We will need a simplified version stated below. In this lemma, ϕ is a hat function, i.e.
a C∞(Rd)-function that is non-negative, radial, non-increasing (as a function of radius), and equals one
0
in a neighborhood of zero.
Lemma 3 (Simplified Lemma 2 in [8]). Let µ be a complex measure of finite variation on Rd. If for
all ξ ∈Rd and r ∈(0,1)
ϕ(r−1ξ+η)dµ(ξ) .rα, (8)
(cid:12)ZRd (cid:12)
then dimµ>α. (cid:12)(cid:12) (cid:12)(cid:12)
(cid:12) (cid:12)
TheclassicalFrostmanlemma3saysthatifµisnon-negativeandsatisfiestheinequalityµ(B (x)).rα
r
for all balls B (ξ) uniformly, then dimµ > α. Lemma 3 deals with a more general case of a signed (or
r
what is the same, complex) measure. We note without proof that (8) does not imply |µ|(B (ξ)).rα.
r
Proof of Theorem 3. By Theorem 5, it suffices to prove that there does not exist a non-zero dis-
tribution ζ ∈ (C∞(Σ))′ such that F−1 Π∗[ζ] ∈ L , q < 2d . Assume the contrary, let ζ be such a
Σ q d−1
distribution. We will come to a contradictionin two steps: firstwe will show that ζ ∈L (Σ), second, we
2
(cid:2) (cid:3)
will apply Lemma 3 to the measure ζdσ (this is a measure since ζ ∈ L (Σ)). But before that we note
2
that the problem is local. Indeed, if we multiply ζ by a smooth function supported in a neighborhood of
a pointonΣ,we do notharmthe condition F−1 Π∗[ζ] ∈L . Thus,wemay assumethatζ is supported
Σ q
in arbitrarily small neighborhood U of some point s∈Σ.
s
(cid:2) (cid:3)
We rewrite the condition F−1 Π∗[ζ] ∈ L in a more convenient form. In the formula below, we
Σ q
identify Rd−1 with some fixed linear hyperplane in Rd:
(cid:2) (cid:3)
q q
F−1 Π∗[ζ] (x) dx= F−1 Π∗[ζ] (Tx˜) |x˜|dx˜dµ(T),
Σ Σ
RZd (cid:12)(cid:12) (cid:2) (cid:3) (cid:12)(cid:12) OZdRdZ−1 (cid:12)(cid:12) (cid:2) (cid:3) (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
here µ stands for the Haar measure on O (we disregard a constant factor in this formula). Since the
d
value on the left is finite,
q
F−1 Π∗[ζ] (Tx˜) |x˜|dx˜<∞ (9)
Σ
RdZ−1 (cid:12)(cid:12) (cid:2) (cid:3) (cid:12)(cid:12)
for almostall T ∈Od. Therefore,there(cid:12)exists an element(cid:12)T ∈Od such that the hyperplane TRd−1 is not
collinear with the normals to Σ at the points in U and the corresponding integral(9) is finite (we recall
s
that U is small). We apply H¨older’s inequality:
s
2 q−2
2 q q q
F−1 Π∗Σ[ζ] (Tx˜) dx˜6 F−1 Π∗Σ[ζ] (Tx˜) |x˜|dx˜ |x˜|−q−22dx˜ . (10)
! !
|x˜Z|>1 (cid:12)(cid:12) (cid:2) (cid:3) (cid:12)(cid:12) |x˜Z|>1 (cid:12)(cid:12) (cid:2) (cid:3) (cid:12)(cid:12) |x˜Z|>1
(cid:12) (cid:12) (cid:12) (cid:12)
3TheeasierpartoftheFrostmanlemma.
6
We note that 2 >d−1 when q < 2d , so, the function F−1 Π∗[ζ] (Tx˜) belongs to L (Rd−1).
q−2 d−1 Σ 2
Since TRd−1 is not collinear with the normals to Σ∩Us,(cid:2)one ca(cid:3)n represent the surface Σ∩Us as
a graph of some function h: TRd−1 → R in the orthonormal coordinates of the hyperplane TRd−1
(here Vs ⊂TRd−1 is a neighborhood of zero):
Σ∩U ={(t,h(t))|t∈V }.
s s
The mapping t→(t,h(t)) is a diffeomorphism onto its image, therefore, there exists a distribution ζ ∈
T
(C∞(V ))′, whose image under this diffeomorphism is ζ. In the chosen coordinates, we have:
0 s
F−1 Π∗Σ[ζ] (t,t)=hζT,e−2πi ht,·i+th(·) i, t∈TRd−1, t∈R.
(cid:0) (cid:1)
In particular, F−1 Π∗[ζ](cid:2)(t,0) i(cid:3)s the (d−1)-dimensional inverse Fourier transform of ζ . By (10), the
Σ T
function t7→F−1[Π∗[ζ]](t,0) is square summable. Thus, by the Plancherel theorem, ζ ∈L . Applying
(cid:2)Σ (cid:3) T 2
the inverse diffeomorphism, we see that ζ is a square summable function with respect to the Lebesgue
measure on Σ.
WearegoingtoapplyLemma3tothe measureν =ζdσ toshowthatits dimensionisstrictlygreater
than d−1 (obviously, this will give us the desired contradiction). It is convenient to use the following
potential:
I(ν,α)d=ef F−1[ν](x) 2 1+|x| α−ddx,
RZd (cid:12) (cid:12) (cid:0) (cid:1)
(cid:12) (cid:12)
where α∈(0,d). The inequality
2 q−2
F−1[ν](x) 2 1+|x| α−ddx6 F−1[ν](x) qdx q 1+|x| q(qα−−2d) dx q
RZd (cid:12) (cid:12) (cid:0) (cid:1) (cid:18) RZd (cid:12) (cid:12) (cid:19) (cid:18) RZd (cid:0) (cid:1) (cid:19)
(cid:12) (cid:12) (cid:12) (cid:12)
shows that I(ν,α) is finite for some α>d−1 (since q >d if 26q < 2d ; the case q <2 is obvious).
q−2 d−1
It remains to show that the assumptions of Lemma 3 follow from the condition I(ν,α)<∞. We use the
notation ϕ (ξ)=ϕ(r−1ξ). First,
r
1
hν,ϕr(·+η)i 6 F−1[ν](x)F−1[ϕr](x) dx6I(ν,α)21 F−1[ϕr](x) 2(1+|x|2)d−2α dx 2.
(cid:12) (cid:12) RZd (cid:12) (cid:12) (cid:16)RZd (cid:12) (cid:12) (cid:17)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
Second,
2 d−α 2 d−α
F−1[ϕ ](x) 1+|x|2 2 dx= rdF−1[ϕ](rx) 1+|x|2 2 dx6
r
RZd (cid:12) (cid:12) (cid:16) (cid:17) RZd (cid:12) (cid:12) (cid:16) (cid:17)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) 2 1 (cid:12) |rx|2 d−α (cid:12) 2 d−α
rdF−1[ϕ](rx) + 2 r−dd(rx)=rα F−1[ϕ](z) 1+|z|2 2 dz,
r2 r2
RZd (cid:12) (cid:12) (cid:16) (cid:17) RZd (cid:12) (cid:12) (cid:16) (cid:17)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
provided r 61.
5 Case d = 2, p = 4
3
In this section, we prove Theorem 4.
7
Proof of Theorem 4. By Theorem 5, it suffices to prove that there does not exists a non-zero distri-
bution ζ ∈ (C∞)′(Σ) such that F−1 Π∗[ζ] ∈ L (R2). We may assume that ζ is supported in a small
Σ 4
ball U centered at a point on Σ (let it be the origin). It is convenient to parameterize Σ by the line:
(cid:2) (cid:3)
Σ∩U = (ξ,h(ξ))∈R2 ξ ∈(−ε,ε) .
Since rotations commute with the Fourier(cid:8)transform, we(cid:12)may assum(cid:9)e that h′(0) = 0. By our assump-
(cid:12)
tions, h is convex. We also assume that h′′ is almost constant on (−ε,ε).
The condition F−1 Π∗[ζ] ∈L (R2) can be rewritten as
Σ 4
(cid:2) (cid:3) Π∗[ζ]∗Π∗[ζ]∈L (R2). (11)
Σ Σ 2
Let Φ∈C∞ U , then the function Φ˜ ∈C∞(R2) is defined as
0 0
(cid:0) (cid:1) Φ˜(s,t)=Φ(s+t,h(s)+h(t)). (12)
In the light of this definition, the action of the convolution on a test function is written as
hΠ∗[ζ]∗Π∗[ζ],Φi=hζ,hζ,Φ˜ii. (13)
Σ Σ
Let us study the change of variables (ξ,η)= s+t,h(s)+h(t) in detail. The mapping (s,t)→(ξ,η) is
injective on the set s<t and C∞-smooth there. Its image is the set (Σ∩U)+(Σ∩U). So, it allows us
(cid:0) (cid:1)
to make the change of variables:
Φ˜(s,t)|h′(s)−h′(t)|dsdt=2 Φ(ξ,η)dξdη. (14)
ZZ ZZ
(−ε,ε)2 (Σ∩U)+(Σ∩U)
Let ϕ∈C∞ (−ε,ε) be a function. Then, by formula (13),
0
(cid:0) (cid:1)
|hζ,ϕi|2 = hζ,hζ,Φ˜ii = hΠ∗[ζ]∗Π∗[ζ],Φi 6 Π∗[ζ]∗Π∗[ζ] Φ ,
Σ Σ Σ Σ L2 L2
(cid:12) (cid:12) (cid:12) (cid:12) (cid:13) (cid:13)
whereΦ˜(s,t)=ϕ(s)ϕ(t)(a(cid:12)(cid:12)ndthefun(cid:12)(cid:12)ction(cid:12)(cid:12) Φisrecoveredby(cid:12)(cid:12)form(cid:13)(cid:13)ula(12)). Usi(cid:13)(cid:13)ngf(cid:13)(cid:13)orm(cid:13)(cid:13)ula(14),weobtain
1
kΦk2 = |ϕ(s)|2|ϕ(t)|2|h′(s)−h′(t)|dsdt.kϕk4 . (15)
L2 2 L2(R)
ZZ
(−ε,ε)2
So, |hζ,ϕi| .kϕkL2(R). Thus, ζ is a square summable function.
Since ζ is a summable function, the convolution can be computed directly using formulas (13)
and (14),
ζ s,h(s) ζ t,h(t)
Π∗[ζ]∗Π∗[ζ](ξ,η)= , s=s(ξ,η),t=t(ξ,η).
Σ Σ 2|h′(s)−h′(t)|
(cid:0) (cid:1) (cid:0) (cid:1)
We are going to prove that the function on the right-hand side does not belong to L (R2). We may
2
assume that the set {s||ζ(s,h(s))|>1} has positive measure. Let I be an interval such that
{s||ζ(s,h(s))|>1}∩I >0.9|I| (16)
(cid:12) (cid:12)
(such an interval exists around an(cid:12)y Lebesgue point of the(cid:12) set {s | |ζ(s,h(s))| > 1}). Consider the
(cid:12) (cid:12)
set SN ⊂ R2 of the points (ξ,η) for which s(ξ,η),t(ξ,η) ∈ I and N1 6 |s − t| 6 1N.1. This set lies
in a N−2-neighborhood of Σ, and by the condition that Σ has non-vanishing curvature, |S | ≍ N−2.
N
Moreover,
|h′(s)−h′(t)|≍N−1, (ξ,η)∈S .
N
8
It is not hard to see that
1
(ξ,η)∈S |ζ(s,h(s))|>1,|ζ(t,h(t))|>1 > |S |. (17)
N N
2
(cid:12) (cid:12)
(cid:12)n (cid:12) o(cid:12)
Indeed,(16)andthefac(cid:12)(cid:12)tthat|h′(s)−(cid:12)(cid:12)h′(t)|isalmostconstantonSN (w(cid:12)(cid:12) ehaveassumedthath′′ isalmost
constant) gives
1 1
(ξ,η)∈S |ζ(s,h(s))|61 6 |S |, (ξ,η)∈S |ζ(t,h(t))|61 6 |S |,
N N N N
5 5
(cid:12)n (cid:12) o(cid:12) (cid:12)n (cid:12) o(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
which lea(cid:12)ds to (17). (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
Therefore, by (17), the function (ξ,η) 7→ ζ(s,h(s))ζ(t,h(t)) is not less than N on the set of measure at
|h′(s)−h′(t)|
least 1|S |≍N−2. This contradicts (11), because the sets S are disjoint.
2 N N
6 One more definition of L
p
Σ
In this section we restrict our attention to a more specific class of surfaces. Let S be a polynomial in d
variables that grows at infinity. Let Σ be its zero set, i.e.
Σ={ξ ∈Rd |S(ξ)=0}. (18)
We assumethat∇SdoesnotvanishonΣ. Thus,Σsatisfiesthe conditionsweimposedonitbefore. The
aim of this section is to give an alternative definition of L in this special case.
Σ p
Theorem 6. If S is a polynomial in d variables that grows at infinity, Σ is given by (18), and ∇S does
not vanish on Σ, then
∂
ΣLp =closLp S 2πi [g] g ∈S(Rd) .
(cid:18)n (cid:16) (cid:17) (cid:12) o(cid:19)
(cid:12)
The notation S(∂) denotes the differential polynomia(cid:12)l, e.g. the case of Σ = Sd−1 corresponding
to S(ξ)= dξ2−1 results in the Helmholtz operator:
1 i
P ∂ ∆
S [g]=− +1 [g].
2πi 4π2
(cid:16) (cid:17) (cid:16) (cid:17)
We need a simple lemma about division of smooth functions.
Lemma 4. Let P be a polynomial in d variables. Suppose that ϕ ∈ C∞(Rd) vanishes on its zero
0
set {ξ |P(ξ)=0} and ∇P does not vanish on this set. In such a case, ϕ ∈C∞(Rd).
P 0
Proof. We may assume that ϕ is supported in a small ball whose center ξ lies on the surface {ξ ∈ Rd |
0
P(ξ)=0}. We may also assume that ∂ P does not vanish on suppϕ.
d
Letus firstconsiderthe easiestcased=1 andP(ξ)=ξ. Itis clearthat the function ϕ(ξ) is infinitely
ξ
differentiable outside zero. Let us provethat it is k times differentiable at zero. Consider the orderk+1
Taylor polynomial of ϕ at 0:
k+1ϕ(i)(0)ξi
ϕ(ξ)= +O(ξk+2).
i!
i=1
X
Surely,
ϕ(ξ) k ϕ(i+1)(0)ξi
= +O(ξk+1),
ξ (i+1)!
i=0
X
9
which shows that ϕ(ξ) is k times differentiable at zero.
ξ
Thegeneralcasemaybederivedfromthe trivialone-dimensionalcaseconsideredwiththehelpofthe
changeofvariablesξ →η,η =ξ forj =1,2,...,d−1andη =P(ξ). Duetothecondition∂ P 6=0,such
j j d d
a change of variables is a C∞-homeomorphism of suppϕ onto its image. Thus, this change of variables
takes ϕ to a function ϕ˜ ∈ C0∞(Rd) and P to a simple polynomial ηd. Applying the already considered
statement for the case d=1 to ϕ˜, we obtain that ϕ˜ ∈C∞(Rd). Consequently, ϕ ∈C∞(Rd).
ηd 0 P 0
Remark 2. In fact, we did not use that P is a polynomial. It could have been a C∞ function. Lemma 4
is a common knowledge, for example, it may be found in [6].
Proof of Theorem 6. By Definition 1, it suffices to prove that
∂
f ∈S(Rd) ∀ξ ∈Σ fˆ(ξ)=0 = S [g] g ∈S(Rd) .
2πi
(cid:8) (cid:12) (cid:9) n (cid:16) (cid:17) (cid:12)(cid:12) o
Clearly, the latter set belongs to(cid:12) the former. Thus, we need to pro(cid:12)ve that for every f ∈ S(Rd) such
that fˆvanishes on Σ, there exists g ∈ S(Rd) such that S ∂ [g] = f. In the Fourier transform world,
2πi
this statement looks like Lemma 4. The only difference is(cid:16)that(cid:17)the class C∞ is replaced by the Schwartz
0
class. To overcome this difficulty, one has to multiply the function fˆ=ϕ by a C∞ function that equals
0
one in a neighborhood of Σ, and use the fact that S grows at infinity.
The definition of L provided by Theorem 6 can be naturally generalized.
Σ p
Definition 5. Let k ∈N, define the space Lk by the formula
Σ p
∂
ΣLkp =closLp Sk 2πi [g] g ∈S(Rd) .
(cid:18)n (cid:16) (cid:17) (cid:12) o(cid:19)
(cid:12)
Conjecture 1. For every d, p, and S fixed, there exists l(cid:12)such that Ll = Ll+1 = Ll+2 = ..., i.e. the
Σ p Σ p Σ p
chain of these spaces stabilzes.
7 Conjectures
The study of L spaces is far from being complete. We list several conjectures the author is interested
Σ p
in.
First,itisnaturalto replacethe sphereinTheorem2bya moregeneralmanifold. Itsufficestoprove
the conjecture below.
Conjecture 2. Let p∈[1,∞) be such that Σ admits the L restriction property. The set
p
g ∈Lq(Rd) ∃ζ ∈C∞(Σ) such that gˆ=Π∗Σ[ζ]
n (cid:12) o
is dense in AnnLq(ΣLp). (cid:12)(cid:12)
We have proved the conjecture for the case Σ = Sd−1, our reasoning heavily relied on the rotational
symmetry of the sphere. A similar reasoning works for the case of a more general quadratic surface.
TheproofsofTheorems3and4arebasedonduality. Infact,weproveanuncertaintyprinciple(inthe
sense of the book [5]): a Fourier transform of a small function (in our case small means fastly decaying)
cannot be small (here small means generatedby a distribution “depending on less number of variables”).
Our proof of Theorem 3 uses two properties of Σ: the continuity of the normal and its dimension. The
author believes that the structure of the distribution ζ may be omitted.
10