HAL Id: hal-03200268
https://hal.archives-ouvertes.fr/hal-03200268
Preprint submitted on 16 Apr 2021
HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
On the continuity of Fourier multipliers on ˙
W
l,1(
R
d )and
˙
W
l,∞ (R
d )Eduard Curcă
To cite this version:
Eduard Curcă. On the continuity of Fourier multipliers on ˙Wl,1
On the continuity of Fourier multipliers on
˙
W
l,1
R
d
and ˙
W
l,∞
R
d
Eduard Curc˘
a
*Abstract
Suppose d ≥ 2. Kazaniecki and Wojciechowski proved in 2013 that every Fourier mul-tiplier on ˙W1,1
Rd is a bounded continuous function on Rd. This is a generalization of a previous result of Bonami and Poornima (1987) concerning homogeneous multipliers of degree zero. We further generalize the result of Kazaniecki and Wojciechowski. We prove that, given an integer l ≥ 1, every multiplier on ˙Wl,1
Rd or on ˙Wl,∞ Rd is a bounded continuous function on Rd. We obtain these results via a substantial simplification of the Riesz products technique used by Kazaniecki and Wojciechowski.
1
Introduction
In this paper, we study the continuity properties of functions which are Fourier multipliers on the homogeneous Sobolev spaces ˙Wl,1
Rd and ˙Wl,∞ Rd, where l ≥ 1 is an arbitrary integer. Given a nonnegative integer l and a parameter 1 ≤ p ≤ ∞, the space ˙Wl,p Rd consists of those distributions f on Rd for which ∇lf ∈ Lp
Rd. This space is endowed with the seminorm given by kf kW˙l,p( Rd) = ∇lf Lp( Rd)=α∈Nmaxd,|α|=lk∇ αf k Lp( Rd) .
Given a function m ∈ L1loc(Rd), we say that m is a Fourier multiplier on ˙Wl,p Rd if, for each Schwartz function f ∈ S(Rd) the distribution m bf is tempered and if, in addition, the following
estimate holds:
kTmf kW˙l,p ≤ Ckf kW˙l,p, ∀ f ∈ S(Rd), (1)
for some constant C < ∞, where Tm is defined by the relation
d
Tmf = m bf , ∀ f ∈ S(Rd).
The least constant C in the above inequality will be called the norm of m and will be denoted by kTmk (which is a quantity depending on p and l).
*Universit´e de Lyon, CNRS UMR 5208, Universit´e Lyon 1, Institut Camille Jordan, 43 blvd. du 11 novembre
1918, F-69622 Villeurbanne cedex, France.
Faculty of Computer Science, University “Alexandru Ioan Cuza”, General Berthelot, 16, Iasi 700483, Romania.
Email address: curca eduard@yahoo.ro
The Fourier transform that we work with is given by the following formula b f (ξ) := Z Rd e−ihx,ξif (x)dx, ∀ f ∈ S(Rd).
Some classical examples of multipliers on ˙Wl,p Rd in the case 1 < p < ∞ (for any l), are the functions mj(ξ) := iξj/ |ξ|, defined for ξ ∈ Rd\ {0} and any j = 1, 2, ..., d. In this case we have
Tmj = Rj, where R1, ..., Rd are the Riesz transforms on R
d. Let us observe that the functions m j
are homogeneous of degree zero, i.e., mj(λξ) = mj(ξ), for all ξ ∈ Rd\ {0} and any λ > 0. Also,
mj are not continuous at zero.
If p = 1, the situation is different. When l = 0, the mj’s fail to be multipliers on ˙W0,1 Rd =
L1 Rd, since the Rj’s are not bounded on L1 Rd. In fact, if m is a multiplier on L1 Rd, then
it is easy to see that m is the Fourier transform of a finite measure and hence m ∈ Cb Rd. The case of the multipliers on L∞ Rd is similar.
Suppose d ≥ 2. Then there exist Fourier multipliers on ˙W1,1
Rd which are not Fourier trans-forms of finite measures (see [7, Proposition 2.2]). In fact, the proof in [7] concerning ˙Wl,1 Rd applies to all the spaces ˙Wl,p Rd, with l ≥ 1 and 1 ≤ p ≤ ∞.
Let us illustrate this when d = 2 and l = 1, via a simple example from [7]. As a consequence of Ornstein’s L1 non-inequality (see [6]), there exists a distribution u, supported in the unit ball,
such that ∂2
1u, ∂22u are L1 functions on R2and ∂1∂2u is not a finite measure. We define m := \∂1∂2u.
Clearly, m is not the Fourier transform of a finite measure, however m is a multiplier on ˙W1,p(R2) for all 1 ≤ p ≤ ∞. Indeed,
Tmf = ∂1∂2u ∗ f ,
for any Schwartz function f . Hence,
∇Tmf = ∂12u ∗ ∂2f, ∂22u ∗ ∂1f
and thus, by Young’s inequality,
k∇Tmf kLp ≤ ∂12u L1k∂2f kLp + ∂22u L1k∂1f kLp = ∂12u L1 + ∂22u L1 k∇f kLp.
Using Ornstein’s L1non-inequality, Bonami and Poornima proved in 1987 that the only Fourier
multipliers on ˙Wl,1
Rd which are homogeneous functions of degree zero and continuous outside the origin are the constant functions. More precisely, they proved the following (see [1, Theorem 2]).
Theorem 1 Suppose d ≥ 2, l ≥ 1 are two integers and let Ω be a continuous function on Rd\ {0}, homogeneous of degree zero. If Ω is a Fourier multiplier on ˙Wl,1
Rd, then Ω is a constant.
When l = 1, this result was generalized by Kazaniecki and Wojciechowski in 2013 as follows (see [3, Theorem 1.1]).
Theorem 2 Suppose d ≥ 2. If m is a Fourier multiplier on ˙W1,1 Rd, then m ∈ Cb Rd.
Since any function homogeneous of degree zero that is continuous on Rd has to be constant,
a relatively difficult lemma of Wojciechowski (see [9, Lemma 1], [10, Lemma 1]) concerning the L1-norm of some trigonometric polynomials.
It seems that, since the above result of Kazaniecki and Wojciechowski, there was no progress in studying the general case of arbitrary multipliers when the regularity of the space is higher than 1.
We follow the ideas in [3] in order to prove a generalisation of Theorem 2 for the case of ˙
Wl,1 Rd, where l ≥ 1. The proof is also based on the Riesz products technique and the con-structions we use are very similar to those in [3]. However, rather than using Wojciechowski’s lemma we rely on much easier facts instead (see Lemmas 14 and 16 below). The other ingredient is Theorem 1. We do not use transference theorems and, apart from Bonami and Poornima’s result, the proof is quite elementary. Another advantage of our approach is that it also applies to multipliers on ˙Wl,∞ Rd. Note that the case of ˙W1,∞ Rd was not considered in [3].
Our results are the following.
Theorem 3 Suppose d ≥ 2 and l ≥ 1 are integers. If m is a Fourier multiplier on ˙Wl,∞ Rd, then m ∈ Cb Rd.
Theorem 4 Suppose d ≥ 2 and l ≥ 1 are integers. If m is a Fourier multiplier on ˙Wl,1 Rd, then m ∈ Cb Rd.
Remark 5 In fact, what we prove in these theorems is that m is a.e. equal to some bounded continuous function.
The proofs go as follows. First, following the idea in the proof of Lemma 3.1 in [3] we show, by simple arguments, that, whenever m is a multiplier on ˙Wl,1
Rd or on ˙Wl,∞ Rd, we need to have m ∈ Cb Rd\ {0}. (More specifically, as a preliminary step in our analysis, we define a
function which equals m a.e. on Rd and is continuous and bounded on Rd\ {0}.) Next, using this conclusion, we prove that, if m is a multiplier on ˙Wl,1
Rd or on ˙Wl,∞ Rd, then m has to be continuous at the origin. For this part of the proof, we use constructions based on Riesz products. While the proofs of these facts are similar, we start by studying the multipliers on ˙Wl,∞ Rd, since the proof is simpler in this setting. For this purpose, we adapt and simplify the ideas in [3] to the case of ˙Wl,∞
Rd, with the help of Lemma 14 below. Next, we study multipliers on ˙Wl,1
Rd. We show by a duality method that the boundedness of Tm implies the existence
of bounded solutions for some underdetermined differential equation. We conclude that such solutions do not exist if m is not continuous. Here, the technique is similar to the one in [2], which was in turn inspired by the one in [10].
2
Continuity outside the origin
Suppose that m is a multiplier of ˙Wl,p Rd for some 1 ≤ p ≤ ∞ and l ≥ 0. Let us notice that the norm of Tm is invariant by dilations and isometries. More precisely, if λ ∈ R\ {0}, respectively
R ∈ O(d) and we set mλ(ξ) := m (λξ), respectively mR(ξ) := m(Rξ), ∀ξ ∈ Rd, then
kTmλk = kTmk and kTmRk ∼l,d kTmk , (2)
Let us justify this when m ∈ L1(Rd); the general case is obtained by approximation. For any
d × d real invertible matrix A and any Schwartz function f on Rd, we have the following identity:
c
fA(ξ) = 1
| det A|f Ab
−1ξ ,
where fA(x) := f (Ax) for any x ∈ Rd. Via a change of variables, we find that
TmAf (x) =(2π)−d Z Rd eihx,ξim (Aξ) bf (ξ) dξ = (2π) −d | det A| Z Rd eihx,A−1ξim (ξ)f Ab −1ξ dξ =(2π)−d Z Rd ei D (A−1)tx,ξE m (ξ) cfA(ξ) dξ = T mfA A−1tx for a.e. x ∈ Rd.
When, A = λI, we get that Tmλf = Tmfλ(·/λ). We obtain
∇lT mλf Lp =λ −l ∇lTmfλ (·/λ) Lp = λ −lλd/p ∇lT mfλ Lp ≤ kTmk λ−lλd/p ∇lf λ Lp = kTmk ∇lf Lp.
Hence, kTmλk ≤ kTmk for any λ 6= 0. This gives the first equivalence in (2).
When, A = R, where R is orthogonal, we get TmRf = TmfR(R·). Since the absolute value of
each entry of R is bounded by 1, we obtain ∇lT mRf Lp .l,d ∇lTmfR (Rx) Lp = ∇lT mfR Lp ≤ kTmk ∇lfR Lp .l,d kTmk ∇lf Lp.
Hence, kTmRk .l,dkTmk for any orthogonal d × d matrix R. This gives the second equivalence
in (2).
Let us first observe that the multipliers of ˙Wl,1
Rd and the multipliers of ˙Wl,∞ Rd are bounded and continuous on Rd\ {0}.
Lemma 6 If m is a multiplier on ˙Wl,1 Rd, then m ∈ Cb Rd\ {0}.
Remark 7 We recall that (see Remark 5) the statement means that m has a bounded continuous representative on Rd\ {0}.
Proof. We recall that l ≥ 1. We essentially follow the argument in [3, Lemma 3.1]. We first prove that m has a continuous representative on Rd\ {0}. Indeed, let ρ be any Schwartz function such that ρ(ξ) 6= 0, ∀ξ (e.g. a standard Gaussian). Set αb j := (δj
kl)1≤k≤d, where δ j
k is the Kronecker
delta. Since ∂αj(T
mρ) ∈ L1, we find that mj := (iξ)α
j
d
Tmρ is a continuous bounded function, and
(iξj)lm(ξ)ρ(ξ) = mb j(ξ) for a.e. ξ ∈ R
d. (3) Set, for ξ 6= 0, e m(ξ) := mj(ξ) (iξj)lρ(ξ)b , if ξj 6= 0.
It is not clear if the above definition is correct, since the result may depend not only on ξ, but also on the choice of the coordinate ξj. However, (3) implies first that this definition is correct
for a.e. ξ, next, using the continuity of mj, the definition is correct for every ξ 6= 0 and that, in
addition m is continuous on Re d\ {0}. Clearly (from (3)), we have that m =m a.e. ande (iξj)lm(ξ)e ρ(ξ) = mb j(ξ) for every ξ ∈ R
Using the fact that each mj is bounded, we find from (4) thatm is bounded on the unit spheree Sd−1. More specifically, we have
max
|ξ|=1|m(ξ)| . kTe mk. (5)
Combining (5) and (2), we find that m is bounded on Re d\ {0}.
Lemma 8 If m is a multiplier on ˙Wl,∞
Rd, then m ∈ Cb Rd\ {0}.
Proof. As in the proof of Lemma 6, we first prove that m has a representative which is continuous on Rd\ {0}. Set m0 defined by m0(ξ) := m (−ξ). Clearly, m0 is also a multiplier on ˙Wl,∞
Rd, with the same norm as m.
It follows that ∂1lTm0ϕ L∞( Rd)≤ kTmk ∇lϕ L∞( Rd), (6) for any ϕ ∈ Cc∞(R2).
Consider now the normed subspace
V :=∇lϕ | ϕ ∈ Cc∞ Rd ⊂ C0 Rd
β ,
endowed with the norm induced by C0 Rd
β
, where β := #α ∈ Nd | |α| = l . Let ρ be as in
the proof of Lemma 6. We consider the linear functional Lρ: V → R defined by
Lρ ∇lϕ := ρ, ∂1lTm0ϕ , ∀ϕ ∈ C∞
c R d .
Thanks to (6), Lρ is well-defined and bounded on V and kLρk ≤ kTmk kρkL1(
Rd).
Using the Hahn-Banach theorem, we obtain a bounded extension eLρ of Lρ to C0 Rdβ. Moreover, we can choose eLρ∈
C0 Rd
β∗
= M Rdβ such that its norm equals kLρk. Let
(µα)|α|=l∈ M Rd
β
be an element representing eLρ. We have that
kµαkM(Rd) ≤ kTmk kρkL1(
Rd) , (7)
for any multiindex α, with |α| = l. Also, we have ∂l 1Tmρ, ϕ = (−1)lρ, ∂l1Tm0ϕ = (−1)lLρ ∇lϕ = (−1)lLeρ ∇lϕ = (−1)lX |α|=l hµα, ∇αϕi = X |α|=l h∇αµ α, ϕi , i.e., ∂1lTmρ = X |α|=l ∇αµ α, (8)
in the sense of tempered distributions on Rd. Taking the Fourier transform in (8), we obtain
(iξ1)lm(ξ)ρ(ξ) =b X
|α|=l
(iξ)αµcα(ξ) := m1(ξ) a.e. on Rd. (9)
Similar identities hold for the partial derivatives ∂l
jTmf , j = 2, . . . , d. Noting that each µcα is a continuous function (since µα is a finite measure), we continue as in the proof of Lemma 6 and
find some m ∈ C(Re d\ {0}) such that m =m a.e.e
Remark 9 In the case where 1 < p < ∞, it is not true that if m is a multiplier on ˙Wl,p
Rd, then m ∈ Cb Rd\ {0}. For example if m(ξ) := isgn (ξ1), then Tm is the Hilbert transform on the
first coordinate and hence m is a multiplier of any space ˙Wl,p
Rd, with 1 < p < ∞. However, m is singular on the whole hyperplane {ξ1 = 0}.
Also, if p = 2, any bounded measurable function is a multiplier. Hence, in this case, the multiplier may be even less regular.
It remains to study, in ˙Wl,1 and ˙Wl,∞, the continuity of the multipliers at the origin.
3
Almost radial limits
Following [3, Section 2], we will say that a function f : Rd\ {0} → C has almost radial limits1 at
the origin if the following condition is satisfied.
If (vn)n≥1, (wn)n≥1 ⊂ Rd\ {0} are two sequences converging to 0
such that (f (vn))n≥1, (f (wn))n≥1 are convergent sequences
and lim
n→∞f (vn) 6= limn→∞f (wn), then lim infn→∞
vn |vn| − wn |wn| > 0. (I)
Note that, if (I) does not hold for f = m, which is bounded, then there exists a sequence (vn)n≥1 ⊂ Rd\ {0}, converging to 0 and such that
vn
|vn| → ν ∈ S d−1
, m(v2n) → b1, m(v2n+1) → b2, with b1, b2 ∈ C, b1 6= b2.
By considering the possible limits (up to subsequences) of (m(−v2n)) and (m(−v2n+1)), we
obtain the following. If m : Rd \ {0} → C is a bounded function which does not have almost radial limits at the origin, then there exists a sequence (vn)n≥1 ⊂ Rd\ {0}, converging to 0, and
such that (at least) one of the two happens: vn |vn| → ν ∈ S d−1, m(v 2n) → b1, m(−v2n) → b1, m(v2n+1) → b2, m(−v2n+1) → b2, with b1, b2 ∈ C, b1 6= b2, (IIs) or vn |vn| → ν ∈ S d−1, m(v n) → b1, m(−vn) → b2, with b1, b2 ∈ C, b1 6= b2. (IIa)
We will refer to the first case as the symmetric case, and to the second as the asymmetric case. The plan of the proofs of Theorems 3 and 4 consists of establishing the desired results separately in cases (I), (IIs) and (IIa). In case (I), the proof relies on Theorem 1 or on its ˙Wl,∞ variant,
Theorem 10 below.
1Actually, the definition given in [3, Section 2] is slightly different. However, this aspect is unimportant in what
4
Proof of Theorems 3 and 4 in case (I)
4.1
The case of ˙
W
l,1R
dFirst, as in [3], we observe that a bounded function having almost radial limits at 0 also has (genuine) radial limits at 0, and therefore we may define the function
Ω (ξ) := lim
n→∞m (ξ/n) . (10)
Clearly, the function Ω is homogeneous of degree 0. Now, one can easily see that Ω is a multiplier on ˙Wl,1
Rd. Indeed, let f be a Schwartz function and let ψ be an arbitrary Schwartz function with kψkL∞ ≤ 1. Thanks to (2) we have, for any multiindex α with |α| = l and any
n ≥ 1, D ∇αT m1/nf, ψ E ≤ kTmk kf kW˙l,1.
On the other hand, the dominated convergence theorem gives, with c = cα,d := ı|α|(2π)−d,
D ∇αT m1/nf, ψ E = c Z Rd ξαm (ξ/n) bf (ξ) bψ (ξ) dξ → c Z Rd ξαΩ (ξ) bf (ξ) bψ (ξ) dξ = h∇αTΩf, ψi , and hence, |h∇αTΩf, ψi| ≤ kTmk kf kW˙l,1.
By Lemma 6, we have that Ω ∈ C Rd\ {0}. We are now in position to apply Theorem 1 and obtain that Ω is constant. From this and condition (I), we deduce that m is continuous at the
origin.
4.2
The case of ˙
W
l,∞R
dAs above, we conclude that the function Ω defined by (10) is a multiplier on ˙Wl,∞
Rd. In particular, by Lemma 8, we have that Ω ∈ C Rd\ {0}. In order to complete (as above) the proof
in this case, it suffices to establish the following analogue of Theorem 1.
Theorem 10 Let d ≥ 2 and l ≥ 1 be some integers and let Ω ∈ C Rd\ {0} ; C be homogeneous of degree zero. If Ω is a multiplier on ˙Wl,∞ Rd, then Ω is a constant.
Proof of Theorem 10. We adapt the arguments from [5]. As in the proof of Lemma 8, for any Schwartz function ρ with integral 1, one can find some finite measures µα such that
∂1lTΩ0ρ = X |α|=l ∇αµ α, (11) where Ω0(ξ) := Ω (−ξ). Now, if ϕ ∈ S Rd and ϕ
ε(x) := ϕ(εx) for some ε > 0, then
∂1lTΩϕε (x) = εl ∂1lTΩϕ (εx), (12)
Combining (11) and (12), we find that εl Z Rd ρ(x) ∂1lTΩϕ (εx) dx = εl X |α|=l Z Rd (∇αϕ) (εx) dµα(x). (13) Since Ω is bounded, ∂l
1TΩϕ is the inverse Fourier transform of an L1 function and hence, ∂l1TΩϕ
is continuous and bounded. Dividing both sides in (13) by εl and taking ε → 0, we get by the dominated convergence theorem,
∂1lTΩϕ (0) =
X
|α|=l
µα(Rd) (∇αϕ) (0),
for any ϕ ∈ S Rd. This implies that
∂1lTΩϕ (x) =
X
|α|=l
µα(Rd) (∇αϕ) (x),
for any ϕ ∈ S Rd and any x ∈ Rd. Hence, by taking the Fourier transform, we get
ξl1Ω (ξ)ϕ (ξ) =b X |α|=l µα(Rd) ξαϕ (ξ) ,b and we have ξ1lΩ (ξ) = X |α|=l µα(Rd) ξα =: p1(ξ) . We can write Ω (ξ) = p1(ξ) ξl 1 , (14)
as an equality of two continuous functions in the domain where ξ1 6= 0. Similarly, there exists a
homogeneous polynomial pd of degree l such that
Ω (ξ) = pd(ξ) ξl
d
, (15)
as an equality of two continuous functions in the domain where ξd 6= 0. From (14) and (15), we
get
ξdl p1(ξ) = ξ1lpd(ξ) everywhere in Rd. (16)
By identifying the coefficients in (16), we see that p1 must be a multiple of ξ1l, thus a constant
multiple of ξ1l (since p1 is of degree l). Going back to (14), we find that Ω is constant in the region
{ξ1 6= 0}. Similarly, Ω is constant in the region {ξd6= 0}, and thus constant (by arguing similarly
in other directions).
From now on, we investigate cases (IIs) and (IIa), which are more involved.
5
Proof of Theorem 3 in case II
Lemma 11 Let m be a multiplier on ˙Wl,∞
Rd for some integer l ≥ 0. Consider the set of functions Pm := ( n X j=1 cjeih·,qji n ∈ N∗, q1, ..., qn ∈ Rd\ {0} and c1, ..., cn∈ C ) . Let Tm0 : Pm → Pm be defined by Tm0 n X j=1 cjeih·,qji ! := n X j=1 cjm(qj)eih·,qji,
for any n ∈ N∗, q1, ..., qn ∈ Rd\ {0} and c1, ..., cn∈ C.
We have that
kTm0 f kW˙l,∞(
Rd) ≤ kTmk kf kW˙l,∞(
Rd) , (17)
for any function f ∈ Pm.
Remark 12 Note that, since the exponentials eih·,qji are linearly independent and P
m is formed
only with (finite) linear combinations of these exponentials, the definition of Tm0 is correct.
Proof. We note that at this point we know that m is continuous and bounded on Rd\ {0}. This
will be used in the proof below.
Consider a function η ∈ Cc∞ Rd whose integral is 1. Fix q ∈ Rd\ {0}. For any small ε > 0,
we set ϕεq(t) := eiht,qiVη
(εt) on Rd. Since ϕεq is a Schwartz function, we have
Tmϕεq V (ξ) = m(ξ)ϕε q V (ξ) ,
in the sense of tempered distributions on Rd. A direct computation gives
c ϕε q(ξ) = Z Rd e−iht,ξ−qibη(εt)dt = 1 εd Z Rd e−iht,ξ−qε i b η(t)dt =1 εdbηb ξ − q ε = (2π) d εd η q − ξ ε . Hence, [ Tmϕεq(ξ) = m(ξ) (2π)d εd η q − ξ ε , (18)
Note that, since m ∈ L∞, the right hand side of (18) is L1; we obtain using the Fourier inversion formula, Tm ϕεq (t) = Z Rd eiht,ξim(ξ)1 εdη q − ξ ε dξ = Z Rd eiht,q−εξim(q − εξ)η(ξ)dξ, (19)
in the sense of tempered distributions. We naturally identify Tm ϕεq with the right hand side of
(19).
Now we can prove (17). Let f ∈ Pm, with
f (t) =
n
X
j=1
Using (19) we have ∇l n X j=1 cj Z Rd
m(qj− εξ)eiht,qj−εξiη(ξ)dξ
L∞ t = ∇lTm n X j=1 cjϕεqj ! (t) L∞ t ≤ kTmk ∇l(f (t) b η(εt)) L∞t .
In other words, for every multiindex α ∈ Nd with |α| = l,
n X j=1 cj Z Rd
(qj− εξ)αm(qj − εξ)eiht,qj−εξiη(ξ)dξ
L∞ t ≤ kTmk ∇lf (t) b η(εt) L∞t + ε kTmk Cf,η,
where Cf,η is a finite constant only depending on f and η. Letting ε → 0 we find that:
∇α n X j=1 cjm(qj)eiht,qji L∞t = n X j=1 cjqαjm(qj)eiht,qji L∞t ≤ kTmk ∇lf L∞.
Here, we use the fact thatη(0) = 1 and the obvious fact thatb ∇lf (t) b η(0) L∞t ≤ lim infε→0 ∇lf (t) b η(εt) L∞t .
The proof of Lemma 11 is complete.
Remark 13 For simplicity, from now on, we will denote both operators Tm and Tm0 by Tm. We
keep this convention even in the case where p = 1. As we will see this will turn out to be convenient in some computations.
5.1
The symmetric case, (IIs)
In what follows we suppose for simplicity that d = 2. Also, we suppose without loss of generality that b1 = 1, b2 = 0 and ν = (1, 0). This is possible thanks to the rotation and dilation invariance
we have discussed.
We will need the following simple lemma (see the Appendix).
Lemma 14 Fix N ∈ N∗. There exists a finite sequence (σk)1≤k≤N in {0, 1} such that
N X k=1 σk k k−1 Y j=1 1 + i j ≥ 1 πln N . (20)
Suppose N ∈ N∗ is fixed and σ1, ..., σN ∈ {0, 1} are some fixed numbers such that inequality
(20) holds. We construct, by backward induction on k, a sequence (ak)1≤k≤N in R2 satisfying the
following properties:
(P1) for each k ∈ {1, ..., N } we have m εkak+ X 1≤j≤k−1 εjaj ! − σk < 1 4N,
(P2) for each k ∈ {1, ..., N − 1} we have
4 |ak(1)| < |ak+1(1)| and 4 |ak(2)| < |ak+1(2)| ;
(Here, ak(1) and ak(2) are the two coordinates of ak. )
(P3) for each k ∈ {1, ..., N } we have
0 < ak(1) + X 1≤j≤k−1 εjaj(1) , ak(2) + X 1≤j≤k−1 εjaj(2) < 1, for all ε1, ..., εk−1 ∈ {−1, 0, 1};
(P4) for each k ∈ {1, ..., N } we have ak(2) + P 1≤j≤k−1εjaj(2) ak(1) + P 1≤j≤k−1εjaj(1) < 1 4N, for all ε1, ..., εk−1 ∈ {−1, 0, 1}.
(A similar construction appears in [3, Subsection 2.2] and in [10].) We proceed by backward induction; we start by constructing aN, then aN −1, ending eventually with a1. At the step N − k
we construct the vector ak taking into account the values aN, aN −1,....,ak+1. More precisely, the
construction goes as follows. We first modify the sequence (vn)n≥1 in (IIs) such that vn(2) 6= 0 for
all n ≥ 1. This is possible, since m is continuous on R2\ {0}. At each step we choose ak to be a
term in the set {vn | n ≡ 0 ( mod 2)} or {vn | n ≡ 1 ( mod 2)} if σk = 1 or σk = 0 respectively
(recall that b1 = 1 and b2 = 0). It remains to see that at each step the term ak can be chosen
sufficiently small in order to satisfy the above conditions. Since vn→ 0, we can choose a vector ak
with both components nonzero and such that |m(εkak) − σk| < (1/2) 4−N, for any εk ∈ {−1, 1}.
Since m is continuous outside the origin, there exists rk > 0 such that |m(ξ) − σk| < 4−N for any
ξ ∈ B(ak, rk) ∪ B(−ak, rk). Hence, if ak−1, ..., a1 are sufficiently small, then (P1) is satisfied. We
have that ν = (1, 0), and hence if ak = vn, for n sufficiently large, and ak−1, ..., a1 are sufficiently
small, then (P4) is satisfied. It is easy to see that the remaining conditions can be satisfied too.
Consider the set
ΛN := ( N X k=1 εkak ε1, ..., εN ∈ {−1, 0, 1} , not all 0 ) . (21)
Thanks to (P2), for each q ∈ ΛN the representation
q =
N
X
k=1
εkak, for some ε1, ..., εN ∈ {−1, 0, 1} ,
is unique. Let us also observe that, for each q ∈ ΛN we have (from (P3) and (P4)),
Define the function RN(t) := −1 + N Y k=1 1 + i kcos ht, aki , t ∈ R2. (24)
By (57) (see the Appendix below),
RN(t) = N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 Y εj6=0 i 2j eiht,ε1a1+...εkak i = X q∈ΛN cqeiht,qi, (25)
for some coefficients cq with |cq| ≤ 1. Thanks to (22) we have that q(1) 6= 0, for any q ∈ ΛN. This
allows us to define the function
hN(t) := X q∈ΛN cq (q(1))le iht,qi , on R2. (26) We claim that ∇lhN L∞(R2)≤ 4. (27) Indeed, we have ∂1lhN L∞(R2)= kRNkL∞(R2)≤ 1 + N Y k=1 1 + 1 k2 1/2 ≤1 + N Y k=1 e1/2k2 ≤ 1 + eπ2/12 ≤ 4. (28)
On the other hand, if l1,l2 are nonnegative integers with l1+ l2 = l and l1 < l, we have (using
(22), (23)), ∂l1 1 ∂ l2 2 hN L∞(R2)= X q∈ΛN (q(2))l2 (q(1))l−l1cqe iht,qi L∞(R2) ≤ X q∈ΛN |q(2)|l2 |q(1)|l−l1 = X q∈ΛN |q(2)| |q(1)| l2 ≤ X q∈ΛN 4−N l2 ≤ |Λ N| 4−N ≤ 3N4−N ≤ 1.
We are now going to estimate kTmhNkW˙l,∞(R2). Since by (26), hN ∈ Pm, with Pm as in Lemma
11, we may define TmhN via Lemma 11 (see Remark 13). More specifically, we will prove that
kTmhNkW˙l,∞(R2) ≥
1
π ln N − 1. (29)
In order to see this, it suffices to prove that
Using (25), we obtain TmRN(t) = N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 m(ε1a1+ ...εkak) Y εj6=0 i 2j eiht,ε1a1+...εkak i . (31)
Introducing the function
Z(t) := N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 σk Y εj6=0 i 2j eiht,ε1a1+...εkak i , (32)
we have (by the identities (57) and (59) in the Appendix),
Z(t) = N X k=1 iσk k cos ht, aki k−1 Y j=1 1 + i j cos ht, aji . Lemma 14 yields kZkL∞(R2)≥ |Z(0)| ≥ 1 πln N . (33)
Also, using property (P1), together with (31) and (32), we get
kTmRN − ZkL∞(R2)≤ |ΛN| 4−N ≤ 3N4−N ≤ 1. (34)
Using (33), (34) and the triangle inequality, we arrive at
kTmRNkL∞(R2)≥ kZkL∞(R2)− kTmRN − ZkL∞(R2) ≥
1
πln N − 1, concluding the proof of (30).
By taking N → ∞, (27) and (29) give us that m is not a multiplier on ˙Wl,∞(R2).
Remark 15 To deal with the case d > 2 we may suppose that ν = (1, 0, ..., 0); we consider constructions like RN ⊗ 1, where RN is defined as above on R2 and the constant function 1 is
defined on Rd−2.
5.2
The asymmetric case, (IIa)
This case is very similar to the previous one. We again suppose without loss of generality that b1 = 1, b2 = 0 and ν = (1, 0). In a similar way we construct a sequence (ak)1≤k≤N satisfying the
above properties (P2)- -(P4) and (P1’) below:
(P1’) for each k ∈ {1, ..., N } we have m εkak+ X 1≤j≤k−1 εjaj ! − 1 + εk 2 σk < 1 4N,
With this new sequence (ak)1≤k≤N we define ΛN as in (21). We again have (22), (23). We also
define RN and hN as in (24) and (26) respectively. The inequality (27) holds in this case too and
it remains to show that
∂1lTmhN L∞(R2) = kTmRNkL∞(R2)≥ 1 πln N − 1. (35) Using (57), we have TmRN(t) = N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 m(ε1a1+ ...εkak) Y εj6=0 i 2j eiht,ε1a1+...εkak i . (36)
Introducing the function
Z(t) := N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 1 + εk 2 σk Y εj6=0 i 2j eiht,ε1a1+...εkaki, (37)
we observe that, by (57) and (59), we have
Z(t) = N X k=1 iσk k e iht,aki k−1 Y j=1 1 + i j cos ht, aji .
Lemma 14 gives us that
kZkL∞(R2)≥ |Z(0)| ≥
1
πln N . (38)
Property (P1’), together with (36) and (37), give
kTmRN − ZkL∞(R2)≤ |ΛN| 4−N ≤ 3N4−N ≤ 1. (39)
Using (38), (39) and the triangle inequality, we get
kTmRNkL∞(R2)≥ kZkL∞(R2)− kTmRN − ZkL∞(R2) ≥
1
πln N − 1,
concluding the proof of (35). (For the case d > 2, see Remark 15.)
6
Proof of Theorem 4 in case II
We prove now Theorem 4. Suppose m is not continuous in 0. As in the preceding section, we may assume that m is in one of the cases (IIs) or (IIa). Again we work under the hypothesis d = 2, b1 = 1, b2 = 0 and ν = (1, 0).
Comparing with the strategy we had in the preceding section, there are two main differences. First, since it turns out it is more convenient to work with L∞ than L1, we reformulate the problem by duality. Secondly, we work with genuine trigonometric polynomials on T2 rather than
exponential sums on R2. This allows us to use some techniques specific to T2 that are not available
6.1
The symmetric case, (IIs)
We need the following analogue of Lemma 14 (see the Appendix):
Lemma 16 Fix N ∈ N∗. There exists a finite sequence (σk)1≤k≤N in {0, 1} such that
N X k=1 σk 2k k−1 Y j=1 1 + i 2j ≥ 1 2πln N .
In what follows (ak)1≤k≤N is a sequence in Q
2 satisfying the properties (P1)- -(P4) for the
sequence (σk)1≤k≤N from Lemma 16 above. It is easy to see that such a sequence exists. Using
this sequence we construct the function RN as in (24).
Suppose that m is a multiplier on ˙Wl,1(R2). Then m0 defined by m0(ξ) := m (−ξ) is also a
multiplier on ˙Wl,1(R2) with the same norm as m. It follows that
∂1lTm0ϕ L1(R2)≤ kTmk ∇lϕ L1(R2), (40) for any ϕ ∈ Cc∞(R2).
Consider now the normed subspace
V :=∇lϕ | ϕ ∈ Cc∞ R2 ⊂ L1 R22
l
,
endowed with the norm induced by (L1(R2))2l. We consider the linear functional L
N : V → R defined by LN ∇lϕ := RN, ∂1lTm0ϕ = Z R2 RN(t)∂1lTm0ϕ(t)dt, ϕ ∈ C∞ c R 2 .
Note that LN is well-defined on V . Thanks to (40), LN is bounded on V . Using the
Hahn-Banach theorem, we get that there exists a bounded extension eLN of LN, on (L1(R2)) 2l
. Moreover, we can choose eLN ∈
(L1(R2))2l∗ = (L∞
(R2))2l such that its norm equals kL
Nk. Note that, by
(28),
kLNk ≤ kTmk kRNkL∞(R2) ≤ 4 kTmk .
Let (uα)|α|=l ∈ (L∞(R2)) 2l
be the element representing eLN, where α ∈ N2 are multiindexes.
We have that
kuαkL∞(R2) ≤ 4 kTmk , (41)
for any multiindex α, with |α| = l. Also, we have (see Remark 13)
∂l 1TmRN, ϕ = (−1) lR N, ∂1lTm0ϕ = (−1)lLN ∇lϕ = (−1)lLeN ∇lϕ = (−1)l X |α|=l huα, ∇αϕi = X |α|=l h∇αu α, ϕi , i.e., ∂1lTmRN = X |α|=l ∇αu α, (42)
As in (31) we have TmRN(t) = N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 m(ε1a1+ ...εkak) Y εj6=0 i 2j eiht,ε1a1+...εkak i . (43)
For each N we fix a positive integer M = M (N ) such that M ak ∈ Z2 for all 1 ≤ k ≤ N .
From (43) we get that TmRN is a component-wise 2πM -periodic function. Hence, TmRN(M t) is
component-wise 2π-periodic.
We will show that uα in (42) can be chosen to be component-wise 2πM -periodic. In order to
prove this we need the following easy lemma.
Lemma 17 Let A > 0 be a real number and suppose u ∈ L∞(R2) is given. We consider the
sequence of functions un(t) := 1 |Bn| X χ∈Bn u (t + Aχ) , t ∈ R2, n ≥ 1,
where Bn := B(0, n) ∩ Z2. Then, there exists g ∈ L∞(R2), component-wise A-periodic, with
kgkL∞(R2)≤ kukL∞(R2) and such that un → g up to a subsequence, in the sense of distributions.
Proof of Lemma 17. Since kunkL∞(R2) ≤ kukL∞(R2) for any n ≥ 1, by the sequential
Banach-Alaoglu theorem, there exists g ∈ L∞(R2) with kgkL∞(R2) ≤ kukL∞(R2) such that un * g in
the w∗-topology of L∞ up to a subsequence. (For simplicity we still denote the subsequence by (un)n≥1.) In particular, un → g in the sense of distributions. Also, we easily get that g is
component-wise A-periodic. Indeed, for ϕ ∈ Cc∞(R2), and any χ0 ∈ Z2, we have
Since |(Bn+ χ0) \Bn|, |Bn\ (Bn+ χ0)| . n and |Bn| ∼ n2, we have rn→ 0. Hence, lim n→∞ Z R2 un(t + Aχ0) ϕ (t) dt = lim n→∞ Z R2 un(t) ϕ (t) dt,
which together with (44) concludes the proof of Lemma 17.
Now, since ∂1lTmRN is component-wise 2πM -periodic, we have ∂1lTmRN = ∂1lTmRN
n (with
A = 2πM ) for any n ≥ 1. From (42) we get
∂1lTmRN = ∂1lTmRN n= X |α|=l ∇α(u α)n,
for any n ≥ 1. Taking n → ∞ and applying Lemma 17 to the functions uα ∈ L∞, with A := 2πM ,
we get ∂1lTmRN = X |α|=l ∇αg α, (45)
for some component-wise 2πM -periodic functions gα ∈ L∞(R2) such that
kgαkL∞(R2) ≤ 4 kTmk (46)
(see (41)).
From now on, for each function ψ on R2, we write ψM for the function ψM(t) := ψ (M t). Consider the function
GN(t) := −1 + N
Y
k=1
(1 + cos ht, aki) , t ∈ R2.
Notice that GMN is component-wise 2π-periodic. (We recall here that each M ak belongs to Z2.)
Also, (TmRN)M and each gMα are component-wise 2π-periodic. From (45) we get
∂1l (TmRN)M =
X
|α|=l
∇αgM α ,
in the sense of distributions on R2and hence in the sense of distributions on T2. Taking convolution
(on the torus T2) with GM
N, we get ∂1l (TmRN)M ∗ GMN = X |α|=l ∇α gMα ∗ GMN . (47)
From the properties of GN it follows that the spectrum of each gαM ∗ GMN and the spectrum of
(TmRN) M
∗ GM
N, as functions on the torus T2, are included in M ΛN and therefore do not touch
the set {0} × Z (see (22)). Hence, we can apply the operator ∂1−l in (47) to obtain
We claim that ∇α∂−l 1 gαM ∗ GMN L∞(T2)≤ 8 kTmk , (49)
for any multiindex α with |α| = l. This estimate is similar to (27). Indeed, for α = (l, 0) we have, using (46),
∂1l∂1−l gαM ∗ GM N L∞(T2)= gαM ∗ GM N L∞(T2)≤ gMα L∞(T2) GMN L1(T2) ≤2 kgαkL∞(R2) ≤ 8 kTmk . (50)
Here, we have used the fact that GM N
L1(T2)≤ 2. This can be justified as follows. We have
N
Y
k=1
(1 + cos ht, M aki) ≥ 0
and hence, thanks to (57) and (P2)2, we obtain N Y k=1 (1 + cos h·, M aki) L1(T2) = 1.
We now turn to the proof of (49) for α 6= (l, 0). Writing gMα ∗ GM N(t) = X q∈ΛN c0qeiht,M qi,
we get that (note that c0q is a Fourier coefficient): c0q ≤ gαM ∗ GM N L∞(T2)≤ 8 kTmk , for all q ∈ ΛN.
Hence, if α = (l1, l2), with l1+ l2 = l and l1 < l, we have (using (22), (23)),
∂l11∂l22 ∂1−l gαM ∗ GMN L∞(T2) = X q∈ΛN (M q(2))l2 (M q(1))l−l1c 0 qe iht,M qi L∞(T2) ≤8 kTmk X q∈ΛN |q(2)|l2 |q(1)|l−l1 = 8 kTmk X q∈ΛN |q(2)| |q(1)| l2 ≤8 kTmk X q∈ΛN 4−N l2 ≤ 8 kT mk |ΛN| 4−N ≤ 8 kTmk 3N4−N ≤8 kTmk . (51)
We see that (50) and (51) imply (49).
We next obtain a contradiction. The starting point is the left-hand side of (48). We claim that (TmRN) M ∗ GM N L∞(T2) ≥ 1 2πln N − 1. (52)
2Note that (P2) implies the fact that the sequence (M a
The method applied to obtain this estimate is similar to the one used to obtain (30). Indeed, by using (57) and (58) (see the Appendix) we have:
(TmRN)M ∗ GMN (t) = N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 m(ε1a1+ ...εkak) Y εj6=0 i 4j eiht,ε1M a1+...+εkM aki. (53)
Introducing the function
Z(t) := N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 σk Y εj6=0 i 4j e iht,ε1M a1+...+εkM aki on T2, (54)
we observe that, by (57) and (59), we have
Z(t) = N X k=1 iσk 2k cos ht, M aki k−1 Y j=1 1 + i 2j cos ht, M aji .
Lemma 16 gives us that
kZkL∞(T2) ≥ |Z(0)| ≥
1
2πln N . (55)
Also, using property (P1), together with (53) and (54), we get (TmRN) M ∗ GM N − Z L∞(T2)≤ |ΛN| 4 −N ≤ 3N4−N ≤ 1. (56)
Using (55), (56) and the triangle inequality, we obtain
(TmRN) M ∗ GMN L∞(T2)≥ kZkL∞(T2)− (TmRN) M ∗ GMN − Z L∞(T2)≥ 1 2πln N − 1, concluding the proof of (52).
Now, (48), (49) and (52) allow us to write 1
2πln N − 1 ≤ 8 kTmk 2
l.
Since N is arbitrary, the last inequality implies that m is not a multiplier on ˙Wl,1(R2).
6.2
The asymmetric case, (IIa)
This case is very similar to the previous one and we skip the proof. We can again suppose by contradiction that m is a multiplier on ˙Wl,1(R2) and use this result to obtain a representation
result similar to the one in (42). The only difference is that now we have to follow the ”asymmetric case” as in the proof corresponding to multipliers on ˙Wl,∞(R2). The functions R
N and GN will
be constructed as above, starting, as in the case of ˙Wl,∞(R2), with a sequence (a
k)1≤k≤N in Q2
7
Appendix
7.1
Some useful identities
We quickly recall here some elementary facts and formulas concerning some trigonometric poly-nomials on the torus.
Fix a finite sequence (ak)1≤k≤N in Zd. For each finite sequence α1, ..., αN of complex numbers
we have the following expansion rule:
N Y k=1 (1 + αkcos ht, aki) = 1 + N X k=1 X ε1,...,εk∈{−1,0,1} εk6=0 Y εj6=0 αj 2 eiht,ε1a1+...+εkak i . (57) A sequence (ak)k=1,N in Z
d will be called dissociated if the only solution to the equation
ε1a1+ ... + εNaN = ε01a1+ ... + ε0NaN,
with ε1, ..., εN, ε01, ..., ε0N ∈ {−1, 0, 1} is the trivial solution ε1 = ε01, ..., εN = ε0N. For example any
sequence (ak)1≤k≤N in Zd which is lacunary on at least one component is dissociated. If (ak)1≤k≤N
is dissociated and α1, ..., αN and β1, ..., βN are complex numbers, by using (57) and the relation
between convolution and the Fourier transform, we obtain that
N Y k=1 (1 + αkcos h·, aki) ∗ N Y k=1 (1 + βkcos h·, aki) = N Y k=1 1 + αkβk 2 cos h·, aki , (58)
as functions on the d-dimensional torus.
The following identity is also useful. We have
N Y k=1 (1 + ck) = 1 + N X k=1 ck k−1 Y j=1 (1 + cj) (59)
for any complex numbers c1,..., cN.
7.2
A selection lemma
The following interesting fact is taken from [8] (Lemma 6.3, p. 118).
Lemma 18 Suppose z1,..., zN are some complex numbers. Then, there exist σ1,..., σN ∈ {0, 1}
such that N X k=1 σkzk ≥ 1 π N X k=1 |zk| .
The proof is elementary and we skip it.
Let us define two sequences (z0k)1≤k≤N and (zk1)1≤k≤N by the expressions
zkβ := 1 2βk k−1 Y j=1 1 + i 2βj for k = 1, ..., N ,
where β = 0, 1 is an index. Here, the product over an empty set is by convention equal to 1. It is easy to see that, using Lemma 18 applied to the sequence (z0
k)1≤k≤N we get Lemma 14.
7.3
Remarks on Wojciechowski’s inequality
We discuss here some inequalities in the spirit of Lemma 14 and Lemma 16. Wojciechowski was the first one to use such inequalities in the proof of non-estimates. In particular, he obtained in [9] the following relatively difficult estimate (see [9, Lemma 1], [10, Lemma 1]):
Lemma 19 There exists a constant C > 0 such that, for any integer N ≥ 2 there exist M = M (N ) and a sequence σ1,..., σN ∈ {0, 1} such that
N X k=1 σkcos h·, aki k−1 Y j=1 (1 + cos h·, aji) L1( Td) ≥ CN , (60)
whenever the sequence (ak)1≤k≤N in Zd satisfies
|ak+1| > M |ak| , for 1 ≤ k ≤ N − 1.
This lemma was already used in conjunction with the Riesz products technique in [10], [3], [4]. Lemma 19 was used in [10] to prove that there exists g ∈ L1(T2) such that there are no f0, f1, f2 ∈ W1,1(T2) with
g = f0+ ∂1f1+ ∂2f2.
It was also used in [4] in order to prove some anisotropic Ornstein-type non-inequalities and in [3] to study the continuity of the multipliers on ˙W1,1
Rd.
Here we want to point out that, in the above-mentioned applications, a weaker form suffices: we only need to know that the lower bound in (60) goes to ∞ when N → ∞. (In the case of the application of Lemma 19 given in [10], this was observed by Wojciechowski [10, Remark 1].) This weaker version can be achieved by much cheaper arguments than the ones used to obtain Lemma 19. In this direction we mention the following.
Lemma 20 For any integer N ≥ 2 there exists a sequence σ1,..., σN ∈ {0, 1} such that
N X k=1 σkcos h·, aki k−1 Y j=1 (1 + cos h·, aji) L1( Td) ≥ 1 2π r N e ,
for any dissociated sequence (ak)1≤k≤N in Z d.
Proof. The proof follows the ideas in [2]. By applying Lemma 18 to the sequence (zk)1≤k≤N,
where zk:= 1 2√N 1 + i 2√N k−1 for k = 1, ..., N , (61)
we obtain a sequence (σk)1≤k≤N in {0, 1} such that
N X k=1 σk 2√N 1 + i 2√N k−1 ≥ 1 2π N X k=1 1 √ N = √ N 2π . (62)
Suppose (ak)1≤k≤N is a dissociated sequence in Zd. Consider the functions
defined on Td. Note that, by (58), we have gN ∗ GN(t) = N Y k=1 1 + i 2√N cos ht, aki . (63)
Also, we consider the set
A := N [ k=1 σk=1 {ε1a1+ ... + εkak| ε1, ..., εk∈ {−1, 0, 1} , εk 6= 0}
and the projection PA defined by dPAf (n) = bf (n) if n ∈ A and dPAf (n) = 0 otherwise, for any
trigonometric polynomial f on Td. Observe that (59) and (63) give
PA(gN ∗ GN) (t) = N X k=1 σk 2√N cos ht, aki k−1 Y j=1 1 + i 2√N cos ht, aji , and thanks to (62), |PA(gN ∗ GN) (0)| ≥ √ N 2π . Since, kgNkL∞( Td) = 1 + 1 N N/2 ≤√e, we obtain √ e kPAGNkL1( Td) ≥ kgNkL∞(Td) kPAGNkL1(Td) ≥ |hgN, PAGNi| = |PA(gN ∗ GN) (0)| ≥ √ N 2π . It remains to observe that,
PAGN(t) = N X k=1 σkcos ht, aki k−1 Y j=1 (1 + cos ht, aji) ,
which concludes the proof.
Remark 21 In fact, it is possible to prove Lemma 20 (up to a multiplicative constant) without using Lemma 18. Indeed, the sequence (zk)1≤k≤N defined in (61) has a quite simple form: the
ar-gument of zk is (k − 1) θN (mod 2π), where θN := arctan
1/2√N. One can choose the sequence (σk)1≤k≤N explicitly: σk = 1, if −π/4 ≤ (k − 1) θN (mod 2π) ≤ π/4, and σk = 0, otherwise.
Acknowledgements
References
[1] Bonami, A., Poornima, S., Nonmultipliers of the Sobolev spaces Wk,1
Rd. J. Funct. Anal. 71(1), 175-181, 1987.
[2] Curc˘a, E., The divergence equation with L∞ source. To appear in Annales de la Facult´e des Sciences de Toulouse.
[3] Kazaniecki, K., and Wojciechowski, M. On the continuity of Fourier multipliers on the homo-geneous Sobolev spaces ˙W1,1
Rd. Annales de l’institut Fourier. 66. 10.5802/aif.3036, 2013. [4] Kazaniecki, K., Wojciechowski, M., Ornstein’s non-inequalities: Riesz product approach. arXiv
preprint arXiv:1406.7319, 2014.
[5] de Leeuw, K., Mirkil, H., A priori estimates for differential operators in L∞ norm. Illinois J. Math. 8, 112–124, 1964.
[6] Ornstein, D., A non-inequality for differential operators in the L1 norm. Arch. Rational Mech. Anal., 11:40–49, 1962.
[7] Poornima, S., On the Sobolev spaces Wk,1 Rd. Harmonic analysis (Cortona, 1982), Lecture Notes in Math., vol. 992, Springer, Berlin, p. 161-173, 1983.
[8] Rudin, W., Real and Complex Analysis, 3d ed., McGraw-Hill Book Company, New York, 1987. [9] Wojciechowski, M., On the strong type multiplier norms of rational functions in several
vari-ables, Illinois J. Math. 42, no. 4, p. 582-600, 1998.
