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Volume 2, Issue 3, Article 37, 2001

Lp−IMPROVING PROPERTIES FOR MEASURES ON R4 SUPPORTED ON HOMOGENEOUS SURFACES IN SOME NON ELLIPTIC CASES

E. FERREYRA, T. GODOY, AND M. URCIUOLO

FACULTAD DEMATEMATICA, ASTRONOMIA YFISICA-CIEM, UNIVERSIDADNACIONAL DECORDOBA, CIUDADUNIVERSITARIA, 5000 CORDOBA, ARGENTINA

eferrey@mate.uncor.edu godoy@mate.uncor.edu urciuolo@mate.uncor.edu

Received 8 January, 2001; accepted 5 June, 2001 Communicated by L. Pick

ABSTRACT. In this paper we study convolution operators Tµ with measures µin R4 of the formµ(E) = R

BχE(x, ϕ(x))dx,whereB is the unit ball ofR2, andϕis a homogeneous polynomial function. Ifinfh∈S1

det d2xϕ(h, .)

vanishes only on a finite union of lines, we prove, under suitable hypothesis, thatTµ is bounded fromLp intoLq if

1 p,1q

belongs to a certain explicitly described trapezoidal region.

Key words and phrases: Singular measures,Lp−improving, Convolution Ooperators.

2000 Mathematics Subject Classification. 42B20, 42B10.

1. INTRODUCTION

It is well known that a complex measure µ on Rn acts as a convolution operator on the Lebesgue spacesLp(Rn) : µ∗Lp ⊂ Lpfor1 ≤p ≤ ∞.If for somepthere exists q > psuch thatµ∗Lp ⊂ Lq, µis calledLp−improving. It is known that singular measures supported on smooth submanifolds ofRnmay beLp−improving. See, for example, [2], [5], [8], [9], [7] and [4].

Letϕ1, ϕ2 be two homogeneous polynomial functions onR2 of degreem ≥ 2and let ϕ = (ϕ1, ϕ2).Letµbe the Borel measure onR4 given by

(1.1) µ(E) =

Z

B

χE(x, ϕ(x))dx,

ISSN (electronic): 1443-5756

c 2001 Victoria University. All rights reserved.

Partially supported by Agencia Cordoba Ciencia, Secyt-UNC and Conicet.

052-00

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whereB denotes the closed unit ball around the origin inR2 anddxis the Lebesgue measure onR2.LetTµbe the convolution operator given byTµf =µ∗f, f ∈S(R4)and letEµbe the type set corresponding to the measureµdefined by

Eµ= 1

p,1 q

:kTµkp,q <∞,1≤p, q ≤ ∞

,

where kTµkp,q denotes the operator norm of Tµ from Lp(R4) into Lq(R4) and where theLp spaces are taken with respect to the Lebesgue measure onR4.

Forx, h∈R2,letϕ00(x)hbe the2×2matrix whosej−thcolumn isϕ00j (x)h,whereϕ00j (x) denotes the Hessian matrix ofϕj atx.Following [3, p. 152], we say thatx ∈ R2 is an elliptic point forϕifdet (ϕ00(x)h)6= 0for allh ∈R2\ {0}.ForA ⊂R2,we will say thatϕis strongly elliptic onAifdet (ϕ001(x)h, ϕ002(y)h)6= 0for allx, y ∈Aandh∈R2\ {0}.

If every pointx∈B\ {0}is elliptic forϕ,it is proved in [4] that form≥3, Eµis the closed trapezoidal regionΣmwith vertices(0,0),(1,1), m+1m ,m−1m+1

and m+12 ,m+11 .

Our aim in this paper is to study the case where the set of non elliptic points consists of a finite union of lines through the origin,L1, ..., Lk. We assume from now on, that forx ∈ R2 − {0}, det (ϕ00(x)h)does not vanish identically, as a function ofh.For eachl= 1,2, ..., k,letπLland πL

l be the orthogonal projections fromR2 ontoLlandLl respectively. Forδ >0,1≤l ≤k, let

Vδl=n

x∈B : 1/2≤ |πLl(x)| ≤1and πL

l (x)

≤δ|πLl(x)|o .

It is easy to see (see Lemma 2.1 and Remark 3.6) that forδsmall enough, there existsαl ∈ N and positive constantscandc0 such that

c πL

l (x)

αl

≤ inf

h∈S1|det (ϕ00(x)h)| ≤c0 πL

l (x)

αl

for all x ∈ Vδl. Following the approach developed in [3], we prove, in Theorem 3.7, that if α= max1≤l≤kαl and if7α≤m+ 1,then the interior ofEµagrees with the interior ofΣm.

Moreover in Theorem 3.8 we obtain that

Eµ =

Σm still holds in some cases where7α >

m+ 1,if we require a suitable hypothesis on the behavior, near the linesL1, ..., Lk,of the map (x, y)→infh∈S1|det (ϕ001(x)h, ϕ002(y)h)|.

In any case, even though we can not give a complete description of the interior ofEµ, we obtain a polygonal region contained in it.

Throughout the papercwill denote a positive constant not necessarily the same at each oc- currence.

2. PRELIMINARIES

Letϕ1, ϕ2 : R2 → Rbe two homogeneous polynomials functions of degree m ≥ 2and let ϕ= (ϕ1, ϕ2). Forδ >0let

(2.1) Vδ=

(x1, x2)∈B : 1

2 ≤ |x1| ≤1and |x2| ≤δ|x1|

.

We assume in this section that, for someδ0 >0,the set of the non elliptic points forϕ inVδ0 is contained in thex1 axis.

Forx ∈ R2,let P = P (x)be the symmetric matrix that realizes the quadratic form h → det (ϕ00(x)h),so

(2.2) det (ϕ00(x)h) =hP (x)h, hi.

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Lemma 2.1. There existδ ∈ (0, δ0), α ∈ Nand a real analytic function g = g(x1, x2)onVδ withg(x1,0)6= 0forx1 6= 0such that

(2.3) inf

|h|=1|det (ϕ00(x)h)|=|x2|α|g(x)|

for allx∈Vδ.

Proof. SinceP(x) is real analytic onVδ andP (x) 6= 0forx 6= 0,it follows that, forδsmall enough, there exists two real analytic functionsλ1(x) andλ2(x) wich are the eigenvalues of P (x). Also, inf|h|=1|det (ϕ00(x)h)| = min{|λ1(x)|,|λ2(x)|} for x ∈ Vδ. Since we have assumed that (1,0)is not an elliptic point forϕ and that P(x) 6= 0for x 6= 0, diminishingδ if necessary, we can assume thatλ1(1,0) = 0and that|λ1(1, x2)| ≤ |λ2(1, x2)|for|x2| ≤ δ.

SinceP (x)is homogeneous in x,we have that λ1(x)and λ2(x)are homogeneous in xwith the same homogeneity degree d. Thus |λ1(x)| ≤ |λ2(x)| for allx ∈ Vδ.Now, λ1(1, x2) = xα2G(x2)for some real analytical function G = G(x2) withG(0) 6= 0and soλ1(x1, x2) = xd1λ1

1,xx2

1

=xd−α1 xα2G

x2

x1

.Takingg(x1, x2) =xd−α1 G

x2

x1

the lemma follows.

Following [3], forU ⊂R2 letJU :R2 →R∪ {∞}given by JU(h) = inf

x, x+h∈U|det (ϕ0(x+h)−ϕ0(x))|,

where the infimum of the empty set is understood to be∞.We also set, as there, for0< α <1 RUα (f) (x) =

Z

JU(x−y)−1+αf(y)dy.

Forr >0andw∈R2,letBr(w)denotes the open ball centered atwwith radiusr.

We have the following

Lemma 2.2. Let wbe an elliptic point for ϕ. Then there exist positive constants candc0 de- pending only on1kC3(B)and2kC3(B)such that if0< r≤cinf|h|=1|det (ϕ00(w)h)|then

(1) |det (ϕ0(x+h)−ϕ0(x))| ≥ 12|det (ϕ00(w)h)|ifx, x+h∈Br(w). (2)

RB1r(w)

2

(f) 6

≤c0r12 kfk3

2 , f ∈S(R4).

Proof. LetF (h) = det (ϕ0(x+h)−ϕ0(x))and letdjxF denotes thej−th differential ofF at x.Applying the Taylor formula toF (h)aroundh= 0and taking into account thatF (0) = 0, d0F (h) = 0and thatd20F(h, h)≡2 det (ϕ00(x)h)we obtain

det (ϕ0(x+h)−ϕ0(x)) = det (ϕ00(x)h) + Z 1

0

(1−t)2

2 d3thF (h, h, h)dt.

LetH(x) = det (ϕ00(x)h).The above equation gives det (ϕ0(x+h)−ϕ0(x)) = det (ϕ00(w)h) +

Z 1 0

dw+t(x−w)H(h)dt

+ Z 1

0

(1−t)2

2 d3thF(h, h, h)dt.

Then, forx, x+h∈Br(w)we have

|det (ϕ0(x+h)−ϕ0(x))−det (ϕ00(w)h)| ≤M|h|3 ≤2M r|h|2

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withM depending onlykϕ1kC3(B)andkϕ2kC3(B).If we choose c≤ 4M1 ,we get, for0< r <

cinf|h|=1|det (ϕ00(w)h)|that

|det (ϕ0(x+h)−ϕ0(x))| ≥ 1

2|det (ϕ00(w)h)|

and that

JBr(w)(h)≥ 1

2|det (ϕ00(w)h)| ≥ 1 2cr|h|2 Thus

RB1r(w)

2

(f) 6

≤ c0r12 kI2(f)k6 ≤ c0r12 kfk3

2 , where Iα denotes the Riesz potential onR4,defined as in [10, p. 117]. So the lemma follows from the Hardy–Littlewood–Sobolev theorem of fractional integration as stated e.g. in [10, p. 119].

Lemma 2.3. Letwbe an elliptic point forϕ.Then there exists a positive constantcdepending only on1kC3(B) and2kC3(B) such that if 0 < r ≤ cinf|h|=1|det (ϕ00(w)h)| then for all h6= 0the mapx→ϕ(x+h)−ϕ(x)is injective on the domain{x∈B :x, x+h∈Br(w)}. Proof. Suppose thatx, y, x+handy+hbelong toBr(w)and that

ϕ(x+h)−ϕ(x) = ϕ(y+h)−ϕ(y). From this equation we get

0 = Z 1

0

0(x+th)−ϕ0(y+th))hdt= Z 1

0

Z 1 0

d2x+th+s(y−x)ϕ(y−x, h)dsdt.

Now, forz ∈Br(w),

d2zϕ−d2wϕ

(y−x, h) =

Z 1 0

d3z+u(w−z)ϕ(w−z, y−x, h)du

≤ M r|y−x| |h|

then

0 = Z 1

0

Z 1 0

d2x+th+s(y−x)ϕ(y−x, h)dsdt

= d2wϕ(y−x, h) + Z 1

0

Z 1 0

d2x+th+s(y−x)ϕ−d2wϕ

(y−x, h)dsdt.

So|d2wϕ(y−x, h)| ≤M r|y−x| |h|withM depending only onkϕ1kC3(B)andkϕ2kC3(B). On the other hand,wis an elliptic point forϕand so, for|u|= 1,the matrixA:= ϕ00(w)u is invertible. AlsoA−1 = (detA)−1Ad(A),then

A−1x

=|detA|−1|Ad(A)x| ≤ Mf

|detA||x|,

whereMfdepends only onkϕ1kC2(B) andkϕ2kC2(B). Then, for |v| = 1andx = Av,we have

|Av| ≥ |detA|/fM .Thus

d2wϕ(y−x, h)

≥ |y−x| |h| inf

|u|=1,|v|=1

d2wϕ(u, v)

= |y−x| |h| inf

|u|=1,|v|=1|hϕ00(w)u, vi|

≥ 1

Mf|y−x| |h| inf

|u|=1|detϕ00(w)u|.

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If we chooser < 1

MMfinf|u|=1|detϕ00(w)u|the above inequality impliesx=yand the lemma

is proved.

For any measurable set A ⊂ B, let µA be the Borel measure defined by µA(E) =R

AχE(x, ϕ(x))dxand letTµA be the convolution operator given byTµAf =µA∗f.

Proposition 2.4. Letw be an elliptic point for ϕ. Then there exist positive constantscand c0 depending only on1kC3(B)and2kC3(B)such that if0< r < cinf|h|=1|detϕ00(w)h|then

TµBr(w)f

3 ≤c0r13 kfk3

2 .

Proof. Taking account of Lemma 2.3, we can proceed as in Theorem 0 in [3] to obtain, as there, that

µBr(w) ∗f

3 3

≤(A1A2A3)13 , where

Aj = Z

R2

Fj(x) Y

1≤m≤3,m6=j

RBr(w)1 2

Fm(x)dx

andFj(x) = kf(x, .)k3 2

Then the proposition follows from Lemma 2.2 and an application of the triple Hölder in-

equality.

For0< a <1andj ∈N let Ua,j =

(x1, x2)∈B :|x1| ≥a, 2−j|x1| ≤ |x2| ≤2−j+1|x1| and letUa,j,i, i= 1,2,3,4the connected components ofUa,j.

We have

Lemma 2.5. Let0 < a < 1. Suppose that there existβ ∈ N, j0 ∈ Nand a positive constant c such that |det (ϕ001(x)h, ϕ002(y)h)| ≥ c2−jβ|h|2 for all h ∈ R2, x, y ∈ Ua,j,i, j ≥ j0 and i= 1,2,3,4.Thus

(1) For allj ≥j0, i= 1,2,3,4ifxandx+hbelong toUa,j,ithen

|det (ϕ0(x+h)−ϕ0(x))| ≥c2−jβ|h|2.

(2) There exists a positive constantc0 such that for allj ≥j0, i= 1,2,3,4

RU1a,j,i

2

(f) 6

≤c022 kfk3 2 .

Proof. We fixiandj ≥j0.Forx∈Ua,j,iwe have det (ϕ0(x+h)−ϕ0(x)) = det

Z 1 0

ϕ00(x+sh)hds

.

For eachh ∈ R2\ {0}we have eitherdet (ϕ001(x)h, ϕ002(y)h) > c2−jβ|h|2 for allx, y ∈ Ua,j,i or det (ϕ001(x)h, ϕ002(y)h) < −c2−jβ|h|2 for allx, y ∈ Ua,j,i.We consider the first case. Let

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F (t) = det Rt

0 ϕ00(x+sh)hds .Then F0(t) = det

Z t

0

ϕ001(x+sh)hds, ϕ002(x+th)h

+ det

ϕ001(x+th)h, Z t

0

ϕ002(x+sh)hds

= Z t

0

det (ϕ001(x+sh)h, ϕ002(x+th)h)ds

+ Z t

0

det (ϕ001(x+th)h, ϕ002(x+sh)h)ds≥c2−jβ|h|2t.

SinceF (0) = 0we getF (1) =R1

0 F0(t)dt ≥c2−jβ|h|2.Thus

det (ϕ0(x+h)−ϕ0(x)) =F (1)≥c2−jβ|h|2.

Then JUa,j,i,(h) ≥ c2−jβ|h|2, and the lemma follows, as in Lemma 2.2, from the Hardy–

Littlewood–Sobolev theorem of fractional integration. The other case is similar.

For fixedx(1), x(2) ∈R2,let Ba,j,ix(1),x(2) =

x∈R2 :x−x(1) ∈Ua,j,i andx−x(2) ∈Ua,j,i , i= 1,2,3,4.

We have

Lemma 2.6. Let0< a <1and letx(1), x(2) ∈R2.Suppose that there existβ ∈N, j0 ∈Nand a positive constantcsuch that|det (ϕ001(x)h, ϕ002(y)h)| ≥c2−jβ|h|2 for allh∈R2, x, y ∈Ua,j,i, j ≥ j0 and i = 1,2,3,4.Then there exists j1 ∈ N independent of x(1), x(2) such that for all j ≥j1, i= 1,2,3,4and all nonnegativef ∈S(R4)it holds that

Z

Ba,j,ix(1),x(2)×R2

f y−ϕ x−x(1)

, y−ϕ x−x(2)

dxdy ≤ m2

JUa,j,i(x(2)−x(1)) Z

R4

f.

Proof. We assert that, ifj ≥j0then for each(z, w)∈R2×R2andi= 1,2,3,4,the set n

(x, y)∈Ba,j,ix(1),x(2) ×R2 :z =y−ϕ x−x(1)

andw=y−ϕ x−x(2)o is a finite set with at mostm2elements. Indeed, ifz =y−ϕ x−x(1)

andw=y−ϕ x−x(2) withx∈Ba,j,ix(1),x(2),Lemma 2.5 says that, forj large enough,

det ϕ0 x−x(1)

−ϕ0 x−x(2)

≥c2−jβ|h|2.

Thus the Bezout’s Theorem (See e.g.[1, Lemma 11.5.1, p. 281]) implies that for each(z, w)∈ R2×R2 the set

n

x∈Ba,j,ix(1),x(2) :ϕ x−x(2)

−ϕ x−x(1)

=z−wo is a finite set with at mostm2 points. Sincexdeterminesy,the assertion follows.

For a fixed η > 0 and fork = (k1, ..., k4) ∈ Z4, let Qk = Q

1≤n≤4[knη,(1 +kn)η].Let Φk,j,i :

Ba,j,ix(1),x(2) ×R2

∩Qk →R2×R2 be the function defined by Φk,j,i(x, y) = y−ϕ x−x(1)

, y−ϕ x−x(2) and letWk,j,i its image. Alsodet Φ0k,j,i

(x, y) = det ϕ0 x−x(2)

−ϕ0 x−x(1) .Thus

(2.4)

det Φ0k,j,i (x, y)

≥JUa,j,i x(2)−x(1)

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for(x, y)∈

Ba,j,ix(1),x(2)×R2

∩Qk.

SinceΦk,j,i(x, y) = Φk,j,i(x, y)implies thatϕ x−x(1)

−ϕ x−x(1)

=ϕ x−x(2)

− ϕ x−x(2)

, taking into account Lemma 2.1, from Lemma 2.3 it follows the existence of ej ∈ N with ej independent of x(1), x(2) such that for j ≥ ej there exists ηe = eη(j) > 0 satisfying that for 0 < η < ηe(j) the map Φk,j,i is injective for all k ∈ Z4. Let Ψk,j,i : Wk,j,i

Ba,j,ix(1),x(2) ×R2

∩Qkits inverse. Lemma 2.5 says that

det Φ0k,j,i

≥c2−jβ|h|2on

Ba,j,ix(1),x(2)×R2

∩Qk.We have Z

Ba,j,ix(1),x(2)×R2

f y−ϕ x−x(1)

, y−ϕ x−x(2) dxdy

= X

k∈Z4

Z

Bxa,j,i(1),x(2)×R2

∩Qk

f y−ϕ x−x(1)

, y−ϕ x−x(2) dxdy

= X

k∈Z4

Z

Wk,j,i

f(z, w) 1

det Φ0k,j,i

k,j,i(z, w))

dzdw

≤ 1

JUa,j,i(x(2)−x(1)) Z

R4

X

k∈Z4

χWk,j,i(v)f(v)dv

≤ m2

JUa,j,i(x(2)−x(1)) Z

R4

f

where we have used (2.4).

Proposition 2.7. Let 0 < a < 1. Suppose that there exist β ∈ N, j0 ∈ N and a positive constantcsuch that|det (ϕ001(x)h, ϕ002(y)h)| ≥c2−jβ|h|2 for allh∈R2, x, y ∈Ua,j,i, j ≥j0, i= 1,2,3,4.Then, there existj1 ∈N, c0 >0such that for allj ≥j1, f ∈S(R4)

TµUa,jf 3

≤c023 kfk3 2 . Proof. Fori= 1,2,3,4,let

Ka,j,i =

x, y, x(1), x(2), x(3)

∈R2×R2×R2×R2×R2 :x−x(s) ∈Ua,j,i, s= 1,2,3 . We can proceed as in Theorem 0 in [3] to obtain, as there, that

µUa,j,i ∗f

3 3 =

Z

Ka,j,i

Y

1≤j≤3

f(xj, y−ϕ(x−xj))dxdydx(1)dx(2)dx(3)

taking into account of Lemma 2.6 and reasoning, with the obvious changes, as in [3], Theorem 0, we obtain that

µUa,j,i ∗f

3

3 ≤m2(A1A2A3)13 with

Aj = Z

R2

Fj(x) Y

1≤m≤3,m6=j

RU1a,j,i 2

Fm(x)dx

andFj(x) = kf(x, .)k3

2 .Now the proof follows as in Proposition 2.4.

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3. ABOUT THE TYPE SET

Proposition 3.1. Forδ > 0 letVδ be defined by (2.1). Suppose that the set of the non elliptic points for ϕ inVδ are those lying in the x1 axis and let α be defined by (2.3).ThenEµ con- tains the closed trapezoidal region with vertices (0,0),(1,1), 7α−1 ,7α−2

, 2 ,1

, except perhaps the closed edge parallel to the principal diagonal.

Proof. We first show that(1−θ) (1,1) +θ 7α−1 ,7α−2

∈Eµ if0≤θ <1.

Ifw= (w1, w2)∈U1

2,jthen2−j−1 ≤ |w2| ≤2−j+1.Thus, from Lemmas 2.2, 2.3 and Propo- sition 2.7, follows the existence ofj0 ∈N and of a positive constantc=c

1kC3(B),kϕ2kC3(B)

such that ifrj =c2−jα,then Tµ

Brj(w)f 3

≤c023 kfk3 2

for somec0 > 0and all j ≥ j0, w ∈ U1

2,j, f ∈ S(R4). For0 ≤ t ≤ 1 let pt, qt be defined by

1 pt,q1

t

=t 23,13

+ (1−t) (1,1).We have also Tµ

Brj(w)f 1

≤πc22−2jαkfk1,thus, the Riesz-Thorin theorem gives

TµBr(w)f qt

≤c2j(3−(1−t)2α)kfkp

t. SinceU1

2,j can be covered withN of such ballsBr(w)withN '2j(2α−1)we get that

TµU

1 2,j

pt,qt

≤c2j(73αt−1).

LetU =∪j≥j0U1

2,j.We have thatkTµUkp

t,qt ≤P

j≥j0

TµU

1 2,j

pt,qt

<∞,fort < 3 . Since for t = 3 we have p1

t = 1− 1 and q1

t = 1− 2 and since every point inVδ\U is an elliptic point (and so, from Theorem 3 in [3],

Tµ\U

3

2,3 <∞), we get that(1−θ) (1,1)+θ 7α−1 ,7α−2

∈ Eµ for0≤θ < 1.On the other hand, a standard computation shows that the adjoint operator Tµ

is given by Tµ

f =

Tµ(f)

,where we write, forg : R4 → C, g(x) = g(−x). ThusEµ is symmetric with respect to the nonprincipal diagonal. Finally, after an application of the Riesz-Thorin interpolation theorem, the proposition follows.

Forδ >0,letAδ ={(x1, x2)∈B :|x2| ≤δ|x1|}.

Remark 3.2. For s > 0, x = (x1, ..., x4) ∈ R4 we set s• x = (sx1, sx2, smx3, smx4). If E ⊂ R2, F ⊂ R4 we setsE ={sx:x∈E}ands•F = {s•x:x∈F}.Forf : R4 → C, s >0, letfsdenotes the function given byfs(x) =f(s•x).A computation shows that (3.1)

Tµ2j Vδf

2−j •x

= 2−2j

Tµf2−j

(x) for allf ∈S(R4), x∈R4.

From this it follows easily that

Tµ

2j Vδ

p,q

= 2−j(2(m+1)q 2(m+1)p +2) Tµ

p,q

.

This fact implies that

(3.2) Eµ

1 p,1

q

: 1 q ≥ 1

p− 1 m+ 1

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and that if 1q > 1pm+11 then

1 p,1q

∈Eµ if and only if

1 p,1q

∈Eµ.

Theorem 3.3. Suppose that for someδ > 0 the set of the non elliptic points forϕ inAδ are those lying on thex1 axis and letαbe defined by (2.3). ThenEµ contains the intersection of the two closed trapezoidal regions with vertices (0,0), (1,1), m+1m ,m−1m+1

, m+12 ,m+11 and (0,0), (1,1), 7α−1 ,7α−2

, 2 ,1

respectively, except perhaps the closed edge parallel to the diagonal.

Moreover, if7α≤m+ 1then the interior ofEµ is the open trapezoidal region with vertices (0,0),(1,1), m+1m ,m−1m+1

and m+12 ,m+11 .

Proof. Taking into account Proposition 3.1, the theorem follows from the facts of Remark 3.2.

For0< a < 1andδ >0we setVa,δ ={(x1, x2)∈B :a≤ |x1| ≤1and |x2| ≤δ|x1|}.We have

Proposition 3.4. Let0< a < 1.Suppose that for some0< a < 1, j0, β∈N and some positive constantcwe have|det (ϕ001(x)h, ϕ001(y)h)| ≥ c2−jβ|h|2 for allh ∈ R2, x, y ∈ Ua,j,i, j ≥ j0 andi= 1,2,3,4.Then, forδpositive and small enough,EµVa,δ contains the closed trapezoidal region with vertices (0,0), (1,1),

β+2 β+3,β+1β+3

,

2

β+3,β+31

, except perhaps the closed edge parallel to the principal diagonal.

Proof. Proposition 2.7 says that there existj1 ∈N and a positive constantcsuch that forj ≥j1

andf ∈S(R4)

TµUa,j,if

3 ≤c23 kfk3

2 . Also, for somec >0and allf ∈S(R4)we have

TµUa,j,if 1

≤c2−jkfk1.Then

TµUa,j,if qt

≤ c2j(tβ3−(1−t))kfkptwherept, qtare defined as in the proof of Proposition 3.1. LetU =∪j≥j1Ua,j. ThenkTµUfkp

t,qt <∞ift < β+33 .Now, the proof follows as in Proposition 3.1.

Theorem 3.5. Suppose that for some0< a <1, j0, β ∈N and for some positive constantcwe have|det (ϕ001(x)h, ϕ001(y)h)| ≥c2−jβ|h|2 for allx, y ∈ Ua,j,i, j ≥j0 andi = 1,2,3,4.Then for δ positive and small enough,Eµ contains the intersection of the two closed trapezoidal regions with vertices(0,0),(1,1), m+1m ,m−1m+1

, m+12 ,m+11

and(0,0),(1,1),

β+2 β+3,β+1β+3

, 2

β+3,β+31

, respectively, except perhaps the closed edge parallel to the diagonal.

Moreover, ifβ ≤ m−2then the interior ofEµ is the open trapezoidal region with vertices (0,0),(1,1), m+1m ,m−1m+1

and m+12 ,m+11 .

Proof. Follows as in Theorem 3.3 using now Proposition 3.4 instead of Proposition 3.1.

Remark 3.6. We now turn out to the case whenϕis a homogeneous polynomial function whose set of non elliptic points is a finite union of lines through the origin,L1,...,Lk.

For eachl,1≤l ≤ k,letAlδ =

x∈R2 : πL

lx

≤δ|πLlx| whereπLl andπL

l denote the orthogonal projections fromR2 intoLl andLl respectively. Thus eachAlδ is a closed conical sector aroundLl.We chooseδsmall enough such thatAlδ∩Aiδ =∅forl 6=i.

It is easy to see that there exists (a unique)αl∈N and positive constantsc0l, c00l such that

(3.3) c0l

πL

lw

αl

≤ inf

|h|=1|det (ϕ00(w)h)| ≤c00l πL

lx

αl

for allw∈Alδ.Indeed, after a rotation the situation reduces to that considered in Remark 3.2.

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Theorem 3.7. Suppose that the set of non elliptic points is a finite union of lines through the origin, L1,...,Lk.Forl = 1,2, ..., k,letαl be defined by (3.3), and let α = max1≤l≤kαl.Then Eµ contains the intersection of the two closed trapezoidal regions with vertices(0,0),(1,1),

m

m+1,m−1m+1

, m+12 ,m+11

and (0,0), (1,1), 7α−1 ,7α−2

, 2 ,1

, respectively, except per- haps the closed edge parallel to the diagonal.

Moreover, if7α ≤m+ 1then the interior ofEµis the interior of the trapezoidal regions with vertices(0,0),(1,1), m+1m ,m−1m+1

, m+12 ,m+11 .

Proof. Forl = 1,2, ..., k,letAlδ be as above. From Theorem 3.3, we obtain thatEµ

Alδ

contains the intersection of the two closed trapezoidal regions with vertices (0,0),(1,1), m+1m ,m−1m+1

,

2

m+1,m+11

and (0,0) , (1,1),

l−1 l ,l−2

l

,

2 l,1

l

respectively, except perhaps the closed edge parallel to the diagonal.

Since every x ∈ B\ ∪l Alδ is an elliptic point for ϕ, Theorem 0 in [3] and a compactness argument give that kTµDk3

2,3 < ∞ where D =

x∈B\ ∪lAlδ : 12 ≤ |x| . Then (using the symmetry of EµD, the fact of that µD is a finite measure and the Riesz-Thorin theorem) EµD is the closed triangle with vertices (0,0), (1,1), 23,13

. Now, proceeding as in the proof of Theorem 3.3 we get that

Tµ

B\∪lAl δ

p,q

< ∞if 1q > 1pm+11 .Then the first assertion of the theorem is true. The second one follows also using the facts of Remark 3.2.

For0< a <1,we set Ua,jl =

x∈R2 :a≤ |πLl(x)| ≤1and2−jLl(x)| ≤

πLl(x)

≤2−j+1Ll(x)|

letUa,j,il , i = 1,2,3,4be the connected components ofUa,jl .

Theorem 3.8. Suppose that the set of non elliptic points forϕis a finite union of lines through the origin,L1,...,Lk.Let0< a < 1and letj0 ∈N such that

Forl = 1,2, ..., k, there existsβl ∈ N satisfying|det (ϕ001(x)h, ϕ001(y)h)| ≥ c2−jβj|h|2 for allx, y ∈Ua,j,il , j ≥j0andi= 1,2,3,4.Letβ = max1≤j≤kβj.ThenEµcontains the intersec- tion of the two closed trapezoidal regions with vertices(0,0),(1,1), m+1m ,m−1m+1

, m+12 ,m+11 and(0,0),(1,1),

β+2 β+3,β+1β+3

,

2

β+3,β+31

, respectively, except perhaps the closed edge par- allel to the diagonal.

Moreover, ifβ ≤ m−2then the interior ofEµis the interior of the trapezoidal region with vertices(0,0),(1,1), m+1m ,m−1m+1

, m+12 ,m+11 .

Proof. Follows as in Theorem 3.7, using now Theorem 3.5 instead of Theorem 3.3.

Example 3.1. ϕ(x1, x2) = (x21x2−x1x22, x21x2+x1x22)

It is easy to check that the set of non elliptic points is the union of the coordinate axes. Indeed, forh = (h1, h2)we havedetϕ00(x1, x2)h = 8x22h21 + 8x1x2h1h2 + 8x21h22 and this quadratic form in(h1, h2)has non trivial zeros only ifx1 = 0orx2 = 0.The associated symmetric matrix to the quadratic form is

8x22 4x1x2 4x1x2 8x21

and for x1 6= 0 and|x2| ≤ δ|x1| withδ small enough, its eigenvalue of lower absolute value isλ1(x1, x2) = 4x21+ 4x22 −4p

(x42−x21x22+x41).Thusλ1(x1, x2) ' 6x22 for such(x1, x2). Similarly, for x2 6= 0and|x1| ≤ δ|x2|withδ small enough, the eigenvalue of lower absolute value is comparable with6x21. Then, in the notation of Theorem 3.7, we obtainα = 2and so Eµcontains the closed trapezoidal region with vertices(0,0),(1,1), 1314,67

and 17,141

except

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perhaps the closed edge parallel to the principal diagonal. Observe that, in this case, Theorem 3.8 does not apply. In fact, forx= (x1, x2),xe= (ex1,ex2)andh= (h1, h2)we have

det (ϕ001(x)h, , ϕ002(ex)h)

= 4h21(x2ex1−ex2x1+ 2x2xe2) + 4h1h2(x1ex2+xe1x2) + 4h22(x1ex2−x2xe1+ 2x2ex1). Takex1 = ex1 = 1and let A = A(x2,xe2)the matrix of the above quadratic form in (h1, h2). Forx2 = 2−j,xe2 = 2−j+1 we havedetA <0forj large enough but if we takex2 = 2−j+1and ex2 = 2−j,we getdetA > 0forj large enough, so, for all j large enough,detA = 0for some 2−j ≤x2,xe2 ≤2−j+1.Thus, for suchx2,ex2,

|(h1inf,h2)|=1det (ϕ001(1, x2) (h1, h2), ϕ002(1,ex2) (h1, h2)) = 0.

Example 3.2. Let us show an example where Theorem 3.8 characterizes

Eµ.Let ϕ(x1, x2) = x31x2−3x1x32,3x21x22 −x42

. In this case the set of non elliptic points forϕis thex1axis. Indeed,

det (ϕ00(x1, x2) (h1, h2)) = 18 x21+x22

(h2x1+x2h1)2 + 2x22h21+ 6h22x22 . In order to apply Theorem 3.8, we consider the quadratic form inh= (h1, h2)

det (ϕ001(x1, x2)h, ϕ002(ex1,ex2)h).

Ifx= (x1, x2)andex= (xe1,xe2),letA=A(x,x)e its associated symmetric matrix. An explicit computation ofAshows that, for a given0< a <1and for alljlarge enough andi= 1,2,3,4, ifxandxebelong toUa,j,i,then

a2 ≤tr(A)≤20

thus, ifλ1(x,ex)denotes the eigenvalue of lower absolute value ofA(x,ex),we have, forx,xe∈ Wathat

c1|detA| ≤ |λ1(x,x)| ≤e c2|detA|

wherec1, c2 are positive constants independent ofj.Now, a computation gives detA= 324 −x21ex21−9x22ex22 −12x1x2xe1xe2+ 2x21ex22

× x22ex21−2x22xe22−4x1x2xe1xe2+x21xe22 . Now we writeex2 =tx2,with 12 ≤t ≤2.Then

detA= 324x22

−x21ex21−9t2x42−12tx1x22ex1+ 2t2x22x21 ex21−2t2x22−4tx1xe1+t2x21 . Note that the the first bracket is negative forx,ex ∈ Wa ifj is large enough. To study the sign of the second one, we consider the functionF (t, x1,xe1) = xe21 −4tx1ex1 +t2x21.SinceF has a negative maximum on{1} × {1} ×1

2,2

,it follows easily that we can choosea such that forx,xe ∈ Wa andj large enough, the same assertion holds for the second bracket. So detA is comparable with2−2j,thus the hypothesis of the Theorem 3.8 are satisfied withβ = 2and such a.Moreover, we have β = m −2, then we conclude that the interior ofEµ is the open trapezoidal region with vertices(0,0),(1,1), 35,45

, 25,15 .

On the other hand, in a similar way than in Example 3.1 we can see that α = 2 (in fact detA(x, x) = 648 (x21+ 9x22) (x21+x22)2x22), so in this case Theorem 3.8 gives a better result (a precise description of

Eµ) than that given by Theorem 3.7, that asserts only that

Eµcontains the trapezoidal region with vertices(0,0),(1,1), 1314,67

and 17,141 .

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