Revista Matematica Iberoamericana
Vol. 14, N.o 3, 1998
Average decay of Fourier
transforms and geometry
of convex sets
Luca Brandolini, Marco Rigoli and Giancarlo Travaglini
Abstract. Let B be a convex body in
R2 ,
with piecewise smooth
boundary and let bB denote the Fourier transform of its characteristic function. In this paper we determine the admissible decays of the
spherical Lp -averages of bB and we relate our analysis to a problem in
the geometry of convex sets. As an application we obtain sharp results
on the average number of integer lattice points in large bodies randomly
positioned in the plane.
1. Introduction.
Given a convex body B; that is, a compact convex set with non
empty interior in R n ; we denote by B its characteristic function. The
study of the decay of the Fourier transform
bB ( ) =
Z
B
e;2ix dx ;
as j j ;! 1, in terms of the geometric properties of B , is a fascinating
and by now classical subject (see [18, Chapter VIII] for basic results,
related problems and references). For instance, it is well known that,
when the boundary is smooth with everywhere strictly positive Gauss519
520
L. Brandolini, M. Rigoli and G. Travaglini
Kronecker curvature, the order of decay of bB in a given direction is
independent of this latter.
This situation is far from being typical, as one can easily check by
considering either a cube or any convex body with a smooth boundary
containing at points. Furthermore, a number of problems requires
some sort of \global information" on the decay of bB ( ) which is not a
direct consequence of the presently known directional estimates.
In this setting, the study of the spherical Lp -averages
Z
n;1
jbB ( )jp d
1=p
turns out to be quite useful.
We point out that the L2 case has been investigated by various
authors, notably [14], [15], [19], [13], [12]; while for general p's and
B a polyhedron, a detailed analysis with applications to problems on
lattice points and on irregularities of distributions can be found in [3].
We note that the L1 case is also naturally related with the summability
of multiple Fourier integrals (see e.g. [2] or [4] ), moreover, F. Ricci and
one of us (G. Travaglini) have recently shown that the general Lp case
is connected to boundedness of Radon transforms (see [16]).
Throughout this paper, unless otherwise explicitly stated, we consider convex bodies B in R 2 with piecewise smooth boundary. More
precisely, we assume that @B is a union of a nite number of regular
arcs, each one of them being C 1 in its interior.
According to a more general result of Podkorytov [13], (see also
[19] ) the L2-average decay of bB satises
(1.1)
Z 2
where, from now on,
0
jbB ( )j2 d
1=2
c ;3=2 ;
2 [0; 2) ;
1 and c; c1; c2; : : : , denote positive constants independent of which
may change from line to line.
It is an easy consequence of a result of Montgomery [12, p. 116]
that (1.1) is sharp. Namely, for any B ,
= (cos ; sin ) ;
(1.2)
lim sup 3=2
!1
Z 2
0
jbB ( )j2 d
1=2
> 0:
Average decay of Fourier transforms
521
We stress that in the L2 case the order of decay is independent of B .
The aim of this paper is to study the general Lp case where the results
turn out to depend on the shape of B .
It is worth to begin with the case of a polygon P . It has been
proved in [3] that
Z 2
(1.3)
1=p
p
jbP ( )j d
0
( ;2
c log (1 + ) ; when p = 1 ;
c ;1;1=p ;
when 1 < p 1 :
Here we prove
Z 2
0
jbP ( )j d c ;2 log (1 + )
and, for each 1 < p 1
(1.4)
lim sup 1+1=p
!1
Z 2
0
jbP ( )jp d
1=p
> 0:
Next, we consider the case when B is not a polygon. We show that
(1.5)
lim sup 3=2
!1
Z 2
1=p
p
jbB ( )j d > 0 ;
0
whenever 1 p 1 (note that, when p = 2, (1.4) and (1.5) agree with
(1.2)). These results, when compared with (1.1) and (1.3), completely
describe the case 1 p 2. As for p 2; an easy interpolation
argument between p = 2 and p = 1 gives
Z 2
0
jbB ( )jp d
1=p
c ;1;1=p ;
for every 2 p 1. Contrary to the case 1 p < 2; in the range
2 < p 1 every order of decay between ;3=2 and ;1;1=p is possible.
More precisely we exhibit, for any 2 p 1 and 1 + 1=p a 3=2,
a corresponding convex body B such that
c1 ;a Z 2
1=p
p
jbB ( )j d c2 ;a :
0
522
L. Brandolini, M. Rigoli and G. Travaglini
When 1 + 1=p < a < 3=2 such examples are constructed so to have,
for a suitable > 2, a piece of the curve of equation y = jxj in its
boundary. As a side-product, we obtain a result on the average decay
of the Fourier transforms of singular measures supported on the above
curves (see Proposition 3.17 below).
The dierent results for p < 2 and p > 2 are due to the following fact. When B is not a polygon, its boundary @B must contain
points with positive curvature and for 1 p 2 they give the relevant
contribution to
Z 2
1=p
p
jbB ( )j d :
0
On the other hand, when 2 p 1 the main contribution is given by
the at points (if any), as one may guess considering the L1 case.
We summarize the main results discussed so far in Figure 1. For
p > 1 and a > 0 the point (1=p; a) is marked black if and only if there
exists B satisfying
Z 2
0
and
lim sup a
!1
jbB ( )jp d
1=p
Z 2
< c ;a
1=p
p
jbB ( )j d > 0 :
0
Figure 1.
Average decay of Fourier transforms
523
It is natural to ask whether (1.4) and (1.5) can be turned into
estimates from below. As a matter of fact, a negative answer is given
by the two simplest examples of convex bodies in R 2 : the square (see
Lemma 3.12) and the disc (because of the zeroes of the Bessel function
J1 ). On the other hand, we show that, for any 1 p 1 and for some
polygons P , we have
Z 2
1=p
p
jbP ( )j d c2 ;1;1=p ;
0
while, if B is neither a polygon nor a body too \similar" (see Denition
3.3) to a disc, then
Z 2
0
jbB ( )j d c ;3=2 :
The above results are organized in our main theorem of Section 2. We
stress that such general Lp estimates hold provided @B is piecewise
smooth. In Section 4 we shall see that in the framework of arbitrary
convex bodies one can nd very \chaotic" situations.
A basic tool in some of our proofs is the following known fact.
Let S = supx2B x . For > 0 suciently small we dene, see
Figure 2, the set
(1.6)
AB (; ) = fx 2 B : S ; x S g :
Figure 2.
524
L. Brandolini, M. Rigoli and G. Travaglini
Then (see Lemma 3.8, [5], or [13])
jbB ( )j c (jAB (;1; )j + jAB (;1; + )j) ;
where jK j denotes the Lebesgue measure of a measurable set K .
As a consequence, for each p 1,
(1.7)
Z 2
1=p
Z 2
1=p
p
;
1
p
jbB ( )j d c
jAB ( ; )j d
0
0
providing a way to estimate the average decay of bB from above. Moreover we shall see (cf. also [5]) that (1.7) can be reversed under additional
assumptions on B .
Observe that the right hand side of (1.7) does not involve any
Fourier transform and the problem of estimating
Z 2
0
jAB (; )jp d
1=p
;
as ;! 0, is indeed a genuine problem in the geometry of convex sets.
To the best of our knowledge, such a problem has never been considered
before and the closest area in the eld is perhaps the study of oating
bodies (see e.g. [17]). In Section 5 we shall investigate the admissible
decays of
Z 2
1=p
p
jAB (; )j d ;
0
as ;! 0, mostly as a consequence of the similar problem for bB .
We end the paper by applying some of the previous results to a
problem on the number of lattice points in a large convex planar body
B.
Elementary geometric considerations show that
card (B \ Z2) 2jB j
and
(1.8)
card (B \ Z2) ; 2jB j = O()
as ;! 1. The improvement of (1.8) and the related problems constitute a whole area of research (see e.g. [11] or [8]), where the pointwise
estimate (1.8) is often substituted by mean estimates.
Average decay of Fourier transforms
525
Here we consider a large convex body B randomly positioned in
the plane. More precisely, for 2 SO(2) and t 2 R 2 we study the
discrepancy
DB (; ; t) = card (( ;1(B ) ; t) \ Z2) ; 2 jB j ;
where ;1(B ) ; t is a rotated, dilated and translated copy of B . Since
this function is periodic with respect to the variable t we restrict this
latter to T2 = R 2 =Z2. Kendall ([10]) has proved L2 estimates related
to the above discrepancy (see also [3]). Here we prove that if B is a
convex planar body with piecewise smooth boundary, dierent from a
polygon, then, for any 1 p 2,
(1.9)
c1 1=2 kDB (; ; )kLp(SO(2)T ) c2 1=2 :
2
We do not know whether (1.9) holds for some p > 2. We point out
that, in general, it is false when p = 1. Indeed, as a consequence of
Hardy's -result for the circle problem (see [7] or [11]) we have, for a
disc D,
lim sup ;1=2(log );1=4 kDD (; ; )kL1(SO(2)T ) > 0 :
2
!1
2. Statement of the main result.
Let 1 be the unit circle in R 2 . For any complex measurable function g on 1 and for any p 1, let
kgkLp( ) =
Z 2
0
1
1=p
p
jg()j d ;
where d is the normalized Lebesgue measure. As usual we set
kgkL1( ) = ess sup jg()j :
1
21
Let B be a convex body in R 2 ; ' : [1; +1) ;! R + a non-increasing
function and let 1 p 1. We say that ' is an optimal estimate of
the p-average decay of bB whenever
i) kbB ( )kLp( ) c '(),
1
526
L. Brandolini, M. Rigoli and G. Travaglini
ii) lim sup
kbB ( )kLp( ) > 0 :
1
'()
Similarly, ' is a sharp estimate of the p-average decay of bB provided
c1 '() kbB ( )kLp( ) c2 '() :
Our main result essentially concerns the case '() = ;a and the following denition will be useful.
!1
1
Denition 2.1. When '() = ;a is an optimal or sharp estimate of
the p-average decay of bB we say that the p-average decay of bB has
optimal order a or sharp order a respectively.
With this preparation we state our main result.
Theorem 2.2. I) Let 1 < p 1 and dene
n 1 3 or a = 1 + 1 o ;
;
a
:
1
<
p
<
2
;
a
=
p
2
p
o
n T = 1p ; a : 2 p 1; 1 + p1 a 23 :
S=
The following are equivalent :
i) There exists a convex body B with piecewise C 1 boundary such
that the p-average decay of bB has optimal order a.
ii) (1=p; a) 2 S [ T .
II) Let p = 1. If P is a polygon then '() = ;2 log (1 + ) is
an optimal estimate for the 1-average decay of bP . If B is any other
convex body with piecewise C 1 boundary, then the 1-average decay of
bB has optimal order 3=2.
Moreover it will be clear from the proof that this result still holds
after substituting the word \optimal" with the word \sharp".
The above theorem will be obtained as a consequence of the following somewhat more informative results.
In the rst Proposition we cover the case 1 p 2 when B is not
a polygon.
Average decay of Fourier transforms
527
Proposition 2.3. Let 1 p 2 and let B be a convex body with
piecewise C 1 boundary. Suppose B is not a polygon, then 3=2 is the
optimal order of the p-average decay of bB . Moreover, 3=2 is the sharp
order of the p-average decay of bB for some, but not for all, bodies B .
The above Proposition follows from Lemma 3.1, Lemma 3.2, Lemma 3.6 and the example of the disc.
We now consider the case of a polygon.
Proposition 2.4. Let P be a compact convex polygon with non empty
interior. Then '() = ;2 log (1 + ) is a sharp estimate of the 1average decay of bP . If 1 < p 1, then 1 + 1=p is the optimal order
of the p-average decay of bP . Moreover, 1 + 1=p is the sharp order of
the p-average decay of bP for some, but not for all, polygons P .
This is a consequence of Lemma 3.9, Lemma 3.10, Lemma 3.11 and
Lemma 3.12.
Finally, for 2 p 1, we have
Proposition 2.5. Let 2 p 1, then the following are equivalent :
j) There exists a convex body B with piecewise C 1 boundary such
that the p-average decay of bB has optimal order a.
jj) 1 + 1=p a 3=2.
The above Proposition follows from Lemma 3.2, Lemma 3.13 and
Lemma 3.16.
3. Lemmas.
The following lemma is contained in [13, p. 63].
Lemma 3.1. Let B be a convex body in R 2 . Then
Z 2
1=2
2
jbB ( )j d c ;3=2 :
0
We now prove the following result.
528
L. Brandolini, M. Rigoli and G. Travaglini
Lemma 3.2. Let B be a convex body in R 2 with piecewise C 1 boundary
@B . Assume B is not a polygon then, for any p 1,
lim sup 3=2
!1
Z 2
0
jbB ( )jp d
1=p
> 0:
Proof. It is enough to prove the lemma when p = 1. Let ; be an arc
in @B where the curvature is strictly positive. We examine two cases.
i) There exists an open interval U of angles such that for every
2 U there is exactly one point () 2 ; whose tangent is orthogonal to
= (cos ; sin ) (this may happen since @B is only piecewise smooth).
ii) There exists an open interval U of angles such that for every
2 U there are exactly two points 1 (); 2() 2 @B whose tangent is
orthogonal to .
We proceed with the proof in case i).
We apply [1, Theorem 1] (see also [13]) to obtain
(3.1)
bB ( ) = ; 21i ;3=2 e;2i()+i=4K ;1=2(()) + E ;
where K (P ) denotes the curvature at P 2 @B and jEj c ;2. We
remark that although [1, Theorem 1] is stated for sets with smooth
boundary, in the bidimensional case it still holds true for sets having a
piecewise smooth boundary. From (3.1) we have
Z
1
jbB ( )j d 2 K ;1=2 (()) d ; c1 ;1=2 c2 > 0 :
U
0
We now turn to ii).
As in the previous case we obtain
3=2
Z 2
2
X
bB ( ) = ; 21i ;3=2 e;2ij ()+i=4K ;1=2(j ()) + E :
j =1
We consider three subcases.
a) Suppose rst there exists a neighborhood Ue U where
K (1()) 6= K (2()) :
Average decay of Fourier transforms
529
Then
3=2
Z 2
0
jbB ( )j d
Z
1
2 e jK ;1=2(1()) ; K ;1=2(2 ())j d ; c1 ;1=2 c2 > 0 :
U
b) Suppose there exists a neighborhood Ue U where the vectors
and 2 () ; 1 () are not parallel. Let Aj () = K ;1=2 (j ()). We
have
3=2
Z 2
0
jbB ( )j d
Z X
2
1
;
2
i
(
)
j
2 e e
Aj () d ; c1 ;1=2
U j =1
Z
1
= 2 jA1() + A2() e;2i( (); ()) j d ; c1 ;1=2
Ue
Z
21 e (A1() + A2 () e;2i( (); ()) ) d ; c1 ;1=2
U
Z
1
;
2
i
(
(
)
;
(
))
M ; 2 e A2 () e
d ; c1 ;1=2 :
U
2
1
2
2
1
1
We claim that the last integral tends to zero as tends to innity.
Observe that j0 () = 0 since is normal to @B at the point j ().
Hence
(3.2)
d ( ( () ; ())) = (; sin ; cos ) ( () ; ())
2
1
2
1
d
is dierent from zero since (; sin ; cos ) is not orthogonal to 2 () ;
1 (). Integration by parts shows that the integral vanishes as ;!
+1.
c) We suppose now that for every 2 U the points 1 (), 2 ()
have the same curvature and that and 2() ; 1() are parallel. In
this case the quantity (3.2) vanishes so that
= (2() ; 1 ())
530
L. Brandolini, M. Rigoli and G. Travaglini
is constant. Let K () = K (1()) = K (2()), then
3=2
Z 2
Z
1
jbB ( )j d 2 e K ;1=2() j1 + e;2i j d ; c1 ;1=2
0
U
Z
1
;
2
i
2 j1 + e
j e K ;1=2 () d ; c1 ;1=2 ;
U
and since
the proof is complete.
lim sup j1 + e;2i j > 0 ;
!+1
The result of the previous lemma can be strengthened under simple
geometric hypothesis on the boundary. The following denition may be
useful.
Denition 3.3. We say that a convex body B is a cut disc if it is not
a polygon and if its boundary @B is the union of a nite number of
segments and of a nite number of couples of antipodal arcs of a given
circle.
We now need a technical lemma.
Lemma 3.4. Let I and J be two neighborhoods of the origin in R
and let f 2 C 2 (I ), g 2 C 2 (J ). Assume f (x) < 0, f 00(x) > 0, for
x 2 I; g(x) > 0, g00 (x) < 0 for x 2 J ; also suppose f (0) = ;1,
g(0) = 1, f 0 (0) = g0(0) = 0. Finally we assume the existence of a
bijection H : I ;! J such that
i) f 0 (x) = g0(H (x)),
ii) the curvature of the graph of f at (x; f (x)) equals the curvature
of the graph of g at (H (x); g(H (x))),
iii) the segment joining the points (x; f (x)) and (H (x); g(H (x))) is
orthogonal to the tangent lines at these points.
Then the graphs of f and g are two (antipodal) arcs of equal length
in the same circle.
Proof. By our assumptions,
i) f 0(x) = g0(H (x)),
Average decay of Fourier transforms
531
f 00 (x)
;g00 (H (x)) ,
=
(1 + (f 0(x))2)3=2 (1 + (g0(H (x)))2)3=2
iii) (x ; H (x)) + (f (x) ; g(H (x)))f 0(x) = 0.
Then i) and ii) imply f 00 (x) = ;g00 (H (x)), while dierentiating i)
one gets f 00 (x) = g00 (H (x)) H 0(x). Because of the other assumptions,
this implies H (x) = ;x and
ii)
f (x) = ;g(;x) :
Then iii) becomes
2 x + 2 f (x) f 0(x) = 0 ;
which gives the equation of a circle.
Lemma 3.4 can be restated in the following, more geometrical, way.
Lemma 3.5. Suppose B is a convex body with piecewise C 1 boundary
which is not a cut disc, then @B contains a regular point P with unit
exterior normal such that either there is no other regular point in
@B with unit exterior normal ;, or, if such a point Q exists, at least
one of the following facts happens: i) P ; Q is not parallel to , ii) the
curvatures of @B at P and at Q dier.
The following is a strengthened version of Lemma 3.2.
Lemma 3.6. Suppose B is a convex body with piecewise C 1 boundary
which is neither a polygon nor a cut disc, then, for 1 p 2,
c1 ;3=2 Z 2
1=p
p
jbB ( )j d c2 ;3=2 :
0
Proof. The estimate from above is contained in Lemma 3.1. On the
other hand the estimate from below holds in cases i), ii)-a, ii)-b of the
proof of Lemma 3.2. Our assumptions and Lemma 3.5 exclude the case
ii)-c. This ends the proof .
The forthcoming lemma is probably known. However, since we
have not found a suitable reference, we provide an elementary argument.
532
L. Brandolini, M. Rigoli and G. Travaglini
Lemma 3.7. Let f : R ;! [0; +1) be supported and concave in
[;1; 1]. Then, for every j j 1,
jfb( )j (3.3)
1 f 1 ; 1 + f ; 1 + 1 :
j j
2 j j
2 j j
Proof. It is enough to prove (3.3) when > 1. The assumption on
the concavity of f allows us to integrate by parts obtaining
1 f (1; ) + 1 f (;1+ ) + 1 Z 1 f 0(t) e;2it dt :
2
2
2 ;1
jfb( )j Let be a point where f attains its maximum. Then f will be nondecreasing in [;1; ] and non-increasing in [; 1]. We can assume 0 1, so that f (;1+) f (;1+1=(2 )). To estimate f (1;) we observe
that when 1 ; 1=(2 ), one has f (1;) f (1 ; 1=(2 )). On the
other hand, since f is concave, in case > 1 ; 1=(2 ) we have
f (1;) f () 2f (0) 2f 1 ; 21 :
To estimate the integral we observe that, by a change of variable,
I=
Z1
;1
f 0 (t) e;2it dt = ;
Z 1+1=(2)
f 0 t ; 21 e;2it dt :
;1+1=(2)
So that
2I =
=
Z1
f 0(t) e;2it dt ;
;1
Z ;1+1=(2)
;1
+
Z1
Z 1+1=(2)
;1+1=(2)
1
0
f t;
e;2it dt
2
f 0 (t) e;2it dt
1
0
0
e;2it dt
f (t) ; f t ;
2
;1+1=(2)
Z 1+1=(2) +
f 0 t ; 21 e;2it dt
1
= I1 + I2 + I3 :
Average decay of Fourier transforms
533
To estimate I1 from above we note that
jI1j Z ;1+1=(2)
;1
f 0(t) dt = f
1
1
+
; 1 + 2 ; f (;1 ) f ; 1 + 2 ;
since 0 1.
The estimate for I3 is similar in case 1 ; 1=(2 ). If >
1 ; 1=(2 ), then
Z +1=(2) Z 1+1=(2) 1
1
0
0
jI3 j f t ; 2 dt ;
f t ; 2 dt
1
+1=(2)
= 2f () ; f 1 ; 1 ; f (1;)
2f ()
4f (0)
4f 1 ; 21 :
2
As for I2, since f 0 is non increasing, we have
jI2j Z1
;1+1=(2)
1
0
0
f t;
; f (t) dt
2
= f 1 ; 21 ; f (;1+ ) ; f (1;) + f ; 1 + 21
1
1
f 1; 2 +f ; 1+ 2
ending the proof. Note that no constant c is missing in (3.3).
Remark. A dierent proof of the above lemma can be modeled on an
argument similar to that of Lemma 3.15 below.
The following result is similar to [5, Theorem 6.1] (see also [13,
Lemma 3]). Our proof is based on the previous lemma.
Lemma 3.8. Let B be a convex body in R 2 , = (cos ; sin ) and
S = supx2B x . For 1 we set (see Figure 2 with ;1 in place of
)
AB (;1; ) = fx 2 B : S ; ;1 x S g :
534
L. Brandolini, M. Rigoli and G. Travaglini
Then
jbB ( )j c (jAB (;1; )j + jAB (;1; + )j) ;
where jE j denotes the Lebesgue measure of a measurable set E .
Proof. Without loss of generality we choose = (1; 0). Then
(3.4)
bB (1; 0) =
Z +1 Z +1
;1
= bh(1) ;
;1
B (x1 ; x2) dx2 e;2ix dx1
1 1
where h(s) is the lenght of the segment obtained intersecting B with
the line x1 = s. Observe that h is concave on its support, say [a; b]. We
can therefore apply Lemma 3.7 to obtain, after a change of variable,
jbh(1)j j1 j h b ; 2 j1 j + h a + 2 j1 j
1
1
1
c (jAB (j1j;1; 0)j + jAB (j1j;1 ; )j) :
We now consider polygons.
The following lemma appears in [3]; here we give a dierent, more
geometric, argument based on the previous lemma.
Lemma 3.9. Let P be a compact polygon in R 2 . Then
(3.5)
Z 2
0
jbP ( )jp d
1=p
(
c ;2 log (1 + ) ; when p = 1 ;
c ;1;1=p ;
when p > 1 :
Proof. Without loss of generality we can assume that the polygon
is convex, lies in the left halfplane and that the points (0; ;1) and
(0; 1) are vertices. By Lemma 3.8 we reduce the problem to estimating jAP (1=; )j in a suitable right neighborhood of zero. A simple
geometric consideration shows that
jAP (;1; )j (
c ;1 ;
for 0 c1 ;1 ;
c2 ;1;1 ; for c1 ;1 c3 ;
which implies (3.5) by integration.
Average decay of Fourier transforms
535
We still have to check sharpness of the estimates in (3.5). This is
not entirely trivial since parallel edges of P (if any) give the same contribution to the decay of bP so that cancellations may occur. Actually
this does not happen for p = 1, but it may happen for p > 1, as shown
in the next three lemmas.
Lemma 3.10. Let P be the characteristic function of a compact convex polygon P in R 2 with non empty interior. Then
Z 2
0
jbP ( )j d c ;2 log (1 + ) :
Proof. Let Lj = [Pj ; Pj +1 ], j = 1; : : :; S , be the edges of the polygon
P and let lj be their lengths. Then, with the aid of the divergence
formula, we obtain
bP ( ) =
Z
P
e;2it dt
Z
1
= ; 2i
e;2it (t) dt0
@P
S
;2iPj
;2iPj
X
1
= 422 j e (P ; ;e P )
lj ;
j +1
j
j =1
+1
where dt0 is the 1-dimensional measure and j is the outward unit normal to Lj . The argument is divided in three cases.
Case 1. Suppose there exists an edge, say L1 , which is not parallel to
any other edge. We can suppose P1 = (0; ;1) and P2 = (0; 1).
Because of these assumptions there exists a right neighborhood
U (0) [0; 2) such that
inf j (Pj+1 ; Pj )j c > 0 ;
2U (0)
for each j 2. Hence
Z 2
Z
c
e;2iP ; e;2iP l d ; c2
1
jbP ( )j d 2
1
1
(P2 ; P1 )
2
0
U (0)
Z
c
cos sin (2 sin ) d ; c2
3
2
sin 2
U (0)
2
1
536
L. Brandolini, M. Rigoli and G. Travaglini
Zc c
sin
(2
u
)
4
2 u du ; c22
0
c ;2 log (1 + ) :
5
Case 2. Suppose there exists a couple of parallel edges of dierent
length. Let M1 = [Q1; R1] and M2 = [Q2; R2] be such a pair.
We can assume Q1 = H1 + (0; ;a1), R1 = H1 + (0; a1), Q2 =
H2 + (0; ;a2), R2 = H2 + (0; a2) with a2 > a1 > 0.
Then, arguing as above,
Z 2
0
jbP ( )j d
Z
2
X
;2iQj ;2iRj c
1
2
1 e (Q;;e R )
lj d ; c22
j
j
U (0) j =1
Z
2
X
c
aj sin ) d ; c2
1
2
e;2iHj sin (2asin
cos 2
j
U (0)
j =1
Zc c
Z c sin (2 a u) sin
(2
a
u
)
c
du ; 2
1
2
3
2
du ;
a1 u
a2 u 2
0
0
c32 a1 log (a1 ) ; a1 log (a2 ) ; c52
1
2
;
2
c log (1 + ) :
4
4
Case 3. Suppose the edges of P are pairwise parallel and with the same
length. Let M1 = [Q1; R1] and M2 = [Q2; R2] be one of these couples.
We can assume Q1 = H +(0; ;1), R1 = H +(0; 1), Q2 = ;H +(0; ;1),
R2 = ;H + (0; 1). Then
Z 2
0
jbP ( )j d
Z
2
X
;2iQj ; e;2iRj c
e
c
1
2
1
d ; 22
(Qj ; Rj )
U (0) j =1
Z
c
cos (2 H ) sin (2 sin ) d ; c2 :
1
2
sin 2
U (0)
Average decay of Fourier transforms
537
Let H = (h1 ; h2) and H = h1 cos + h2 sin . We choose ' so that
H = jH j sin ( + '). Since h1 6= 0 we have ' 6= 0 and for symmetry
reasons we can restrict ourselves to the case 0 < ' =2.
We obtain
Z 2
0
jbP ( )j d
Z
c
cos (2 jH j sin( + ')) sin (2 sin ) d ; c2 :
1
2
sin 2
U (0)
Observe that choosing a sequence n so that n jH j sin ' is close to an
integer we immediately get
Z
sin (2 sin ) c
d c ;2 log (1 + ) ;
n
jbP (n )j d 2
n
n
sin 0
n U (0)
that is, we have proved that ;2 log (1+ ) is an optimal estimate of the
1-average decay of bP . To get the full statement of the lemma we must
deal with the values of close to those annihilating cos (2 jH j sin ').
We begin with the case 0 < ' < =2. Let 0 < " < =2 ; '
such that [0; "] U (0) and let f[aj ; bj ]g be the collection of intervals
determined by the choice
j + 1 + j + 3 ; aj = arcsin 22jH j ; ' ; bj = arcsin 22jH j ; '
Z 2
and j = [2 jH j sin '] + 1; : : :; [2 jH j sin (' + ")] for some suciently
small > 0. We observe that on each [aj ; bj ] we have
j cos (2 jH j sin( + '))j 0 > 0 :
As a consequence
Z
U (0)
(3.6)
cos (2 jH j sin ( + ')) sin (2 sin ) d
sin X 1 Z bj
0
j sin (2 sin )j d
sin b
j
a
j
j
X 1 Z sin bj
0
j sin (2 u)j du :
c sin b
j
sin
a
j
j
538
L. Brandolini, M. Rigoli and G. Travaglini
Using the elementary inequality
sin (bj + ') ; sin (aj + ') sin bj ; sin aj
and the above denition of aj and bj we see that the quantity
sin bj ; sin aj
is bounded away from zero and therefore
Z sin bj
sin aj
j sin (2 u)jdu > c > 0 :
Now the choice of bj implies
X
1 c log (1 + ) :
j sin bj
Indeed, let k = j ; [2 jH j sin '], so that we have to estimate
c
X
1
k=1 sin bk+[2jH j sin ']
from below. The choice of bj shows that
sin bk+[2jH j sin '] 2k+jH2j
and therefore the last term in (3.6) is greater than
c1
c
X
1
k=1 sin bk+[2jH j sin ']
c2
c
X
1 c log (1 + ) :
3
k=1 k + 2
The case ' = =2 is similar. We x " > 0 so that [0; "] U (0). Next,
we consider the collection of intervals f[aj ; bj ]g with
j + 1 + aj = 2 ; arcsin 2 2jH j ;
j + 3 ; bj = 2 ; arcsin 22jH j
Average decay of Fourier transforms
539
and j = [2 jH j sin (=2 + ") + 1]; : : :; [2 jH j] for some suciently small
> 0. As before on each [aj ; bj ] we have
j cos (2 jH j sin( + '))j 0 > 0 :
Using the fact that
r
; arcsin x = 2 arcsin 1 ; x
2
2
one deduces the estimates
v
v
u
u
1
u 2 jH j ; j ; 3 + u 2 jH j ; j ; ; t
t
2
2
aj ;
b
j
jH j
jH j
and consequently the required result.
Lemma 3.11. Let P be the characteristic function of a compact polygon P in R 2 . For any p > 1
lim sup 1+1=p
!1
Z 2
0
jbP ( )jp d
1=p
> 0:
Proof. We can suppose that one of the sides of P is vertical. We
assume the following facts, which will be proved in the sequel:
(3.7)
there exists k ;! +1 so that jbP (k ; 0)j c ;
k
jrbP ( )j j j c+ 1 :
(3.8)
Next we consider
Z 2
Z "=k
p
jbP (k )j d jbP (k )jp d :
0
0
By choosing " suciently small we can make (k cos ; k sin ) close to
(k ; 0) so that (3.8), (3.7) and the mean value theorem imply
jbP (k )j c1 :
k
540
L. Brandolini, M. Rigoli and G. Travaglini
Hence,
Z 2
Z "=k
p
jbP (k )j d c
;k p d c ;k p;1 :
0
0
We now prove (3.7).
First we recall, see (3.4), that
bP (1; 0) = bh(1) ;
where h(t) is the length of the chord given by the intersection of P
with the line x1 = t. Observe that h(t) is a piecewise linear function,
continuous at any point except at least one of the extremes of the
support. Split
h(t) = b(t) + g(t) ;
where b(t), g(t) and h(t) share the same support, b(t) is linear inside
the support and g(t) is continuous on R . Our choice forces b(t) to be
discontinuous in at least one of the extremes (recall that at least one
side of P is ortogonal to (1; 0)), while g(t) must be piecewise linear.
An explicit computation gives a sequence k ;! +1 such that
b
jb(k )j c ;k 1, while
jgb(1)j c 1 +1 2 :
1
This proves (3.7).
In order to prove (3.8) we observe that, for any unit vector u,
@ b ( ) = @ Z e;2ix dx
@u P
@u P
Z
= ;2i (u x) e;2ix dx
P
= ;2i
Z
R2
((u x) P (x)) e;2ix dx ;
and (3.8) follows since the function x ;! (u x) P (x) has bounded
variation.
The following lemma is taken from [3]. We reproduce the short
proof.
Average decay of Fourier transforms
541
Lemma 3.12. i) Let P be a polygon having an edge not parallel to any
other. Then, if 1 p 1,
Z 2
1=p
p
jbQ ( )j d c ;1;1=p :
0
be the unit square [;1=2; 1=2]2. If 1 < p +1 and if
ii) Let Q
is a positive integer, then
Z 2
0
jbQ (k )jp d
1=p
k
c k;3=2;1=(2p) :
Proof. i) Arguing as in the rst case in the proof of Lemma 3.10 we
are reduced to bounding
1 Z c sin (2 u) p du
2p 0 u from below. A computation ends the proof of this case.
ii) We have
1
Z 2
Z =4 p
sin
(
k
cos
)
sin
(
k
sin
)
d
p
jbQ (k )j d = 8
k cos k sin 0
0
Z =4 sin (k cos ) p d
;
2
p
ck
sin 0
Z =4 sin 2k sin2 p ;p d
;
2
p
(3.9)
ck
2 0
Z k;1=2
Z =4
;
2
p
p
p
;
2
p
ck
k d + c k
;p d
;
1
=
2
0
k
c k;3p=2;1=2 :
2.5.
The forthcoming results will be used in the proof of Proposition
Lemma 3.13. Let 2 p +1 and let s < 1 + 1=p. Then the paverage decay of bB has optimal order s for no convex body B with
piecewise C 1 boundary.
542
L. Brandolini, M. Rigoli and G. Travaglini
Proof. Lemma 3.1 and the theorem on the decay of the Fourier trans-
form of a function of bounded variation imply this lemma when p = 2
and p = 1 respectively. When 2 < p < 1 we have
Z 2
1=p Z 2
1=p
p
2
p
;
2
jbB ( )j d =
jbB ( )j jbB ( )j d
0
0
Z 2
1=p
c
jbB ( )j2 d ;1+2=p
0
c ;1;1=p :
Lemma 3.14. Let P = (s0 ; s0 ) be a given point in the
graph of the
;1
function t = s , with 0 < < 1. Let ' = arctan ( s0 ) be the slope
of the corresponding tangent line and let, for a small positive ,
t = s0 ;1 (s ; s0) + s0 ; cos '
be parallel to the above tangent line, at distance : Here we assume that
this last line and the curve t = s intersect in two points A = (s1 ; s1 )
and B = (s2 ; s2 ) (see Figure 3). We denote by d() the distance between
A and B . Then d0 () is a convex function of .
Figure 3.
Average decay of Fourier transforms
543
Proof. Since d( ) = (s2 ; s1 )= cos ' it is enough to check that the
functions h() = s2 ; s0 and
0 ; s1 + tan '
k() = scos
'
have convex derivatives.
We start with h(). By the denition of the point B we have
s2 = s0 ;1 (s2 ; s0) + s0 ; cos ' ;
that is
(h() + s0 ) = s0 ;1 h() + s0 ; cos ' :
Dierentiating the above with respect to we get
(h() + s0 );1h0 () = s0 ;1 h0 () ; cos1 ' ;
which implies h0 () > 0 since 0 < < 1. Further dierentiations show
that h00 () < 0 and h000 () > 0.
We now turn to k(), which is the distance between the points A
and C in Figure 3. In order to prove that the negative function k00()
increases with we observe that
(3.10)
k00 () = ;K (A) (1 + (k0 ())2)3=2 ;
where K (A) denotes the curvature at the point A. Now it is easy to
check that K (A) decreases as A moves towards O (that is as grows).
On the other hand, by convexity, k0 () decreases too. Therefore, by
(3.10), k00 () increases and this ends the proof of the lemma.
The following result is related to [5, Lemmas 6.2 and 6.3].
Lemma 3.15. Let f : R ;! R + be supported in [;1; 1], such that
f 2 C 1 (R nf1g), f 2 C (R ) , f and f 0 are concave in [b; 1) and f 0 (b) = 0,
f 0(1; ) = ;1. Then, for j j 1,
(3.11)
jfb( )j c
1 f 1 ; 1 :
j j
6 j j
544
L. Brandolini, M. Rigoli and G. Travaglini
The constant c depends only on the supremum of jf (t)j on R and on
the variation of f 0 (t) outside a neighborhood of t = 1.
Proof. We write
fb( ) =
Z1
;1
f (t) e;2it dt
Z1
1
f 0(t) e;2it dt
= 2i ;1
Zb
Z1
1
1
0
;
2
it
= 2i f (t) e
dt + 2i f 0(t) e;2it dt
;1
b
= I1 ( ) + I2( ) :
Since f 0 is of bounded variation on [;1; b] we have jI1( )j c j j;2
where c depends only on the variation of f 0. Morover, f concave on
[b; 1] and f 0 (1;) = ;1 imply
j j;2 = o j j;1f 1 ; 6 1j j
so that
jI1( )j = o j j;1f 1 ; 6 1j j :
To analyze I2 ( ) we proceed as follows: we assume > 0 (the case
< 0 is similar) we write = [ ] + and let = (1 ; 6 )=(6 ) (this
choice will be appreciated later on, while estimating I5( )). Then
Z 1
1
jI2( )j = 2 (;f 0 (t)) e2it dt
b
Z1
= 21 (;f 0 (t)) e2i(t+) dt
b
Z1
1
0
2 (;f (t)) cos (2 (t + ) ) dt
b
= 21 jI3( ) + I4 ( ) + I5 ( )j ;
Average decay of Fourier transforms
where
I3 ( ) =
I4( ) =
Z j0 =(4);
b
4[X
];1 Z (j +1)=(4);
545
(;f 0 (t)) cos (2 (t + ) ) dt ;
4[X
];1
0
(;f (t)) cos (2 (t + ) ) dt =
Aj ;
j =j0
j =j0 j=(4);
Z1
I5( ) =
(;f 0 (t)) cos (2 (t + ) ) dt ;
1;1=(6)
with j0 the smallest even integer such that j0=(4 ) ; b. First we
observe that jI3( )j c= and therefore its contribution is negligible.
We consider I4 ( ) and we show that
I4( ) =
(3.12)
4[X
];1
j =j0
Aj 0 :
Indeed,
i) A4[];1 > 0, A4[];2 < 0, A4[];3 < 0, A4[];4 > 0, A4[];5 > 0,
A4[];6 < 0; : : :
ii) jAj j jAj+1j so that A4[];1 +A4[];2 > 0, A4[];3 +A4[];4 < 0,
A4[];5 + A4[];6 > 0; : : :
iii) jA4[];1 + A4[];2j jA4[];3 + A4[];4j jA4[];5 + A4[];6j
The validity of i) is obvious, while ii) depends on the monotonicity
of f 0 . As for iii) we note that the concavity of f 0 implies
jA4[];1j ; jA4[];3j jA4[];2j ; jA4[];4j By i), ii), iii) it follows that the sum
(Aj + Aj +1 ) + (Aj +2 + Aj +3 ) + + (A4[];2 + A4[];1)
shares the sign of its last term (A4[];2 + A4[];1), thereby proving
(3.12).
Hence,
Z1
I4( )+I5( ) I5( )=
(;f 0(t)) cos (2 (t+) ) dt 12 f 1; 61 ;
1;1=(6)
0
0
0
0
546
L. Brandolini, M. Rigoli and G. Travaglini
since cos (2 (t + ) ) 1=2 on the domain of integration.
Lemma 3.16. For each
1 1 < a < 3o
;
a
:
2
p
1
;
1
+
p
p
2
there exists a convex body B with piecewise C 1 boundary such that the
p-average decay of bB has sharp order a.
p; a 2 T =
n 1 Proof. Let B be a convex body symmetric with respect to the vertical
axis and assume that its boundary @B satises the following conditions.
i) @B passes through the origin and it is of class C 1 in any other
point.
ii) @B coincides with the graph of the function y = jxj in a neighborhood of the origin (the exponent = (p; a) > 2 will be chosen
later).
iii) @B has strictly positive curvature out of the above neighborhood.
We rst prove that jbB ( )j c ;1;1= for any 2 1 . This
bound seems to be quite obvious since j2j;1;1= is the order of decay
of bB (0; 2), that is, the decay associated to the attest point in @B .
However, a proof seems to be necessary (in order to check that the
constant does not depend on ), and the argument will be needed in
the sequel.
Let = + =2. We choose " > 0 suciently small and we assume
" j j ; ". Since @B has strictly positive curvature away from the
origin, by Lemma 3.8 we have,
jbB ( )j c AB (;1; ; 2 + AB ;1; + 2 c ;3=2 ;
for " j j ; ".
Symmetry enables us to consider only the case 0 ". The
assumptions on the curvature of @B show that the contribution of
jAB (;1; + =2)j is not larger than c ;3=2 so that it suces to consider AB (;1; ; =2) (which is a cap close to the origin).
We set more notation. For any 0 ", we consider the straight
line with slope and tangent to the curve y = x at a point (x0; x0 ).
Then AB (;1; ; =2) is the set enclosed between the line y = r(x) =
Average decay of Fourier transforms
547
x0 ;1 (x ; x0) + x0 + ( cos );1 and the curve y = x . Let us call x1
and x2 the abscissae of the two points where they intersect (see Figure
4).
Figure 4.
Since tan = x0 ;1 we have
(3.13)
c1 x0 ;1 c2 :
We further split the interval 0 " into 0 c ;1+1= " for some suitable constant c.
Assume
(3.14)
c ;1+1= and
0 c ;1+1= :
Since is positive, jAB (;1; ; =2)j c ;1 x2 . We recall that x2 is
the largest solution of the equation
x = x0 ;1 (x ; x0 ) + x0 + ( cos );1 :
We now estimate x2. This gives a bound for jAB (;1; ; =2)j since
the assumption 0 yields x2 jx1 j. To do this we observe that
(3.14) implies that the above equation has no solutions for x > k ;1=
548
L. Brandolini, M. Rigoli and G. Travaglini
for k suciently large. Indeed, (3.13) and (3.14) imply x0 c3 ;1=
and therefore
x ; x0 ;1 (x ; x0 ) ; x0 ; ( cos );1
> x ; c4 ;1+1= x ; c4 ;1 ; ( cos );1
> ;1 ((1= x) ; c4 1= x ; c4 ; (cos );1)
> 0;
when 1= x is larger than a suitable k. Then x2 k ;1= and
(3.15) AB ;1; ; 2 c ;1;1= ;
for 0 c ;1+1= :
Next, let c ;1+1= ". Then (3.13) and a suitable choice of the
constant c imply x1 > 0. We want to show that
x ; x0 ;1 (x ; x0 ) ; x0 ; ( cos );1
becomes positive whenever jx ; x0j > c5 ;1=2 x10;=2 . Towards this
aim one checks the inequality
(3.16)
(1 + u) ; 1 ; u 2 u2 ;
which holds true for > 2 and u ;1. Then
x ; x0 ;1 (x ; x0 ) ; x0 ; ( cos );1
= (x0 + (x ; x0 )) ; x0 ;1 (x ; x0) ; x0 ; ( cos );1
= x0 1 + x ;x x0 ; x ;x x0 ; 1 ; ( cos );1
0
0
2
x0 2 x ;x x0 ; ( cos );1
0
2 c25 ;1 ; ( cos );1
> 0;
for a suitably large c5 . Consequently
(3.17)
jx ; x0j c5 ;1=2 x10;=2 ;
Average decay of Fourier transforms
549
for any x1 x x2 . This and (3.13) show that
;
1
(3.18)
AB ; ; c6 ;3=2 (2; )=(2( ;1)) ;
2
for c ;1+1= ". Then (3.15), (3.18), the assumptions on the
curvature of @B and Lemma 3.8 yield
jbB ( )j c7 ;1;1= :
(3.19)
for any .
We now study the estimates of the Lp-norm, 2 p < +1. Because
of the symmetry of B it is enough to bound
Z =2
1=p Z ;=2+c(1;)=
1=p
p
jbB ( )j d jbB ( )jp d
;=2
;=2
Z ;=2+"
1=p
p
+
jb ( )j d
;=2+c(1;)= B
Z =2
1=p
+
jbB ( )jp d
;=2+"
= I1 + I2 + I3 :
By the assumptions on the curvature of @B we have I3 c8 ;3=2.
Furthermore, by (3.19),
I1 c7 ;1;1=
Z c(1;)=
0
d
1=p
c9 ;1;1=p;1=+1=(p) :
In order to estimate I2 we observe that Lemma 3.8, (3.18), the assumptions on the curvature of @B and the choice > 2 give
I2 c10 ;3=2
Z "
p(2; )=(2 ;2) d
c(1;)=
8
>
>
c11 ;3=2 ;
>
>
>
>
>
<
> c11 ;3=2 (log )(;2)=(2;2) ;
>
>
>
>
>
>
: c11 ;1;1=p;1= +1=(p) ;
1=p
for p < 2;;22 ;
for p = 2;;22 ;
for p > 2;;22 :
550
L. Brandolini, M. Rigoli and G. Travaglini
In particular, for p > (2 ; 2)=( ; 2),
(3.20)
Z 2
0
jbB ( )jp d
1=p
c12 ;1;1=p;1=+1=(p) :
Observe that (3.20) cannot be obtained interpolating between L2 and
L1 . Moreover, we shall see in a moment that the above estimates are
sharp and therefore
(3.21)
kbB ( )kLp( )
1
8
>
>
;3=2 ;
>
>
>
>
>
<
> ;3=2(log )(;2)=(2;2) ;
>
>
>
>
>
>
: ;1;1=p;1= +1=(p) ;
for p < 2;;22 ;
for p = 2;;22 ;
for p > 2;;22 :
When p < (2 ; 2)=( ; 2) the estimate from below follows from Lemma
3.6. We shall now prove the estimates from below in (3.21) when p (2 ; 2)=( ; 2). Indeed, (3.17) can be reversed so that, by (3.13),
x 2= (x1; x2) implies
jx ; x0j c13 ;1=2 x10;=2 ;
whence
1
A
B ;
;
c13 ;3=2 (2; )=(2( ;1)) ;
2
for c ;1+1= ". To prove this, we argue as for the estimate from
above, after substituting (3.16) with the inequality
(1 + u) ; 1 ; u 2 2 u2 ;
valid for > 2 and ;1 u 1: The restriction u 1 causes no
troubles since the monotonicity of the curvature of y = x implies
x2 ; x0 x0 ; x1 for c ;1+1= ", so that
;1 x ;x x0 1
0
if x1 x x2.
Average decay of Fourier transforms
551
To estimate bB , let be xed in c ;1+1= " and recall
that = ; =2. By Lemma 3.8 the decay of bB ( ) depends on the
shape of @B at the point (x0; x0 ), see Figure 4, and at the \opposite"
point. The latter will turn out to give a negligible contribution because
of our assumption on the curvature of @B outside the origin. Then, if
2 C01 (R 2 ), (x) = 1 in a neighborhood of the point (x0; x0 ), Lemma
3.14 allows us to apply Lemma 3.15 so to obtain
Z 2
1=p
p
jbB ( )j d
0
Z "
jbB ( )jp d
=
1=p
c;1+1
Z
"
j
[ B ]^ ( )jp d
=
;
c;1+1
Z "
c14
c;1+1
Z
"
1=p
j[(1 ; ) B ]^ ( )jp d
=
1
A
B ;
;
2
c;1+1=
c16 ;1;1=p;1=+1=(p) :
1=p
p 1=p
; c15 ;3=2
d
We recall that the above holds whenever p > (2 ; 2)=( ; 2). This
ends the proof once we observe that when p > (2 ; 2)=( ; 2) we have
(2 p ; 2)=(p ; 2) < < 1 and therefore the range of the exponent
1 + p1 + 1 ; 1p
is the open interval (1 + 1=p; 3=2).
The proof of the previous lemma can be used to get a result for
singular measures supported on the curve y = jxj , > 2.
Proposition 3.17. Let d be the measure on the curve y = jxj , > 2,
induced by the Lebesgue measure on R 2 . Let 2 C01 (R2 ); (t) = 1 in
a neighborhood of the origin and let d = (t) d. Let `(;1; ) be the
552
L. Brandolini, M. Rigoli and G. Travaglini
length of the chord as in Figure 4. Let 1 p 1, then
c ( )kLp( )
k`(;1; )kLp([0;2)) kd
1
8
>
>
;1=2 ;
for
>
>
>
>
>
<
> ;1=2 (log )(;2)=(2;2) ; for
>
>
>
>
>
>
: ;1=p;1= +1=(p) ;
for
p < 2;;22 ;
p = 2;;22 ;
p > 2;;22 :
4. A remark on the average decays associated to arbitrary
convex sets.
Let C be the space of convex bodies in R 2 endowed with the Hausdor metric H dened by
H (C; D) = max sup yinf
jx ; yj; sup xinf
jx ; yj ;
x2C 2D
y2D 2C
for C; D 2 C . A weak version of Blaschke selection theorem (see [9])
shows that (C ; H ) is locally compact and therefore of second category
(not meager) by Baire theorem. We x n 2 N . On C we consider the
functional
Z 2
1=p
p
n (B ) =
jbB (n )j d
and we observe that
0
j kbC (n )kLp( ) ; kbD (n )kLp( ) j kbC (n ) ; bD (n )kLp( )
jC D j ;
1
1
1
implies continuity of n .
Next, let 1 < p < 2 and 3=2 < < 1 + 1=p. Let B be a convex set
with piecewise smooth boundary. The results in the previous section
show that the family fn g satises
n (B ) = o (n; ) ;
when B is a polygon and
n; = o (n(B )) ;
Average decay of Fourier transforms
553
if B is not a polygon nor a cut disc. Therefore, the sets
A1 = fB 2 C : n (B ) = o (n; )g
and
A2 = fB 2 C : n; = o (n(B ))g
are dense in C . A similar argument also applies when p > 2.
We now use the following result due to Gruber, [6].
Lemma 4.1. Let T be a second category topological space.
i) Let 1; 2; 2 R + and let 1 ; 2; : T ;! R + be continuous
functions such that
A = fx 2 T : n (x) = o (n ) as n ;! +1g
is dense in T . Then for all, but a meager subset of x's belonging to T ,
the inequality n (x) < n holds for innitely many n.
ii) Let 1 ; 2; 2 R + and let 1 ; 2; : T ;! R + be continuous
functions such that
B = fx 2 T : n = o ( n (x)) as n ;! +1g
is dense in T . Then for all, but a meager subset of x's belonging to T ,
the inequality n < n (x) holds for innitely many n.
By way of summary we have.
Proposition 4.2. Let 1 < p < 2 and 3=2 < 1 2 < 1 + 1=p or let
2 < p < 1 and 1 + 1=p < 1 2 < 3=2. Then there exists a meager
set E C such that for all B 2 C n E there exist two sequences nk , mk
satisfying
and
Z 2
0
Z 2
0
1=p
n;k 1=p
m;k :
jbB (nk )jp d
jbB (mk )jp d
1
2
554
L. Brandolini, M. Rigoli and G. Travaglini
5. A result on the geometry of convex sets.
At this point, little eort is needed to prove the following result,
which may be of independent interest.
Theorem 5.1. Let AB (; ) be as in (1:6) and let 1=2.
If B is a polygon, then
1 Z 2
2
c1 log AB (; ) d c2 2 log 1 ;
0
while if B is not a polygon
c1 3=2 Z 2
0
AB (; ) d c2 3=2 :
Let 1 < p 1. Then the following are equivalent.
i) There exist a > 0 and a convex body B with C 2 boundary such
that
Z 2
1=p
a
p
c1 AB (; ) d c2 a :
0
ii) The pair (1=p; a) belongs to the set S [ T , where
n 1 o
3
1
S = p ; a : 1 < p < 2; a = 2 or a = 1 + p ;
o
n T = 1p ; a : 2 p 1; 1 + p1 a 23 :
The proof of this theorem is largely a consequence of results in the
previous section. Actually, the present problem is simpler since AB (; )
is positive and no cancellation can arise. We sketch the argument for a
reader specically interested in this result.
Proof. We split the proof into several steps. We assume > 0
suciently small.
Step 1. Upper bound when 1 p 2
Z 2
0
1=2
2
AB (; ) d c 3=2 ;
Average decay of Fourier transforms
555
for any B .
This has been proved by Podkorytov in [13, p. 60].
Step 2. If P is a polygon, then
1 Z 2
1
2
2
c1 log AP (; ) d c2 log ;
0
Z 2
1=p
1+1
=p
c1 AP (; )p d c2 1+1=p ; for 1 < p 1 :
0
These estimates are easy consequences of the argument in Lemma 3.9.
Step 3. Upper bound when 2 p 1
Z 2
0
AB (; )p d
1=p
c2 1+1=p ;
for any B .
The case p = 1 is obvious; the case 2 < p < 1 follows as in
Lemma 3.13.
Step 4. Admissible decays when 2 p 1.
For any 2 p 1 and any 1 + 1=p a 3=2 there exists B such
that
Z 2
1=p
a
c1 AP (; )p d c2 a :
0
This is precisely the content of Lemma 3.16.
Step 5. Lower bound for 1 p 1 when B is not a polygon
Z 2
0
AB (; )p d
1=p
c 3=2 :
Indeed, if B is not a polygon, there exists a regular arc in @B which
does not coincide with its chord. Then, at any point in this arc one
can apply the following elementary observation. Let f 2 C 2 [;1; 1] be
a real function satisfying f (0) = f 0(0) = 0 and 0 f 00 (x) 2 c for
any x 2 [;1;p
1]. Writing f (x) = for x = x1 and x = x2, we have
jx1 ; x2 j 2 =c .
556
L. Brandolini, M. Rigoli and G. Travaglini
6. Lattice points in large convex planar sets.
From the Introduction we recall the following:
Denition 6.1. Let 2 SO(2) and t 2 T2 . The discrepancy function
DB (; ; t) is dened by
DB (; ; t) = card ((;1(B ) ; t) \ Z2) ; 2 jB j
X
=
; (B);t (m) ; 2 jB j :
1
m2Z2
We prove the following result.
Theorem 6.2. Assume B is a convex body in R 2 with piecewise C 1
boundary, which is not a polygon. Let 1 p 2, then
c1 1=2 kDB (; ; )kLp(SO(2)T ) c2 1=2 :
2
Proof. The estimate from above is easy (and essentially known).
Indeed a computation gives
DB (; ; )^(m) = 2 bB ( (m)) ;
for any m 2 Z2, m 6= 0 (please note that the hat symbol in the left
hand side and in the right hand side refer to the Fourier transform on
T2 and on R 2 respectively). Hence, by Lemma 3.1,
Z
Z
SO(2) T
2
jDB (; ; t)j2 dt d = 4
=
Z
X
SO(2) m6=0
Z
4X
jbB ( (m))j2 d
jbB ( (m))j2 d
m6=0 SO(2)
X
4 j mj;3
m6=0
= c:
Therefore, whenever 1 p 2,
kDB (; ; )kLp(SO(2)T ) kDB (; ; )kL (SO(2)T ) c2 1=2 :
2
2
2
Average decay of Fourier transforms
557
On the other hand, for any m 2 Z2, m 6= 0,
kDB (; ; )kLp(SO(2)T ) kDB (; ; )kL (SO(2)T )
2
Z
Z
1
2
jDB (; ; t)j dt d
SO(2) T2
Z
jDB (; ; )^(m)j d
SO(2)
Z
2
=
jb ( (m))j d :
SO(2) B
=
(6.1)
We split the argument for the estimate from below into three cases.
First case. Suppose B is not a cut disc (see Denition 3.3). Then,
making use of Lemma 3.6, (6.1) implies
kDB (; ; )kLp(SO(2)T ) c1 1=2 :
2
Second case. Suppose we have a disc D. First assume
2 ;
min
n2Z
1 ; n 1 :
4 10
Let m = (1; 0); then, by (6.1), and the asymptotic of Bessel functions,
kDD (; ; )kLp(SO(2)T ) J1(2)
2
= ;1 1=2 cos 2 ; 43 + O(1)
c 1=2 :
On the other hand, when
1
min
2 ; ; n n2Z
4
1
10
we choose m = (2; 0), then
kDD (; ; )kLp(SO(2)T ) ;1 1=2 cos 4 ; 34 + O(1) c 1=2 :
2
558
L. Brandolini, M. Rigoli and G. Travaglini
Third case. Suppose B is a cut disc, coming from a given disc D.
Without loss of generality we can assume
f(cos ; sin ) : jj or j ; j g @B ;
for a small > 0. Let cos
sin
U = ; sin cos : jj < 2 :
Then, for m = (1; 0) or m = (2; 0),
kDB (; ; )kLp(SO(2)T )
Z
2
2 jbB ( (m))j d
U
Z
Z
2
2
jbD ( (m))j d ; jbDnB ( (m))j d :
U
U
Now the third case isZ a consequence of the second one if we prove that
jbDnB ( (m))j d c ;2 :
U
Indeed DnB looks like in the following picture and therefore, by applying Lemma 3.8 to each one of the connected components of DnB , we
get
jbDnB ( (m))j c ;2 ;
uniformly in 2 U .
Figure 5.
Average decay of Fourier transforms
559
Acknowledgements. We wish to thank S. Campi and C. Schutt for
some interesting comments. We are grateful to R. Schneider for suggesting us the argument in Section 4. Finally we wish to thank L.
Colzani for several suggestions he gave us during the preparation of the
paper.
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L. Brandolini, M. Rigoli and G. Travaglini
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Recibido: 28 de enero de 1.997
Revisado: 30 de octubre de 1.997
Luca Brandolini, Marco Rigoli and Giancarlo Travaglini
Dipartimento di Matematica
Universita di Milano
Via Saldini 50
20133 Milano, ITALY
[email protected]
[email protected]
[email protected]