A NNALI
DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA
Classe di Scienze
E. B OMBIERI
H. DAVENPORT
Some inequalities involving trigonometrical polynomials
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 3e série, tome 23,
no 2 (1969), p. 223-241
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SOME INEQUALITIES
INVOLVING TRIGONOMETRIOAL POLYNOMIALS
E. BOMBIERI and H. DAVENPORT
1.
Introduction.
integer and let
Let ~Y be a positive
complex numbers. Define
he any real
be any real numbers which
Let Xs ’
satisfy
denotes the difference between 8 and the nearest
positively
positively,
and 0
’
or
integer, taken
! . 20131.
d
In a recent paper
1)
proved
that
and that
the latter
about the
represents
au
improvement
are
of
same
Pervenuto alla Redazione il 96
tbft
(t)
large
Narrn.
Ciiuguo }968,
method,
Landaii,
An
I.
the former if N and 3-1
on
V0P18g
deo
Wissenschaften,
U1Jd
Horlil1 1968.
~ur
224
In the present paper we investigate more deeply the two cases in
which N 6 is either large or small. The factor on the right of (3) is a little
greater than N in the first case, and a little greater than 6-1 in the second case. Our object is to determine the order of magnitude of the term
that must be added to N or ~-1, as the case may be, to ensure the validity of the inequality.
For the case
large, we prove :
‘
‘
THEOREM 1.
then
If
On the other hand, if c is a constant less than 1 there exist 8utns S (x)
with 6 arbitrarily 8mall and N 6 arbitrarily large for which
r
features in the proof of (4), as compared with the
previous paper. The first of these is a maximization
argument (Lemma 1), which has the effect of allowing us to limit ourselves
to sums S (x) in which the
I have a measure of approximate
of the function 0, (t), defined in
The
feature
the
use
second
is
equality.
in
of
the
simpler function 4Y (t) of our previous paper.
place
(13),
For the
small, we prove:
There are two
proof of (3) in our
THEOREM 2.
’
1
1 p then
N 6 G
IIff N
On the otlte1’ hand there exist
sums
..
(x) with N 6 arbitrarily 81nall for
whic4
This case represents a problem which is entirely different from that
of the first case; it has much in common with the problem of approximating to a Riemann integral by a finite sum. The arguments used in the
proof of (6) are quite delicate.
225
The present paper is self-contained, except for the general inequality
(27), due to Davenport and Halberstam, which is needed for the proof of (4).
2. The ikiaxiiiiizatioii
Let 0
G (N) by
argument.
be fixed. For each
...,
positive integer N,
define
n
and the maximum is taken
over
all
complex numbers at... , aN satisfying
The maxiinum is obviously attained, and
which it is attained a maximal set.
for
an
LEMMA 1.
Tlten, for
we
set
a
whereI 9 !I,
Proof.
N is such that G
Suppose
1 and G
Write
e
(0)
=
=
of coefficients,
G
(N) ;
e2ni9,
(N) ~
we
and this
we
call
a
set of coefficients
0 and
have
holds
all
M,
H
satisfying
and
where A 7~ &#x3E; 0 and a~~ is real. For a maximal set of coefficients we have a
value
Hence we must have
subject
stationary
226
for
some
we see
A.
Writing
---
that the conditions
If this is
Ss,
and
noting
tbat
are
multiplied by A,~
and summed for
m
= 1, ... , N,
it
gives
whence
If however the
sum
is restricted to
in
= M + 1,
... ,
M
+ H,
we
get
where
Hence
Now
Also, by
Hence
the definition of G
(H)
and the
hypothesis (11),
we
have
227
that
is,
N
This is
on
the
hypothesis
that
NE I a, 12 = 1.
Piainly
that
1
can
be omitted if
We
can
we
obtain
modify
a
the
inequality
so
that it reads
complementary inequality by applying this
sums over
and
and
subtracting
The two
3. A
from the
complete
inequalities together
sum.
prove
particular Fourier series.
LEMMA 2. Let
Suppose
that 1
hypothesis
) ~I exI +
1. Then
We obtain
(12).
to the
228
Proof. (14)
To prove
is almost
(15)
we
start from the relations
we
if A
= n (1 + a -1 )
where
immediate, for since sin 03C0t
have
Hence
nt we have
229
To estilnate .I (x)
we
; in the complex plane,
2
contribution of the
We
the line of
rotate
so
quadrant
integration through
that it becomes the line
at
infinity
an
1 + iu, u &#x3E;
angle
0. The
vanishes.
get
Hence
Putting this,
for
that If
3.
teger. There exists
1l,ith real
in
a
it.
(16),
ive
obtain (I
&#x3E; 2. Let
satisfies
No
be any
positive
in-
Fourier series
b-,,
coefficients bn
=
bn,
such that
and, for
P,’oof.
We define A
=
(No -~- K) ð,
and
R~e
define 1p (x)
forx I
c ~ by
230
We define y (x) for other real x hy periodicity with period l. Then
11’ (.v) is an even function of x and satisfies (18). The Fourier coefficients bn
of 1f’ (.1’) are given by
where
a
By
~z~.
Parseval’s
=
formula,
Hence, by (14) of Lemma 2,
It remains
By (15)
only
ca
prove
(20).
Forn ~
1V’, ,
we
have
of Lemma 2,
Now
[ 1,
and for 0 (y ;
2013
46
we
bave
have
Hence
Since 4.2 03C0-3
1/77
this
gives (20),
on
recalling
that
231
4.
Proof of the first
part
of Theorem 1.
We observe first tha,t there is no loss of generality in taking M = 0
in the definition
in (1), for we can reduce the general case to this
= 3i
we can take S (.r) to be defined by (9).
Thus
by hutting n
-~For fixed 6 and fixed xl ,,...,.r~let N be the least positive integer
for which G (N), defined in § 2, satisfies
if there is
no
such integer the desired conclusion holds. For this N, the
of Lemma 1 is satisfied. For a maximal set of coefficients,
of course we also have
hypothesis (11)
(12) holds ; and
that N is
Suppose first
odd, say N
=
2NO + 1.
We define
and have
(24)
where
By (12),
if
that N is even, say N = 2No .
above in (23) for 2013
and put aNo = 0.
is still valid, with the same definition (25). Also (26) is still va-
Suppose next
We define a’n
Then
(24)
lid, provided 0
Let 1p (x) be
of Lemma 3. It
13
as
any
was
(2) « The values
(1966), 91-96, and
of
14
function of
period 1 satisfying the condition (18)
proved by Davenport and Halberstam (2) that
even
trigonometrical polynomial
(1967), 229-232.
a
at well
spaced points
Mathematika
232
With the
of Lemma
particular
3, this gives
where
In view of (22)
we can
write the result
as
where
Applying partial summation,
Since
the inner
0, vca
the right, This gives
c,n
sum on
&#x3E; 0
and using the fact that C-n
can
apply
Since
+ 1 ~ ~, the last term in the bracket
Substitution in (28) gives
the
can
we
=
obtain
inequality (26)
be omitted.
where
By
the
inequality
of the arithmetic and
geometric
means,
we
have
in
233
-I 2 N
and since
this
implies
Since the function Cn, for
second
derivative,
We
now
we
take If
that
a
continuous
variable n,
has
a
positive
have
~
26-1 . and have
Also
Hence
From
(30)
we now
ohtain
which gives a contradiction to (21). This contradiction proves the first part
of Theorem 1.
VVe have not used the hypothesis that
1, but the result (4)
1.
becomes of little value if Nb
234
5. Proof of the second
We give a simple
large, for which
part of Theorem
example,
with 3
1.
arbitrarily
small and N3
arbitrarily
-
This suffices for (5), since ð-l - 1 &#x3E;
Let h and L be arbitrarily large
For this sum,
Take 6
regarded
as a case
=1/(2h + 1),
At each of these
of
if 6 is
sufficiently small.
positive integers, and take
(1),
and take the
points
we
have
we
have
points
S (x)
=
x, to be
2Z
+ 1,
and therefore
Also
This proves
(31).
6. Lemmas for Theorem 2.
As observed at the
that
S (x)
is defined
beginning of § 4, we
by (9). We suppose that
can
take M = 0 in
(1),
so
235
by
We note that in every
term 16 (111 - n)I
!~ ~ -.- .
LEMMA 4. lve have
Since
we
have
The result
now
5. Let
zafcere the Ck
ai,e
follows from
F (z) =S (x) 12. Then
= 2-,
and, J*o)constants, and c
positive
T4
For any 9
withI 9 I 2~,
we
have
0
0
2n,
236
where the Ck
ting
0
=
2n 6
positive
are
(itt
-
constants.
1
Also
it) and substituting
in
24
(33),
we
from
26
00 -2
03A3n-1
=Put1
obtain
Now
These two
equations yield (35).
LEMMA G. For any
Let T
(z)
=
positive integer k,
X
(z),
so
that F (z)
=
S
(z) ~’ (z), By Leibniz’s formula,
By Cauchy’s inequality,
Now
whence
There is
a
similar result for the
integral containing
T.
Finally
237
.for
Let E denote the set
LEMMA 7.
1vlticll
of
real
considered modulo 1,
Then
Lemma 4 the left hand side is
1’roof. By
and
by (2)
the intervals ot’
the latter being
terms with ~o =
a
n
integration
are
disjoint. We have
consequence of the definition of 0 (z) in (33), since the
in the double sum have coethcients 0. It follows that
where ~’ (z) = ( S (z) ~2 as before, and where ~’ denotes the complement of
the set of intervals (xr
!/2, xr + 8/2), The integral in the last expression
can only be increased if ve replace E’ by E, since jE7 comprises all z for
which the integrand is positive.
-
7, Proof of the Crst
part of Theorem 2.
represent numbers z by points
on the circumference of a circle
of
7
set
~’
the
Lemma
consists of a finite number of open
perimeter 1,
intervals. Each interval has length less than 6, for by Lemma 4 it is impossible for (38) to hold throughout any interval of length !.
We divide the intervals of E into two types. The first type are those
for which F’ (z) does not vanish in the interval, and we denote the union
of these by Et The second type are those for which ~" (z) vanishes at some
point of the (open) interval, and we denote the union of these by .E~.
If
we
of
8
238
Let I be
points of I,
of the intervals of
have
one
we
for any real number y. Since F’
in I so that
V’e
now
(z)
E1.
Since 03A6 (z) - F (z) = 0 at the end
is of constant
sign,
we
can
choose y
have
Hence
By lemmas
We
now
5 and
6,
the
right
turn to the set
By Lemmas 5
and
6, this
hand side is
Since
is
F (z) ~ 0 always,
we
have
239
It remains to estimate the
of only one or two
consists
~;4
less than 6. Thus in this case
integral over E2 of F’’ (z). Suppose first that
intervals; as noted earlier, each has length
Now
and therefore
Ilence7
in the
present case,
Now suppose that
.~’2
consists of at least 3
intervals,
~1), ···
say
... ,
By
the definition of
E2 )
each interval
(aj, Pj) contains some $;
By Rolle’s theorem, there exists some qj between 03BEj and
We make the obvious convention that
We have
pj
Both qj-l and
6,
- (Xj
2. Annali della Scuola Norm.
z are
Sup..
$s+1 = ~1’
contained in the interval
Pi8a.
and
for which
for which
so on.
(f~-i,
Hence,
since
240
The intervals
Hence
(~1-1, $j+,)
cover
Lemma- 6. Comparing this witit
the two cases.
It follows from (41) that
by
Adding to this the estimate
stituting in (39), we obtain
where
1
24
Since c1=
el
By (36) with 0
NG
f’or N6
Hence, for
This proves
(6).
this
=
implies
n, we have
I
41
in
the whole interval of
(42),
(40)
we see
for the
that
(43)
integral
length
1 twice.
is valid in both
over
and sub-
241
8. Proof of the second
We take N =
so
that (x) ~2
=
V1’e take the
2,
part of Theorem 2.
and
4 cos2
We have
points xr
at
where
and 0
1 . Note
~
1 - 2 1n ð &#x3E; ð.
that the
gap
(mod 1) between m 6
sin
(2m+ 1);76=sinnC6.
We have
Now (2~a
H ence
-E- 1) b = 1 - C ð,
so
where
For
and
on
fixed (,
taking ~
Since -
as 6
--&#x3E;
= IjV3
0,
we
we
have
get
this example satisfies
(7).
and - m6 is