Ife'"Xd~(z) l I e"Xdz By

A R K I V F O R M A T E M A T I K B a n d 4 n r 15

Communicated 10 February 1960 by O. FROSTMAN and L. CARLESON

O n linear d e p e n d e n c e in closed sets By

I N G E M A R W I K

1. K r o n e c k e r ' s t h e o r e m can be f o r m u l a t e d as follows: A n e c e s s a r y a n d sufficient c o n d i t i o n for 22 - 21, 43 - 21 . . . h e - 21 to be l i n e a r l y i n d e p e n d e n t (rood 2 z ) is

k k

inf s u p l ~ a , e ' % l = ~ [ a , [ .

a v n > . O

T h r o u g h o u t the p a p e r n r e p r e s e n t s integers F o r a n a r b i t r a r y closed set on (0,2~) we i n d e p e n d e n c e :

(a) Pc

(E) = inf sup

,ueU* n~>O

(b) P ~ ( E ) = i n f sup

pe 1~* t>~O

(c) Pn

(E) = inf sup

peP* n

(d) P ~ (E) = inf sup

,u~F* t

where r ~ is t h e class of functions ~u which

a n d t a r b i t r a r y real numbers.

define t h e following indices of l i n e a r

If e'"x dr (x) I

I] e"Xdz (zll Ife'"Xd~(z) l

E

a r e c o n s t a n t o u t s i d e E a n d s a t i s f y

fld~,l=l.

E

A n i m m e d i a t e consequence of t h e definitions is:

O<.Pc<- ( p n < ~ P ~ l .

I n K r o n e c k e r ' s t h e o r e m t h e c o n d i t i o n n~> 0 can be c h a n g e d to t~> 0, - c ~ <

< n < ~ or - c~ < t < oo.

I f E consists of a finite n u m b e r of l i n e a r l y i n d e p e n d e n t p o i n t s we t h u s h a v e

Pc = P~ = Pn= P * =

1. W e shall p r o v e t h a t this p r o p e r t y holds for all sets E for which

P*H(E)=I

(Theorem 1) a n d t h a t P ~ ( E ) > 0 implies P c ( E ) > 0 (Theo- r e m 2).

A set E is called a

Kronecker set if Pc

( E ) = 1 a n d a

weak Kronecker set

if

Pc (B) > O.

13:'2 2 0 9

An example of a perfect Kronecker set of the Cantor t y p e was given in R u d i n [4]. Theorem 2 provides a positive answer to the question, raised b y Kahane-Salem in [3], concerning the equivalence of Carleson-sets and Helson-sets.

B o t h notions are equivalent to weak Kronecker sets.

The following theorem b y Carleson [1] is f u n d a m e n t a l for the proof of Theo- rem 2:

Theorem A. A necessary and sufficient condition that PH (E)> 0 is that for each

> 0 every continuous function q~ (x) has a representation

r

r (x)= ~ a,e ~=, x f E,

where l a" l "< G + 1

2. To prove Theorem 1 below we use

L e m m a 1. I f P * ( E ) = 1 we have for all IX f F ~ lira Ij eUZdix (x) l = 1.

I t l ~ o r

Proof. Suppose t h a t the lemma is false. Then there exists a function IX f i f ~ so t h a t for t = t o we have

f e'"= dix (~)= d~..

E

IX cannot be a step function. I n t h a t case the lemma follows immediately from Kroneeker's theorem. W i t h o u t loss of generality we p u t ~ 0 = 0 and we have d#=e-tt~ dp(x)>~O.

Choose ~ r E , a point of continuity of IX, so t h a t e~t*~# 1 and form d Ixx = e-tt'x d p (x) + d (~ ( x - ~). [d} (x) = 0 for x < 0, = 1 for x >~ 0].

There is no v satisfying

Ife'=d ,ll=2.

E

Thus lim [f e 't~ dixl (x) l = 2 a n d lira I f ei t~dl x (x) l = 1 which is a contradiction t h a t

Itt ~ * r ~ ltl ~ r E

proves L e m m a 1.

L e m m a 2. I f P*n(E)= 1 ue have ]or all IX f F ~

~m Ifen~dix(~)l= 1.

I n l ~

Proof. The same as for L e m m a 1 if t is changed to n.

Cor. 1. If P*H(E)=I we have for all IXEI "~

lim fie"=- llldix[=O.

t - - ~ E

210

ARKIV FOR MATEMATIK. B d 4 nr 15

Proo/.

Choose an a r b i t r a r y e > 0. B y L e m m a 1 there is a sequence {t,}i ~ so t h a t

[ f e % ~ + ~ d # - l l < ~ , r>~v o.

E

P u t d/z2 =

e%'. z +i~ d be = d tt~ + i d #~'.

t d H ~2

Then

^{w e}

have f(Id~,t-dm)<~ ~ and fl

~'~ < - - -

E E 6

B y the triangle inequality we obtain for v >~ ro

e 2

6 g

whence

E E E

/32 E2

f 1 - - c o s

( ' . - t , o ) X l d l ~ l < ~ + ~ + . . . .

E 6 2

a n d b y Schwarz's inequality

9 s i n I !_t~ - t.0) x

f i e,(t _,,~ 11 [ d/~ i = 2~

(t,~t~~

i d # l ~ < 2 ~ sin 2 2 dja]

fie~(e.-t.o)X-ll[djal<e

for v~>r o a n d Cor. 1 is shown.

Cor.

2. I[ PH(E)= 1 we have /or all # E F ~

lira

fl~'"=- llld/,[=o.

n---~oO E

Proo].

The same as for Cor, 1 if t is changed to n and L e m m a 2 is used instead of L e m m a 1.

Cor. 3.

Ph,(E)= 1 implies P c ( E ) = 1

Proo[.

Suppose t h a t P c ( E ) < I . Then there exists a function # E F ~ with

I / e ' n X d / a ( x ) l < r < l , n>~O.

According to L e m m a 2 we choose n o so t h a t

E

l + r

B

and b y Cor. 2 there exists n x with the properties: n I + n o > 0 a n d I - r

f I~'"'~- Illet'l< 2

E

T h e triangle i n e q u a l i t y gives

r>l f e'<~'+~"~d:, (x) I > I f e'~'~d# (~) I- f I e'(~'+~')~- e'~'~ I I d. (x) I >

E E E

l + r 1 - r

2 2 ^{r .}

This c o n t r a d i c t i o n proves Cor. 3.

L e m m a 3. From a set o/ rational numbers with denominator m, it is su//icient to remove m - 1 to get an integer as the s u m o / t h e others.

Proo/. E n u m e r a t e t h e n u m b e r s I P ' ! N a n d from t h e sums ( m J1

~ P ' = n ~ + a k ~ k = 1,2 . . . m.

1 m m

A t least two of t h e gk:s are identical, e.g. ~k, a n d ~ , . T h u s k, p~

These P , are r e m o v e d a n d t h e o t h e r s are e n u m e r a t e d a g a i n from 1 t o N - (k I - k2).

W e o b t a i n a similar sum:

k ' t + l lf~

A n a l o g o u s l y we can continue u n t i l th~ n u m b e r of t e r m s is less t h a n m - 1. The sum of t h e o t h e r s is t h e n a n integer, which was t o b e p r o v e d .

T h e o r e m 1. I / P ~ (E) = 1, then E is a Kronecker set.

Proo/: Suppose t h a t P * ( E ) = 1 a n d P x ( E ) < 1. T h e n t h e r e e x i s t s a f u n c t i o n /z E F ~ satisfying

lfe'"=d/~(~)l

< P I , all n, P l < 1. (1)

E

a n d P u t

F o r j>~ Jo we h a v e

B y L e m m a 1 t h e r e is a sequence (tj}~ w i t h t h e properties:

E

t, - [t,]-+~, [t,] = n,.

~ = 1 l a n d m = ~ + 1 .

l - If ⁽²⁾

212

ARKIV FOR MATEMATIK. B d 4 nr 15 /~ is c o n t i n u o u s on E c a n d d i s c o n t i n u o u s on E s. E = E c U E s a n d f [ d ~ [ = a .

E

a ~= 0 for otherwise t h e t h e o r e m follows d i r e c t l y from K r o n e c k e r ' s t h e o r e m . Before p r o v i n g the t h e o r e m we m a k e t h e following s t a t e m e n t : To each n u m b e r fl a n d each ~p t h e r e exist a n i n t e g e r N~ a n d a c o n s t a n t ~@ w i t h t h e p r o p e r t i e s

f I e ' " ~ + ~ . - ?~",~11dgl < 2 ~

a n d

f I?"~+~,- 11 Idol < ~.

E,.g

To show t h i s we d i v i d e E c i n t o d i s j o i n t subsets E . U E , = E c, r e s t r i c t e d b y t h e c o n d i t i o n

a(~ .< fl a(~

9 ~

W e c o n s t r u c t new f u n c t i o n s / ~ of b o u n d e d v a r i a t i o n on E b y choosing

d~t, = ] d g (x) l otherwise.

f l d ~ , l = l a n d t h u s b y L e m m a I t h e r e exists a sequence t'k a n d a c o n s t a n t ~0,

N

so t h a t

f

ett~z+~% dl~,-->l (3)

E

a n d t" - k r~.'. 1 L v ~ j ~ a , rt" ~ L kj - - n" k.

H o w e v e r , ~ is a r a t i o n a l n u m b e r w i t h d e n o m i n a t o r less t h a n m. F o r s u p p o s e t h a t t h e c o n t r a r y is true. T h e n t h e r e are two integers ql a n d qu satisfying

q l ~ + z c - q 2 = h , where [ h i < 1 . m F o r k>~k o we h a v e according t o (3) a n d t h e definition of / ~

v _ I s,

a n d thus, i f q l / I , k -~- q 2 - k ,

w h e n c e

E - E p E - E v

m l E

This gives

If e'~'x e'~ d ~ (~) - f e'('~+ ~' ,~+'~'~, d ~ (~)1 < 4 ~.

E E

F o r 7">~?'0 we have b y (2)

] f ei('~ +'q)~ d,u ] > 1 - 5 ~ > P1

E

contrary to (1), a n d we have proved t h a t all ~ are rational numbers with de- n o m i n a t o r m !.

a52

We p u t e l = 4 m !. If k ~ k 1 we have b y (3) for all v

< 1

E v E - E ~

Applying Schwarz's inequality as in the proof of Cor. 1 we get:

Ev

and ^{E 1}

E - E v

Since all ~ are rational numbers with d e n o m i n a t o r m! we have by L e m m a 3

~ = N, where the s u m m a t i o n runs over all v except at m o s t m!. The sets E , corresponding to these indices are removed a n d we get a new set E c'.

m ! a ~

f ]d~l<~q-.t =a~.

E C _ EC"

B y the triangle inequality

EC, I' E v

< 2 r a ! . ~ , ~ < 8 for k > k 1,

a ~ 2

P u t ~ ' n~, = Nk, a n d ~ ' ~o, = Ta.

N o w we eonclude t h a t

EC

The triangle inequality also gives 214

ARKIV F6R MATEMATIK. B d 4 nr 15

ES

P u t N + N ~ , = 2Va and our statement is proved.

Returning to the proof of the theorem we assume t h a t / z is continuous except at the points 2s. 2 s - 2 1 are linearly independent since P* ( E ) = 1. There are by Kronecker's theorem an integer M and a constant W so that

fle-'~'~+'~-~'~lld~,l<,~.

ES;

According to the preeeeding statement there exist an integer NM and a constant

~M with the properties

EC

and

fl,'~+'~,-llld~,l<O.

ES

By the triangle inequality

EC

E$

(4)

There also exist an integer N~ and a constant q0~ so t h a t

(4) and (5) imply

EC

ES

E

P u t N ~ + N M - M = N ' and qJ~q-q)Mq-~p=q~ '.

For ]~>?o we have b y (1) and (2)

(5)

E E

whence 8~ > 1 - P 1 , which is a contradiction.

Thus we have proved t h a t P * ( E ) = 1 implies P H ( E ) = 1 but b y Cor. 3 this implies P c ( E ) = 1 and E is a Kronecker set which was to be proved.

T h e o r e m 2. I / P ~ ( E ) > 0, then E is a weak Kronecker set.

215

Proo/:

Carleson has shown in [1] that P~ (E) > 0 implies

PH

(E) > 0. I t is therefore sufficient to prove that

PH(E)>

0 implies Pc (E)> 0. We give an indirect proof and assume that Pc (E)= 0 while

PH(E)> O.

Choose 0 < ~ < Pn. Since Pc = 0 there exists a function /~ 6 F ~ satisfying

:

ePH

By the Radon-Nikodym theorem we obtain

E E E

where [/0 (x)[= 1 and [0 is measurable (#).

Then we can approximate /o (x) with a continuous function ~ (x), [~ (x)[ ~< 1 in the sense that

f [ ~ ( x ) - / o ( X ) [ [ d r ( x ) [ < i "

E

Since

PH > O, ~ (x)

can be represented

oo

cp(x)=~a'e'~'-oo

where _~or ~ ~ + e . 1

Choose N so that ~ a, < e I,l>~ ~. By the triangle inequality we obtain for

n>~N

E E E

N " 8

<4 +'fe'nX~-a'e"~dl~(X)'+l.~N]a~'<i+ 2+ 4

i.e. Iftnxd~(x)l<~ I ~ 1 ~ . (6>

E

If e *nz d/~ (x)] assumes its greatest value, which is called P0, 1 >~ Po >/P~, (7)

E

when

n=no,

]noi~<N.

Hence we have

[! e'"~:(l+eu'~)d~(x)]-

{ ~ P ~ + e' In] < 3 N < 2 e ,

In[>~ 3N (8)

We form the functions % (x) and az(x).

z x

% (x) = ] (1 + e u'Nx) d/~ (x), ~2 (x) = f (1

- e 2'~z) d/~ (x),

0 0

(9)

216

ARKIV FOR MATEMATIK, Bd 4 nr 15 Choose ~ = ~ ( P o - e) and suppose t h a t

E E

and This gives

fld~:l = f l l -e2'~: I Id~ (x)] < 1 +~

E E

E

whence

f 211 + e~'~l [ 1 -e~'~ I I d/x (x) I I1 + ~ ' ~ 1 + 1 1 - ~ ' ~ I +2

E

< 2 ~

and But by (6)

fll-e"~lJd#(x) l<2(l + V~)O<5o.

E

E E E

This is a contradiction and thus

m a x [ [ d ~ ] ~ > l + P ~ - - e ' ' , = l . z

Suppose that the maximum is assumed by ~1.

The function

z

0

s

then belongs to F ~ and by (8)

E

(7')

l fetnXdl~1(x)[

assumes its greatest value, which is called P1 for

n = n 1,

E

n 1 ~< 3 N , P I < - P o + e l + P o - e "

5

In the same way as (8) and (9) were formed from (6) and (7) we now form (8') and (9') and obtain a new function p z ( x ) E r ~ w i t h the property

where

~ <P2, [ n [ < 3 z ' N , = P z for n = n 2

P l + 2 e P~

1-~ P 1 - 2 e "

5 F r o m /~ (x) we construct /z a (x) etc.

We get functions /~k(x)EF ~ satisfying [ / e lnz d/zk (x)[~< Pk, where Pk ~<

g

P k - 1 + 2 k-1 "

P k - 1 - - 2 ~-1" s "

I +

5 a k - 1

However, the sequence given b y % = 1, ak tends to zero. F u r t h e r ak-1

1 + - - 5

when e---~O.

w o o oo o soo

If we start with ~ = e 0 we obtain a function /zk. satisfying s u p I f c n x e ~ . ( x l l < - ~

2P~

- .

n

This contradicts our assumption and the theorem is proved.

R E F E R E N C E S

1. CARLESOI~, L,: Representations of continuous functions. Math. Zeitschr. Bd. 66, pp. 447-451 (1957).

2. - - , Sets of u n i q u e n e s s for f u n c t i o n s regular in t h e u n i t circle. A c t a Math. 87, pp. 325-345 (1952).

3. KAHANE, J.-P.--SALEM, R.: Construction de pseudomeasures sur los ensembles parfaits sym6- triques. C. R. Acad. Sci. Paris 243. pp. 1986-1988 (1956).

4. RUDI~, W.: The role of perfect sets in harmonic analysis. S y m p o s i u m on harmonic analysis.

CorneU University. J u l y 23-27 (1956).

2 1 8

Tryckt den 10 december 1960

Uppsala 1960. Almqvist & Wiksells Boktryckeri AB