bochner’s theorem on the fourier transform on r

7/23/2019 Bochners Theorem on the Fourier Transform on R

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Bochners Theorem on the Fourier Transform on R

Yitao Lei

October 2013

1 Introduction

Typically, the Fourier transformation sends suitable functions on R to functions on R.This can be defined on the space L1(R) +L2(R), i.e. functions which can be writtenas the sum of a function in L1(R) and a function in L2(R). A celebrated result (theHausdorffYoung inequality) states that the Fourier transform takes functions in Lp(R)to Lq(R) for 1 p 2, where 1

p+ 1

q = 1.

However, this does not extend to the case whenp >2. In addition, the maps F :Lp(R)Lq(R) are not surjective when p < 2. Therefore it seems natural to try to extend theFourier transform to ob jects other than functions. The most complete method of doingthis is extending the Fourier transform to the space of tempered distributions, i.e. thespace of linear functionals on the Schwartz functions.

Instead, we will study the FourierStieltjes transform, a slight generalisation of theFourier transform. We now transform complex finite Borel measures rather than func-tions, and output a function. Bochners Theorem answers the question of which functions are the FourierStieltjes transform of some positive Borel measure. It states that thefunction is continuous and positivedefinite is a necessary and sufficient condition for itto be a FourierStieltjes transform.

We shall first explore the analogous situation on the torus (or the circle here, when thedimension is one). FourierStieltjes coefficients will be examined, and are related toFourier coefficients. There is much similarity between FourierStieltjes coefficients and

the FourierStieltjes transform. However, the theory building up to a Bochnertyperesult on the torus is clearer and simpler than going directly to Bochners theorem on R.

2 Preliminaries

Let T be the torus [0, 2] with the points 0 and 2 identified, i.e. the same point. In thispaper, we shall be working with two spaces of continuous functions, both equipped withthe supremum norm: C(T), continuous functions on T, and C0(R), continuous functionsthat vanish at infinity.

We will also consider two spaces of finite complex Borel measures on T and R, namelyM(T) and M(R). For clarity, will denote a measure in M(T), while will denote

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a measure in M(R). The norm on both spaces are given by M(T) = ||(T) andM(R)=||(R).

Recall that ifVis a Banach space, then the dual spaceV is the set of linear functionals : V C which are continuous/bounded. The following theorem is very useful in

multiple ways; the proof can be found in [3].

Theorem 1 (Riesz representation theorem). (1) Any linear functional(C(T)) canbe identified with a unique measure M(T) such that (f) = 1

2

T

f(t) d(t). Inaddition, C(T) = M(T).

(2) Any linear functional (C0(R)) can be identified with a unique measureM(R)

such that(g) =R

g(x) d(x). Furthermore, C(R) = M(R).

Recall that the Fourier series on T of an integrable function f is given by f(n) =

T f(t)eint dt. Trigonometric polynomialsare functions of the form P(t) =

N

n=Naneint.

It can be easily verified thatP(t) =Nn=N P(n)eint. A basic result in Fourier analysis isthat the partial sums

Nn=N f(n)e

int do not necessarily converge to the function itself.Nevertheless, we have convergence in a related series. A proof of this statement can befound in[2].

Lemma 2 (Fejer). For positive integerN, define theFejer kernelby

FN(x) = 1

N+ 1

sin2[(N+ 1)x/2]

sin2[x/2] .

Then for a continuous functionf onT, we have the following convolution:

(FN f)(t) =N

n=N

1

|n|

N+ 1

f(n)eint.

Furthermore, (FN f)(t) converges uniformly to f asN .

As the convergents in the previous lemma are trigonometric polynomials, any functioncan be approximated by trigonometric polynomials. Hence trigonometric polynomialsare dense in C(T).

3 FourierStieltjes Coefficients

FourierStieltjes coefficients are an extension of Fourier coefficients, defined with measuresin M(T):

Definition 3. The FourierStieltjes coefficientsof a measure M(T) is a functionon Zgiven by the expression

(n) =

1

2 T eint d(t).2


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Note that if we have an integrable function f, we can identify it with a measure d =f(t) dt (dt represents the usual Lebesgue measure). Computing the FourierStieltjescoefficients gives (n) = 1

2

T

eint d = 12

T

eintf(t) dt = f(n), so we really do havean extension of Fourier coefficients.

For Fourier transforms of functions in L2

(R), an important result is Parsevals formula:R

f(x)g(x) dx= 12R

f()g() d. An analogue holds in the context of FourierStieltjescoefficients in the following sense:

Proposition 4 (Parsevals Formula). If M(T) andfC(T), then

1

2

T

f(t)d(t) = limN

Nn=N

1

|n|

N+ 1

f(n)(n).

Proof. Suppose fis any continuous function. Then by using lemma2,we can approximate

f uniformly as limN1 |n|N+1 f(n)eint. As is a finite measure, we can exchange thelimit and integral to yield

T

f(t)d(t) = limN

T

Nn=N

1

|n|

N+ 1

f(n)eint

d(t)

= limN

Nn=N

1

|n|

N+ 1

f(n)

T

eint d(t)

= limN

N

n=N1

|n|

N+ 1f(n)(n).

We remark that without the

1 |n|N+1

factor, the right hand side may not necessarily

converge.

Bochners theorem is about trying to determine which sequences are the FourierStieltjescoefficients of a measure. As a first step, we give a necessary and sufficient condition fora sequence to the FourierStieltjes coefficients of a measure :

Proposition 5. Let {an}n= be a sequence of complex numbers. Then the following

are equivalent:

(a) There exists M(T) with C and(n) =an for alln.

(b) For all trigonometric polynomialsP, P(n)an Csup

tTP(t).

Proof. (a) =(b): Given any trigonometric polynomial P, use Parsevals formula to get

N

n=NP(n)an

=

T

P(t) d

sup

tTP(t)

where we used the Riesz representation theorem to show that |(f)| fC(T) = suptT f(x).

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(b) = (a): The linear map P P(n)an defines a linear functional on the space

of trigonometric polynomials. As trigonometric polynomials are dense in C(T), we canextend our linear functional to C(T).

By the Riesz representation theorem our linear functional is of the form f 12 T

f d

for a suitable measure M(T) of norm C. Substituting f = eint

into both P P(n)an and f 12 T f dimmediately gives (n) =an or (n) =an.Consider a sequence {an}nZ which eventually vanishes, i.e. an= 0 when |n|> K. Thenit can be checked that the measure given by dN :=

Kn=Kane

int dtsatisfiesN(n) =anfor all n. This allows us to make sense of the following corollary. The proof involvesalgebraic manipulations and Parsevals formula only, and hence will not be presentedhere.

Corollary 6. A sequence {an}nZ is the FourierStieltjes coefficients of some with C, if, and only if, NM(T) C for allN. Here, N is the measure such thatN(n) = (1 |n|N+1)an whenever |n| N, and zero when |n|> N.4 Hergoltzs Theorem

Hergoltzs theorem is the analogue of Bochners theorem on the torus, as in it givesnecessary and sufficient conditions for a sequence to be the FourierStieltjes coefficientsof a positive measure. To prove this, we first need the following lemma:

Lemma 7. A sequence{an}n

Z

is the FourierStieltjes series of a positive measure if,and only if, for allN andt T,

N(t) :=N

n=N

1

|n|

N+ 1

ane

int 0.

Proof. First suppose (n) = an for some positive measure M(T). Let f C(T) bean arbitrary non-negative function. Then

1

2 T f(t)N(t) dt= 12 T f(t) N

n=N1 |n|N+ 1 aneint dt=

Nn=N

1

|n|

N+ 1

f(n)(n)

=N

n=N

1

|n|

N+ 1

f(n)

T

eint d(t)

=

T

1

|n|

N+ 1

f(n)eint d

= T(FN f)(t) d (from lemma2).4


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Now note that the Fejer kernel is non-negative. As f is non-negative, the convolution(FN f) is also non-negative. From the positivity of the measure , our quantity isnon-negative. As this is true for any non-negative function f, we get that N(t) 0.

Now supposeN

n=N1 |n|N+1 ane

int 0. As before, denote Nas the measure with

FourierStieltjes coefficients 1 |n|N+1

an. Then we want to compute NM(T). Bythe Riesz representation theorem we just need to compute the norm of the functional

f 12

T

f dN. However, from earlier discussion, dN=

1 |n|N+1

an dt. Therefore

NM(T) = supfC(T)=1

1

2

T

f(t) dN = supf=1

1

2

T

f(t) N(t) dt.

AsN(t) is positive,Nis a positive measure and the above quantity is maximised whenf(t) = 1. This means that

NM(T)= 1

2

T

Nn=N

1

|n|

N+ 1

ane

int dt= a0.

This must be true for all N. Therefore, NM(T) are uniformly bounded by a0. Bycorollary6, we can find a M(T) such that (n) =an for all n. To show that is apositive measure, take an arbitrary non-negative function f. Then

T

f d= limN

T

f dN0

as this convergence comes from taking the limit in Parsevals formula. Thus is positive.

We remark in the above proof that one can replace the condition N(t) 0 for all Nwith N(t) 0 for infinitely many N. The proof remains the same, except that we onlytake a subsequence Nj where Nj (t) 0.

Definition 8. A sequence {an}nZ is positive definite if for all sequences of complexnumbers{zn}nZ which have all but a finite number of terms zero, we have

n,mZ anmznzm 0.The work from the last few pages can now be combined to prove Herglotzs theorem:

Theorem 9 (Herglotz). A sequence{an}nZ is the FourierStieltjes transform of a pos-itive measure M(T) if, and only if, the sequence is positive definite.

Proof. Ifis a positive measure with (n) =an, then

n,mZ

anmznzm = n,mZ

einteimtznzm d= nZ

zneint2

d 0.

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On the other hand, suppose {an} is positive definite. Fix values t, N. Define numbers

zn =

eint, |n| N

0, else. By direct substitution, one can verify that

n,m anmznzm =

jCj,Naje

ijx, where Cj,N= max(0, 2N+ 1 |j|). Therefore by positivedefiniteness,

2N(t) =2N

j=2N

1

|j|

2N+ 1

aje

ijx = 1

2N+ 1

2Nj=2N

Cj,Najeijx

= 1

2N+ 1

n,mZ

anmznzm0.

We proved that N(t) 0 for even N. The result then follows from lemma 7.

5 The FourierStieltjes Transform

We define the main object of interest here, the FourierStieltjes transform.

Definition 10. The FourierStieltjes transform of a measure M(R) is a functionon Rgiven by the expression

() =

R

eix d(x).

Like FourierStieltjes coefficients, there is consistency between the definition here, andthe Fourier transform ofL1 functions. Ifg is an integrable function, identify it with ameasure d=g dx. Hence () =

R

eixg(x) dx= g(), which shows consistency in thetransforms.

The RiemannLebesgue lemma on the ordinary Fourier transform states that if g L1(R),then g() is uniformly continuous and goes to 0 as || . The uniform continuity stillholds for the FourierStieltjes transform, but () does not necessarily go to 0. A simpleexample is=0, the measure with mass at 0. It can be easily verified that () = 1 forall .

We can deduce another version of Parsevals formula in this new context:

Proposition 11 (Parsevals Formula). IfM(R) and bothg, g are inL1(R), thenR

g(x)d(x) = 1

2

R

g()() d.

Proof. If both g and g are integrable, then the Fourier inversion formula holds: g(x) =12

R

g()eix d. Therefore,

Rg(x)d(x) =

1

2 Rg()eix d(x) d=

1

2 Rg()(),

where we used the integrability of g() to justify the usage of Fubinis theorem.

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The following proposition gives a necessary and sufficient statement for a function to bethe FourierStieltjes transform of a measure, and is the first step to Bochners theoremon R.

Proposition 12. If is a continuous function defined on R, then it is the Fourier

Stieltjes transform of a measure if, and only if, there exists a constantCsuch that 12R

g()() d

CsupxR

|g(x)|

for every continuous functiongL1(R) such thatg has compact support.

Proof. Firstly, suppose that = . Then the statement follows directly from Parsevalsformula by setting C=M(R).

On the other hand, suppose our inequality is valid. Then the linear functional which

maps g to 12 Rg()() d is a bounded, continuous linear functional on the set ofcontinuous functions g such that g is compactly supported. This is a dense subset ofC0(R). Therefore we can extend to a bounded functional on C0(R).

By the Riesz representation theorem, can be represented by a measure M(R),where M(R) C. Therefore maps g to

R

g(x) d(x). Using Parsevals formulahere yields that 1

2

R

g()() d = 12

R

g()() d. This must hold for all g, so= .

6 Connection to FourierStieltjes CoefficientsWe have a natural covering map p: R Tsending xto xmod 2. This can be used topushforward a measure M(R) to a measure M(T). If we extend a continuousfunction f on T to a 2-periodic function g on R (which do not necessarily go to zero),then we get that

R

g(x) d(x) =

T

f(t) d(t).

An immediate consequence is that (n) = (n) for all integers n. This allows us to relatethe original problem on R to the problem on T. The following theorem is a key step in

doing this:

Theorem 13. Ifis a continuous function defined onR, then it is the FourierStieltjestransform if, and only if, there exists a constantC >0 such that for any choice of >0,{(n)}n= are the FourierStieltjes coefficients of a measure of normC onT.

Proof. First suppose = . Then (n) = (n) = (n), where is the pushforwardmeasure ofwith respect to the covering. Note that M(T) M(R). Denote(x/)the measure on Rwhich satisfies the following equation for all g:

R g(x)d(x) = R g(x) d(x).7


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We get that (x/)M(R) =M(R) and (x/)() = (). Setting=n yields that(n) =(x/)(n), so {(n)}nZform the FourierStieltjes coefficients of norm at mostM(R).

To show the converse, we shall use theorem12. Letg be continuous such that gis contin-

uous and compactly supported. Then by those conditions the integral 12 Rg()() d

can be approximated by its Riemann sums (where the width of each rectangle is ). Forarbitrary , choose sufficiently small to obtain 12

R

g()() d

< 2

n

g(n)(n)

+.Note that

2g(n) is the Fourier coefficient for the function G(t) =

mZ

g((t+ 2m)/)

on T. If is sufficiently small, then by the decay ofg, we have

suptT

|G(t)| supxR

|g(x)| +.

By assumption,(n) =(n) for someM(T), with M(T)C. Then Parsevalsformula for Tgives 2

n

g(n)(n)

=

n

G(n) (n)

CsuptT |G(t)|.Combining all inequalities gives

2 n

g(n)(n)< CsupxR |g(x)| + (C+ 1).As >0 was arbitrary, the condition for proposition12is satisfied.

Now here is a necessary and sufficient condition for a function to be a FourierStieltjestransform of a positive measure. The proof is very similar to theorem13, so it will notbe presented here.

Proposition 14. If is a continuous function defined on R, then it is the Fourier

Stieltjes transform of a positive measure if, and only if, there exists a constant C > 0such that for any choice of >0, {(n)}n= are the FourierStieltjes coefficients ofa positive measure onT.

7 Bochners Theorem

Definition 15. A function defined on Ris positive definiteif for all 1, . . . , N Rand z1, . . . , z N C, we have

Nj,k=1

(j k)zjzk 0.

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Finally, we have the machinery to prove the main subject of this paper.

Theorem 16. A function defined onR is the FourierStieltjes transform of a positivemeasure if, and only if, it is continuous and positive definite.

Proof. First assume that = for a positive measure M(R). Given numbers1, . . . N Rand z1, . . . z N C, we get

Nj,k=1

(j k)zjzk =

R

Nj,k=1

eijxzjeikxzkd(x)

=

R

N

j=1

zjeijx

2

d(x) 0.

For the other direction, if is positive definite, then by definition, {(n)} is a positivedefinite sequence for all. Herglotzs theorem guarantees a measure, such that(n) =(n). By proposition14, there must exist such that =.

References

[1] Katznelson, Y. (2004). An Introduction to Harmonic Analysis(3rd edn), CambridgeMathematical Library.

[2] Grafakos, L. (2003).Classical and Modern Fourier Analysis, Prentice Hall.

[3] Rudin, W. (1987).Real and Complex Analysis (3rd edn), McGrawHill Book Com-pany.

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bochner’s theorem on the fourier transform on r

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