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Fourier Transform and Plancherel Theorem

Oliver DazUniversity of Texas at AustinEmail: odiaz@math.utexas.edu

September 27, 2000

Here we present the L1(Rn) and L2(Rn) theory of the Fourier transform. We make

a brief treatment of tempered distributions and use it to classify the operators in

L1(Rn) and in L2(Rn) that commute with translations.

1 L1(Rn) theory of the Fourier transform

We begin by defining the fourier transform in L1(Rn).

Definition 1 If f L1(Rn) , the Fourier transform of f is the function fdefined by letting

f(t) = n

f(x)e2ixt dx

The following properties of the Fourier transform are easy to obtain.

Proposition 1 Suppose f L1(Rn) and h Rn, R. Then,(a) If g(x) = f(x)e2ixh then g(t) = f(t h).(b) If g(x) = f(x h) then, g(t) = f(t)e2iht.(c) If g L1(Rn) and = f g then (t) = f(t)g(t).

(d) If g(x) = f(x), then g(t) = f(t).(e) If g(x) = f(x/) and > 0, then g(t) = nf(t).

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Proof: We only prove (c). The rest are straight forward.

(t) =

n

e2ixt

n

f(x y)g(y) dy dx

=

n

e2iytg(y)

n

f(x y)e2i(xy)t dx dy

=

n

e2iytg(y)

n

f(x)e2ixt dx

dy

=

f(t)g(t)

For any positive number a and any vector h we define the dilation by a, a,and the translation by h, h, as the operators mapping any function g(x)into g(ax) and g(x h) respectively. Proposition 1 formulated in terms ofthese operators says

(a) (e2ixhf(x))(t) = (hf)(t).

(b) (hg)(t) = e2ithf(t).

(e) (af)(t) = anf(a1t).

The following lemma will be needed:

Lemma 1 If 1 p < , then the mapping from R to Lp(Rn)given byh hf is uniformly continuous.Proof: We first prove tihs lemma for continuous functions of compact sup-port. Suppose that g is such a function and that supp(g) B(0, a) then,g is uniformly continuous. Given > 0, by uniform continuity of there is a0 < < a such that |s t| < implies

|g(t)

g(t)

|< (m(B(0, 3a)))1/p

Thus n

|g(x t) g(x s)|p dx < p

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so that

tg

s

g

p =

tsg

g

|p < . The conclusion follows from the

density ofCc(Rn) in Lp(Rn).

Another useful result of this nature, which we will use until the next section,has to do with convolution of functions.

Theorem 1 If 1/p + 1/q = 1 and f L1(Rn) and g Lq(Rn) then f g isuniformily continuous. If 1 < p < then we also have lim

|x|f g(x) = 0.

Proof: Without lost of generality, we might assume that 1 p < , thenby Holders inequality and translation invariance of Lebesgue measure wehave

|(f g)(x + h) (f g)(x)| h(f g) f gpgqand lemma 1 implies uniform continuity. To prove that f g vanishes atinfinity, we use sequences fk and gk of compact supported functions aprox-imating f and g. If supp fk supp gk Bak(0), then fk gk is continuousand of compact support, supp (fk gk) B2ak(0). Using Holders inequalitywe get that

f g fk gk f fkpgq + fkpg gkqThis clearly finishes the proof.

It is easy now to establish and prove the following result:

Theorem 2 (Riemann -Lebesgue) (a) The mappingf f is a boundedlinear transformation from L1(Rn) to L(Rn). In factf f1. (b)If f L1(Rn) then f is uniformly continuous and f(t) 0 as |t| .Proof: We only prove (b). Note that

|f(t) f(s)| n

f(x)e2ixs

e2ix(ts) 1

dx

Dominated convergence does the rest. To prove that f vanishes at infinity,

note that since ei = 1 then

f(t) =

f(x)e2i x+

t2|t|2

tdx =

f

x t2|t|2

e2ixt dx

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Hence

2f(t) =

f(x) fx t2|t|2 e2ixt dxso that

2f(t) f hf1with h = t2|t|2 . Now, lemma 1 implies that f(t) 0 as |t| .

Differentiation and fourier transformation are related in the following way:

Theorem 3 Suppose f L1(Rn) and xkf(x) L1(Rn) , where xk is ther-th coordinate function. Then f is differentiable with respect to tk and

ftk

(t) = (2ixkf(x))(t).

Proof: Letting h = (0, . . . , hk, . . . , 0) be a nonzero vector along the k-thaxis. Since |e2itkhk 1| tkhk, we have that

f(t + h) f(t)hk

=

e2ith 1

hkf(x)

(t) (2itkf(x))(t)

as hk 0 by dominated convergence.

It is also true that we can take fourier transforms of partial derivatives of

funtions. To make precise this statement, we introduce the following

Definition 2 We say that f Lp(Rn) is differentiable in the Lp(Rn) normwith respect to xk, if there exists g Lp(Rn) such thatf(x + h) f(x)hk g(x)

p

0

as hk 0

Applying proposition 1 and part (a) of theorem 2 leads to

e2ith 1hk f(t) g(t) f(x + h) f(x)hk g(x)

1

and taking hk 0 proves the following

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Theorem 4 If f

L1(Rn) and g is the partial derivative of f with respect

to xk in the L1(Rn) norm then

g(t) = 2itkf(t)

everywhere.

For n = 1 we can use integration by parts to get a simpler result.

Proposition 2 If f and f belong to L1(R) and f is the indefinite integralof f, then (f)(t) = 2itf(t)

Proof: Since f(x) =x

f(t) dt, it follows that lim

xf(x) = 0 = lim

xf(x).

Integration by parts does the rest.

The last two theorems can be extended to higher derivatives. Without goinginto the details, we note the following formulas:

(i) P(D)f(t) = (P(2ix)f(x))(t)(ii) (P(D)f)(t) = P(2it)f(t)

(1)

where, for a n n-tuple = (1, , n) of nonnegative integers we letx = x11 x

22 xnn , D = 1+2+...+n/x11 x22 xnn , P is a polyno-

mial in the n variables x1, x2, . . . , xn and P(D) is the associated differentialoperator.

The main problem in the L1(Rn) thoery of Fourier transform is to obtaina function f back from its Fourier transform f. Our next task is to answerthis question. Here we introduce certain summabilitymethods for integrals.

Definition 3 For each > 0 and f a locally integrable function, we defindethe Abel mean Af to be the integral

Af =

n

f(x)e|x| dx (2)

In the same way, we define the Gauss-Weierstrass mean Wf as

Wf =n

f(x)e

|x|2

dx. (3)

Whenever the limit lim0

Af or lim0

Wf exists, we say that

n f(x) dx is

Abel or Weierstrass summable.

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Clearly, if f is integrable, then lim0

Af = n f(x) dx.It is noteworthy to observe that a function f might be Abel or Weierstrasssummable even if f is not integrable itself, as the next example shows:

Example 1 Let f be the function

f(x) =sin x

x1(0,)(x)

We claim that A f = lim0

Af = /2 For any > 0 there exist P > 0 such

that

p

0

sin x

x

dx

2 0 we have thatn

e2iyte|y|2

dy = n/2e|t|2/ (5)

and n

e2iyte2|y| dy = cn

(2 + |t|2)(n+1)/2 (6)

where cn = [(n + 1)/2]/((n+1)/2).

Proof: By a change of variables, it suffices to consider the case = 1.Equality (5) follows easily from lemma 2. Since

n

e2iyte|y|2

dy =

nj=1

e2iyj tjey2

j dyj

=n

j=1

et2

k = e|t|2

Equation (6) is harder to obtain. Using equation (4) we get

n

e2iyte2|y| dy = n

1

0

euu

e2|y|2/u du

e

2iyt dy

=1

0

euu

n

e2|y|2/ue2iyt dy

du

=1

0

euu

u

n/2eu|t|

2

du

=1

(n+1)/2

0

euu(n1)/2eu|t|2

du

= 1(n+1)/2

1(1 + |t|2)(n+1)/2

0

ess(n1)/2 ds

=[(n + 1)/2]

(n+1)/21

(1 + |t|2)(n+1)/2

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For > 0, let us denote the Fourier transform of the functions (x) =

e4|x|2 and (x) = e2|x| by W(t, ) and P(t, ) respectively. That is,

W(t, ) =1

(4)n/2e|t|

2/4

P(t, ) = cn

(2 + |t|2)(n+1)/2

W(t, ) is called the GaussWeierstrass kernel. P(t, ) is called the Poissonkernel.We will show among other things that the Abel and Gaussian means of theintegral

n

f(t)e2itx dt (7)

converges almost everywhere to f(x). The idea will be to expressed thosemeans in tearms of convolutions with the Poisson and Gauss kernels and thenuse the theory of aproximations to the identity. for the first step towardsthis, we will use the following

Theorem 6 If f anf g belong to L1(Rn) thenn

f(t)g(t) dt =

n

f(t)g(t) dt

Proof: Applying Fubinis theorem, we get

n

f(t)g(t) dt =

n

n

f(x)e2ixt dx

g(t) dt

=

n

n

g(t)e2ixt dt

f(x) dx

=

n

f(x)g(x) dx

For any given function we let (x) = n(1x). In this notation,

proposition 1 says that ()(t) = (t).

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Example 2 Consider the function

(x) = e42|x|2

its Fourier transform is given by

(t) = (4)n/2e|t|2/4

Thus()(x))

(t) = (42)n/2e|t|2/42 = W(t, 2)

Example 3 Consider the function

(x) = e2|x|

its Fourier transsform is given by

(t) = cn1

(1 + |t|2)(n+1)/2

Thus()(x))

(t) = cn

(2 + |t|2)(n+1)/2 = P(t, )

These examples together with theorem 6 lead to the following

Theorem 7 If f and belong to L1(Rn) thenn

f(x)e2itx(x) dx =

n

f(x)(x t) dx

for all > 0. In particular,n

f(x)e2itxe2|x| dx =

n

f(x)P(x t, ) dx

and n

f(x)e2itxe42|x|2 dx =n

f(x)W(x t, ) dx

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We have almost everything set up to use the theory of aproximations to

the identity to show that integral (7) is sumable to f for a large class ofmethods that includes both the Able and Guass summability. The task willbe to prove that the means

n f(t)e2itx(t) dt converge to f in L1(Rn)

provided that both and are integrable and

n (t) dt = 1. The followinglemma shows that the Gauss and Poisson kernels satisfy that condition.

Lemma 3 For all > 0 n

W(x, ) dx = 1

and n

P(x, ) dx = 1

Proof: For > 0, a simple change of variable gives thatn

W(x, ) dx =

n

W(x, 1/4) dx =

n

e|x|2

dx = 1

Similarly, we have thatn

P(x, ) dx =

n

P(x, 1) dx

. The following geometrical facts will be at handy in our calculations:

(i) The surface element dn1 of the sphere Sn1 on the space Rn is givenby:

dn1 = sinn2 1 d1 dn2

= sinn2 1 sinn3 2 sin n2 d1 d2 dn1

where 0 k < for 1 k n 2 and 0 n1 < 2(ii) The volume n of the unit ball in R

n and the surphace n1 of itsboundary are:

n = |x|1

dx =n1

n

n1 =

Sn1

dn1(u) =2n/2

[n/2]

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With this in mind, if we use polar coordinates and then the change of variable

r = tan and we will have

n

1

(1 + |x|2)(n+1)/2 dx =0

Sn1

rn1

(1 + r2)(n+1)/2dn1(u) dr

= n1

0

rn1

(1 + r2)(n+1)/2dr

= n1

/20

sinn1 d

=1

2n =

(n+1)/2

[(n + 1)/2]

This finishes the proof.

Now we state and proof a theorem concerning approximations to the identity

Theorem 8 (Aproximation to the Identity) Suppose L1(Rn) witha =

n (x) dx, and for > 0 let (x) = n(1x). If f Lp(Rn) or

f Co(Rn) L(Rn), then f a fp 0 as 0. In particular,u(x, ) =

nf(t)P(x t, ) dt and s(x, ) =

nf(t)W(x t, ) dt converge

to f on Lp(Rn) norm as

0.

Proof: By a simple change of variables we haven

(s) dx =

n

n(1x) dx =

n

(x) dx = a

Hence,

(f g)(x) af(x) =n

[f(x t) f(x)](t) dt

=n

[f(x t) f(x)](t) dt

For 1 p < let S = {h Lq(Rn):hq = 1}, where 1/p + 1/q = 1. UsingMinkowskis duality equation, Fubinis theorem and Holder inequality we

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get

f g afp = supS

n

h(x)

n

[f(x t) f(x)](t)dt dx

supS

n

|h(x)|n

|f(x t) f(x)||(t)|dt dx

supS

n

|(x)|

n

|f(x t) f(x)||h(t)|dx

dt

n

|(t)|hqtf fp dt

=

n

|(t)|tf fp dt

For f C0(Rn), the same relationship takes place with p = . We in-troduce the function modulus of continuity of f, f,p, defined as f,p(h) =(h) = hf fp. It is clear that |(h)| 2fp for all h. Continuity oftranslations, lemma 1, implies that (h) 0 as h 0. Therefore f converges to f in Lp(Rn) norm as 0.

The same argument applies to give a more general result:Corollary 1 Suppose A() is a familly of invertible matrices such thatlim0

A() = 0. Let us define A()(x) = (det A())n (A1()x). Then

f A() a fp 0.

The strength of theorems 7 and 8 is reflected in the solution to the Fourierinverse problem

Theorem 9 If and its Fourier transform are integrable and

n (x) dx =

1, then the means of the integral

n f(t)e

2ixt dt converges to f(x) inL1(Rn) norm. In particular, the Abel and Gauss means of this integral

converges to f in L1(Rn).

Since s(x, ) =

n f(t)e2ixte42|t|2 dt converges to f(x) in L1(Rn) as

0 there is a sequance k 0 such that s(x, k) f(x) for a.e. x. If

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f happens to be integrable as well, then by dominated convergence we get

the following result:

Corollary 2 If both f ad f are integrable then

f(x) =

n

f(t)e2itx dt

for almost everyx.

Remember that f is continuous, and if f is integrable then the integral

n f(t)e2itx dt will be continuous as well (in fact, it is (f)(x)). Hence,

by changing f on a set of measure zero we can obtain equality in corolary 2.In other words, f can be make into a continuous function by changing itsvalues on a set of measure zero.If f(t) = 0 then f(x) = 0 a.e. Applying this to f1 f2 we obtain thefollowing result on uniqueness of the Fourier transform.

Corollary 3 If f1 and f2 belong to L1(Rn) and f1(t) = f2(t) for all t Rn

then f1(x) = f2(x) a.e.

The Fourier inversion problem has a pointwise solution as well. The nextresult is a pointwise version of the theorem of aproximations to the identity.

for any given measurable function , we defined the least decreasing radialmajorant of as the function

(x) = ess. sup|y||x|

|(y)|

Theorem 10 Suppose L1(Rn) with a = n (x) dx. If L1(Rn)and f Lp(Rn) 1 p < , then(i) lim

0(f )(x) = af(x) whenever x is a Lebesgue point of f.

(ii) sup>0

|(f)(x)| 1 (M(f))(x), where M(f) is the Hardys maximalfunction of f

In particular

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(a) The Poisson integral of f

u(t, ) =

n

f(x)P(t x, ) dx

and the GaussWeierstrass integral of f

s(t, ) =

n

f(x)W(t x, ) dx

converge to f(x) as 0 for almost every x Rn.(b)

sup>0

|u(t, )| (M f)(t)sup>0

|s(t, )| (M f)(t)

Proof: Let x be a Lebesgue point of f. This means that for any > 0, onecon find > 0 such that r < implies that

1

m(Br(x))

Br(x)

|f(y) f(x)| dy <

Changing to polar coordinates we can reinterpret the former condition as:

if r < then

G(r) =

r0

sn1g(s) ds < nrn (8)

where

g(s) =

Sn1

|f(x s u) f(x)| dn1(u)

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On the other hand, for all > 0 we have

|f (x) a f(x)| =n

[f(x t) f(x)](t) dt

|t|

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observations on equation 8 we get

I1

|y|

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But

=

n(1) = n(/)n0(/)

0 as

0 as pointed

out before, and (i) follows.As for part (ii), note that |(f )(x)| (|f| ||)(x) (|f| )(x). Thuswe only need to consider f 0 and show that

sup>0

|(f )(x)| 1 (M(f))(x)

Also, since

((xf) )(0) = (f )(x)M(xf)(0) = (M(f))(x)

(f )(0) = ((f) )(0)it suffices to prove that

(f )(0) 1(M(f))(0) for all f Lp(Rn)Let

(r) =

Sn1

f(ru) dn1(u)

(r) =

ro

(s)sn1 ds =

|x|r

f(x) dx

(f )(0) = f(y)(y) dy = f(y)(y) dy = f(y)(y) dy. Thus,using polar coordinates, the radial property of and integration by partswe obtain

(f )(0) = lim 0

N

0(r)(r)]N +

N

(r) d(0(r))

For the first term converges to 0 as the following estimate shows it:

|0(N)(N) 0()()| n(M f)(0)(0(N)Nn + 0()n)For the second term we have the following estimate

N

(r) d(0(r)) n(M f)(0)0

rn d(0(r))

= 1(M f)(0)

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This concludes our proof.

Any point of continuity of f is of course a Lebesgue point off. Thus iff iscontinuous at 0, then by theorems (7) and (10) we have that

lim0

n

f(x)e2|x| dx = lim0

n

f(x)P(x, ) dx = f(0)

If we further assume that f 0 it follows from monotone convergence thatf L1(Rn). In this way we obtain the following simple and neat result:Corollary 4 Suppose f L1(Rn) and f 0. If f is continuous at 0 thenf L1(Rn) and

f(x) =n

f(t)e2itx dt

almost everywhere (i.e. at every Lebesgue point of f). In particular,

f(0) =

n

f(t) dt

Applying this result ot the Poisson ans GaussWeierstrass kernels we get:

Corollary 5

n

W(x, )e2itx dx = e42|t|2

n

P(x, )e2itx dx = e2|t|

for all > 0.

We inmidiately obtain the semigroup properties of the Poisson and GaussWeierstrass kernels:

Corollary 6 If 1 and 2 are positive real numbers then

(a) W(x, 1 + 2) = n

W(x

t, 1)W(t, 2) dt

(b) P(x, 1 + 2) =

n

P(x t, 1)P(t, 2) dt

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2 L2(Rn) Theory and the Plancherel Theorem

We know that the space L1(Rn)L2(Rn) is dense in L2(Rn). Here we willextend the fourier transform from the former space to the latter. One of thenice properties of this extension is that it turns out to be a unitary lineartranformation.

Theorem 11 If f L1(Rn)L2(Rn) then f2 = f2Proof: Let g(x) = f(x). Then by theorem 1 h = f g is conitnuous.Being the convolution of functions in L1(Rn) h is also integrable. Since

g = f then h = fg = |f|2. By corollary 4 it follows that h is integrable andh(0) = n h(t) dt. Thus

n

|f(t)|2dt =n

h(t) dt = h(0) =

n

f(y)g(0 y) dy

=

n

f(y)f(y) dy =

n

|f(y)|2 dy

This theorem asserts that the Fourier transform is a unitary operator definedon the dense subspace L1L2 ofL2(Rn) into L2(Rn). Thus by Caratheodoryextension, there exists a unique bounded extension,

Fon the whole space

L2(Rn). We will also use the notation f = Ff. Clearly, for any functionf L2(Rn)

f(t) = Ff(t) = limk

|x|k

f(x)e2ixt dx

It is also clear that F is a unitary operator from L2(Rn) into L2(Rn). Infact this operator is onto:

Theorem 12 (Plancherel)

(i) The Fourier transform is a unitary operator on L2(Rn).

(ii) The inverse Fourier transform, F1, can be obtained by letting(F1g)(x) =(Fg)(x) for all g L2(Rn).

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Proof: Since

Fis a unitary operator,

F(L2(Rn)) is a closed subspace of

L2(Rn). Let g F(L2(Rn)). A simple density argument extends theo-rem 6 to functions in L2(Rn). Thus

n f(t)g(x) dx =

n f(x)g(x) dx = 0

for all f L2(Rn). This implies that g2 = g2 = 0.

For the second part of the proof, note that since F is a unitary operator,then it preserves the inner product: (u|v) = n uvdx. Let g be any functionin L1 L2 and f L2(Rn). Let fk be the sequence given by

fk(x) =

|t|k

f(t)e2ixt dt

Clearly each fk L1L2, and fk(x) converges in L2 to the function f givenby f(x) = F(f)(x).Then

(g|f) = limk

(g|fk) = limk

n

g(x)

|t|k

f(t)e2itx dt

dx

= limk

|t| k

f(t)

n

g(x)e2itx dx

dt

= limk

|t|k

g(t)f(t) dt = (g|f) = (g

|f)

This clearly implies that f(x) = (F1f)(x) = f(x) = (Ff)(x) for allf L2(Rn).

3 Tempered Distributions and a Characterization

of Operators that Commute with Translations

The basic idea in the theory of distributions is to consider them as linearfunctionals on a space of regular functionsoften called testing functions.

This space is assumed to be well behave with respect to operations wehave studied so far, namely: differentiation, Fourier transform, convolution,translation, dilation, etc. This will be reflected, in return, on the propertiesof distributions. We are after a space of testing functions Son whichthese

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operations are defined and preserve the space. Motivated by theorems 3, 4

and formulas 1, the space Smust consist of infinitely differentiable functionssuch that when they or their derivatives are multiplied by polinomials, theymust still be integrable.

Definition 4 The space Sof testing functions is defined to be the class ofall functions C onRn such that

supx n

|x(D)(x)| <

for all ntuples = (1, . . . , n) and = (1, . . . , n) of nonnegative inte-gers.

Clearly that space is not void: the family of functions (x) = e|x|2

belongs to S. Salso contains the space D of all C functions with compactsupport. The later space is not void. To see this, we first consider the casen = 1. Let f(t) = e1/t if t > 0 and f(t) = 0 when t 0. It is an easyexercise to see that f C and that f and all its derivatives are bounded.Let (t) = f(1 + t)f(1 t); then (t) = e2/(1|t|2) if |t| < 1 and zerootherwise. Thus D(R). We can easily obtain ndimensional variantsfrom :

(a) For x

R

n define (x) = (x1)

(xn); then

D(Rn)

(b) For x Rn define (x) = e2/(1|x|2), if |x| < 1 and zero otherwise.Then D(Rn)

(c) If C and is the function in (b) then (x)(x) defines a functionin D(Rn).

The spaces Co (continuous functions that vanish at infinity with the supnorm) and Lp(Rn), with 1 p < contain S. Using the desity of Cc onboth spaces and the StoneWeierstrass theorem we can show that D is infact dense in both spaces and therefore so would S. Another interestingobservation is given by the following

Lemma 4 The Lp norm of a function Sis bounded by a linear combi-nation of L norms of the form x

(x), The coefficients and exponents ofthese terms depend only on p and the dimension n.

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Proof: Let A =

and B = sup

x n |x

|2n

|(x)

|then

n

|(x)|p dx

1/p

|x|1

|(x)|p dx

1/p

+

|x|>1

|(x)|p dx

1/p

An1

n

1/p+ B

|x|>1

|x|2np dx

1/p

= An1

n 1/p

+ Bn1

(2p

1)n

1/p

Theorem 3 implies that C whenever S. Combinig this withtheorem 4 gives that S. In other words,Theorem 13 If Sthen S.Remember that ( ) = . Since Sis closed under multiplication, thenthe inverse Fourier transform theorem shows that

Theorem 14 If and are inSthen so is .

We introduce now a metric on Sthat makes this sapce a topological vectorspace. There is a natural countable family of norms {} in S, indexed byordered pairs (, ) of n-tuples of nonnegative integers, given by:

() = supx n

|xD(x)|

These norms induce the distances d(, ) = (). Let d1, . . . , dn, . . .be an ordering of these metrics. Thus, we define a metric d in Sas

d(, ) =

k=1

2kdn(, )

1 + dn(, )

It is clear that m

with respect to d, if and only ifm

with respectto each dk (as m ). From this observation, it follows that the spaceoperations (, ) + and (a, ) a are continuous. Thus Swith thetopology induced by d becomes a topological vector scape. The followingproperties ofSare not difficult to be checked:

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Theorem 15 (

S, d) has the following properties:

(1) The mapping (x) xD(x) is continuous(2) If Sthen lim

h0h =

(3) Suppose Sandh = (0, . . . , hi, . . . , 0) lies on the i-th coordinate axisofRn, then the difference quotient [ h]/hi tend to /partialxias h 0

(4) Sis a complete metric space(5) The Fourier transform is a homeomorphism of Sonto itself.(6) D is dense subset of S(7) Sis separable.

The collection S of all continuous linear functionals on Sis called the spaceof tempered distributions. Thus to show that a linear functional L on S isa tempered distributionm it suffices to show continuity at the origin, i.e,lim

kL(k) = 0 whenever k 0 in S. Here we present some examples:

Example 4 Let f Lp(Rn), 1 p , and let L = Lf defined as

L() = Lf() =

n

f(x)(x) dx

for S. Lemma 4 implies that kq 0 if k 0 in S. By Holdersinequality

|L()| fpqThus, L is a tempered distribution

Example 5 If is a finite Borel measure, then the linear functional L = Ldefined by

L() = L() =n

(x) d(x)

is a tempered distribution.

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Example 6 Let f be a measurable function such that f(x)(1 +

|x

|2)k

belongs to Lp(Rn), 1 p for some integer k. Define L = Lf just as inexample 4: L() =

n f(x)(x) dx. Since

L() =

n

(1 + |x|2)k(x)

f(x)/(1 + |x|2)k

dx

it follows that L is a tempered distribution. f is called a tempered Lp func-tion. When p = , f is said to be slowly increasing.

Example 7 A Borel measure is called a tempered measure ifn

(1 + |x|2)k d||(x) <

for some integer k. If we define L() = L() =

n (x) d(x), then weget a tempered distribution.

Example 8 For a fixed point x0 Rn, and an n-tuple of nonnegativeintegers, we define the linear functional L() = D(x0). The continuity ofthe metric 0 in Simplies that L is a tempered distribution. In particular,for x0 = 0 and = (0, . . . , 0) we get the Diracs delta distribution: L() =(0). The latter can be obtained also, by considering the Borel measure of

mass 1 concentrated at the origin.

There is a simple characterization of tempered distributions.

Theorem 16 A linear functional L on S is a tempered distribution if andonly if there exists a constant C > 0 and integers m and l such that

|L()| C

||l,||m

()

for all S.Proof: The existence of C, l, m clearly implies the continuity of L.

For the converse, note that the family of sets of the form

N,l,m =

:

||l,||m

() <

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where > 0, and l and m are nonnegative integers, forms a basis of neigh-

borhoods arround 0 in S. To see that, let k = lm and consider the sets

U, = { : () < /k}Clearly, these are open sets in S. Also

||l,||m

U N,l,m.

For the ball B(0, ) in S, pick K > 0 such that

n=K2n < /2. Take l

and m big enough such that the norms in the list {1, . . . , K} are includedamong the norms {} with || l, || m. Then

N/2,l,m K

n=1

{ : n() < /2} B(0, )

Ith follows that there is a set N,l,m such that |L()| < 1 for all N,l,m.For = 0, let

=

||l,||m

()

it is easy to see now that C = 2/ will do the job:

|L()| = 2

L 2 2

Many important operations on functions, i.e: diferentiation, translation,dilation, reflections, Fourier transform, convolution, etc, can be extended ina natural way to distributions. Here we motivate some of these extension.

Derivative:If f Lp(Rn) admits an Lp(Rn) derivative g = Df, then by integrationby parts we have that

Lg() =

n

(Df(x))(x) dx = (1)||n

f(x)D(x) dx = (1)||Lf(D)

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for all

S. Thus, for any distribution u, we define Du as the linear

functionalv on Ssuch thatv() = (1)||u(D)

Clearly Du is a distribution.

Translation:If f Lp(Rn) then

Lhf() =

n

f(x h)(x) dx =n

f(x)(x + h) dx = Lf(h)

Thus, for any distribution u, we define h

u as the linear functional onSsuch that

hu() = u(h)

It is clear that hu is a distribution since it is the composition of continuousfunctions on S.

Fourier transform:If f L1(Rn) then from theorem 6 we have

Lf() =

n

f(x)(x) dx =

n

f(x)(x) dx = Lf()

thus, for any given distribution u we define u by

u() = u()

Multiplication by a test function:If Sand u S then we define u as the linear functional

(u)() = u()

for all S. Clearly this is a distribution on S.

Convolution:If f

Lp(Rn) and and belong to

S. Let (x) = (

x). A simple

application of Fubinis theorem shows that

Lf() =

n

(f )(x)(x) dx =n

f(x)( )(x) dx = Lf( )

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Thus, for any distribution u and test function we define u

as the linear

functional(u )() = u( )

The right side of the equation above is the composition of continuous func-tions in S. Therefore u is a distribution.The following theorem gives a better characterization of the convolutiondistribution just described:

Theorem 17 If u S and Sthen the convolution u is the functionf(x) = u(x). Moreover, f C and it is slowly increasing.Proof: First we show that (u

)() =

n

f(t)(t) dt:

(u )() = u( ) = u

n

(x t)(t) dt

= u

n

(t)(x)(t) dt

The Reimann sums of the last integral converge in the topology of S, thus

un

(t)(x)(t) dt = n

u(t)(t) dt = n

f(t)(t) dt

Continuity follows from theorem 15. Let h = (0, . . . , hj , . . . , 0), then

x+h xhj

x

xj

in the Stopology. Thus, by continuity of u,f(x + h) f(x)

hj u

x

xj

By an iterative argument we have (Df)(x) = (1)||u(xD). The factthat f is slowly increasing follows from theorem 16: There exist C > 0 andnonnegative integers l and m such that

|f(x)| = u(x) C

||l,||m

(x)

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But

(x) = supw n

|(w + x)(D)(w)|

whis is clearly a polynomial in x.

In what follows, we will apply the theory of distributions to the study of aclass of operators that occur in the theory of singular integrals: those thatcommute with translations. Suppose B is an operator mapping a linearspace V of functions in Rn into another space of the like. We say that Bcommutes with translations if hB = bh for all h Rn.

Example 9:

Fix f Lp

(R

n

) and let be B the operator on L

1

(R

n

) to L

p

(R

n

) definedby convolution: g f g. We have that Bgp fpg1. Clearly itcommutes with translations.

Example 10:Consider the Banach space of all complex measures on Rn with norm .The Fourier transform off as a distribution concides with that of a measure:

(t) =

n

e2ixt d(x)

Let be the map:

: Rn Rn Rn(x, y) x + y

Let be the measure on Rn induced by , i.e:

( )(E) = ( ) 1(E)It follows stright forward that

From the Rieszrepresentation theorem, it follows that

is the uniquemeasure such that

n

f(x) d(x) =

n n

f(x + y) d( )(x, y)

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for every f

Cc(R

n). Application of Fubunis theorem implies

( )(E) =n

(E t) d(t) =n

(E s) d(s)

and if d = f dx, where f L1(Rn) then d( )(y) = (f )(y) dy:

( )(E) =

n n

1E(x + y) d( )(x, y)

=

nn

1E(x + y)f(y) dy

d(x)

=

n

n

1E(y)f(y x) dy d(x)

=

E

n

f(y x) d(x) dy =

E

(f )(y) dy

It is clear that the map on L1(Rn) given by f f is continuous:f 1 = f1 = . Clearly it also commutes withtranslation.

We will show that all bounded operators in Lp(Rn)are of the similar convo-lution type. The following lemma will imply the former statement:

Lemma 5 If f Lp(Rn) has derivatives in the Lp norm of all orders n + 1, then f equals almost everywhere a continuous function g satisfying

|g(0)| Cn,p

||n+1

Dfp

where Cn,p depends only on the dimension n and the exponent p.

Proof: Let x

R

n, then

(1 + |x|)(n+1)/2 1 + n

j=1

|xj |

n+1

C

||n+1

|x|

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For p = 1 we will prove that f is integrable and use corollary 2:

|f(t)| C(1 + |t|2)(n+1)/2

||n+1

|t|||f(t)|

= C(1 + |t|2)(n+1)/2

||n+1

|((2)||Df)(t)|

C(1 + |t|2)(n+1)/2

||n+1

Df1

Therefore

f1 Cn2

||n+1

Df1

thus f is equal to a continuous function g almost everywhere and

|g(0)| g = f f1 Cn2

||n+1

Df1

For p > 1, choose 0 1 in D such that (x) = 1 if|x| 1 and (x) = 0if |x| 2. Then f satisfies the hypothesis of p = 1. Thus varphif equalsa continuos function h such that

|h(0)| C||n+1

Df1

Since D(f) =

+=C(Df)(D) and using Holders inequality we

have

D(f)1

|x|2

+=

C|D| |Df| dx

K

+=

0()

|x|2

|Df| dx

K2nn1

n1/q

+=

0()

Df

p

K

2nn1n

1/qmax||||

{0()}

||||

Dfp

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Thus, there is a constant Cn,p depending only on n and p such that

h(0) Cn,p

||n+1

Dfp (9)

Since (x) = 1 on the unit ball we see that f is equal to a continuousfunction h almost everywhere on the unit ball. Choosing k(x) = (x/k)shows that f is almost everywhere a continuous function hk on the ball ofradius k arround 0. Note that hk(0) = h(0), hence equation (9) is preserved.This finishes the proof of the lemma.

The following theorem characterizes all continuous operators on Lp(Rn) thatcommute with translations:

Theorem 18 Suppose B : Lp(Rn) Lq(Rn), 1 p, q , is a linearbounded operator that commutes with translations; then, there exists a uniquetempered distribution u such that B = u for all S.Proof: Let h = (0, . . . , hj , . . . , 0) be a nonzero vector along the jth

axis. Since hhj

xjin S, it converges in Lp as well. Therefore,

B

hhj

= hBBhj converges to B

xj

= Bxj in Lq by continuity

of B. By an iteration argument, it follows that B has derivatives in Lq

of all orders and D(B) = B(D). Hence by the previous lemma, Bequals a continuos function g almost everywhere and

|g(0)| Cn,q

||n+1

D(B)q

= Cn,q

||n+1

B(D)q

BCn,q

||n+1

Dq

This shows that the linear functional g(0) is in fact a distributionu1 on S. Letting u = u1 ge obtain the distribution we are looking for:(u)(x) = u(x) = u((x)) = u(x) = u1(x) = (B(x))(0) =(xB)(0) = (B)(x). Uniqueness follows inmidiately from the construc-tion.

Let us denote by (Lp, Lq) the space of those tempered distributions u forwhich there exists A > 0 such that u q Ap for all S. Last

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theorem shows that there is a onetoone relationship between this space,

and that of continuous linear operartors from Lp(Rn) to Lq(Rn) that com-mute with translations. For the case p = 2 = q there is a much bettercharacterization:

Theorem 19 The distribution u belongs to (L2, L2) if and only if thereexists b L(Rn) such that u = b. In this case, b is the norm ofthe operator B : L2 S L2 defined by putting B = u ; moreover(u ) = u.Proof: By the Fourier inverse theorem, we have that is the Fouriertransform of . Then, by definition of Fourier transform and convolutionin the distributional sence we have that

(u )() = (u )() = u( ) = u(()) = (u)()This shows that (u ) = u in general.

Consider the function 0(x) = e|x|2 in S. Since u (L2, L2), Plancherels

theorem implies that 0 = (u 0) = u0 = u0 belongs to L2(Rn). Letb(x) = e|x|

2

0(x) = 0(x)/0(x). Let Sand D, then(u)() = u() = u( 0/0) = (u)(/0)

=

n

0(x)(x)e|x|2(x) dx

=n

b(x)(x)(x) dx

= (b)()

Taking such that the support of its Fourier transform contains the supportof we get that u() = b() for all D. The density ofD in Simpliesthat u = b.Since u (L2, L2) there exists a constant A > 0 such that

b2 = (u )2 = u 2 A2for all

S. Since

Sis dense in L2(Rn), it is now easy to show that

b L(Rn) with b A.The converse is easy: if u = b L(Rn) then

u 2 = u2 = b2 b2 = b2

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This finishes the proof.

For p = 1 = q there is a nice characterization as well.

Theorem 20 The distribution u belongs to (L1, L1) if and only if it is afinite Borel measure. In this case, the total variation of u equals the normof the operator B : L1 S L1 defined by B = u for SProof: Consider the space Co(Rn) of continuous functions that vanishat infinity with the sup norm. Its dual space, according to the Rieszrepresentation theorem, is the normed space of finite Borel measures M =M(Rn) on Rn: mapping the measure to the linear functional that assings the value

n (x) d(x). Note that L

1(Rn) is naturally embeded into Mby the map f

f dx.

On the other hand, we can topologize M with the weaktopology and byAlaoglus theorem, the unit ball B = { : 1} in M is compact withrespect this topology; moreover, since C0 is separable, the unit ball is in factmetrizable (with the weaktopology), thus sequantially compact.Consider the family of L1 functions u = u W(, ), > 0, where W isthe GaussWeierstrass kernel. By hypothesis, there exists A > 0 such thatu1 AW(, )1 = A. Thus, by the previous observations, there exists M and a sequence {k} coverging to 0 such that uk as k inthe weaktopology. That is

limk

n

(x) duk (x) = n

(x) d(x)

for any S. Let (x) =

n (x t) W(t, ) dt. By dominated conver-gence we have that (D)(x) =

n(D

)(x t)W(t, ) dt = (D) forany ntuple of nonnegative numbers. By the theorem of aproximations tothe identity, (case p = ) we have that D(x) converges uniformily in xto D(x) as 0. Thus in Sas 0. Hence, by continuity ofu, u() u(). Since W(, ) = W(, ) we have

u() = u(W(, ) ) = (u W(, ))() =n

(x)u(x) dx

Taking = k and letting k

0 we obtain

u() =

n

(x) d(x)

This shows that u = M(Rn). The rest of the proof is easy.

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