Appel W. Mathematics for Physics and Physicists

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Properties of the Fourier transform 287

In particular, it will be u seful to remember that



′

(x)



= 2iπν

f (ν) and F



−2iπx f (x)



dν

f (ν).

Proof. For any ﬁxed x ∈ R, the function ν 7→ f (x) e

−2πiν x

is of C

∞

class, and

its k-th derivative is bounded in modulus by |(2πx)

f (x)|, which is integrable by

assumption. Applying the theorem of differentiation under the integral sign, we obtain

′

(ν) =

dν

f (ν) =

dν

f (x) e

−2πiν x

dx =

(−2πi x) f (x) e

−2πiν x

dx,

and then, by an immediate inducti on, the ﬁrst formula stated.

Now for the second part, recall that for any integrable function ϕ, we have

ϕ(x) dx = lim

R→+∞

−R

ϕ(x) dx.

Since f

′

is assumed to be integrable, an integration by parts yields

F [ f

′

] (ν) = lim

R→+∞

−R

′

(x)

|{z}

′

−2πiν x

|{z}

= lim

R→+∞

f (x) e

−2πiν x

−R

(2πiν) f (x) e

−2πiν x

As f and f

′

are integrable, f tends to zero at inﬁnity (see Exercise 10.7 on page 296).

The previous formula then shows, by letting R tend to inﬁnity, that

′

(x) e

−2πiν x

dx =

(2πiν) f (x) e

−2πiν x

dx,

which e stablishes the second formula for k = 1. An induction on k concludes the

proof.

Noticing that if f is integrable and has bounded suppor t then x 7→ x

f (x)

is also integrable for any k ¾ 0, we get the following corollary:

COROLLARY 10.2 5.1 If f ∈ L

is a function w ith bounded support, then its Fourier

transform is of C

∞

class.

The formula of Th eorem 10.25 on the preceding page also implies a very

important result:

COROLLARY 10.2 5.2 Let f and g be measurable func tions, and let p, q ∈ N.

Assume that the derivatives of f , of order 1 to p, exist and are integrable, and that

x 7→ x

g(x) is also integrable for k = 0, . . ., q. Then, for any ν ∈ R, we have



f (ν)



¶ |2πν|

−p



(p)

(x)



and



(q)

(ν)



∞

−∞

|2πx|



g(x)



dx,

288 Four i er transform of functions

These inequalities give very strong bounds: if f is ﬁve times differentiable, for

instance, a nd f

(5)

is integrab le, then

f decays at least as fast as 1/ν

So th ere e xists a link between the regularity of f and the rate of decay of

f at inﬁnity, and similarly between the rate of decay of f and the regularity

f .

10.2.d Rapid ly dec aying functi ons

DEFINITION 10.26 A rapidly decaying function is any function f such that

lim

x→±∞



f (x)



= 0 for any k ∈ N.

In ot her words, a rapidly decaying function is one that decays at inﬁnity

faster than any power of x.

Example 10.27 The functions x 7→ e

−x

and x 7→ x

(log x) e

−|x|

are rapidly decaying.

This is a useful notion, for instance, because if f is rapidly decaying

and is locally integrable, then x

f (x) is integrable for all k ∈ N. From

Theorem 10.25, we deduce:

THEOREM 10.28 If f is an integrable rapidly decaying function, then its Fourier

transform is of C

∞

class.

Still using Theorem 10.25, one can show (exercise!) that

THEOREM 10.29 Let f be an integrable f unction of C

∞

class. If f

(k)

is integrable

for all k ∈ N, then

f is a rapidly decaying function.

10.3

Fourier transform of a function in L

The are some functions which are not necessarily integrable, but whose

square is. Such is the “sine cardinal” function:

x 7−→ sinc(x)

def

sin x

(extended to R by continuity by putting sinc(0) = 1), which belongs to L

but not to L

It t urns out that in physics, square integrable functions are of paramount

importance and occur frequently:

Fourier transform of a function in L

289

• in quantum mechanics, the wave function ψ of a particle is a square

integrable function such that

|ψ(x)|

dx = 1;

• in optics, the square of th e modulus of the light amplitude represents a

density of energy; the total energy

|A(x)|

dx is ﬁnite;

• similarly in electricity, or more generally in signal theory,

|f (t)|

is the total energy of a temporal signal t 7→ f (t).

It is therefore ad visable to ex tend the deﬁnition of th e Fourier transform

to the class of square integrable functions. (It will be seen in Cha pter 13,

Section 13.6, how to extend this further to functions with inﬁnite total energy

but ﬁnite power.) For this purpose, we start by introducing the Schwartz space,

a technical device to reach our goal.

10.3.a The s pace SS

DEFINITION 10.30 The Schwartz space , denoted SS

SS , is the space of functions

of C

∞

class which are rapidly decaying along with all their derivatives.

Example 10.31 Let f ∈ D. Then

f is deﬁned and

f ∈ S .

Indeed, f has bounded support, so, according to C orollary 10.25.1,

f is inﬁnitely differ-

entiable. M oreover, f can be differentiated p times and f

(p)

is integrable for any p ∈ N, so

that Corollary 10.25.2 implies that

f decays at least as fast as 1/ν

at inﬁnity, for any p ∈ N.

Moreover (still by Theorem 10.25), the derivatives of

f are also Fourier transforms of functions

in D, so the same reasoning implies that

f ∈ S .

The most important result concerning S is the following:

THEOREM 10.32 The Fourier transform deﬁnes a continuous linear operator from

S into itself. In other words, for f ∈ S , we have

f ∈ S , and if a sequence ( f

)

n∈N

tends to 0 in S then (

)

n∈N

also tends to 0 in S .

Proof that S is stable. We ﬁrst show that if f ∈ S , we have

f ∈ S .

The function f is rapidly decaying, so its Fourier transform

f is of C

∞

class. For

all k ∈ N, the function f

(k)

is also rapidly decaying (and integrable). By Theorem 10.25

on page 286, we deduce that

f is rapidly decaying.

There only remains to show that the derivatives of

f are also rapidly decaying. But

by Theorem 10.25, the derivatives of

f are Fourier transforms of functions of the type

x 7→ (−2iπx)

f (x), and it is clear that by deﬁnition of S , these are also in S . So

applying the previous argument to those functions shows that

(k)

is rapidly decaying

for any k ∈ N.

To speak of continuity of the Fourier transform F [·] on S , we must

make precise what the notion of convergence is in S .

The Schwartz space will also be useful to deﬁne the Fourier transform in the sense of

distributions.

290 Fourier transform of functions

DEFINITION 10.33 A sequence ( f

)

n∈N

of elements of S converges in SS

to 0 if, for any p, q ∈ N, we have

lim

n→∞

sup

x∈R



(q)

(x)



= 0.

This notion of convergence in S is very strong; in particular, it implies the

uniform convergence on R of ( f

)

n∈N

together with that of all its der ivatives



(k)



n∈N

To end this discussion of the most important proper ties of the Fourier

transform on the space S , it is natural to ask what happens for the inversion

formula. We know that if f ∈ S , its Fourier transform

f is in S and, since

f is continuous and integrable, we will have (by T heorem 10.13) the relation

F [F [ f ]] = f . In fact even more is true:

THEOREM 10.34 (Inversion in SS

SS ) The Fourier transform is a bicontinuous linear

isomorphism (i.e., bijective and c ontinuous along with its inverse) from S into S .

Its inverse is given by F

−1

[·] = F [·].

10.3.b The Fourier transform in L

To deﬁne the Fourier transform on the space L

of square integ rab le functions,

we will use the results of the previous section concerning the Fourier transform

on S , together with the following lemma, which is given without proof.

LEMMA 10.35 The space S is a dense subspace of the space L

(R).

The next step is to show that, for two functions f and g in S , we have

∞

−∞

f (x) g(x) dx =

∞

−∞

f (ν) eg(ν) dν.

This is an easy consequence of Fubini’s theorem and some manipulations with

complex conjugation.

We can guess that a Hilbert space structure is in the

ofﬁng!

Then we use the properties

• S is dense in L

• L

is complete,

to show that the Fourier transform deﬁned on S can b e extended to a con-

tinuous linear operator on L

. This step, essentially technical, is explained in

Sidebar 5 on page 294.

We end up with an operator “Fourier transform,” deﬁned for any function

f ∈ L

(R) which satisﬁes the identities of the nex t theorem:

See, for instance, the proof of Lemma 10.14.

Fourier transform of a function in L

291

THEOREM 10.36 (Parseval-Plancherel) The Fourier transform F [·] is an isome-

try on L

: for any two square integrable f unctions f and g, the Fourier transforms

and eg also in L

and we h ave

∞

−∞

f (x) g(x) dx =

∞

−∞

f (ν) eg(ν) dν.

In particular, taking f = g, we have

∞

−∞



f (x)



dx =

∞

−∞



f (ν)



dν

or, in other words,

kf k

= k

f k

THEOREM 10.37 (Inversion in L

) For any f ∈ L

, we have



F [ f ]



= F



F [ f ]



= f almost everywhere.

The inverse Fourier transform is thus deﬁned by

−1

[·] = F [·].

It is natural to wonder if there is a relation between the Fourier transform

thus deﬁned on L

(R) (after much wrangling and sleight of hand) and the

original Fourier transform on L

(R).

PROPOSITION 10.38 The Fourier transform on L

and that on L

coincide on the

subspace L

∩ L

In other words, if f is square integrable on the one hand, but is also integrable

itself on the other hand, then its Fourier transform deﬁned in th e section is

indeed given by

F [ f ] : ν 7−→

f (x) e

−2πiν x

dx.

But t hen, how does one express the Fourier transform of a square integrable

function which is not integrable? One can show that, in practice, its Fourier

transform in L

can be deﬁ ned by the following limit:

f (ν) = lim

R→+∞

−R

f (x) e

−2πiν x

dx for almost all ν ∈ R

(which is again a special case of improper integral, deﬁned in the sense of a

principal value).

292 Fourier transform o f functions

10.4

Fourier transform and convolution

10.4.a Convolution formula

Recall ﬁrst the d eﬁnition of the convolution of two functions:

DEFINITION 10.39 Let f and g be two locally integrable functions. Their

convolution, when it exists, is the function denoted h = f ∗ g s uch that

h(x)

def

+∞

−∞

f (t) g(x − t) dt.

Fubini’s theorem (page 79) provides the proof of the following fundamen-

tal theorem, also known as the Faltung theorem (Faltung being the G erman

word for “convolution”):

THEOREM 10.40 Let f and g be two func tions such that their Fourier transforms

exist and the convolution f ∗ g is deﬁned and is integrable. Then we have

F [ f ∗ g] = F [ f ] ·F [g] ,

f ∗ g(ν) =

f (ν) · eg(ν).

Similarly, when those expressions are deﬁned, we have

F [ f · g] = F [ f ] ∗F [g] ,

f · g(ν) =

f ∗ eg(ν).

Proof. We consider the special case where f and g are integrable. Then the Fourier

transforms of f and g are deﬁned and f ∗ g is indeed integrable. Deﬁne ϕ(x, t) =

f (t) g(x − t) e

−2πiν x

. Then

F [ f ∗ g] (ν) =

Z Z

f (t) g(x − t) dt



−2πiν x

dx =

Z Z

ϕ(x, t) dt



dx.

Since ϕ i s integrable on R

by Fubini’s theorem (and the integrability of f and g), we

obtain

F [ f ∗ g] (ν) =

Z Z

ϕ(x, t) dx



Z Z

f (t) g(x − t) e

−2πiν x



f (t)

Z

g(x − t) e

−2πiν(x−t)



−2πiν t

f (t) eg(ν) e

−2πiν t

dt =

f (ν) eg( ν).

The second formula is in general more delicate than the ﬁrst; if we assume that

and eg are both integ rable, it follows from the ﬁrst together with the Fourier inversion

formula.

Fourier transform and convolution 293

Example 10.41 We want to compute without a problem th e Fourier transform of the function

Λ deﬁned by

Λ(x) =







1 + x for −1 ¶ x ¶ 0,

1 − x for 0 ¶ x ¶ 1,

0 for |x| ¾ 1.

Λ(x)

1−1

First we show that Λ(x) =



Π ∗Π



(x) and then we deduce that



Λ(x)





sin πν

πν



Note that the Parseval-Plancheral formula gives

Λ(x) dx =

Π(x) dx ×

Π(x) dx = 1,

which is the expected result.

10.4.b Cases of the c onvolution formula

One can show, in particular, that the following cases of the convolution for-

mula are valid:

i) if f , g ∈ L

, th en

f ∗ g(ν) =

f ·eg(ν) for any ν ∈ R;

ii) if f , g ∈ L

and their Fourier transforms are also in L

, then

f · g(ν) =

f ∗ eg(ν) for any ν ∈ R;

iii) if f and g are in L

, one can take the inverse Fourier transform of the

ﬁrst formula and write f ∗ g(t) = F



f ·eg



(t) for all t ∈ R, which is

a good way to compute f ∗ g (see Exercise 10.2 on page 295); moreover,

the second formula is valid:

f · g(ν) =

f ∗ eg(ν) for all ν ∈ R;

iv) if f ∈ L

and g ∈ L

, then f ∗ g(t) = F



f · eg



(t) for almost all

t ∈ R.

294 Fourier transform o f functions

Sidebar 5 (Extension of a continuous linear operator) Let E be a normed

vector space and V a dense subspace of E. If F is a complete normed vector

space and ϕ : V → F is a continuous linear map, then

a) there exists a unique continuous linear extension of ϕ, say bϕ : E → F (i.e.,

bϕ satisﬁes bϕ(x) = ϕ(x) for any x ∈ V );

b) moreover, the norm of bϕ and the norm of ϕ are equal.

Let x ∈ E. We want to ﬁnd the vector which will be called bϕ(x). Using the

fact that V is dense in E, there exists a sequence (x

)

n∈N

with values in V which

converges to x in E. To show that the sequence (ϕ(x

))

n∈N

converges, we star t by

noting that (x

)

n∈N

is a Cauch y sequence. Since ϕ is linear and continuous, we

have, denoting by |||ϕ||| the norm of ϕ (see Appendix A, page 583), the inequality



ϕ(x

) −ϕ(x

)



¶ |||ϕ|||



− x



for any p, q ∈ N. The sequence



ϕ(x

)



n∈N

is therefore also a Cauchy sequence,

and since it takes values in the complete space F , it is convergent. There only

remains to deﬁne

bϕ(x) = lim

n→∞

ϕ(x

It is easy to check that ϕ(x) does not depend on the chosen sequence (x

)

n∈N

converging to x: indeed, if ( y

)

n∈N

also converges to x, we have

lim

n→∞

ϕ(x

) − lim

n→∞

ϕ( y

) = lim

n→∞

ϕ(x

− y

) = 0

by continuity of ϕ.

The map bϕ thus constructed is linear and satisﬁes



bϕ(x)



lim

n→∞

ϕ(x

)



= lim

n→∞



ϕ(x

)



¶ |||ϕ||| lim

n→∞

k = |||ϕ|||·kxk,

showing that |||bϕ||| ¶ |||ϕ|||.

To prove the converse inequality, note that for y ∈ V , we ha ve bϕ( y) = ϕ( y)

and therefore

|||bϕ||| = sup

x∈E\{0}



bϕ(x)



kxk

¾ sup

x∈V \{0}



bϕ(x)



kxk

= |||ϕ|||.

Hence the equality |||bϕ||| = |||ϕ||| holds as stated.

Exercises 295

EXERCISES

 Exercise 10.1 (Fourier transform of the gaussian) Using the link bet ween the Fourier trans-

form and derivation, ﬁnd a differential equation satisﬁed by the Fourier transform of the

function x 7→ ex p(−πx

). Solve t h is differential equation and deduce from this the Fourier

transform of this function. Recall that

∞

−∞

−πx

dx = 1 .

Deduce from this the Fourier transform of the centered gaussian with standard deviat ion σ:

(x) =

2π

exp



−

2σ



 Exercise 10.2 Using the convolution theorem for functions f , g ∈ L

and the preceding

exercise, compute the convolution product of a centered gaussian f

with standard deviation

σ and a centered gaussian f

′

with standard deviation σ

′

 Exercise 10.3 Let α ∈ C such that Re(α) > 0, and let n ∈ N

∗

. Compute the Fourier

transform of the function

x 7−→ f (x) =

(2πi x −α)

n+1

In particular, show that

f (ν) is zero for ν < 0.

 Exercise 10.4 Compute the following integrals, using the known properties of convolution

and the Parseval-Plancheral formula, and by ﬁnding a relation with the functions Π and

Λ = Π ∗Π:

+∞

−∞

sin πx

πx

dx,

+∞

−∞



sin πx

πx



dx,

+∞

−∞



sin πx

πx



dx.

 Exercise 10.5 (sine and cosine Fourier transforms) Let f : R → R be an odd integrable func-

tion. Deﬁne

(ν)

def

= 2

+∞

f (t) sin(2πνt) dt.

Show th at

f (t) = 2

+∞

(ν) sin(2πν t) dν.

The f u nction

thus deﬁned is called the sine Fourier transform of f , and f is the

inverse sine Fourier transform of

. These transforms can be extended to arbitrary (integrable)

functions, the values of which are only of interest for t > 0 or ν > 0.

Similarly, show that if f is even and if th e cosine Fourier transform of f is deﬁned by

(ν)

def

= 2

+∞

f (t) cos(2πνt) dt,

then we have

f (t) = 2

+∞

(ν) cos(2πνt) dν.

Compute the cosine Fourier transform of x 7→ e

−αx

, where α is a strictly positive real

number.

296 Fourier transform of functions

Deduce then the identity

∀t > 0 ∀α > 0

+∞

cos 2πν t

+ α

dν =

2α

−2παt

 Exercise 10.6 Given functions a : R → R and b : R → R, and assuming their Fourier

transforms are deﬁned, solve the integral equation

y(t) =

+∞

−∞

y(s) a(t − s) ds + b(t).

 Exercise 10.7 (proof of Theorem 10.25) Let f ∈ L

(R) be a function which is differentiable

with a derivative which is Lebesgue-integrable: f

′

∈ L

(R). Show that lim

x→±∞

f (x) = 0.

SOLUTIONS

 Solution of exercise 10.1. Deﬁne f (x) = e

−πx

, which is an integrable function, and let

be its Fourier transform. From

∀x ∈ R f

′

(x) + 2πx f (x) = 0,

we deduce, evaluating the Fourier transform,

∀ν ∈ R 2πν

f (ν) +

dν

f (ν) = 0 ,

which has solutions given by

f (ν) = A e

−πν

for some constant A to be determined.

Since

f (0) =

+∞

−∞

f (x) dx = 1, it follows that A = 1.

By a simple change of variable, it also follows that

2π

exp



−

2σ



F.T.

−−−→ e

−2π

a result which will be ver y useful in probability theory.

Compare this with the method given in Exercise 4.9 on page 125, which uses the Cauc hy

formula in the complex plane.

 Solution of exercise 10.2. Using the convolution formula, we get F



∗ f

′



′

, and since f

and f

′

are both square integrable, their Fourier transforms are also square

integrable. Using the Cauchy-Schwarz inequality, it follows that

′

is integrable; there its

conjugate Fourier transform is deﬁned and

∗ f

′

(t) = F



′



(t) = F



−2π

·e

−2π

′2



(t)

= F



−2π(σ

+σ

′2

)ν



(t) = f

+σ

′2

(t).

The squares of the standard deviations (i.e., the variances) add up when taking the convolution

of gaussians.

 Solution of exercise 10.3. We ﬁrst notice that f is integrable: f ∈ L

. So its Fourier

transform is deﬁned by the integral

f (ν) =

−2πiν x

(2πi x −α)

n+1

dx.