Appel W. Mathematics for Physics and Physicists
Подождите немного. Документ загружается.
Solutions of exercises 297
We c ompute this using the meth od of residues. The only pole of t h e integrand in C is at
x = iα/2π.
• I f ν < 0, we close the c o ntour in the l ower half-plane to apply Jordan’s second lemma,
and no residue appears; therefore the function
e
f vanishes identical ly for ν < 0.
• For ν > 0, we close the contour in the upper half-plane and, by applying the residue
formula, we get
e
f (ν) = (−1)
n+1
ν
n
e
αν
/n !.
To conclude, the Fourier transform of f can be written
e
f (ν) = (−1)
n+1
ν
n
n!
e
αν
H (ν).
Solution of exercise 10.5. An immediate calculation shows that if f is an odd function,
we have
e
f
s
= (i/2)
e
f . Moreover,
e
f
s
is odd; hence the Fourier inversion formula yields
f (t) =
Z
+∞
−∞
e
f (ν) e
2πiνt
dν =
Z
+∞
−∞
e
f
s
(ν) sin(2πν t) dν = 2
Z
+∞
0
e
f
s
(ν) sin(2πν t) dν.
By definition, the cosine Fourier transform of x 7→ e
−αx
is given by
2
Z
+∞
0
e
−αt
cos(2πν t) dt =
2α
α
2
+ 4π
2
ν
2
.
The inverse cosine Fourier transform leads to x 7→ e
−α|x|
, whic h is the original input function,
after it has been made even. The formula follows.
Solution of exercise 10.6. Assume that the Fourier transform of y also exists. De note by
Y , A, and B the Fourier transforms of y, a, and b, respectively. Then, by the convolution
theorem, we have
Y (ν) = Y (ν) A(ν) + B(ν) and hence Y (ν) =
B(ν)
1 − A(ν)
.
So, under the assumption that it exists, the solution is given by the inverse Fourier transform
of the function ν 7→ Y (ν) thus defined.
Solution of exercise 10.7. Since f
′
is integrable, we can write
Z
+∞
0
f
′
(t) dt = lim
x→+∞
Z
x
0
f
′
(t) dt.
Moreover, for any x ∈ R we have
Z
x
0
f
′
(t) dt = f (x) − f (0),
which shows that f admits a limit at +∞, and similarly at −∞. If those limits were nonzero,
f would certainly not be integrable on R.
❧
Chapter
11
Fourier transform
of distributions
11.1
Definition and properties
The next objective is to define the Fourier transform for distr ibutions.
This is interesting for at least two reasons:
1. it makes it possible to define the Fourier transform of distr ibutions
such as δ or X;
2. possibly, it provides an extension of t he Fourier transform to a larger
class of functions; in particular, functions which are neither in L
1
(R),
nor in L
2
(R), but which are of constant use in physics, such as the
Heaviside function.
In order to define the Fourier transform of a distribution, we begin, as
is our custom, by restricting to the special case of a regular distr ibution.
Consider therefore a locally integrable function. But then, unfortunately,
we realize that “being locally integrable” is not, for a function, a sufficient
condition for the Fourier transform to be defi ned.
Restricting even more, consider a function f ∈ L
1
. Being integrable,
it is also locally integrable and h as an associated regular distribution, also
300 Fourier transform of distributions
denoted f . Its Fourier transform
e
f is continuous, hence a lso locally integrable,
and defi nes therefore a distribution
e
f , which acts on a test function ϕ by the
formula
〈
e
f , ϕ〉 =
Z
e
f (t) ϕ(t) dt =
Z Z
f (x) e
−2πi x t
dx
ϕ(t) dt,
which is the same, using Fubini’s theorem to exchange the order of integration
(this is allowed because f is integrable and ϕ is also), as
〈
e
f , ϕ〉 =
Z
f (x)
Z
e
−2πi x t
ϕ(t) dt
dx =
Z
f (x) eϕ(x) dx = 〈f , eϕ〉.
This computation justifies the following definition:
DEFINITION 11.1 The Fourier transform of a distribution T is defined by
its action on compactly supported infinitely differentiable test functions ϕ,
which is given by
F [T ] , ϕ
def
=
T , F [ϕ]
.
Sometimes the notation
e
T is used instead of F [T ]:
e
T , ϕ
=
T , eϕ
.
Remark 11.2 It is known that if ϕ has compact support and is nonzero, F [ϕ] does
not have c o mpact support; this means that the quantity
T , eϕ
is not always defined.
Hence not all distribution have an associated Fourier t ransform. For this reason, the
notion of tempered distribution is introduced.
11.1.a Tempered distrib utions
DEFINITION 11.3 (SS
S
S
SS
′
) Recall tha t a rapidly decaying f unction is a function
f such that lim
x→±∞
x
k
f (x)
= 0 for all k ∈ N, and that th e Sch wartz space (of
functions which are infinitely differentiable and rapidly decaying along with
all their derivatives) is denoted S .
The topological dual of S (i.e., the space of continuous linear forms on
S ) is denoted S
′
. A tempered distribution is an element of S
′
.
Hence, to say t hat T is a tempered distribution implies that T is continu-
ous, that is, that
if ϕ
n
−−−→
n→∞
0 in S , then 〈T , ϕ
n
〉 −−−→
n→∞
0 in C.
PROPOSITION 11.4 Any tempered distribution is a “classical” distribution. In other
words, we have S
′
⊂ D
′
.
Definition and properties 301
Proof. Note first that D ⊂ S . Let now T ∈ S
′
be a tempered distribution; it acts
on any test functi on ϕ ∈ S and hence, a fortior i, on any function ϕ ∈ D, so T can
be restricted to a linear functional on D.
Moreover, if (ϕ
n
)
n∈N
converges to 0 in D, it converges also to 0 in S by the
definition of those two convergences, so 〈T , ϕ
n
〉 tends to 0 in C. This shows that T is
continuous, and hence that T ∈ D
′
.
Now let us look for examples of tempered distributions.
Let f be a locally integrable function, which increases at most like a power
function |x|
k
, where k is some fixed integer, i.e., we have f (x) = O(|x|
k
) in
the neighborhood of ±∞. T hen for any function ϕ which decays rapidly, the
integral
R
f ϕ exists. The map
ϕ ∈ S 7−→
Z
+∞
−∞
f (x) ϕ(x) dx
is also linear, of course, and continuous — which t he reader is invited to check.
Therefore, it defines a tempered distribution, also denoted f .
DEFINITION 11.5 (Slowly increasing function) A function f : R → C is
slowly increasing if it increases at most like a power of |x| at infinity
1
.
PROPOSITION 11.6 Any loca lly integrable slowly increasing function defines a
regular tempered distribution.
Exercise 11.1 Let T be a tempered distribution. Show that for any k ∈ N, x
k
T and T
(k)
are also tempered distributions.
Exercise 11.2 Show that th e distribution exp(x
4
) X(x) is not tempered.
(Solution page 326)
11.1.b Fourier transform of tempered distributions
Let T be a tempered distribution. For ϕ ∈ S , that is, for a C
∞
function
which is rapidly decaying, and with all its derivatives also rapidly decaying,
the Fourier transform F [ϕ] is also in the space S (Theorem 10.32). H ence
the quantity
T , F [ϕ]
is defined and, consequently, it is possible to define
the Fourier transform of T .
THEOREM 11.7 Any tempered distribution has a Fourier transform in the sense of
distributions, which is also tempered.
Remark 11.8 Does the Fourier transform of tempered distributions defined in this manner
coincide with the Fourier transform of functions, in the case of a regular distribution?
1
A slowly increasing function, multiplied by a rapidly decaying function, is therefore inte-
grable; the terminology is coherent!
302 Fourier transform of distributions
Yes indeed. Let f be an integrable function, and denote this time by T ( f ) the associated
regular distribution (to make the distinction cle arly). The Fourier transform of g (in the sense
of functions) is a bounded function
e
f , which therefore defines a regular tempered distribution.
The associated distribution, denoted T (
e
f ), satisfies therefore for any ϕ ∈ S
T (
e
f ), ϕ
=
Z
e
f (x) ϕ(x) dx =
Z
f (x) eϕ(x) dx =
T ( f ), eϕ
=
F
T ( f )
, ϕ
.
This implies that T (
e
f ) is indeed equal to th e Fourier transform, in the sense of distributions,
of T ( f ).
Similar reasoning holds for f ∈ L
2
(R) (Exercise!) and shows that the Fourier transform
defined on S
′
coincides with that defined on L
2
(R).
The proper ties of the Fourier t ransform with respect to scaling, translation,
and differentiation remain valid in the setting of tempered distributions:
THEOREM 11.9 Let T be a tempered distribution. Then
F
T
′
= (2πiν) F [T ] , F
T
(n)
= (2πiν)
n
F [T ] ,
F
T (x −a)
= e
−2πiνa
F [T ] , F
e
2πiax
T
=
e
T (ν − a).
Proof. Take the first property, for instance. Before starting, remember that the
letters x and ν, usually used to denote the variable for a function and its Fourier
transform, respectively, are in fact interchangeable. So, it is perfectly legitimate to write
F [T ] (ν), ϕ( ν)
as well as
F [T ] (x), ϕ(x)
.
This being said, let T be a tempered distribution and ϕ ∈ S . Then we have
F [T
′
] , ϕ
=
T
′
, F
ϕ
= −
T ,F
−2iπx ϕ(x)
= −
F [T ] , −2πi x ϕ(x)
=
2πi xF [T ] (x), ϕ(x)
,
which is the stated formula (with the variable x replacing ν).
The other statements are proved in an identical manner and are left as warm-up
exercises.
Finally, here is the theorem linking convolution and Fourier transform.
THEOREM 11.10 Let S be a tempered distribution and T a distribution with
bounded support. T hen the convolution product S ∗ T makes sense and is a tem-
pered distribution, and therefore it has a Fourier transform which is given by
F [S ∗ T ] =
e
S ·
e
T .
If S is a regular distribution associated to a function of C
∞
class, and T a
distribution such that the product S · T exists and is a tempered distribution, then its
Fourier transform is given by
F [S · T ] =
e
S ∗
e
T .
Definition and properties 303
Remark 11.11 The preceding formulas need some comment. Indeed, the product of t wo
distributions is generally not defined (Jean-François Colombeau worked on the possibility of
defining such a product, but his results [23] are far from the scope of this book). However,
it is defined wh en one of the two distributions is regular and associated with a f u nct ion of
class C
∞
.
In the first formula (F [S ∗ T ] =
e
S ·
e
T ), the distribution
e
T is really a regular distribution,
associated with th e function f , of class C
∞
, defined by
f (ν) =
T (t), e
−2πiν t
(the right-hand side is well defined for all ν because T has bounded support).
11.1.c Examples
The first important example of the Fourier transform of a tempered d istribu-
tion is that of a constant function.
THEOREM 11.12 (FF
F
F
FF [1] = δ) The Fourier transform of the constant function 1 is
the Dirac distribution.
1
F.T.
−−−→ δ(ν).
Proof. Let ϕ ∈ S be a Schwartz function. We have, by definition and the Fourier
inversion formula,
e
1, ϕ
def
=
1, eϕ
=
Z
eϕ(ν) dν.
Furthermore, as eϕ is in S , the inversion formula for the Fourier transform yiel ds
Z
eϕ(ν) dν = F [eϕ] (0) = ϕ(0) = 〈δ, ϕ〉.
This th eorem indicates that for a constant f unction, th e “signal” contains
a s ingle frequency, namely the zero frequency ν = 0. Similarly, we have:
THEOREM 11.13 The Fourier transforms of the trigonometric functions a re given by
F
e
2πiν
0
x
= δ(ν −ν
0
),
F [cos(2πν
0
x)] =
1
2
h
δ(ν −ν
0
) + δ(ν + ν
0
)
i
,
F [sin(2πν
0
x)] =
1
2i
h
δ(ν −ν
0
) −δ(ν + ν
0
)
i
for all ν
0
∈ R.
This result is quite intuitive: for the functions considered, only the frequency
ν
0
occurs, but it may appear with a sign ±1 (see Section 13.4, page 365).
THEOREM 11.14 (FF
F
F
FF [δ] = 1) The Fourier transform of the Dirac distribution is the
constant function 1 : x 7→ 1.
δ(x)
F.T.
−−−→ 1 .
304 Fourier transform of distributions
The Dirac distribution is so “spiked” that all frequencies are represented.
Proof. Let ϕ ∈ S be a Schwartz function. We have
e
δ, ϕ
def
=
δ, eϕ
= eϕ(0).
But, from the definition of eϕ, it follows that
eϕ(0) =
Z
ϕ(x) e
−2πi0·x
| {z }
1(x)
dx = 〈1, ϕ〉.
Using Theorem 11.9 for differentiation and translation, we deduce:
THEOREM 11.15 Let m ∈ N and a ∈ R. Th en we have
F
δ
′
= 2πiν, F
δ
(m)
= (2πiν)
m
,
F
δ(x −a)
= e
2πiνa
.
We now compute the Fourier transform of the Heaviside distribution H
(which is tempered). From H
′
= δ, we deduce, using the differentiation
property of the Fourier transform, that 2πiν ·
e
H (ν) = 1. But, according to
Theorems 8.6 on page 229 and 7.28 on page 191, the distribution solutions of
the equation ν · T (ν) = 1 are of the form
αδ(ν) +
1
2πi
pv
1
ν
, with α ∈ C.
To determine the constant α, note that H −
1
2
is a real-valued function which
is odd, so that its Fourier transform, namely,
e
H −
1
2
δ, must also be odd. S ince
the δ distribution is even, it follows that α =
1
2
. Denoting by “sgn” the sign
function, which is given by sgn(x) = 2H (x) −1, we further obtain:
THEOREM 11.16 (Fourier transform of H and pv
1
x
) The Fourier transform of
the Heaviside distribution is
F
H (x)
=
1
2πi
pv
1
ν
+
1
2
δ.
Similarly,
F
H (−x)
=
i
2π
pv
1
ν
+
1
2
δ.
Moreover, the distribution pv
1
x
satisfies
F
pv
1
x
= −iπ sgn ν, F [sgn x] =
1
iπ
pv
1
ν
.
Definition and properties 305
11.1.d Higher-dimensi onal Fourier trans forms
Consider now distributions defined on R
n
. For a function f : R
n
→ R (with
suitable conditions), its Fourier transform is defined on R
n
by
e
f (νν
ν
ν
νν)
def
=
Z
. . .
Z
R
n
f ( x) e
−2πiνν
ν
ν
νν·x
d
n
x
for n-uples x and νν
ν
ν
νν
The vector νν
ν
ν
νν is a “frequency vector.” It is of ten advantageous to replace it
by the “wave vector” k = 2π νν
ν
ν
νν. Then the conventions defining the Fourier
transform are
F ( k)
def
=
Z
. . .
Z
R
n
f ( x) e
−i k·x
d
n
x, f ( x) =
1
(2π)
n
Z
. . .
Z
R
n
F ( k) e
i k·x
d
n
k.
Those definitions provide a way of defining the Fourier transform of a
tempered distribution on R
n
, imitating the process in Defi nit ion 11.1. T he dif-
ferentiation theorem is easily generalized, and has the following consequence,
for instance:
THEOREM 11.17 (Fourier transform of the laplacian) Let T ( x) be a distribu-
tion in R
3
, which admits a Fourier transform. Then the Fourier transform of its
laplacian △T ( x) is −4πνν
ν
ν
νν
2
e
T (νν
ν
ν
νν).
Consider now the following par tial differential equation (coming from par-
ticle physicis, plasma physics, or ionic solutions, and explained in Problem 5
on page 325):
△−m
2
f ( x) = −4π δ( x).
Taking the Fourier transform of this relation, it follows that
(−4πνν
ν
ν
νν
2
−m
2
)
e
f (νν
ν
ν
νν) = −4π.
This is now an algebraic equation, the solution of which is obvious
2
:
e
f (νν
ν
ν
νν) =
4π
m
2
+ 4πνν
ν
ν
νν
2
. (11.1)
2
But is that really true? Not entirely! Since we are looking for distribution solutions, we
must remember that there are many distribution solutions to the equation
(m
2
+ 4πνν
ν
ν
νν
2
) · T (νν
ν
ν
νν) = 0:
any distribution which is p roportional to δ(m
2
+4πνν
ν
ν
νν
2
) will do. B u t, if the function m
2
+4πνν
ν
ν
νν
2
does not vanish on R
3
, on the other hand it does vanish for complex values of νν
ν
ν
νν. The fact
that
e
f (νν
ν
ν
νν) takes nonzero values for co mplex values of νν
ν
ν
νν is more easily interpreted in terms of
a (bilateral) Laplace t ransform than in terms of a Fourier transform. For instance, a solution
is given by f (x, y, z) = e
mx
, whic h does indeed satisfy (∆ −m
2
) f ( x) = 0, but this f has no
Fourier transform. The solution (11.1) is, in fact, the only one wh ich vanishes at infinity. It is
therefore the one we will keep.
306 Fourier transform of distributions
There only remains to compute the inverse Fourier transform of this func-
tion to obtain a solution to the equation we started with. For this, the method
of residues is still t he most efficient. This computation is explained in detail
in t he physics problem on page 325. In addition, the whole of Chapter 15 is
devoted to the same subject in various physical cases. The result is
f ( r) =
e
−mkr k
krk
.
Letting m to 0, we deduce
3
the following result, of great use in electromag-
netism:
THEOREM 11.18 (Fourier transform of the Coulomb potential)
In R
3
, the Fourier transform of the Coulomb potential r 7→ 1/ krk is given by
F ( k) =
ZZZ
R
3
1
kxk
e
−i k·x
d
3
x =
4π
k
2
.
Remark 11.19 The function k 7→ 1/ kkk
2
is locally integrable (in R
3
), and the right-hand side
is therefore a regular distribution.
11.1.e Inversion formula
The inversion formula, known for functions that belong to L
1
, has a gener-
alization for tempered distrib ut ions. First, let us define the conjugate Fourier
transform of a tempered distribution T ∈ S
′
by
∀ϕ ∈ S
¬
F [T ] , ϕ
¶
=
¬
T , F [ϕ]
¶
=
T (ν), eϕ(−ν)
.
Then F is the inverse of F, that is to say,
THEOREM 11.20 (Inversion formula in SS
S
S
SS
′
) Let T ∈ S
′
be a tempered distri-
bution. I ts Fourier transform U = F [T ] is also a tempered distribution and
T = F [U ].
3
With no rigorous justification. Here i s how to proceed to check this. Start by regularizing
the Coulomb potential by putting, for R > 0,
f
R
( r) =
¨
1/r if r ¶ R,
1/R if r ¾ R.
This regularized Coulomb potential converges to f : r 7→ 1/r in the sens of distributions (the
verification of this is immediate). Since the Fourier transform is a continuous operator, the
Fourier transform of f
R
tends to
e
f as R tends to infinity. After a few lines of computation
(isolating the constant part 1/R), the Fourier t ransform of f
R
is computed explicitly and gives
f
R
( k) =
(2π)
3
R
δ( k) +
4π
k
2
−
4π
R k
3
sin(kR).
The first and third terms tend to 0 in the sense of distributions as [R → +∞], as the reader
can easily show.