Computational Methods and advanced Statistics Tools
I. FOURIER SERIES,
FOURIER TRANSFORM.
1. UNIFORM CONVERGENCE
A series of functions is a series where the general term is a sequence of
functions:
∞
X
fn (x), x ∈ I
n=1
where I is some common domain ⊂ R.
The series is pointwise convergent to a sum S(x) in I if
∀x ∈ I, ∀ǫ > 0, ∃n̄(ǫ, x) > 0 : |
n
X
fk (x) − S(x)| < ǫ, ∀n > n̄(ǫ, x).
n
X
gk (x) − S(x)| < ǫ, ∀n > n̄(ǫ)
k=1
The series is uniformly convergent to S(x) in I if
∀ǫ > 0, ∃n̄(ǫ) > 0 : ∀x ∈ I, |
k=1
If the functions fn (x), n ∈ N , are continuous for x ∈ I, the point
convergence is not sufficient to ensure that the sum S(x) is continuous.
But if the series of continuous functions is uniformly convergent to S(x)
for x ∈ I, then S(x) is a continuous function in I.
In turn, a sufficient condition for the uniform convergence of a series of
functions is the ”total convergence” criterion: the series of functions is
uniformly convergent in I if the modulus of the general term is bounded
by the general term of a convergent numerical series.
In other words, if
∞
X
|fn (x)| ≤ αn , ∀x ∈ I, with
αn < +∞
n=1
then the series of functions
P∞
n=1
fn (x) is uniformly convergent.
When such a condition is satisfied one can prove that:
i) the sum of the series of function S(x) is continuous in I;
ii) the definite integral on [a, b] ⊂ I of the sum of the series is equal to
the sum of the integrals of fn (x);
1
2
iii) if, moreover, the general term fn (x) is differentiable and, in the
interval I, |fn′ (x)| is bounded by the general term of convergent numerical series, the sum S(x) of the series of functions is differentiable and
the sum of the derivatives converges to S ′ (x).
For example, consider the series of continuous functions
+∞
X
n=0
fn (x) =
+∞
X
1
sin(nx)
3
n
n=0
Since the general term is bounded by the general term of a convergent
numeric series,
X 1
1
1
|fn (x)| = 3 | sin(nx)| ≤ 3 , with
< +∞
n
n
n3
n
then the sum S(x) of the series is finite and continuous, ∀x ∈ R.
Moreover, since the series of the derivatives enjoys the same property,
X 1
1
1
< +∞
|fn′ (x)| = 3 |n cos(nx)| ≤ 2 , with
2
n
n
n
n
then S(x) admits a continuous derivative S ′ (x), which coincides with
the sum of the series of the derivatives ∀x ∈ R. In such a case we say
that the series can be derived term by term.
Besides, for any finite a, b with a < b,
Z bX
XZ b
fn (x)dx
fn (x)dx =
n
a
a
n
3
for fn (x) = sin(nx)/n , as well as for fn (x) = cos(nx)/n2 . In such
cases we say that the series can be integrated term by term.
2. FOURIER SERIES
A function f is periodic if it is defined for all real x and there is a
positive number T such that
(1)
f (x + T ) = f (x), ∀x
According to this definition, a constant function is periodic with arbitrary period. Trigonometric functions, such as sin x, sin 2x, ..., sin nx,
or cos x, ..., cos nx, are periodic with period 2π.
Besides exponential functions of imaginary arguments, too, such as
eix , e2ix , ..., einx , or e−ix , e−2ix , ..., e−inx , are periodic functions of period
2π. Indeed e2πi = 1.
If T is a period for f then f (x + nT ) = f (x), ∀n ∈ N : therefore f is
periodic with periods 2T, 3T, ..., too.
3
A trigonometric polynomial is a finite sum of the type
k
a◦ X
+
(an cos nx + bn sin nx)
Sk (x) =
2
n=1
(2)
Any trigonometric polynomial is periodic with period 2π.
Any trigonometric polynomial can be written in exponential form:
(3)
Sk (x) =
k
X
cn einx .
n=−k
Indeed, for n = 0, c◦ = a◦ /2. For each n ≥ 1, in the equality
an cos nx + bn sin nx = cn einx + c−n e−inx
we can use Euler’s formulae
iθ
e = cos θ + i sin θ
e−iθ = cos θ − i sin θ
⇐⇒
cos θ = 21 (eiθ + e−iθ )
sin θ = 2i1 (eiθ − e−iθ )
It suffices to write cos nx, sin nx in terms of exponentials. Thus we
obtain the coefficients cn in terms of the an , bn ’s:
(4)
c◦ =
1
1
a◦
, cn = (an − ibn ), n ≥ 1; c−n = (an + ibn ), n ≥ 1.
2
2
2
In particular, if the an , bn ’s are real, then c−n is the complex conjugate
of cn : c−n = c̄n , ∀n 6= 0.
THEOREM I.2A - Let f (x + 2π) = f (x), ∀x ∈ R, and let us assume
that f (x) is the sum of a trigonometric series
+∞
(5)
a◦ X
f (x) =
+
(an cos nx + bn sin nx)
2
n=1
or, in equivalent manner,
(6)
f (x) =
+∞
X
cn einx .
n=−∞
Moreover, assume that the series is uniformly convergent for x ∈ R.
Then the coefficients of the series in exponential form are
Z 2π
1
f (x)e−inx dx, ∀n ∈ Z
(7)
cn =
2π 0
and the coefficients of the series in trigonometric form are
(8)
Z
Z
1 2π
1 2π
an =
f (x) cos(nx)dx, n ∈ N◦ , bn =
f (x) sin(nx)dx, n ∈ N.
π 0
π 0
4
Proof To fix ideas consider the exponential form (??) and let us
integrate both sides of (??). By virtue of the uniform convergence, we
can integrate term by term:
Z 2π
X einx 2π
f (x)dx = 2πc◦ +
= 2πc◦
in
0
0
n6=0
so that c◦ is the mean of the periodic function.
Multiplying both sides by e−imx , m 6= 0, and integrating we find:
Z 2π
X ei(n−m)x 2π
−imx
+ 2πcm = 2πcm
f (x)e
dx =
cn
i(n
−
m)
0
0
n6=m
R
2π
so that cm = (2π)−1 0 f (x)e−imx dx as stated.
By inversion of (??)
a◦
(9)
= c◦ ; an = cn + c−n , bn = i(cn − c−n ), ∀n ≥ 1,
2
and, by Euler’s formula,
1
1
cos x = (eix + e−ix ), sin x = (eix − e−ix )
2
2i
Thus, for n ≥ 1,
Z 2π
Z 2π
1
1
−inx
inx
f (x)(e
+ e )dx =
f (x)2 · cos(nx)dx
an =
2π 0
2π 0
and
Z 2π
Z 2π
1
1
−inx
inx
f (x)i · (e
− e )dx =
f (x)i · (−2i) sin(nx)dx.
bn =
2π 0
2π 0
△
The above calculation assumes that the period is 2π. In fact any period
T > 0 can be handled so as to obtain similar formulae for the coefficients cn , an , bn . Above all, since the integrals (??), (??) exist under
wide conditions, we can associate Fourier coefficients to any periodic
function in a large class:
DEF. I.2B (Fourier series) Let f (x + T ) = f (x)∀x ∈ R, and let f be
integrable in the interval [0, T ]. The Fourier coefficients and the formal
Fourier series, in exponential form, associated with f are defined by
Z
+∞
X
2π
1 T
nx
−i 2π
T
, f (x) ∼
cn ei T nx
(10)
cn =
f (x)e
T 0
n=−∞
The Fourier coefficients and the formal Fourier series, in trigonometric
form, associated with f are defined by
Z
Z
2 T
2 T
2π
2π
(11) an =
f (x) cos( nx) dx, bn =
f (x) sin( nx) dx
T 0
T
T 0
T
5
(12)
+∞
+∞
X
2π
2π
a◦ X
+
an cos( nx) +
bn sin( nx)
f (x) ∼
2
T
T
n=1
n=1
REMARK I.2C - a) An equivalent expression of coefficients makes use
of integrals from −T /2 to T /2, since the length of the interval is T :
Z
2 T /2
2π
(13)
an =
f (x) cos( nx) dx, n ∈ N◦
T −T /2
T
(14)
2
bn =
T
Z
T /2
f (x) sin(
−T /2
2π
nx) dx, n ∈ N
T
Actually the same numbers are obtained by integrating on any interval
[c, c + T ], with fixed c ∈ R.
b) The symbol ∼ reminds that the trigonometric expansion is still
formal: in fact nothing is said about convergence, and, if the series is
convergent, the sum is not necessarily f (x).
RT
c) The constant term in (??), i.e. c◦ ≡ a◦ /2 = T1 0 f (x) dx, is the
mean value of f in one period.
3. THEOREMS ON FOURIER SERIES
THEOREM I.3A (Uniformly convergent expansion if the periodic function is regular enough)
Let the T −periodic function f admit continuous r−th derivative in
[0, T ] with r ≥ 1. Then the Fourier coefficients of the series in exponential form satisfy
(15)
|cn | ≤ K|n|−r , ∀n ∈ Z − {0},
where the constant K is independent of n. Similarly the Fourier coefficients an , bn of the series in trigonometric form
(16)
|an | ≤ K̃n−r ,
|bn | ≤ K̃n−r , ∀n ∈ N
for some K̃ > 0. As a consequence, any periodic function f of class C r
with r ≥ 2, is the sum of a uniformly convergent Fourier series, both
in exponential form and in trigonometric form.
Proof. - Let f ∈ C r ([0, T ]). The derivative of a periodic function is
still periodic, so integrating by parts r times the expression (?? )
!
T Z T
2π
2π
T
T
1
f (x)
e−i T nx −
e−i T nx dx
f ′ (x)
cn =
T
−i2πn
−i2πn
0
0
6
1 T
=+
T i2πn
Therefore
Z
T
′
f (x)e
0
1
dx = +
T
T
i2πn
r Z
T
2π
f (r) (x)e−i T
nx
dx.
0
r Z T
2π
T
|f (r) (x)||e−i T nx | dx
2π|n|
0
r Z T
T
K
1
|f (r) (x)| dx ≤
≤
r
T |n| 2π
|n|r
0
1
|cn | ≤
T
where
nx
−i 2π
T
r
T
K=
max |f (r) (x)|.
x∈[0,T ]
2π
Since an = cn + c−n and bn = i(cn − c−n ), for the an ’s and the bn ’s it
suffices to choose K̃ = 2K.
Then the uniform convergence is obvious: indeed
K
, ∀x ∈ R
|cn ||einx | = |cn | ≤
|n|r
and
K̃
, ∀x ∈ R
|n|r
In both cases the general term of the Fourier expansion is bounded by
the general term of a convergent numerical series, that is a gerneralized
harmonic series with exponent r > 1:
X K
|a◦ | X 2K̃
|c◦ | +
<
+∞,
+
< +∞. △
r
r
|n|
2
|n|
n≥1
n6=0
|an cos nx + bn sin nx| ≤ |an | + |bn | ≤ 2
DEF. A function f (x) is said piecewise continuous in (a, b) ⊂ R if f
is defined and continuous except in a finite number of points xj ∈ (a, b),
j = 1, ..., N , where there exist the finite limits:
f (xj + 0) = limx→xj + (x), f (xj − 0) = limx→xj − f (x).
Such type of regularity is part of the Dirichlet condition, which is in
the hypothesis of the convergence criterion:
THEOREM I.3B (Dirichlet’s Criterion) Let the function f (x) be periodic (with period T ). Let both f (x) and f ′ (x) be piecewise continuous.
Then the trigonometric series associated with f is convergent to f (x)
in each point where f is continuous; while in each point x◦ where f is
discontinuous, the series is convergent to the half-sum
f (x◦ + 0) + f (x◦ − 0)
.
2
7
Figure 1. Graph of the Fourier series calculated by the
first 25 harmonics, i.e. for n from 0 to 12.
EXAMPLE I.3C (Fourier expansion of a periodic step function) - Let
1 x ∈ [2nπ, (2n + 1)π]
f (x) =
0 elsewhere.
which satisfies the Dirichlet conditions, with period T = 2π. The
Fourier coefficients a◦ , an , bn are directly calculated:
Z π
Z π
1
1
1
a◦
=
f (x) dx =
1 dx = .
2
2π −π
2π 0
2
For n ∈ N,
π
Z
Z
1 π
1 π
1 sin(nx)
an =
= 0,
f (x) cos(nx) dx =
cos(nx) dx =
π −π
π 0
π
n
0
π
Z
Z
1 π
1
cos(nx)
1 π
f (x) sin(nx) dx =
sin(nx) dx =
−
=
bn =
π −π
π 0
π
n
0
1
0,
n even
n
=
[1 − (−1) ] =
2
,
n odd .
nπ
nπ
2
Thus bn vanishes for even n, it equals nπ
for odd n. The Fourier series
is:
∞
∞
2
1 X
1 X
+
sin [(2n + 1)x] .
[an cos(nx) + bn sin(nx)] = +
2 n=1
2 n=0 (2n + 1)π
8
REMARK I.3D - (On the computation of Fourier coefficients)
a) In order to compute the coefficients of the Fourier series, several
remarks are useful.
A function f (x) is said to be even if f (−x) = f (x), ∀x; it is odd if
f (−x) = −f (x), ∀x. If the function f (x) is even, then all the Fourier
coefficients bn vanish: indeed, under a substitution y = −x,
Z
Z
2π
2π
2 T /2
2 T
f (x) sin( nx) dx =
f (x) sin( nx) dx =
bn =
T 0
T
T −T /2
T
2
T
Z
T /2
2
2π
f (−y) sin(− ny) dy = −
T
T
−T /2
Z
T /2
f (y) sin(
−T /2
2π
ny) dy = −bn
T
Thus 2bn = 0, so that bn = 0. Similarly, it turns out that a◦ =
an = 0 if the function is odd. Therefore an even periodic function
admits a series expansion with only cosine terms; while an odd periodic
function admits a series expansion with only sine terms. b) The next
remark regards an arbitrary function f (x), defined in some interval, for
example in (0, π). We can extend it both to an even function and to
an odd function. Indeed, we can define it in the interval (−π, 0) so
that it becomes even:

 f (x), x ∈ (0, π)
limx→0+ f (x), x = 0
f◦ (x) =
 f (−x), x ∈ (−π, 0)
Another choice is to define an odd extension of f (x):

x ∈ (0, π)
 f (x),
0,
x=0
f1 (x) =
 −f (−x), x ∈ (−π, 0)
Finally, one can extend the functions f◦ (x) and f1 (x) to the whole of
R, so that the resulting function is periodic with period 2π. Assuming
the Dirichlet conditions to hold, it follows that
∞
X
1
an cos(nx)
f◦ (x) = a◦ +
2
n=1
(17)
where
1
an =
π
Z
π
2
f◦ (x) cos(nx) dx =
π
−π
Z
π
f (x) cos(nx) dx
0
and
(18)
f1 (x) =
∞
X
n=1
bn sin(nx)
9
where
1
bn =
π
Z
π
2
f1 (x) sin(nx) dx =
π
−π
Z
π
f (x) sin(nx) dx
0
Indeed f1 (x)sin(nx) is even, as a product of two odd functions. Now
under a restriction of (??), (??) to the interval (0, π), it follows that a
function with domain (0, π) can be expanded in a cosine-series as well
as in a sine-series.
4. APPLICATION OF THE FOURIER SERIES TO
DIFFERENTIAL EQUATIONS
EX. I.4A - The damped harmonic oscillator with a periodic force
Let us compute a particular solution of the damped harmonic oscillator
with a forcing term
ẍ + µẋ + ω 2 x = f (t)
(19)
when f (t) is a real valued periodic function with period T . Although
not real-valued, it is convenient first to consider
f (t) = ρ · eiΩt
where Ω = 2π
is the pulsation of the forcing term. In this case a
T
particular solution in the form const. · eiΩt is easily found:
x∗ (t) =
ρeiΩt
ω 2 − Ω2 + iΩµ
As a second step let us consider f (t) given by a finite Fourier expansion
in exponential form:
N
X
f (t) =
cn eiΩnt
n=−N
where cn = c̄−n since f (t) is real valued. A particular solution with
period T is given by
∗
x (t) =
N
X
x∗n (t),
n=−N
∗
x∗n (t) =
cn eiΩnt
ω 2 − n2 Ω2 + inΩµ
Notice that x is real since the addends of the sum with opposite indices
n, −n are complex conjugate. Moreover x∗n (t) is a particular solution
of the differential equation
ẍ + µẋ + ω 2 x = cn eiΩnt .
10
Thus we can immediately verify:
∗
∗
2 ∗
ẍ + µẋ + ω x =
N
X
(ẍ∗n
+
µẋ∗n
+
ω 2 x∗n )
n=−N
=
N
X
cn eiΩnt = f (t).
n=−N
Finally, let f (t) admit a Fourier expansion with infinitely many nonzero
coefficients:
∞
X
f (t) =
cn eiΩnt
n=−∞
A particular solution is looked for in the form:
∞
X
cn
∗
eiΩnt
(20)
x (t) =
2
2
2 + inΩµ
ω
−
n
Ω
n=−∞
To this end we have to check the convergence of (??); then, provided
its convergence, to check whether is it a solution of the differential
equation. Now a sufficient condition to derive k times term by term
is the convergence of the series of the k−th order derivatives:
∞
X
cn (iΩn)k
(21)
eiΩnt
2 − n2 Ω2 + inΩµ
ω
n=−∞
uniformly with respect to t. On the other hand we know that f ∈ C r
implies the bound on the Fourier coefficients |cn | ≤ cn−r . Therefore
the general term of ??) is bounded by
cΩk
nk
p
≤ Cnk−r−2
nr (ω 2 − n2 Ω2 )2 + n2 µ2 Ω2
for some constant C > 0 independent of n. So the series (??) is uniformly convergent with respect to t if
r + 2 − k > 1;
In particular, the series (??) is a solution of the differential equation if
r + 2 − 2 > 1 (k = 2), i.e. if the function f (t) is at least in C 2 .
Notice: for µ small enough, the particular solution x∗ (t) amplifies the
harmonics n = ±[ω/Ω], (where [·] denotes the integral part), of the
function f (t) for which ω 2 − n2 Ω2 attains its minimum, while the other
harmonics are damped.
EXAMPLE I.4B Let us consider the one-dimensional heat equation
with boundary and initial conditions:
(22)
2
 ∂u
= 2 ∂∂xu2
∂t
u(0, t) = u(3, t) = 0,
∀t ≥ 0

u(x, 0) = 5 sin(4πx) − 3 sin(8πx) + 2 sin(10πx), x ∈ [0, 3].
Besides, the solution is required to remain bounded ∀(x, t) ∈ [0, 3]×R+ .
11
Let us solve the equation by separation of variables. By such a method,
we look for a solution (if possible) of the form
u(x, t) = f (x)g(t)
. Then the equation reads f (x)ġ(t) = 2f ′′ (x)g(t). After separating
variables,
ġ(t)
f ′′ (x)
=
.
2·
f (x)
g(t)
Since the left hand side depends only on x, the right hand side depends
only on t, and the equation holds for all (x, t) where x, t are independent
variables,
f ′′ (x)
ġ(t)
∃λ ∈ R : 2
= 2λ =
.
f (x)
g(t)
Therefore we obtain two ordinary differential equations depending on
an arbitrary parameter λ:
f ′′ = λf,
ġ = 2λg
with general solution
(23)
f (x) = c1 e
√
λx
+ c2 e −
√
λx
g(t) = ce2λt
,
f (x) = c1 x + c2 ,
g(t) = c
if λ 6= 0
if λ = 0.
Now, the boundedness requirement ∀t ∈+ implies λ < 0. Therefore,
setting λ = −ω 2 , by Euler’s formulae the general solution (??) can be
written:
f (x) = a cos(ωx) + b sin(ωx),
2
g(t) = ce−2ω t .
Hence a solution of the heat equation turns out to be
2
u(x, t) = f (x)g(t) = ce−2ω t [a cos(ωx) + b sin(ωx)] =
2
= e−2ω t [A cos(ωx) + B sin(ωx)]
with constants A, B, ω to be determined by using the boundary conditions in (??):
2
u(0, t) = 0 =⇒ A = 0, i.e. u(x, t) = e−2ω t B sin(ωx)
u(3, t) = 0 =⇒ B sin(3ω) = 0
This last equality, in turn, implies
either B = 0, arbitrary ω,
or arbitrary B, ω = nπ/3, n ∈ .
Now B = 0 is immediately rejected since it gives an identically vanishing solution. Therefore a nontrivial solution of the diffusion equation
in (??) with zero boundary condition has the form:
nπ
2 2
(24)
u(x, t) = Be−2n π t/9 sin( x), n ∈
3
12
Figure 2. Graph of the function u(x, t) for some values of
t: t = 0 (continuous line), t = 0.001 and t = 0.01
By the superposition principle (which holds for linear differential equations), a linear combination of solutions is still a solution of the homogeneous equation; so for any choice of n1 , n2 , n3 ∈ the function
n1 π
n2 π
n3 π
2 2
2 2
2 2
B1 e−2n1 π t/9 sin(
x)+B2 e−2n2 π t/9 sin(
x)+B3 e−2n3 π t/9 sin(
x)
3
3
3
is still a solution of the equation in (??). By comparing this type of
solution with the initial condition in (??) we find:
B1 = 5, n1 π/3 = 4π, B2 = −3, n2 π/3 = 8π, B3 = 2, n3 π/3 = 10π
i.e. n1 = 12, n2 = 24, n3 = 30. Thus the unique (bounded) solution
of the diffusion problem (??) is
2
2
2
u(x, t) = 5e−32π t sin(4πx) − 3e−128π t sin(8πx) + 2e−200π t sin(10πx)
EXAMPLE I.4C - We solve the same diffusion equation with a different
initial value function:


∂2u
 ∂u
=
2
2
∂t
∂x

u(0, t) = u(3, t) = 0,
∀t ≥ 0

u(x, 0) = 25x(3 − x), x ∈ [0, 3]
Again by separation of variables we obtain a solution of the heat equation, but it is no more sufficient to make a finite superposition of such
functions to solve the problem. We try by a superposition of infinitely
many functions:
∞
X
π
2 2
u(x, t) =
Bn e−2n π t/9 sin(n x),
3
n=1
13
Figure 3. Graph of the function u(x, t), for (x, t) ∈ (0, 3) × (0, 0.01)
where we require u(x, 0) = 25x(3 − x), that is
25x(3 − x) =
∞
X
π
Bn sin(n x), x ∈ (0, 3).
3
n=1
This amounts to expand in a sine Fourier series the function 25x(3−x).
According to Remark I.3D, this is possible if we regard the function
as a restriction of an odd periodic function defined for all x ∈, with
period 6:

 25x(3 − x), x ∈ (0, 3)
25x(3 + x), x ∈ (−3, 0)
f1 (x) =
 ... and so on.
Then we have the usual Fourier coefficients for an odd function of
period T :
Z
Z
2 T /2
4 T /2
2π
2π
Bn =
f1 (x) sin(n x) dx =
f1 (x) sin(n x) dx =
T −T /2
T
T 0
T
Z
2 3
900
=
25x(3 − x) sin(nπx/3) dx = 3 3 [1 − (−1)n ]
3 0
nπ
Finally
∞
1800 X
2 2
(2m + 1)−3 e−2(2m+1) π t/9 sin((2m + 1)πx/3).
u(x, t) = 3
π m=0
The graph of u(x, t) is first represented for some values of t (Fig. 4),
then for (x, t) ∈ (0, 3) × (0, 1) (Fig. 5).
14
Figure 4. For initial shape u(x, 0) = 25x(3 − x), graph of
the solution u(x, t) when t = 0, t = 0.01, t = 0.1
Figure 5. For initial shape u(x, 0) = 25x(3 − x), graph of
the solution u(x, t), (x, t) ∈ (0, 3) × (0, 1)
5. FOURIER TRANSFORM
DEFINITION I.5A (Direct Fourier transform) Let f = f (t) be a real
valued (or a complex valued) function of a real variable. The Fourier
transform of f is the function:
Z +∞
ˆ
(25)
F (f ) = f (ω) =
f (t)e−iωt dt, ω ∈ R
−∞
15
whenever such an integral exists ∀ω ∈ R.
The operator F , which transforms f (t) into fˆ(ω), is said ”Fourier
transform”.
PROPOSITIONS I.5B
Let f (t) be an absolutely integrable function in R, that is:
Z +∞
|f (t)|dt < +∞
(26)
−∞
Then
i) the Fourier transform fˆ(ω), for ω ∈ R, exists;
ii) the modulus of fˆ(ω) is an even function if f is real valued;
iii) limω→±∞ fˆ(ω) = 0 (Riemann’s lemma);
iv) fˆ is a continuous function ∀ω ∈ R.
Let f (t), f ′ (t) be piecewise continuous in any bounded interval, then
v) the inversion formula
Z +∞
Z +R
1
1
iωt
ˆ
(27) f (t) =
f (ω)e dω ≡
PV
lim
fˆ(ω)eiωt dω
R→+∞
2π
2π
−∞
−R
holds in all points t where f (t) is continuous, while the formula
Z +∞
1
1
fˆ(ω)eiωt dω
[f (t◦ + 0) + f (t◦ − 0)] =
PV
(28)
2
2π
−∞
holds in all discontinuity points t◦ .
Remarks.
(i) Since ω and t are real, |eiωt | = 1, and
Z +∞
Z +∞
−iωt
|
f (t)e
dt| ≤
|f (t)|dt < +∞
−∞
−∞
so the existence of fˆ is immediate if f is absolutely integrable. (ii) If
f = f¯ (real valued function), the complex conjugate of fˆ(ω) is equal
to the Fourier transform of f calculated in −ω:
Z +∞
Z +∞
¯ˆ
−iωt
¯
f (ω) = (
f (t)e
dt) =
f (t)eiωt dt = fˆ(−ω).
−∞
−∞
Now writing fˆ(ω) in exponential notation
fˆ(ω) = A(ω)eiφ(ω)
we notice that A(ω)e−iφ(ω) = A(−ω)eiφ(−ω) , so that A(ω) = A(−ω).
The proof of (iii)-(v) is omitted. Only remark that the integral in
(??) is computed as a Cauchy principal value. The Cauchy Principal
R +R
Value integral is convergent if the limit limR→+∞ −R (...) exists and is
16
finite. Recall that the usual generalized integral of a function g(t) is
convergent if the double limit
Z R
g(t) dt
lim
R,S→+∞
−S
exists and is finite with R, S diverging to +∞ independently of each
other. Of course if the integral is convergent in the generalized sense,
then it is convergent in the sense of the Cauchy principal value. But
the opposite implication is not true in general. An example is the
function g(t) = t, which is integrable in the Cauchy principal value
sense:
Z +R
Z +∞
t dt = 0.
P
t dt = lim
R→+∞
−∞
−R
The same function is not integrable in the generalized sense, since many
different
the double limit prescription: for examR R values are attainedRby
2R
ple −2R tdt → −∞, while −R tdt → +∞. A similar situation regards
any odd, piecewise continuous function.
Finally, in some textbooks the direct ad the inverse Fourier transforms
are defined in
√ a slightly different way, attributing to both the same
coefficient 1/ 2π (instead of 1 and 1/2π, respectively). Actually any
other choice is good, provided that the product of the two coefficients
is 1/2π. △
PROPOSITION I.5C Symmetry property If fˆ is the Fourier transform of f , and if both satisfy the inversion
conditions I.5B, (v), then
(29)
F [fˆ](ω) = 2πf (−ω)
Proof
R +∞
1
fˆ(t)eiωt dt = f (ω), then
Since the inversion formula ensures 2π
−∞
Z +∞
Z +∞
1
−iωt
ˆ
ˆ
f (t)e
dt = 2π
F (f ) =
fˆ(t)e−iωt dt = 2πf (−ω).
△
2π −∞
−∞
PROPOSITION I.5D (Computation of Fourier transforms)
(30)
(31)
(32)
1)
If pa (t) =
2) If f (t) =
1,
0,
sin(ω◦ t)
t
3) If f (t) = e
−αt2
t ∈ [−a, a]
|t| > a
then
then p̂a (ω) = 2

 π,
π/2,
fˆ(ω) =
 0,
then fˆ(ω) =
r
π − ω2
e 4α .
α
sin ωa
ω
|ω| < ω◦
ω = ±ω◦
|ω| > ω◦
17
Proof
1) Let pa (t) = I[−a,a] (t), for each a > 0. The Fourier transform (??)
is nothing else than
−iωt a
Z a
e
1
sin ωa
−iωt
e
dt =
p̂a (ω) =
= (eiωa − e−iωa ) = 2
−iω −a iω
ω
−a
for ω 6= 0, while p̂a (0) = 2a.
respectively.
Fig. 6 and Fig. 7 show pa (t) and p̂(ω),
2) Now consider
sin(a t)
.
t
By using (30), the given function f (t) can be seen as a Fourier transform:
1
sin(a t)
= pˆa (t),
f (t) =
t
2
where pa is the indicator function of [−a, a]. By applying the Fourier
transform to both sides and using the symmetry property I.5C,
1
1
fˆ(ω) ≡ F [p̂a ](ω) = 2π pa (−ω) = π · pa (−ω)
2
2
in all points where pa is continuous. The half-sum rule (??) holds in
the discontinuity points, so the explicit result is:


|ω| < a
 π,
π/2, ω = ±a 
(33)
fˆ(ω) =

0,
|ω| > a
f (t) =
3) Finally consider the Gauss integral and the Fourier transform of
the Gaussian
function.
R +∞
2
a) Let I = −∞ e−αx dx. By Fubini’s theorem
Z +∞
2 Z +∞
Z +∞
2
2
−αx2
−αx2
I =
e−αy dy
e
dx
e
dx =
−∞
−∞
−∞
=
Z
+∞ Z
−∞
+∞
e−α(x
2 +y 2 )
dxdy
−∞
Under a change of variables from cartesian to polar coordinates,
(
x = r cos θ
, dxdy → rdrdθ
y = r sin θ
the integral becomes
Z 2π Z +∞
π
2
2
2
e−αr rdrdθ = 2π[e−αr /(−2α)]+∞
= .
I =
0
α
0
0
p
So I = π/α.
18
Figure 6. Graph of pa (t).
b)
F [e
−αt2
]=
Z
e
R
−iωt −αt2
e
dt =
Z
e
R
√
√ )2
−( αt+ 2iω
α
ω2
· e− 4α dt
Now by Cauchy’s theorem of complex analysis, the integral performed
√ (parallel to the real line in the complex plane) is
on the line R + 2iω
α
p
equal to the integral performed on the real line, which is απ :
r
Z
√ 2
ω2
π − ω2
e 4α .
=
e−( αt) · e− 4α dt =
α
R
Therefore, up to a coefficient, the Fourier transform of a Gauss function
is still a Gauss function, with α replaced by −1/(4α). △
6. PROPERTIES OF THE FOURIER TRANSFORM
THEOREM I.6A (Linearity ) Let the functions f1 , f2 admit Fourier
transform F (f1 ) = fˆ1 and F (f2 ) = fˆ2 , respectively. Then for arbitrary
coefficients c1 , c2 ∈ there exists the Fourier transform of c1 f1 (t)+c2 f2 (t)
and
F (c1 f1 + c2 f2 ) = c1 fˆ1 + c2 fˆ2 .
Proof. A straightforward consequence of the linearity property of the
integral. F and F −1 are linear operators, too. △.
19
Figure 7. Graph of p̂a (ω).
THEOREM I.6B (Frequency shifting) Let the function f (t) admit
Fourier transform fˆ(ω). For any constant ω◦ ∈,
F [eiω◦ t f (t)] = fˆ(ω − ω◦ ).
Proof.
F [e
iω◦ t
f] =
Z
+∞
−∞
f (t)e−i(ω−ω◦ )t dt = fˆ(ω − ω◦ )
△
EX. I.6C (Fourier transform of a truncated cosine function)
Is it possible to transform, in a generalized sense, the cosine (sine)
function, which does not admit the integral transform I.5A in the strict
sense? An approximation to the right answer (I.8H iv) is clear if we
consider a truncated cosine, supported in intervals (−a, a) for arbitrarily large a > 0 :
f (t) = pa (t) · cos(ω◦ t),
pa (t) = I(−a,a) (t)
where pa (t) is the indicator of (−a, a) and ω◦ is nonzero and fixed. By
Euler’s formula, linearity, frequency shifting and by the above examples
I.5D,
1
fˆ(ω) = F [pa (t) cos(ω◦ t)] =
F (pa (t)eiω◦ t ) + F (pa (t)e−iω◦ t )
2
1
[p̂a (ω − ω◦ ) + p̂a (ω + ω◦ )]
2
sin[a(ω − ω◦ )] sin[a(ω + ω◦ )]
+
. △
=
ω − ω◦
ω + ω◦
=
20
Figure 8. Graph of the Fourier transform of the modulated
indicator function pa (t) cos(ω◦ t), with ω◦ = 7, a = 5.
As a → +∞, Fig. 8 shows that the Fourier transform appears more
and more concentrated on the frequency ω◦ (and, of course, −ω◦ since
this Fourier transform is an even function). △
Let us prove that the Fourier operator transforms derivatives in the
time representation into multiplication by corresponding monomials
in the frequency representation, and vice-versa. By this property the
Fourier is a powerful tool for differential equations.
THEOREM I.6D (Transformation of derivatives into multiplications
and vice-versa )
Let f ∈ C n () and let f (r) (t) be absolutely integrable on for each r ≤ n.
Then
F [f (n) ] = (iω)n fˆ(ω).
Similarly, assuming that tn f (t) is absolutely integrable,
dn fˆ(ω)
F [(−it) f (t)] =
.
dω n
Proof. The proof is based on integration by parts and the fact that
n
(34)
lim f (r) (t) = 0,
t→±∞
r = 0, 1, ..., n − 1
since f (r) (t), r = 0, 1, . . . , n is absolutely integrable on . We can realize
this fact by writing
Z t
(r)
(r)
f (t) = f (0) +
f (r+1) (τ )dτ.
0
21
Since f (r+1) (τ ) is absolutely integrable, then the following limit exists
and is finite:
Z +∞
(r)
(r)
f (r+1) (τ )dτ.
L = lim f (t) = f (0) +
t→+∞
0
Now, either L = 0 or L 6= 0 : we prove that the second case is impossible. Ab absurdo, let L 6= 0. Then there is an interval [T, +∞) in which
|f (r) (t)| > L/2, so that f (r) (t) cannot be absolutely integrable. Thus
(??) is proved. Consider
Z +∞
(n)
F (f )(ω) =
f (n) (t)e−iωt dt
−∞
= f
(n−1)
+∞
(t)e−iωt −∞
+ iω
Z
+∞
f (n−1) (t)e−iωt dt
−∞
(n−1)
= iω F (f
)(ω)
since limt→±∞ f
(t) = 0. By iterating such relation the statement
follows. Finally, by direct inspection the trasform of (−it)f (t) is equal
to dfˆ(ω)/dω, and the iteration on n concludes the proof. △
(n−1)
7. CONVOLUTION PRODUCT
DEFINITION I.7A -(Convolution product) The convolution product
of two functions f1 (t), f2 (t) defined for t ∈, is the generalized integral
(provided it exists)
Z +∞
f1 (x)f2 (t − x)dx.
(35)
(f1 ∗ f2 )(t) =
−∞
REMARK I.7B - For the convolution product the usual associative
and distributive (with respect to sum) properties hold. Moreover, the
commutative property can be easily proved: by a change of variable
t − x = y,
Z +∞
Z +∞
f1 ∗ f2 (t) =
f1 (x)f2 (t − x)dx =
f1 (t − y)f2 (y)dy = f2 ∗ f1 (t).
−∞
−∞
THM. I.7C - (Convolution theorem in time domain, convolution theorem in frequency domain)
If the functions f1 , f2 are sufficiently regular 1 then
(36)
F (f1 ∗ f2 )(ω) = fˆ1 (ω) · fˆ2 (ω)
1A
generic assumption of regularity so as to ensure the existence of the Fourier
transforms, the change in the order of integration, etc.
22
F −1 (fˆ1 ∗ fˆ2 )(t) = (f1 ∗ f2 )(t).
(37)
Scheme of proof.
Let us consider the Fourier transform of a convolution product:
Z +∞
Z +∞
−iωt
f1 (x)f2 (t − x)dx
e
dt
F (f1 ∗ f2 )(ω) =
−∞
−∞
=
Z
+∞
−∞
Z
+∞
e−iωt f1 (x)f2 (t − x)dt dx
−∞
By the change of variables (t, x) → (y, x), where x = x and y = t − x,
we obtain
Z +∞ Z +∞
e−iω(x+y) f1 (x)f2 (y)dy dx
F (f1 ∗ f2 )(ω) =
−∞
=
Z
+∞
e−iωx f1 (x) dx
−∞
and vice-versa
F
−1
Z
−∞
+∞
−∞
e−iωy f2 (y) dy = fˆ1 (ω) · fˆ2 (ω)
ˆ
ˆ
f1 · f2 (t) = (f1 ∗ f2 ) (t) △
EXAMPLE I.7D - (The diffusion equation on a line) Consider the
heat equation with initial condidion f (x) for ∈:

2
 ∂u
= k ∂∂xu2
∂t
u(x, 0) = f (x), −∞ < x < +∞

|u(x, t)| < M, −∞ < x < +∞, t > 0
In this problem it is useful to consider the Fourier transform of both
sides with respect to x, so that the second order derivative (with respect
to x) is transformed into a multiplication by the square of the conjugate
variable (Thm. I.6F). Denoting p the Fourier variable conjugate with
x, we obtain:
dF [u]
= −kp2 · F [u]
dt
where the Fourier transform of u is denoted by F [u]. Here we deal with
a first order ordinary differential equation, since the unknown function
F [u] no more depends on x. The general solution of such ordinary
equation is:
(38)
F [u(x, t)] = Ce−kp
2t
Setting t = 0 in (??) we see that
F [u(x, 0)] = F [f (x)] = C
so that
2
F [u] = fˆ(p) · e−kp t .
23
Now we can apply the convolution theorem (Thm. 7.1):
2
u(x, t) = f (x) ∗ F −1 [e−kp t ]
From (??) we know that
r
r
π −p2 /4α
α −αx2
−1 −p2 /(4α)
−αx2
e
, or F [e
]=
e
F [e
]=
α
π
With α = 1/4kt, it implies
F
−1
[e
−kp2 t
]=
r
1 −x2 /4kt
e
.
4kπt
Then the solution is the convolution product
Z +∞
1
1
2
−x2 /4kt
u(x, t) = f (x) ∗ √
=
f (z) √
e
e−(x−z) /4kt dz.
4πkt
4πkt
−∞
Notice that in this formula the well known ”heat kernel” appears:
G(z; t) = √
1
2
e−z /4kt
4πkt
which coincides with a Gaussian probability density with mean zero
and variance 2kt. So we have three facts: a) the diffusion equation on
the line with initial condition u(x, 0) = f (x) is solvable with a known
explicit kernel:
Z +∞
f (z)G(x − z; t)dz.
u(x, t) =
−∞
b) The physical interpretation of u(x, t) as the temperature at time t
suggests that a choice of a compact support initial temperature f (x)
gives rise to a temperature which is at once everywhere nonzero (!):
u(x, t) 6= 0 ∀x ∈, ∀t > 0. An instantaneous diffusion to all points of
space takes place. Setting σ 2 = 2kt, the convolution of any function
2
2
f (x) with the probability density σ√12π e−z /2σ is convergent to f as
σ → 0.
24
Figure 9. Graph of the solution of the heat equation, starting from an initial function u(x, 0) with compact support,
the indicator of (0, 10). Its diffusion is represented at times
t = 0.01, t = 1, t = 10.
Figure 10. Graph of the solution of the heat equation,
starting from an initial function u(x, 0) with compact support, the indicator of (0, 10). Evidence of its ”diffusion” is
given at times t = 0.01, t = 1, t = 10.
8. TEST FUNCTIONS AND DIRAC’S DELTA
Let ϕ be a function of one real variable. We recall that the support
of a function is the closure of the set where the function is nonzero:
25
supp(ϕ) = {x ∈ R : ϕ(x) 6= 0}. Also, we recall that a compact subset
of R is a closed and bounded set.
A test function is an infinitely differentiable function with compact
support in R. The set of test functions is denoted C◦∞ (R).
REMARK I.8A - Such functions do exist: an example is
exp(−1/(1 − |x|2 ), |x| < 1
ϕ(x) =
0,
|x| ≥ 1.
Although the derivatives of 1/(1 − |x|2 ) are divergent as x → 1− (as
x− → −1+ ) , we see that all derivatives of ϕ exist and are zero in 1− (in
−1+ ) because they contain the exponentially small factor exp(−1/(1 −
|x|2 ). So ϕ ∈ C◦∞ (R) and supp(ϕ) = {x ∈ Rn : |x| ≤ 1} = B1 (0),
i.e. the support is the unitary ball centered at the origin. Similar test
functions can be written with support coinciding with a given arbitrary
compact set of R.
DEF. I.8B - (The space of test functions) In the set C◦∞ (R), of infinitely
differentiable functions with compact support, a sequence {ϕn } is said
to converge to ϕ ∈ C◦∞ if
(i) there is a compact set K such that supp(ϕn ) ⊂ K, ∀n ∈ N
(k)
(ii) the sequence {ϕn }n∈N of the k−th order derivatives is uniformly
convergent to ϕ(k) as n → ∞, for k = 0, 1, 2... The vector space C◦∞ (R),
when endowed with such notion of convergence, is said ”the space of
test functions”’ and is denoted by D(R).
Notice that the space D(R) is not normalizable, i.e. it is impossible to
introduce a norm compatible with the above type of convergence.
DEF. I.8C - (The space of distributions) Let D(R) be the space of test
functions on R. A map T : D(R) → R is said a linear functional if
< T, (λ1 ϕ1 + λ2 ϕ2 ) > = λ1 < T, ϕ1 > +λ2 < T, ϕ2 > .
T is continuous if
ϕn → ϕ
in D(R),
=⇒ < T, ϕn >→< T, ϕ > in R.
The space D∗ (R) of linear and continuous functionals on D(R) is said
the ”space of distributions” and each element T ∈ D∗ (R) is said a
”distribution”.
D∗ (R) is endowed with a topology by means of the notion of weak
convergence:
26
EXAMPLE I.8D - (Regular distributions). Let f be a ”locally integrable”’ real function on R:
Z
1
f ∈ Lloc (R), i.e. ∃
|f (x)| dx < +∞, ∀ compact K ⊂ R.
K
The map Tf : D(R) → R is defined by
Z
< Tf , ϕ >=
f · ϕ dx, ϕ ∈ D(R).
Ω
Such a map is well defined since ϕ has compact support in R. Moreover
Tf is linear and continuous: indeed, if ϕn → ϕ in D(R), there exists a
compact K ⊂ R such that all ϕn vanish out of K. Since the convergence
is uniform in K,
Z
Z
| < Tf , ϕn > − < Tf , ϕ > | = | (f ϕn − f ϕ)dx| ≤ ǫ
|f |dx
R
K
if only n is larger than some nǫ . Therefore < Tf , ϕn >→< Tf , ϕ >, i.e.
Tf is continuous. Therefore Tf is a distribution, Tf ∈ D∗ (R). Distributions of this kind are said ”regular”. All the remaining distributions,
which do not admit a representation by means of L1loc functions, are
said ”singular”.
DEFINITION I.8E - Dirac’s delta - Let a ∈ R. Dirac’s delta, in the
point a, is the linear functional by which the value of ϕ in a is associated
to any test function ϕ ∈ D(R):
< δa , ϕ >:= ϕ(a).
REMARK I.8F - Such a linear functional on D(R) is a distribution in
the sense of Def. I.8C. Indeed it is continuous: if ϕn → ϕ in D(R), a
fortiori the point convergence holds, so that
< δa , ϕn >≡ ϕn (a) → ϕ(a) ≡< δa , ϕ >
as n → ∞
In case a = 0, it is simply said ”Dirac’s delta”:
< δ, ϕ >= ϕ(0), ∀ϕ ∈ D(R)
. It is a ”singular” distribution in Rthe sense that there exists no function
f ∈ L1loc (R) such that < δ, ϕ >= R f (x)ϕ(x). Nevertheless it is usual
to write δ(x) , just as if it were a function, in a conventional notation:
Z ∞
Z ∞
δ(x − a)ϕ(x)dx = ϕ(a).
δ(x)ϕ(x)dx = ϕ(0),
−∞
−∞
This notation is currently accepted, but in the sense of definitions I.8C,
I.8E. Such a notation can be understood by approximants, too:
THM. I.8G - Sequence of approximants of Dirac’s delta -
27
Let fRn be a sequence of probability densities, i.e. fn ∈ L1loc , with fn ≥ 0,
and R fn (x)dx = 1, ∀n ∈ N . Let, moreover, fn (x) → 0 ∀x 6= 0. Then
Z
lim
fn (x)ϕ(x)dx = ϕ(0), as n → ∞.
n→+∞
R
The same is true for other sequences, such as:
fn = i
e−inx
πx
.
Proof. For the proof we restrict to a simple example of approximants:
n
n
, x ∈ [− n1 , n1 ]
2
fn (x) = I[−1/n,1/n] (x) =
0, elsewhere.
2
Indeed
n
min
ϕ(x)
=
m
n ≤
1
1
2
[− n
,+ n
]
Z
1
+n
1
−n
ϕ(x)dx ≤ Mn = max
ϕ(x)
1
1
[− n ,+ n ]
where, by continuity in 0,
lim mn = ϕ(0),
n→∞
whence the formula.
lim Mn = ϕ(0),
n→+∞
△
In the proper sense (I.5A) we know that the Fourier transform of functions such as constants or cosine functions do not exist. But such
transforms can make sense in the wider framework of distributions. To
this end the formal use of usual properties (linearity, frequency shifting,
etc. ) gives the right results:
EXAMPLES I.8H i) F [δ] ≡
Z
Z
e−iωt δ(t)dt = 1,
F −1 [1] = δ.
R
e−iωt dt ≡ F [1] = 2πδ(ω), F −1 [2πδ(ω)] = 1.
R
Z
iω◦ t
iii)F [e ] =
e−i(ω−ω◦ )t dt ≡ F [1](ω − ω◦ ) = 2πδ(ω − ω◦ )
ii)
R
1
1
iv)F [cos(ω◦ t)] = F (eiω◦ t ) + F (e−iω◦ t )
2
2
= π[δ(ω − ω◦ ) + δ(ω + ω◦ )].
(Compare (iv) with the Fourier transform of the truncated cosine function
sin[a(ω − ω◦ )] sin[a(ω + ω◦ )]
+
,
F [pa (t) cos(ω◦ t)] =
ω − ω◦
ω + ω◦
which tends to zero, as a → ∞, ∀ω 6= ±ω◦ ).
28
In order to understand how to generate a derivative of a distribution,
let us consider the simple case of a differentiable function f , at least
with continuous derivative in an open connected set R. We consider
the distribution Tf ′ associated with f ′ (x). For any ϕ ∈ D(R), take
a, b such that supp(ϕ) ⊂ [a, b] ⊂ R; integrating by parts we obtain:
Z
Z b
′
b
< Tf ′ , ϕ >=
f (x)ϕ(x) dx = [f (x)ϕ(x)]a −
f (x)ϕ′ (x) dx
R
a
′
Taking into account that ϕ ∈ D(R), it follows that
< Tf ′ , ϕ > = − < f, ϕ′ >
is the distribution which is determined by f ′ .
DEFINITION I.8I - The derivative of a distribution. Let r = 1, 2, 3, .... The linear map from D∗ (R) to D∗ (R) defined by
T → Dr T, with
< Dr T, ϕ, >= (−1)r < T, Dr ϕ >
is said r−th derivative, and Dr T is the r-th order derivative of the
distribution T.
EXAMPLE I.8L- The derivatives of the Heaviside function, of δ and
δ′. Let us consider the Heaviside function
1,
x>0
H(x) =
0, elsewhere.
The corrspoinding distribution
< Tf , ϕ >=
has the derivative:
′
< DTf , ϕ >= − < Tf , ϕ >= −
Z
∞
ϕ(x) dx
0
Z
∞
ϕ′ (x) dx = ϕ(0) =< δ, ϕ >
0
Therefore the derivative of the Heaviside function is the Dirac’s delta
distribution.
The derivative of δ is given by:
< Dδ, ϕ >= − < δ, ϕ′ >= −ϕ′ (0).
The second derivative of delta is:
< D2 δ, ϕ >= (−1)2 < δ, ϕ′′ >= ϕ′′ (0).
and so on.
EX. I.8M - Let f : R → R be continuous except in the points
a1 < a2 < ... < an . For sake of simplicity, fix ideas with n = 2. Let
f admit a continuous derivative in the intervals (−∞, a1 ), (a1 , a2 ),
29
+
−
(a2 , +∞). Let, moreover, exists the finite limits f (a−
1 ), f (a1 ), f (a2 ),
f (a+
2 ). f is locally integrable, so the distribution Tf makes sense:
< Tf , ϕ >=
Z
f (x)ϕ(x) dx =
R
s Z
X
k=0
ak+1
f (x)ϕ(x) dx
ak
The derivative is:
′
< DTf , ϕ >= − < Tf , ϕ >= −
=
1
−[f ϕ]a−∞
Z
R−{a1 ,a2 }
+
Z
Z
a1
′
−∞
ϕf −
[f ϕ]aa21
+
a1
′
−∞
Z
fϕ −
Z
a2
′
a1
fϕ −
a2
′
a1
ϕf −
[f ϕ]+∞
a2
+
Z
Z
∞
f ϕ′
a2
+∞
ϕf ′ =
a2
−
+
−
ϕf ′ dx + ϕ(a1 )[f (a+
1 ) − f (a1 )] + ϕ(a2 )[f (a2 ) − f (a2 )].
Therefore
n
X
d
−
Tf = Tf ′ +
[f (a+
k ) − f (ak )]δak .
dx
k=1
In probability, for example, this means that the derivative of a discrete
distribution function F (x) is a ”probability density” in the distributional sense:
n
X
d
F (x) = F ′ (x) · IR−{a1 ,...,an } +
P (X = ak ) · δ(x − ak ).
dx
k=1
30
fˆ(ω)
f (t)
c1 f1 (t) + c2 f2 (t) c1 fˆ1 (ω) + c2 fˆ2 (ω)
f ′ (t)
iω fˆ(ω)
f (n) (t)
(iω)n fˆ(ω)
tn f (t)
in d dωf (ω)
n
f (t − t◦ )
e−iωt◦ fˆ(ω)
eiω◦ t f (t)
(f1 ∗ f2 )(t)
n
ˆ
fˆ(ω − ω◦ )
fˆ1 (ω) · fˆ2 (ω)
f1 · f2
1 ˆ
(f
2π 1
∗ fˆ2 )(ω)
sin(ω◦ t)/t
pa (t) cos(ω◦ t)
fˆ(ω) = π · I(−ω◦ ,ω◦ ) (ω)
p π −ω2 /(4α)
e
α
δ(t)
1
1
2πδ(ω)
δ(t − a)
e−iωa
e−iω◦ t
2πδ(ω + ω◦ )
pa (t) = I(−a,a) (t) 2 sin(aω)/ω
e−αt
2
eiω◦ t
cos(ω◦ t)
H(t) = I(0,∞) (t)
sin[a(ω−ω◦ )]
ω−ω◦
+
sin[a(ω+ω◦ )]
ω+ω◦
2πδ(ω − ω◦ )
π[δ(ω − ω◦ ) + δ(ω + ω◦ )]
πδ(t) + 1/(iω)