Chapter 2

Fourier Integrals
2.1

L1-Theory

Repetition: R = (, ),
1

f L (R)
2

f L (R)

|f (t)|dt < (and f measurable)

|f (t)|2dt < (and f measurable)

Definition 2.1. The Fourier transform of f L1 (R) is given by

Z

f () =
e2it f (t)dt, R

Comparison to chapter 1:

f L1 (T) f(n)
defined for all n Z

f L1 (R) f()
defined for all R
Notation 2.2. C0 (R) = continuous functions f (t) satisfying f (t) 0 as
t . The norm in C0 is

kf kC0 (R) = max|f (t)| (= sup|f (t)|).

tR

tR

Compare this to c0 (Z).

Theorem 2.3. The Fourier transform F maps L1 (R) C0 (R), and it is a
C (R) kf kL1 (R) , i.e.,
contraction, i.e., if f L1 (R), then f C0 (R) and kfk
0
36

CHAPTER 2. FOURIER INTEGRALS

37

i) f is continuous
ii) f() 0 as
iii) |f()|

|f (t)|dt, R.

Note: Part ii) is again the Riemann-Lesbesgue lemma.

Proof. iii) The same as the proof of Theorem 1.4 i).
ii) The same as the proof of Theorem 1.4 ii), (replace n by , and prove this
first in the special case where f is continuously differentiable and vanishes outside
of some finite interval).
i) (The only new thing):
+ h) f()|

|f(

=
=
-ineq.

ZR




e2i(+h)t e2it f (t)dt




e2iht 1 e2it f (t)dt

|e2iht 1||f (t)| dt 0 as h 0

(use Lesbesgues dominated convergens Theorem, e2iht 1 as h 0, and

|e2iht 1| 2).

Question 2.4. Is it possible to find a function f L1 (R) whose Fourier trans-

form is the same as the original function?

Answer: Yes, there are many. See course on special functions. All functions
which are eigenfunctions with eigenvalue 1 are mapped onto themselves.
Special case:
2
0 () = e2 , R
Example 2.5. If h0 (t) = et , t R, then h

Proof. See course on special functions.

Note: After rescaling, this becomes the normal (Gaussian) distribution function.
This is no coincidence!
Another useful Fourier transform is:
Example 2.6. The Fejer kernel in L1 (R) is
F (t) =

sin(t)
2
.
t

38

CHAPTER 2. FOURIER INTEGRALS

The transform of this function is

(
1 || ,
F () =
0
,

|| 1,

otherwise.

Proof. Direct computation. (Compare this to the periodic Fejer kernel on page
23.)
Theorem 2.7 (Basic rules). Let f L1 (R), , R

g() = e2i f()

g() = f( )

a) g(t) = f (t )

b) g(t) = e2i t f (t)

g() = f()
g() = f()

c) g(t) = f (t)
d) g(t) = f (t)
e) g(t) = f (t)
f ) g L1 and h = f g
g)
h)

g(t) = 2itf (t)

and g L1

f is absolutely continuous

and f = g L (R)

g() = f( ) ( > 0)

g ()
h()
= f()
(
f C 1 (R),
and

f () = g()

g() = 2i f().

Proof. (a)-(e): Straightforward computation.

(g)-(h): Homework(?) (or later).
The formal inversion for Fourier integrals is
Z

f () =
e2it f (t)dt
Z

?
f (t) =
e2it f()d

This is true in some cases in some sense. To prove this we need some
additional machinery.
Definition 2.8. Let f L1 (R) and g Lp (R), where 1 p . Then we
define

(f g)(t) =

f (t s)g(s)ds

for all those t R for which this integral converges absolutely, i.e.,
Z
|f (t s)g(s)|ds < .
R

39

CHAPTER 2. FOURIER INTEGRALS

Lemma 2.9. With f and p as above, f g is defined a.e., f g Lp (R), and

kf gkLp (R) kf kL1 (R) kgkLp (R) .
If p = , then f g is defined everywhere and uniformly continuous.
Conclusion 2.10. If kf kL1 (R) 1, then the mapping g 7 f g is a contraction
from Lp (R) to itself (same as in periodic case).

Proof. p = 1: same proof as we gave on page 21.

p = : Boundedness of f g easy. To prove continuity we approximate f by a

function with compact support and show that kf (t) f (t + h)kL1 0 as h 0.

p 6= 1, : Significantly harder, case p = 2 found in Gasquet.

Notation 2.11. BUC(R) = all bounded and continuous functions on R. We

use the norm

kf kBU C(R) = sup|f (t)|.

tR

Theorem 2.12 (Approximate identity). Let k L1 (R), k(0)

=

1, and define

t R, > 0.

k (t) = k(t),
If f belongs to one of the function spaces

a) f Lp (R), 1 p < (note: p 6= ),

b) f C0 (R),
c) f BUC(R),
then k f belongs to the same function space, and
k f f

as

in the norm of the same function space, i.e.,

kk f f kLp (R) 0 as if f Lp (R)
(
if f BUC(R),
suptR |(k f )(t) f (t)| 0 as
or f C0 (R).
It also conveges a.e. if we assume that

R
0

(sups|t| |k(s)|)dt < .

k(t)dt =

40

CHAPTER 2. FOURIER INTEGRALS

Proof. The same as the proofs of Theorems 1.29, 1.32 and 1.33. That is,
the computations stay the same, but the bounds of integration change (T R),
and the motivations change a little (but not much). 
Example 2.13 (Standard choices of k).
i) The Gaussian kernel
2
2

k(t) = et , k()
= e .

This function is C and nonnegative, so

Z
Z

k(t)dt = k(0)
= 1.
kkkL1 = |k(t)|dt =
R

ii) The Fejer kernel

F (t) =

sin(t)2
.
(t)2

It has the same advantages, and in addition

F () = 0 for || > 1.
The transform is a triangle:
F () =

1 ||, || 1
0,

|| > 1

F( )

iii) k(t) = e2|t| (or a rescaled version of this function. Here

k()
=

1
, R.
1 + ()2

Same advantages (except C )).

41

CHAPTER 2. FOURIER INTEGRALS

Comment 2.14. According to Theorem 2.7 (e), k () k(0)

= 1 as

, for all R. All the kernels above are low pass filters (non causal).

It is possible to use one-sided (causal) filters instead (i.e., k(t) = 0 for

t < 0). Substituting these into Theorem 2.12 we get approximate identities,
which converge to a -distribution. Details later.
Theorem 2.15. If both f L1 (R) and f L1 (R), then the inversion formula
Z
f (t) =
e2it f()d
(2.1)

is valid for almost all t R. By redefining f on a set of measure zero we can

make it hold for all t R (the right hand side of (2.1) is continuous).
Proof. We approximate
R

e2it f()d by

2
2
(where > 0 is small)
e2it e f()d
R 2it2 2 R 2is
= Re
e
f (s)dsd
(Fubini)
R
Z
R
2
2
= sR f (s)
e2i(st) e|{z} dds (Ex. 2.13 last page)

k( 2 )

{z

()

() The Fourier transform of k( 2 ) at the point s t. By Theorem 2.7 (e) this

is equal to

1 s t
)=
= k(

1 t s
k(
)

(since k()
= e is even).

The whole thing is

Z

1
f (s) k

sR

ts

ds = (f k 1 )(t) f L1 (R)

as 0+ according to Theorem 2.12. Thus, for almost all t R,

Z
2
2
f (t) = lim e2it e f()d.
0

On the other hand, by the Lebesgue dominated convergence theorem, since

2
2
|e2it e f()| |f()| L1 (R),
Z
Z
2it 2 2

e
f ()d =
e2it f()d.
lim e

42

CHAPTER 2. FOURIER INTEGRALS

Thus, (2.1) holds a.e. The proof of the fact that

Z
e2it f()d C0 (R)
R

is the same as the proof of Theorem 2.3 (replace t by t).

The same proof also gives us the following approximate inversion formula:
Theorem 2.16. Suppose that k L1 (R), k L1 (R), and that
Z

k(t)dt = 1.
k(0)
=
R

If f belongs to one of the function spaces

a) f Lp (R), 1 p <
b) f C0 (R)
c) f BUC(R)
then

e2it k()
f()d
f (t)

in the norm of the given space (i.e., in Lp -norm, or in the sup-norm), and also
R
a.e. if 0 (sups|t| |k(s)|)dt < .
Proof. Almost the same as the proof given above. If k is not even, then we

end up with a convolution with the function k (t) = 1 k(t/) instead, but we
can still apply Theorem 2.12 with k(t) replaced by k(t).

Corollary 2.17. The inversion in Theorem 2.15 can be interpreted as follows:

If f L1 (R) and f L1 (R), then,

f(t) = f (t) a.e.

Here f(t) = the Fourier transform of f evaluated at the point t.

Proof. By Theorem 2.15,
f (t) =

e2i(t) f()d
|R
{z
}

Fourier transform of f at the point (t)

a.e.

43

CHAPTER 2. FOURIER INTEGRALS

Corollary 2.18. f(t) = f (t) (If we repeat the Fourier transform 4 times, then
we get back the original function). (True at least if f L1 (R) and f L1 (R). )
As a prelude (=preludium) to the L2 -theory we still prove some additional results:
Lemma 2.19. Let f L1 (R) and g L1 (R). Then
Z
Z

f (t)
g (t)dt =
f(s)g(s)ds
R

Proof.
Z

f (t)
g (t)dt =

=
=

f (t)
tR
Z Z
ZsR

e2its g(s)dsdt (Fubini)

sR

2ist
f (t)e
dt g(s)ds

tR

f(s)g(s)ds.


sR

L1 (R). Then
Theorem 2.20. Let f L1 (R), h L1 (R) and h
Z
Z

f (t)h(t)dt =
f()h()d.
R

(2.2)

Specifically, if f = h, then (f L2 (R) and)

kf kL2 (R) = kfkL2 (R) .
R

Proof. Since h(t) = R e2it h()d

we have
Z
Z
Z

f (t)h(t)dt =
(Fubini)
f (t)
e2it h()ddt
R
tR
R

Z Z
2ist

=
f (t)e
dt h()d
sR
tR
Z


=
f()h()d.
R

2.2

Rapidly Decaying Test Functions

(Snabbt avtagande testfunktioner).

Definition 2.21. S = the set of functions f with the following properties
i) f C (R) (infinitely many times differentiable)

(2.3)

44

CHAPTER 2. FOURIER INTEGRALS

ii) tk f (n) (t) 0 as t and this is true for all

k, n Z+ = {0, 1, 2, 3, . . . }.
Thus: Every derivative of f 0 at infinity faster than any negative power of t.

Note: There is no natural norm in this space (it is not a Banach space).
However, it is possible to find a complete, shift-invariant metric on this space (it
is a Frechet space).
2

Example 2.22. f (t) = P (t)et S for every polynomial P (t). For example,
the Hermite functions are of this type (see course in special functions).

Comment 2.23. Gripenberg denotes S by C (R). The functions in S are called

rapidly decaying test functions.

The main result of this section is

Theorem 2.24. f S f S
That is, both the Fourier transform and the inverse Fourier transform maps this
class of functions onto itself. Before proving this we prove the following
Lemma 2.25. We can replace condition (ii) in the definition of the class S by
one of the conditions
R
iii) R |tk f (n) (t)|dt < , k, n Z+ or
iv) |

d n k
t f (t)|
dt

0 as t , k, n Z+

without changing the class of functions S.

Proof. If ii) holds, then for all k, n Z+ ,
sup|(1 + t2 )tk f (n) (t)| <
tR

(replace k by k + 2 in ii). Thus, for some constant M,

Z
M
k (n)
=
|tk f (n) (t)|dt < .
|t f (t)|
1 + t2
R
Conversely, if iii) holds, then we can define g(t) = tk+1 f (n) (t) and get
g (t) = (k + 1)tk f (n) (t) + tk+1f (n+1) (t),
|
{z
} |
{z
}
L1

L1

45

CHAPTER 2. FOURIER INTEGRALS

so g L1 (R), i.e.,

This implies

|g(t)| |g(0) +

|g(0)| +
|g(0)| +

|g (t)|dt < .
t

g (s)ds|

Z0

|g (s)|ds

Z0

|g (s)|ds = |g(0)| + kg kL1 ,

so g is bounded. Thus,
1
tk f (n) (t) = g(t) 0 as t .
t
The proof that ii) iv) is left as a homework.

Proof of Theorem 2.24. By Theorem 2.7, the Fourier transform of

 k
d
k (n)

(2it) f (t) is
(2i)n f().
d

Therefore, if f S, then condition iii) on the last page holds, and by Theorem
2.3, f satisfies the condition iv) on the last page. Thus f S. The same argument with e2it replaced by e+2it shows that if f S, then the Fourier
inverse transform of f (which is f ) belongs to S.

Note: Theorem 2.24 is the basis for the theory of Fourier transforms of distributions. More on this later.

2.3

L2-Theory for Fourier Integrals

As we saw earlier in Lemma 1.10, L2 (T) L1 (T). However, it is not true that

L2 (R) L1 (R). Counter example:

f (t) =
(too large at ).

L (R)

1
6 L1 (R)

1 + t2

C (R)

So how on earth should we define f() for f L2 (R), if the integral

Z
e2int f (t)dt
R

CHAPTER 2. FOURIER INTEGRALS

46

does not converge?

Recall: Lebesgue integral converges converges absolutely
Z
|e2int f (t)|dt < f L1 (R).
We are saved by Theorem 2.20. Notice, in particular, condition (2.3) in that
theorem!
Definition 2.26 (L2 -Fourier transform).
i) Approximate f L2 (R) by a sequence fn S which converges to f in

L2 (R). We do this e.g. by smoothing and cutting (utjamning och

2

klippning): Let k(t) = et , define

kn (t) = nk(nt), and
 
t
(kn f )(t)
fn (t) = k
n | {z }
| {z }

|
{z
}

the product belongs to S

() this tends to zero faster than any polynomial as t .

() smoothing by an approximate identity, belongs to C and is bounded.


By Theorem 2.12 kn f f in L2 as n . The functions k nt tend
to k(0) = 1 at every point t as n , and they are uniformly bounded
by 1. By using the appropriate version of the Lesbesgue convergence we
let fn f in L2 (R) as n .
ii) Since fn converges in L2 , also fn must converge to something in L2 . More
about this in Analysis II. This follows from Theorem 2.20. (fn f
fn Cauchy sequence fn Cauchy seqence fn converges.)
iii) Call the limit to which fn converges The Fourier transform of f , and
denote it f.
Definition 2.27 (Inverse Fourier transform). We do exactly as above, but replace e2it by e+2it .
Final conclusion:

47

CHAPTER 2. FOURIER INTEGRALS

Theorem 2.28. The extended Fourier transform which we have defined above
has the following properties: It maps L2 (R) one-to-one onto L2 (R), and if f is the
Fourier transform of f , then f is the inverse Fourier transform of f. Moreover,
all norms, distances and inner products are preserved.
Explanation:
i) Normes preserved means
Z
Z
2
|f (t)| dt = |f()|2d,
R

L2 (R) .
or equivalently, kf kL2 (R) = kfk

ii) Distances preserved means

kf gkL2(R) = kf gkL2 (R)
(apply i) with f replaced by f g)
iii) Inner product preserved means
Z
Z
g ()d,
f()
f (t)g(t)dt =
R

which is often written as

hf, giL2(R) = hf, giL2 (R) .

This was theory. How to do in practice?
One answer: We saw earlier that if [a, b] is a finite interval, and if f L2 [a, b]

f L1 [a, b], so for each T > 0, the integral

Z T

fT () =
e2it f (t)dt
T

is defined for all R. We can try to let T , and see what happens. (This
resembles the theory for the inversion formula for the periodical L2 -theory.)
Theorem 2.29. Suppose that f L2 (R). Then
Z T
lim
e2it f (t)dt = f()
T

in the L -sense as T , and likewise

Z T
lim
e2it f()d = f (t)
T

in the L2 -sense.

48

CHAPTER 2. FOURIER INTEGRALS

Proof. Much too hard to be presented here. Another possibility: Use the Fejer
kernel or the Gaussian kernel, or any other kernel, and define


R

f()
= limn R e2it k nt f (t)dt,
R


f (t) = limn e+2it k f()d.
n

We typically have the same type of convergence as we had in the Fourier inversion
formula in the periodic case. (This is a well-developed part of mathematics, with
lots of results available.) See Gripenbergs compendium for some additional
results.

2.4

An Inversion Theorem

From time to time we need a better (= more useful) inversion theorem for the
Fourier transform, so let us prove one here:
Theorem 2.30. Suppose that f L1 (R) + L2 (R) (i.e., f = f1 + f2 , where
f1 L1 (R) and f2 L2 (R)). Let t0 R, and suppose that
Z t0 +1 

 f (t) f (t0 ) 
dt < .

t t0
t0 1

Then

f (t0 ) = lim

where f()
= f1 () + f2 ().

S
T

e2it0 f()d,

(2.4)

(2.5)

Comment: Condition (2.4) is true if, for example, f is differentiable at the point
t0 .
Proof. Step 1. First replace f (t) by g(t) = f (t + t0 ). Then

g() = e2it0 f(),

and (2.5) becomes
g(0) = lim

S
T

and (2.4) becomes

g()d,
S

Z 1

 g(t t0 ) g(0) 

dt < .
t t0
1

Thus, it suffices to prove the case where t0 = 0 .

49

CHAPTER 2. FOURIER INTEGRALS

Step 2: We know that the theorem is true if g(t) = et (See Example 2.5 and
Theorem 2.15). Replace g(t) by
2

h(t) = g(t) g(0)et .

Then h satisfies all the assumptions which g does, and in addition, h(0) = 0.
Thus it suffices to prove the case where both () t0 = 0 and f (0) = 0 .
For simplicity we write f instead of h but assume (). Then (2.4) and (2.5)
simplify:
Z 1

 f (t) 

dt < ,
t
1
Z T

lim
f()d
= 0.

S
T

(2.6)
(2.7)

Step 3: If f L1 (R), then we argue as follows. Define

g(t) =

f (t)
.
2it

Then g L1 (R). By Fubinis theorem,

Z T
Z TZ

f ()d =
e2it f (t)dtd
S
S
Z Z T
=
e2it df (t)dt
=
=

1
e2it
2it

T

f (t)dt

e2iT t e2i(S)t

= g(T ) g(S),

 f (t)
dt
2it

and this tends to zero as T and S (see Theorem 2.3). This proves

(2.7).

Step 4: If instead f L2 (R), then we use Parsevals identity

Z
Z
h()d

f()
f (t)h(t)dt =

in a clever way: Choose

h()
=

1, S t T,

0, otherwise.

50

CHAPTER 2. FOURIER INTEGRALS

is
Then the inverse Fourier transform h(t) of h
Z T
h(t) =
e2it d
=

1 2it
e
2it

so Parsevals identity gives

Z T
Z

f ()d =

T


1  2iT t
e
e2i(S)t
2it


1  2iT t
e
e2i(S)t dt
2it

= (with g(t) as in Step 3)

Z
 2iT t

=
e
e2i(S)t g(t)dt

(
T ,
= g(T ) g(S) 0 as
S .

f (t)

Step 5: If f = f1 + f2 , where f1 L1 (R) and f2 L2 (R), then we apply Step 3

to f1 and Step 4 to f2 , and get in both cases (2.7) with f replaced by f1 and f2 .

Note: This means that in most cases where f is continuous at t0 we have
Z T
e2it0 f()d.
f (t0 ) = lim
S
T

(continuous functions which do not satisfy (2.4) do exist, but they are difficult
to find.) In some cases we can even use the inversion formula at a point where
f is discontinuous.
Theorem 2.31. Suppose that f L1 (R) + L2 (R). Let t0 R, and suppose that

the two limits

f (t0 +) = lim f (t)

tt0

f (t0 ) = lim f (t)

tt0

exist, and that

Z

t0 +1 


 f (t) f (t0 +) 

dt < ,
t t0
t0
Z t0 

 f (t) f (t0 ) 

dt < .
t t0
t0 1

51

CHAPTER 2. FOURIER INTEGRALS

Then

e2it0 f()d
= [f (t0 +) + f (t0 )].
T T
2
RT
RT
Note: Here we integrate T , not S , and the result is the average of the right
lim

and left hand limits.

Proof. As in the proof of Theorem 2.30 we may assume that

Step 1: t0 = 0 , (see Step 1 of that proof)
Step 2: f (t0 +) + f (t0 ) = 0 , (see Step 2 of that proof).
Step 3: The claim is true in the special case where
(
et , t > 0,
g(t) =
et , t < 0,

because g(0+) = 1, g(0) = 1, g(0+) + g(0) = 0, and

Z T
g()d = 0 for all T,
T

since f is odd = g is odd.

Step 4: Define h(t) = f (t) f (0+) g(t), where g is the function in Step 3. Then
h(0+) = f (0+) f (0+) = 0 and
h(0) = f (0) f (0+)(1) = 0,

so

h is continuous. Now apply Theorem 2.30 to h. It gives

Z T

0 = h(0) = lim
h()d.
T

Since also

0 = f (0+)[g(0+) + g(0)] = lim

we therefore get
0 = f (0+) + f (0) = lim

g()d,

[h()
+ g()]d = lim

f()d.


Comment 2.32. Theorems 2.30 and 2.31 also remain true if we replace
Z T
lim
e2it f()d
T

by

lim

e2it e() f()d

(2.8)

(and other similar summability formulas). Compare this to Theorem 2.16. In

2

the case of Theorem 2.31 it is important that the cutoff kernel (= e() in
(2.8)) is even.

52

CHAPTER 2. FOURIER INTEGRALS

2.5
2.5.1

Applications
The Poisson Summation Formula

P
1

Suppose that f L1 (R) C(R), that

n= |f (n)| < (i.e., f (Z)), and
P
that
n= f (t + n) converges uniformly for all t in some interval (, ). Then

f (n) =

n=

f(n)

(2.9)

n=

Note: The uniform convergence of

possible way out is: If we define

f (t + n) can be difficult to check. One

n = sup |f (t + n)|,
and if

n=

n < , then

<t<

f (t + n) converges (even absolutely), and

n=

the convergence is uniform in (, ). The proof is roughly the same as what we

did on page 29.
We first construct a periodic function g L1 (T) with the
Fourier coefficients f(n):
Z

f (n)
=
e2int f (t)dt

Proof of (2.9).

=
t=k+s

e2int f (t)dt

k= k
Z 1
X
k=

Thm 0.14

k+1

e2ins f (s + k)ds

2ins

e
0

g(n),

f (s + k) ds

k=

where g(t) =

f (t + n).

n=

(For this part of the proof it is enough to have f L1 (R). The other conditions

are needed later.)

So we have g(n) = f(n). By the inversion formula for the periodic Fourier
transform:
g(0) =

n=

e2in0 g(n) =

n=

g(n) =

n=

f(n),

53

CHAPTER 2. FOURIER INTEGRALS

provided (=forutsatt) that we are allowed to use the Fourier inversion formula.
This is allowed if g C[, ] and g 1 (Z) (Theorem 1.37). This was part of
our assumption.

In addition we need to know that the formula

X
g(t) =
f (t + n)
n=

holds at the point t = 0 (almost everywhere is no good, we need it in exactly

P
this point). This is OK if
n= f (t + n) converges uniformly in [, ] (this
also implies that the limit function g is continuous).

Note: By working harder in the proof, Gripenberg is able to weaken some of the
assumptions. There are also some counter-examples on how things can go wrong
if you try to weaken the assumptions in the wrong way.

2.5.2

\
1 (R) = C (R) ?
Is L
0

That is, is every function g C0 (R) the Fourier transform of a function f

L1 (R)?

The answer is no, as the following counter-example shows. Take

, || 1,

ln 2

1
g() =
, > 1,
ln(1+)

1
, < 1.
ln(1)

Suppose that this would be the Fourier transform of a function f L1 (R). As

in the proof on the previous page, we define
h(t) =

f (t + n).

n=

Then (as we saw there), h L1 (T), and h(n)

= f(n) for n = 0, 1, 2, . . ..
P 1
However, since n=1 n h(n) = , this is not the Fourier sequence of any h
L1 (T) (by Theorem 1.38). Thus:

Not every h C0 (R) is the Fourier transform of some f L1 (R).

But:
f L1 (R) f C0 (R) ( Page 36)
f L2 (R) f L2 (R) ( Page 47)
f S
f S
( Page 44)

54

CHAPTER 2. FOURIER INTEGRALS

2.5.3

The Euler-MacLauren Summation Formula

Let f C (R+ ) (where R+ = [0, )), and suppose that

f (n) L1 (R+ )
for all n Z+ = {0, 1, 2, 3 . . .}. We define f (t) for t < 0 so that f (t) is even.

Warning: f is continuous at the origin, but f may be discontinuous! For exam-

ple, f (t) = e|2t|

f(t)=e 2|t|

We want to use Poisson summation formula. Is this allowed?

(n) = (2i)n f(),

By Theorem 2.7, fd
and f(n) is bounded, so

sup|(2i) ||f()|
< for all n
n

By the note on page 52, also

n=

|f(n)|
< .

n=

f (t + n) converges uniformly in (1, 1). By

the Poisson summation formula:

1
1 X
f (n) =
f (0) +
f (n)
2
2
n=0
n=

1 X
1
f(n)
f (0) +
=
2
2 n=

i
1
1
1 Xh

=
f (0) + f(0) +
f(n) + f(n)
2
2
2 n=1
Z
X


1
1 2int
1
e
+ e2int f (t)dt
f (0) + f (0) +
=
2
2
2
{z
}
n=1 |

1
f (0) +
2

f (t)dt +

Z
X
n=1

cos(2nt)

cos(2nt)f (t)dt

Here we integrate by parts several times, always integrating the cosine-function

and differentiating f . All the substitution terms containing odd derivatives of

55

CHAPTER 2. FOURIER INTEGRALS

f vanish since sin(2nt) = 0 for t = 0. See Gripenberg for details. The result
looks something like

f (n) =

n=0

2.5.4

1
1
1
1
f (t)dt + f (0) f (0) +
f (0)
f (5) (0) + . . .
2
12
720
30240

Schwartz inequality

The Schwartz inequality will be used below. It says that

|hf, gi| kf kL2 kgkL2
(true for all possible L2 -spaces, both L2 (R) and L2 (T) etc.)

2.5.5

Heisenbergs Uncertainty Principle

For all f L2 (R), we have

Z
 Z
2
2
t |f (t)| dt

|f()|2d
2

1
f 4 2
2
L (R)
16

Interpretation: The more concentrated f is in the neighborhood of zero, the

more spread out must f be, and conversely. (Here we must think that kf kL2 (R)
is fixed, e.g. kf kL2 (R) = 1.)

In quantum mechanics: The product of time uncertainty and space uncer-

tainty cannot be less than a given fixed number.

56

CHAPTER 2. FOURIER INTEGRALS

Proof. We begin with the case where f S. Then

Z
Z
Z
Z

16 |tf (t)|dt | f()|d

= 4 |tf (t)|dt |f (t)|dt
R

() = 2i f()

(f[
and Parsevals iden. holds). Now use Scwartz ineq.
Z


4
|tf (t)||f (t)|dt
R
Z


= 4
|tf (t)||f (t)|dt
R
Z


Re[tf (t)f (t)]dt

4
R
Z  
 2
1

f (t)f (t) + f (t)f (t) dt

= 4
t
2
R
Z
d
t (f (t)f (t)) dt (integrate by parts)
=
{z }
R dt |
=|f (t)|

Z



|f (t)|dt
[t|f (t)|]
| {z }

=0
Z

|f (t)|dt

This proves the case where f S. If f L(R), but f S, then we choose a

sequence of functions fn S so that
Z
Z
|fn (t)|dt
|f (t)|dt and

Z
Z
|tfn (t)|dt
|tf (t)|dt and

Z
Z

| fn ()|d
| f()|d

(This can be done, not quite obvious). Since the inequality holds for each fn , it
must also hold for f .

2.5.6

Weierstrass Non-Differentiable Function

Define (t) =

k
k=0 a

cos(2bk t), t R where 0 < a < 1 and ab 1.

Lemma 2.33. This sum defines a continuous function which is not differentiable at any point.

57

CHAPTER 2. FOURIER INTEGRALS

Proof. Convergence easy: At each t,

X
k=0

|a cos(2b t)|

ak =

k=0

1
< ,
1a

and absolute convergence convergence. The convergence is even uniform: The

error is

X
 X
X
aK
k
k
k
k 

0 as K
ak =
|a cos(2b t)|
a cos(2b t)
1a
k=K

k=K

k=K

so by choosing K large enough we can make the error smaller than , and the
same K works for all t.
By Analysis II: If a sequence of continuous functions converges uniformly, then
the limit function is continuous. Thus, is continuous.
Why is it not differentiable? At least does the formal derivative not converge:
Formally we should have

(t) =

X
k=0

ak 2bk (1) sin(2bk t),

and the terms in this serie do not seem to go to zero (since (ab)k 1). (If a sum
converges, then the terms must tend to zero.)

To prove that is not differentiable we cut the sum appropriatly: Choose some
function L1 (R) with the following properties:
i) (1)

=1
ii) ()

= 0 for
iii)

1
b

and > b

|t(t)|dt < .

()
0

1/b

We can get such a function from the Fejer kernel: Take the square of the Fejer
kernel ( its Fourier transform is the convolution of f with itself), squeeze it
(Theorem 2.7(e)), and shift it (Theorem 2.7(b)) so that it vanishes outside of

58

CHAPTER 2. FOURIER INTEGRALS

( 1b , b), and (1)

= 1. (Sort of like approximate identity, but (1)

= 1 instead of
(0)

= 1.)
Define j (t) = bj (bj t),

t R. Then j () = (b
j ), so (b
j ) = 1 and

()

= 0 outside of the interval (bj1 , bj+1 ).

b4

Put fj = j . Then
fj (t) =
=

b3

b2

1/b

(t s)j (s)ds

Z X

k1

k=0

i
h
2ibk (ts)
2ibk (ts)
j (s)ds
e
+e

(by the uniform convergence)

k
Xa

2ibk t
k
2ibk t
=
j (bk )
|e {z } j (b ) + e
{z
}
|
2
k=0

=jk

=0

1 j 2ibk t
ae
.
2

(Thus, this particular convolution picks out just one of the terms in the series.)
Suppose (to get a contradiction) that can be differentiated at some point t R.

Then the function

(s) =

(t+s)(t)
s

(t) , s 6= 0
, s=0

is (uniformly) continuous and bounded, and (0) = 0. Write this as

(t s) = s(s) + (t) s (t)

59

CHAPTER 2. FOURIER INTEGRALS

i.e.,
fj (t)

=
=

ZR
R

(t s)j (s)ds
s(s)j (s)ds + (t)

j (s)ds (t) sj (s)ds

| R {z }
| R {z
}
=
j (0)=0

=
bj s=t

s(s)bj (bj s)ds

RZ
s
( j ) s(s)ds
bj
b } | {z }
R | {z
L1

(0)
j
=0
2i

0 pointwise

by the Lesbesgue dominated convergence theorem as j .

Thus,

1  a j 2ibj t
b fj (t) 0 as j
e
0 as j .
2 b

j
j
As |e2ib t | = 1, this is ab 0 as j . Impossible, since ab 1. Our
j

assumption that is differentiable at the point t must be wrong (t) is not

differentiable in any point!

2.5.7

Differential Equations

Solve the differential equation

u(t) + u(t) = f (t), t R

(2.10)

where we require that f L2 (R), u L2 (R), u C 1 (R), u L2 (R) and that u

is of the form

u (t) = u (0) +

v(s)ds,
0

where v L2 (R) (that is, u is absolutely continuous and its generalized

derivative belongs to L2 ).

The solution of this problem is based on the following lemmas:

Lemma 2.34. Let k = 1, 2, 3, . . .. Then the following conditions are equivalent:
i) u L2 (R)C k1 (R), u(k1) is absolutely continuous and the generalized
derivative of u(k1) belongs to L2 (R).

60

CHAPTER 2. FOURIER INTEGRALS

ii) u L2 (R) and

| k u(k)|2 d < .

Proof. Similar to the proof of one of the homeworks, which says that the same
P
2
result is true for L2 -Fourier series. (There ii) is replaced by |nf(n)|
< .)
Lemma 2.35. If u is as in Lemma 2.34, then

d
(k) () = (2i)k u
u
()

(compare this to Theorem 2.7(g)).

Proof. Similar to the same homework.

Solution: By the two preceding lemmas, we can take Fourier transforms in (2.10),
and get the equivalent equation
(2i)2 u()+
u() = f(), R (4 2 2)
u() = f(), R (2.11)
Two cases:
Case 1: 4 2 2 6= 0, for all R, i.e., must not be zero and not a positive

number (negative is OK, complex is OK). Then

u() =

f()
, R
4 2 2

so u = k f , where k = the inverse Fourier transform of

k()
=

1
.
4 2 2

This can be computed explicitly. It is called Greens function for this problem.
Even without computing k(t), we know that
k C0 (R) (since k L1 (R).)
k has a generalized derivative in L2 (R) (since

| k()|
d < .)

k does not have a second generalized derivative in L2 (since

| 2 k()|
d = .)

How to compute k? Start with a partial fraction expansion. Write

= 2

for some C

61

CHAPTER 2. FOURIER INTEGRALS

( = pure imaginary if < 0). Then

1
4 2 2

Now we must still invert

1
1
1
=

2 4 2 2
2 + 2
A
B
=
+
2 + 2
A + 2A + B 2B
=
( 2)( + 2)
)
(A + B) = 1
1
A=B=

2
(A B)2 = 0
=

1
+2

and

1
.
2

This we do as follows:

Auxiliary result 1: Compute the transform of

(
ezt , t 0,
f (t) =
0 , t < 0,
/ C(R) because of the jump at
where Re(z) > 0 ( f L2 (R) L1 (R), but f

the origin). Simply compute:

f()
=

e2it ezt dt
 0 (z+2i)t 
e
1
=
=
.
(z + 2i) 0
2i + z

Auxiliary result 2: Compute the transform of

(
ezt , t 0,
f (t) =
0 , t > 0,
where Re(z) > 0 ( f L2 (R) L1 (R), but f
/ C(R))
f() =
=
Back to the function k:

Z


e2it ezt dt

e(z2i)t
(z 2i)t

0

1
.
z 2i



1
1
1
+
2 2 + 2


i
i
1
.
+
=
2 i 2i i + 2i

k()
=

62

CHAPTER 2. FOURIER INTEGRALS

We defined by requiring 2 = . Choose so that Im() < 0 (possible because

is not a positive real number).
1

Re(i) > 0, and k()

=
2

i
i
+
i 2i i + 2i

The auxiliary results 1 and 2 gives:

k(t) =

i it
e
2
i it
e
2

and
u(t) = (k f )(t) =

, t0
, t<0

k(t s)f (s)ds

Special case: = negative number = a2 , where a > 0. Take = ia

i = i(i)a = a, and

k(t) =

1 at
e
, t0
2a

1 at
2a
e

k(t) =

, t<0

i.e.

1 |at|
e
, tR
2a

Thus, the solution of the equation

u(t) a2 u(t) = f (t),

t R,

where a > 0, is given by

u=kf
k(t) =

where

1 a|t|
e
,
2a

tR

This function k has many names, depending on the field of mathematics you are
working in:
i) Greens function (PDE-people)
ii) Fundamental solution (PDE-people, Functional Analysis)
iii) Resolvent (Integral equations people)

63

CHAPTER 2. FOURIER INTEGRALS

Case 2: = a2 = a nonnegative number. Then

f() = (a2 4 2 2 )
u() = (a 2)(a + 2)
u().
As u() L2 (R) we get a necessary condition for the existence of a solution: If
a solution exists then

2
f()
 d < .
(a 2)(a + 2)

(2.12)

a
a
, this forces f to vanish at 2
,
(Since the denominator vanishes for = 2

and to be small near these points.)

If the condition (2.12) holds, then we can continue the solution as before.
Sideremark: These results mean that this particular problem has no eigenvalues and no eigenfunctions. Instead it has a contionuous spectrum consisting
of the positive real line. (Ignore this comment!)

2.5.8

Heat equation

This equation:

u(t, x)
t

u(0, x)

2
x2

u(t, x) + g(t, x),

= f (x) (initial value)

t>0
xR

is solved in the same way. Rather than proving everything we proceed in a formal
mannor (everything can be proved, but it takes a lot of time and energy.)
Transform the equation in the x-direction,
Z
u(t, ) =
e2ix u(t, x)dx.
R

Assuming that

e2ix t
u(t, x) =

(
(

u(t, )
t

u(0, )

u(t, )
t

u(0, )

e2ix u(t, x)dx we get

= (2i)2 u(t, ) + g(t, )

= f()

= 4 2 2 u(t, ) + g(t, )

= f()

64

CHAPTER 2. FOURIER INTEGRALS

We solve this by using the standard variation of constants lemma:

2 2
u(t, ) = f()e4 t +
|
{z
}

We can invert e4

22t

u1(t, )

= e(2

t)2

|0

e4

= e(/)

2 2 (ts)

g(s, )ds
{z
}
u2 (t, )

where = (2 t)1 :

According to Theorem 2.7 and Example 2.5, this is the transform of

x2
1
1 ( x )2
k(t, x) = e 2 t = e 4t
2 t
2 t

We know that f()k()

= k[
f (), so
u1(t, x) =

2
1 e(xy) /4t f (y)dy,
2 t

(By the same argument:

s and t s are fixed when we transform.)

Rt
u2(t, x) = 0 (k g)(s)ds
RtR
2
= 0 1
e(xy) /4(ts) g(s, y)dyds,
2

(ts)

u(t, x) = u1 (t, x) + u2 (t, x)

The function

x2
1
k(t, x) = e 4t
2 t
is the Greens function or the fundamental solution of the heat equation on the

real line R = (, ), or the heat kernel.

Note: To prove that this solution is indeed a solution we need to assume that
- all functions are in L2 (R) with respect to x, i.e.,
Z
Z
Z
2
2
|u(t, x)| dx < ,
|g(t, x)| dx < ,

|f (x)|2 dx < ,

- some (weak) continuity assumptions with respect to t.

2.5.9

Wave equation

2
t2

u(t, x) =

2
x2

u(t, x) + k(t, x),

u(0, x) = f (x),

u(0, x)
t

= g(x),

t > 0,
x R.

xR
xR

65

CHAPTER 2. FOURIER INTEGRALS

Again we proceed formally. As

t2 u(t, )
u(0, )

u(0, )

above we get
),
= 4 2 2 u(t, ) + k(t,

= f(),
= g().

This can be solved by the variation of constants formula, but to simplify the
) 0. Then the solution is
computations we assume that k(t, x) 0, i.e., h(t,
(check this!)

sin(2t)
u(t, ) = cos(2t)f() +
g().
2

(2.13)

To invert the first term we use Theorem 2.7, and get

1
[f (x + t) + f (x t)].
2
The second term contains the Dirichlet kernel , which is inverted as follows:
Ex. If
k(x) =

then k()
=

1
2

1/2, |t| 1

0,

otherwise,

sin(2).

Proof.
1

k()
=
2

e2it dt = . . . =
1

1
sin(t).
2

Thus, the inverse Fourier transform of

sin(2)
2

is k(x) =

1/2, |x| 1,

0,

|x| > 1,

(inverse transform = ordinary transform since the function is even), and the
inverse Fourier transform (with respect to ) of
sin(2t)
sin(2t)
= t
is
2
2t
(
1/2, |x| t,
x
k( ) =
t
0,
|x| > t.
This and Theorem 2.7(f), gives the inverse of the second term in (2.13): It is
Z
1 x+t
g(y)dy.
2 xt

66

CHAPTER 2. FOURIER INTEGRALS

Conclusion: The solution of the wave equation with h(t, x) 0 seems to be

1
1
u(t, x) = [f (x + t) + f (x t)] +
2
2

x+t

g(y)dy,

xt

a formula known as dAlemberts formula.

Interpretation: This is the sum of two waves: u(t, x) = u+ (t, x) + u (t, x), where
1
1
u+ (t, x) = f (x + t) + G(x + t)
2
2
moves to the left with speed one, and
1
1
u (t, x) = f (x t) G(x t)
2
2
moves to the right with speed one. Here
Z x
G(x) =
g(y)dy,
0

x R.