MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018
T. PRZEBINDA
Contents
1. Complex numbers. 1 1.1. The field of complex numbers 1 1.2. Multiplication in polar coordinates 3 1.3. Some geometry and some analysis 5 2. The finite Fourier Transform. 7 2.1. The L2 Theory 7 2.2. The L1 Theory 9 3. The Fourier series. 14 3.1. The L1 Theory 14 4. The Exam 1, due Wednesday 10/3/2018 in class. We DO NOT meet
on Friday in class. 17 References 20
1. Complex numbers.
1.1. The field of complex numbers.
C ∈ z = x+ iy ↔ (x, y) ∈ R2 .
The addition of complex numbers is defined coordinate-wise:
(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)
and multiplication by
(x1 + iy1)(x2 + iy2) = x1x2 − y1y2 + i(x1y2 + x2y1) .
Then i2 = −1 and the multiplicative inverse of x+ iy is
(x+ iy)−1 = x
x2 + y2 − i y
x2 + y2 .
With these operations C is a field, i.e. behaves like the field R of real numbers.
Date: September 26, 2018.
1
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Problem 1. Check that
(x+ iy)
( x
x2 + y2 − i y
x2 + y2
) = 1 .
The left hand side is equal to
(x+ iy)(x− iy) x2 + y2
= x2 + y2
x2 + y2 = 1 .
In problems 2 and 3 we realize complex numbers as matricesof size two with real entries.
Problem 2. Let
I =
( 1 0 0 1
) and J =
( 0 1 −1 0
) .
Show that for any real numbers x1, x2, y1 and y2,
(x1I + y1J) (x2I + y2J) = (x1x2 − y1y2)I + (x1y2 + x2y1)J. In particular
(x1I + y1J) (x2I + y2J) = (x2I + y2J) (x1I + y1J)
and J2 = −I.
We notice first that J2 = −I.
Then
(x1I + y1J) (x2I + y2J) = x1Ix2I + x1Iy2J + y1Jx2I + y1Jy2J
= x1x2I + x1y2J + y1Jx2 − y1y2I = (x1x2 − y1y2)I + (x1y2 + x2y1)J.
Problem 3. With the notation of Problem 2 show that the matrix
xI + yJ
is invertible if and only if x2 + y2 6= 0. Also, show that
(xI + yJ)−1 = x
x2 + y2 I − y
x2 + y2 J.
If x2 + y2 6= 0, then the matrix on the right hand side is well defined and we see from the formula of previous problem that
(xI + yJ)
( x
x2 + y2 I − y
x2 + y2 J
) = I.
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 3
Thus the matrix xI + yJ is invertible. If x2 + y2 = 0, then xI + yJ = 0 is not an invertible matrix.
Complex conjugation x+ iy = x− iy
is an automorphism of C. In other words the conjugation of the sum is the sum of conjugates
(x1 + iy1) + (x2 + iy2) = x1 + iy1 + x2 + iy2
and the conjugation of the product is the product of conjugates
(x1 + iy1)(x2 + iy2) = x1 + iy1 x2 + iy2 .
Also, {z ∈ C; z = z} = R .
Problem 4. Let a, b, c be complex (or real) numbers. Show that if z solves the equation
az2 + bz + c = 0
then z solves the equation a(z)2 + b(z) + c = 0.
This is clear because az2 + bz + c = a(z)2 + b(z) + c.
1.2. Multiplication in polar coordinates. Eoler’s formula
eiθ = ∞∑ n=0
(iθ)n
n!
= ∞∑ k=0
(iθ)2k
(2k)! + ∞∑ k=0
(iθ)2k+1
(2k + 1)!
= ∞∑ k=0
(−1)k θ 2k
(2k)! + i
∞∑ k=0
(−1)k θ 2k+1
(2k + 1)!
= cos θ + i sin θ .
Hence the multiplication in polar coordinates looks as follows:
0 6= z = x+ iy = reiθ , x = r cos θ , y = r sin θ , r > 0 , θ ∈ R , r1e
iθ1r2e iθ2 = r1r2e
i(θ1+θ2) .
The fundamental Theorem of Algebra.
Theorem 1. Let n be a positive integer. Then for any complex numbers a0, a1, …, an with an 6= 0 there is a z ∈ C such that
anz n + …+ a0 = 0.
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Example. The equation az2 + bz + c = 0, with a 6= 0 has the solutions
z = −b±
√ b2 − 4ac
2a .
Example. The N-th roots of 1 are the solutions of the equation zN = 1. Here is a complete list of them:
zk = e 2πik N (k = 0, 1, 2, …, N − 1) .
Problem 5. Find all the solutions of the equation
z6 = 8
and draw them on the complex plane.
The solutions are zk =
√ 2e2πik/6 (k = 0, 1, 2, 3, 4, 5).
Problem 6. Find all the solutions of the equation
z6 = −8 and draw them on the complex plane.
The solutions are zk =
√ 2eiπ/6e2πik/6 (k = 0, 1, 2, 3, 4, 5).
Problem 7. Solve for z z2 + z + 1 = 0 .
z = −1±
√ 3i
2 .
Problem 8. Find all the solutions of the equation
z3 = 27
and draw them on the complex plane.
Recall the general formula for the solutions of the equation
zn = Reiθ (1)
where R > 0 and θ ∈ R: zk = R
1/nei(θ/n+k2π/n) (k = 0, 1, 2, …, n− 1). (2) In the particular case of this problem
z0 = 3, z1 = 3e i(2π/3), z2 = 3e
i(4π/3).
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 5
Problem 9. Solve for z
(2− 3i)z + 4 = −i. Write the answer in the form z = x+ iy where x and y are real numbers.
z = −i− 4 2− 3i
= (−i− 4)(2 + 3i)
22 + 32 = −5− 14i
13 = −5 13
+ i −14i
13 .
Problem 10. Find all the solutions of the equation
e2z − ez = 2.
Let w = ez. Then the above equation may be written as
w2 − w − 2 = 0.
The solutions are w = −1 and w = 2. Hence we need to find all the z such that ez = −1 and all the z such that ez = 2. The first equation has solution z = iπ + i2πk, k ∈ Z, and the second one z = ln(2) + i2πl, l ∈ Z. Thus the answer is
{iπ + i2πk; k ∈ Z} ∪ {ln(2) + i2πl; l ∈ Z}.
1.3. Some geometry and some analysis. The absolute value:
|z| = |x+ iy| = √ x2 + y2 =
√ zz
Taking the absolute value preserves multiplication:
|z1z2| = |z1||z2| .
Triangle inequality
|z1 + z2| ≤ |z1|+ z2| .
Problem 11. Sketch the sets described by
|z| = 1 , |z| ≤ 1 , |z| ≥ 1 ,
z = z , z = −z ,
The first equation describes the real line and the second one the imaginary line.
Problem 12. Sketch the set of points z ∈ C determined by the condition
|z − 1 + i| ≥ 1.
This is the set of points outside the open disc or radius 1, centered at 1− i.
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Problem 13. Prove that for z ∈ C \ {1},
1 + z + z2 + · · ·+ zn = 1− z n+1
1− z (n = 1, 2, 3, . . . ) .
Notice that
(1 + z + z2 + · · ·+ zn)(1− z) = 1 + z + z2 + · · ·+ zn
− (z + z2 + · · ·+ zn + zn+1) = 1− zn+1 .
Problem 14. Deduce from previous Problems that ∞∑ n=0
zn = 1
1− z (z ∈ C, |z| < 1) .
Since |z| < 1, we have lim n→∞ |z|n = 0. Therefore
1 + z + z2 + · · ·+ zn + … = lim n→∞
(1 + z + z2 + · · ·+ zn) = lim n→∞
1− zn+1
1− z =
1
1− z .
The exponential function
ez = ∞∑ n=0
zn
n! (z ∈ C) .
The main properties:
ez+w = ezew (z, w ∈ C) , e0 = 1 ,
eiy = cos y + i sin y (y ∈ R) , ex+iy = exeiy (x, y ∈ R) ,∣∣ex+iy∣∣ = ex (x, y ∈ R) , ex+iy = ex−iy (x, y ∈ R) .
Problem 15. Check that the set of the two trigonometric formulas
cos(α + β) = cosα cos β − sinα sin β (3) sin(α + β) = sinα cos β + cosα sin β
are equivalent to the single formula
ei(α+β) = eiαeiβ . (4)
This is straightforward:
ei(α+β) = cos(α + β) cos(α + β)− i sin(α + β) sin(α + β)
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 7
and by explicit multiplication of complex numbers
eiαeiβ = (cosα cos β − sinα sin β) + i(sinα cos β + cosα sin β) . Hnece the claim follows.
2. The finite Fourier Transform.
2.1. The L2 Theory. The group Z/NZ = {0, 1, 2, …, N − 1} with addition modulo N . Characters
χy(x) = e 2πi N xy (y, x ∈ {0, 1, 2, …, N − 1}) . (5)
The space L2(Z/NZ) consists of all functions v : Z/NZ → C. It is equipped with the scalar product
(u, v) = 1√ N
N−1∑ x=0
u(x)v(x) (u, v ∈ L2(Z/NZ)) . (6)
Basic orthogonality equations
(N− 1 4χk, N
− 1 4χl) =
{ 1 if k = l , 0 otherwise
(k, l ∈ Z/NZ) . (7)
The Fourier transform
F : L2(Z/NZ)→ L2(Z/NZ) , (8)
Fv(y) = (v, χy) = 1√ N
N−1∑ x=0
v(x)χy(x) = 1√ N
N−1∑ x=0
v(x)e− 2πi N xy .
Plancherell formula
(Fu,Fv) = (u, v) (u, v ∈ L2(Z/NZ)) . (9)
Problem 16. Check that for N = 2
Ff(n) = 1√ 2
(f(0) + (−1)nf(1)) (n = 0, 1).
This is clear from the definition
Ff(n) = 1√ 2
1∑ j=0
f(j)e−i 2π 2 jn =
1√ 2
1∑ j=0
f(j)(−1)jn.
Problem 17. Give an example of a non-zero function f ∈ L2(Z(2)) such that Ff = −f . (This is an eigenvector corresponding to the eigenvalue −1.)
f(0) = 1, f(1) = −(1 + √
2).
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Problem 18. Give an example of a non-zero function f ∈ L2(Z(4)) such that Ff = −if . (This is an eigenvector corresponding to the eigenvalue −i.)
We see from the definition of F that
Fδk = 1
2 (δ0 + i
−kδ1 + i −2kδ2 + i
−3kδ3).
Hence
F(δ1 − δ3) = −i(δ1 − δ3).
Problem 19. Let P,Q > 1 be integers and let N = PQ. Let A = {0, P, 2P, . . . , (Q−1)P} and let B = {0, Q, 2Q, . . . , (P − 1)Q}. Denote by 1A the indicator function of A:
1A(n) =
{ 1 if n ∈ A, 0 if n /∈ A.
Show that
F √ P1A =
√ Q1B.
We compute
F √ P1A(n) =
√ P√ N
Q−1∑ j=0
e−i 2π N jPn =
1√ Q
Q−1∑ j=0
e−i 2π Q jn =
1√ Q Q1B(n) =
√ Q1B(n).
Problem 20. Give an example of a non-zero function f ∈ L2(Z(144)) such that Ff = f . (This is an eigenvector corresponding to the eigenvalue 1.)
Since 144 = 12× 12, we may take f = 1A in terms of Problem 19.
Recall that the Fourier transform applied four times is the identity:
F4 = I .
Problem 21. Show that the only possible eigenvalues of the Fourier transform F : L2(Z(N))→ L2(Z(N)) are ±1 and ±i (Here i =
√ −1.)
Recall that λ ∈ C is an eigenvalue of F if and only if there is a non-zero function f ∈ V (N) such that
Ff = λf.
Since F4 = I we see that λ4 = 1. Hence λ ∈ {1,−1, i,−i}.
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 9
The matrix of F with respect to the basis δ0, δ1, …, δN−1 is equal to
F = FN = 1√ N
1 1 1 … 1 1 z z2 … zN−1
1 z1·2 z2·2 … z(N−1)·2
1 z1·3 z2·3 … z(N−1)·3
.. .. .. … .. 1 zN−1 z2·(N−1) … z(N−1)·(N−1)
, (10)
where z = e− 2πi N .
Problem 22. Show that FF
t = I .
(FF t )j,k =
1
N
N−1∑ l=0
zl·jzl·k = 1
N
N−1∑ l=0
zl·jz−l·k = 1
N
N−1∑ l=0
zl·(j−k) = δ(j − k) = Ij,k .
Problem 23. Show that F−1 = F
t .
This is obvious from Problem 22.
Problem 24. Show that F 4 = I.
Since F is the matrix of F and since F4 is the identity, the formula follows.
Problem 25. Find the eigenvalues and the corresponding eigenspaces for F3.
Use wolframalpha.com. to find approximate solutions. Then guess the exact solutions and check. There are three eigenvalues and the corresponding three one-dimensional eigenspaces:
λ1 = √ −1 , C(0− 1, 1)t
λ2 = 1 , C( √
3 + 1, 1, 1)t
λ2 = 1 , C( √
3− 1, 1, 1)t . Here the superscript t stands for the transpose. Thus the above are column vectors.
2.2. The L1 Theory. Here L1(Z(N)) is the vector space of all functions v : Z(N)→ C, with the norm
‖ v ‖1= 1
N
N−1∑ x=0
|v(x)| .
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As a vector space L1(Z(N)) = L2(Z(N)), but the norm is different. We also have the norm
‖ v ‖∞= max{|v(0)|, |v(1)|, …, |v(N − 1)|} . Here the Fourier transform is defined as
v̂(y) = 1
N
N−1∑ x=0
v(x)e− 2πi N xy (y ∈ Z(N)) .
It differs from F , defined in previous section, only by a constant
v̂ = 1√ N Fv .
Then ‖ v̂ ‖∞≤‖ v ‖1 .
Furthermore we define the translations
Tzv(x) = v(x− z) (x, z ∈ Z(N)) and the modulations
Mzv(y) = v(y)e − 2πi
N zy (y, z ∈ Z(N)) .
A straightforward computation shows that
(Tzv)̂ = Mzv̂ .
In other words, the Fourier transform transforms translations to modulations.
Theorem 2. [MS73] For any non-zero function v ∈ L1(Z(N)) we have the following inequality
| supp v̂|| supp v| ≥ N .
Proof. We follow [GGI05, page 4019]. Since v 6= 0, the support supp v is a non-empty subset of Z(N). Hence all the translates
supp v + n = suppTnv, n ∈ Z(N), cover the whole group Z(N):
Z(N) = ⋃
n∈Z(N)
suppTnv . (11)
Consider the vector subspace V ⊆ L1(Z(N)) consisting of all the linear combinations of the translates of v:
c(0)v + c(1)T1v + …+ c(N − 1)TN−1v . (12) (Here c(0), c(1), …, c(N − 1) are arbitrary complex numbers.) Since the translates Tnv generate V, one may choose a basis of V consisting of such translates:
Tn1v, Tn2v, …, Tnmv .
(Here m = dimV and n1, n2, …, nm ∈ {0, 1, 2, …, N − 1} are distinct numbers.) Fix n ∈ Z(N) and write Tnv as a linear combination of the basis elements:
Tnv = c(n1)Tn1v + c(n2)Tn2v + …+ c(nm)Tnmv .
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 11
We see from the above equation that
suppTnv ⊆ suppTn1v ∪ suppTn2v ∪ … ∪ suppTnmv . Notice that there is no n on the right hand side. In other words, the right hand side does not depend on n. Therefore⋃
n∈Z(N)
suppTnv ⊆ suppTn1v ∪ suppTn2v ∪ … ∪ suppTnmv .
By combining this with (11) we see that
Z(N) ⊆ suppTn1v ∪ suppTn2v ∪ … ∪ suppTnmv . Hence the cardinality
|Z(N)| ≤ | suppTn1v|+ | suppTn2v|+ …+ | suppTnmv| . But
| supp v| = | suppTn1v| = | suppTn2v| = … = | suppTnmv| . Therefore
N ≤ m| supp v| . (13) It remains to compute m, i.e. to show that
m = | supp v̂| . (14)
Let V̂ be the space of the Fourier transforms of all the elements of V. Since the Fourier transform is a linear bijection, it does not change the dimension of any subspace. Thus
dimV = dim V̂ .
Since (Tkv)̂ = Mkv̂, we see from (12) that the space V̂ consists all the functions of the form
ĉ(y)v̂(y) , (15)
where c ∈ L1(Z(N)) is arbitrary. The Fourier transform is a linear bijection, hence the space V̂ consists all the functions of the form
f(y)v̂(y) , (16)
where f ∈ L1(Z(N)) is arbitrary. Let supp v̂ = {y1, y2, …, ys} .
By choosing
f(y) =
{ 1
v̂(y1) if y = y1
0 otherwise
we see from (16) that the Dirac delta at y1, δy1 belongs to V̂. Similarly, by choosing
f(y) =
{ 1
v̂(y2) if y = y2
0 otherwise
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we see from (16) that the Dirac delta at y2, δy2 belongs to V̂. Thus
δy1 , δy2 , …, δys ∈ V̂ . (17)
The Dirac deltas are clearly linearly independent. Also they span V̂. This is because for any f ∈ L1(Z(N)),
f(y)v̂(y) = (f(y1)δy1 + f(y2)δy2 + …+ f(ys)δys)(y)v̂(y) .
Thus the set (17) forms a basis of V̂ . Therefore
s = dim V̂ .
Since s = | supp v̂|, the equality (14) follows and we are done. �
The convolution u ∗ v of two functions u, v ∈ L1(Z(N)) is defined as follows,
u ∗ v(y) = 1 N
N−1∑ x=0
u(y − x)v(x) (y ∈ Z(N)) .
Fourier transform transforms convolution into multiplication:
(u ∗ v)̂ = ûv̂ . (18)
Problem 26. Verify the equality (18).
We compute
(u ∗ v)̂(z) = 1 N
N−1∑ y=0
(u ∗ v)(y)e− 2πi N yz =
1
N
N−1∑ y=0
1
N
N−1∑ x=0
u(y − x)v(x)e− 2πi N yz
= 1
N
N−1∑ y=0
1
N
N−1∑ x=0
u(y)v(x)e− 2πi N
(y+x)z
= 1
N
N−1∑ y=0
1
N
N−1∑ x=0
u(y)e− 2πi N yzv(x)e−
2πi N xz
= û(z)v̂(z) .
Problem 27. For two subsets A,B ⊆ Z(N) let A+B = {a+ b; a ∈ A, b ∈ B}.
(This is the addition modulo N .) Show that for any two functions f, g ∈ L1(Z(N)) supp(f ∗ g) ⊆ supp(f) + supp(g). (19)
Let c /∈ supp(f) + supp(g). This means that c ∈ Z(N) cannot be expressed as the sum c = a+ b where a ∈ supp f and b ∈ supp g. In other words, c ∈ Z(N) cannot be expressed
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 13
as the sum c = a+ b where f(a) 6= 0 and g(a) 6= 0. Thus for any decomposition c = a+ b we have f(a) = 0 or g(a) = 0. Therefore
f ∗ g(c) = 1 N
∑ b∈Z(N)
f(c− b)g(b) = 1 N
∑ a,b∈Z(N), a+b=c
f(a)g(b) = 0 .
Thus the complement of the right hand side of the inclusion (19) is contained in the complement of the left hand side. This verifies the inclusion (19)
Problem 28. Is there a function v ∈ L1(Z(8)) such that | supp v| = 3 and | supp v̂| = 2?
No, because 3 · 2 < 8, which contradicts the
Problem 29. Give an example of a function v ∈ L1(Z(8)) such that | supp v| = 2 and | supp v̂| = 4.
See Problem 19 for a more general case. Here
v(0) = 1, v(1) = 0, v(2) = 0, v(3) = 1, v(4) = 0, v(5) = 0, v(6) = 0, v(7) = 0 ,
will do.
Problem 30. For v ∈ L1(Z(N)) let v∗(x) = v(−x). Show that
(v∗)̂ = v̂ .
We compute
(v∗)̂(y) = 1
N
N−1∑ x=0
v∗(x)e− 2πi N xy =
1
N
N−1∑ x=0
v(−x)e− 2πi N xy =
1
N
N−1∑ x=0
v(x)e 2πi N xy
= 1
N
N−1∑ x=0
v(x)e− 2πi N xy = v̂(y) .
Problem 31. For v ∈ L1(Z(N)) be such that
v̂(y) ≥ 0 (y ∈ Z(N)) .
Show that there is u ∈ L1(Z(N)) such that
u ∗ u = v .
Take u to be the inverse Fourier transform of √ v̂. Then ûû = v̂, so u ∗ u = v.
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3. The Fourier series.
3.1. The L1 Theory.
Problem 32. Let T > 0 be a positive number. Suppose f : R→ R is a T -periodic function integrable on any bounded interval. Show that for any a ∈ R∫ T
0
f(x) dx =
∫ T+a a
f(x) dx =
∫ T 0
f(x+ a) dx .
There is an integer n and a real number b, 0 ≤ b < T , such that a = nT + b. Therefore, by a change of variables,∫ T+a
a
f(x) dx =
∫ T 0
f(x+ a) dx =
∫ T 0
f(x+ nT + b) dx =
∫ T 0
f(x+ b) dx,
where the last equality follows from the fact that f is T -periodic. Again, by a change of variables ∫ T
0
f(x+ b) dx, =
∫ T+b b
f(x) dx =
∫ T b
f(x) dx+
∫ T+b T
f(x) dx
and ∫ T+b T
f(x) dx =
∫ b 0
f(T + x) dx =
∫ b 0
f(x) dx,
where the last equality follows from the fact that f is T -periodic. Altogether∫ T+a a
f(x) dx =
∫ T b
f(x) dx+
∫ b 0
f(x) dx =
∫ T 0
f(x) dx.
The space L1(R/Z) consists of all functions f : R→ C such that
f(x+ 1) = f(x) (x ∈ R) , (20)
‖ f ‖1 := ∫ 1 0
|f(x)| dx <∞ .
The space L∞(Z) consists of all functions φ : Z→ C such that
‖ φ ‖∞ := sup{|φ(n)|; n ∈ Z} <∞ . (21)
The Fourier transform is defined by the formula
Ff(n) = ∫ 1 0
f(x)e−2πixn dx (n ∈ Z) . (22)
Problem 33. Check that for any f ∈ L1(R/Z)
‖ Ff ‖∞≤‖ f ‖1 .
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 15
In particular the Fourier transform is a map from L1(R/Z) to L∞(Z).
|Ff(n)| ≤ ∫ 1 0
|f(x)e−2πixn| dx =‖ f ‖1 .
Define the translations of the 1-periodic functions f : R→ C
Tyf(x) = f(x− y) (x, y ∈ R) (23)
and modulations of functions φ : Z→ C by
Myφ(n) = e −2πiynφ(n) (y ∈ R , n ∈ Z) . (24)
Problem 34. Check that
FTy = MyF . (25) More explicitly, for any f ∈ L1(R/Z),
F(Ty(f)) = My(F(f)) .
We compute
F(Ty(f))(n) = ∫ 1 0
Tyf(x)e −2πinx dx =
∫ 1 0
f(x− y)e−2πinx dx
=
∫ 1 0
f(x− y)e−2πinx dx = ∫ 1 0
f(x)e−2πin(x+y) dx
= e−2πiny ∫ 1 0
f(x)e−2πinx dx = e−2πinyFf(n) ,
where in the fourth equation we used Problem 32.
The convolution in L1(R/Z) is defined by
u ∗ v(y) = ∫ 1 0
u(y − x)v(x) dx (u, v ∈ L1(R/Z), y ∈ R) . (26)
Problem 35. Check that
‖ u ∗ v ‖1≤‖ u ‖1‖ v ‖1 .
Done in class.
Problem 36. Check that
u ∗ v = v ∗ u .
16 T. PRZEBINDA
By the change of variables t = y − x we get
u ∗ v(y) = ∫ 1 0
u(y − x)v(x) dx = ∫ 1 0
u(t)v(y − t) dt = v ∗ u(y) .
Problem 37. Check that
F(u ∗ v) = F(u)F(v) .
F(u ∗ v)(n) = ∫ 1 0
(u ∗ v)(y)e−2πiyn dy = ∫ 1 0
∫ 1 0
u(y − x)v(x) dx e−2πiyn dy
=
∫ 1 0
∫ 1 0
u(y − x)v(x) dx e−2πi(y−x)yne−2πixn dy = F(u)(n)F(v)(n) .
Problem 38. For v ∈ L1(R/Z) let v∗(x) = v(−x). Check that
Fv∗ = Fv .
Fv∗(n) = ∫ 1 0
v∗(x)e−2πixn dx =
∫ 1 0
v(−x)e−2πixn dx == ∫ 1 0
v(−x)e2πixn dx
=
∫ 1 0
v(−x)e2πixn dx = Fv(n) .
Problem 39. For 0 < � < 1 2
define h� ∈ L1(R/Z) by
h�(x) =
{ 1 2�
if |x| ≤ � , 0 if � < |x| ≤ 1
2 .
Check that
Fh�(n) = {
sin(�2πn) �2πn
if n ∈ Z , n 6= 0 , 1 if n = 0 .
Done in class.
Problem 40. Compute the limit
lim �→0 Fh�(n) (n ∈ Z).
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 17
The limit is equal to 1 because
lim t→0
sin t
t = 1 .
Problem 41. Show that for a continuous function g ∈ L1(R/Z) lim �→0
g ∗ h�(0) = g(0).
g ∗ h�(0)− g(0) = 1
2�
∫ � −� g(x) dx− g(0) = 1
2�
∫ � −�
(g(x)− g(0)) dx.
Hence,
|g ∗ h�(0)− g(0)| ≤ 1
�
∫ � −� |g(x)− g(0)| dx ≤ 1
2�
∫ � −�
( max |x|≤� |g(x)− g(0)|
) dx
= max |x|≤� |g(x)− g(0)|.
Since g is continuous
lim �→0
( max |x|≤� |g(x)− g(0)|
) = 0
which proves the claim.
Problem 42. Show that for a continuous function f ∈ L1(R/Z) and for any y ∈ R lim �→0
f ∗ h�(y) = f(y).
This follows from Problem 41 with g(x) = f(y − x).
4. The Exam 1, due Wednesday 10/3/2018 in class. We DO NOT meet on Friday in class.
Problem 43. Find an error in the proof of Theorem 2.1, page 39, in the text and correct it
Problem 44. For any two integers x, y ∈ Z define I(x, y) = {ax+ by; a, b ∈ Z} .
(This is the set of sums of all multiples of x and all multiples of b.) Check that for any m,n ∈ I(x, y) and k ∈ Z we have
m+ n ∈ I(x, y) and mk ∈ I(x, y) .
18 T. PRZEBINDA
Recall division with remainder. More precisely, if d is a non-zero integer, then for any integer m there are integers q and r such that
m = dq + r , (27)
where either r = 0 or 0 < |r| < |m|. For example
7 = 2 · 3 + 1 , −7 = 2 · (−3) + (−1) .
This process is useful in solving the following Problem.
Problem 45. With the notation of Problem 44, let d be the smallest element of I(x, y), in the sense that
d = min{|m|; m ∈ I(x, y)} . Assume that d 6= 0. Show that d divides every element of I(x, y), i.e.
I(x, y) = dZ = {dq; q ∈ Z} .
In other words, I(x, y) consists of all the multiples of d.
Problem 46. With the notation of Problem 44, suppose that the greatest common diviser of x and y is 1. (In other words x and y are relatively prime, like 3 and 7.) Show that there are a, b ∈ Z such that
ax+ by = 1 .
Problem 47. Deduce from Problem 46 that for any prime number p, Z(p) is a field, i.e. every non-zero element x ∈ Z(p) is invertible. In other words, for any non-zero x ∈ Z(p), there is a ∈ Z(p) such that ax = 1 in Z(p). (Remember that in Z(p) we count modulo p.)
Problem 48. Deduce from Problem 47 that for any prime number p, and any non-zero x ∈ Z(p), the map
Z(p) 3 y → xy ∈ Z(p) is bijective (one to one and onto).
MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 19
Problem 49. Let f : Z(p)→ C be a function such that f(0) = 0, f(1) = 1 and
f(xy) = f(x)f(y) , f(z) = f(z−1) (x, y, z ∈ Z(p), z 6= 0) .
Show that
|Ff(1)| = 1 .
Problem 50. Assume that p be a prime different that 2. Let G = Z(p) \ {0} and let H = {x2; x ∈ G}. Show that G has twice as many elements as H, i.e. |G| = 2|H|.
Problem 51. With the notation of Problem 50 show that for any y ∈ G such that y /∈ H,
G = H . ∪ yH .
In other words G is the disjoint union of H and yH = {yx; x ∈ H}.
Problem 52. With the notation of Problem 50 show that product of two non-squares is a square. In other words
yH · yH ⊆ H .
Problem 53. Let p be a prime different that 2. Define
f(x) =
1 if x is a non-zero square, i.e. x ∈ H ,0 if x is a non-zero non-square, i.e. x ∈ yH ,0 if x = 0 . Show that f satisfies the assumptions of Problem 49.
Problem 54. Show that if p is an odd prime then∣∣∣∣∣ p−1∑ x=0
e− 2πi p x2
∣∣∣∣∣ = √p .
20 T. PRZEBINDA
Problem 55. Find out in the Internet which of the function defined above is called the Legendre symbol and which expression is called a Gauss sum.
References
[GGI05] D. Goldstein, R. Guralnick, and I. M. Isaacs. Inequalities for Finite Group Permutation Modules. Trans. Amer. Math. Soc., 357:4017–4042, 2005.
[MS73] Matolcsi and J. Szücs. Intersections des mesures spectrales conjugees. C. R. Acad. Sci. Paris, 277:841–843, 1973.
