Fourier Transform

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 3.2. Exam 1, due Wednesday 10/3/2018 in class. 17 3.3. The L1 Theory continued 22 3.4. The L2 Theory 23 4. The Fourier Transform on the Schwartz space S(R) 27 4.1. Basic properites of The Fourier Transform 27 4.2. Eigenvectors of Fourier Transform and Quantum Oscillator 30 4.3. The Heisenberg Uncertainty Principle 35 5. Exam 2, due Monday 11/26/2018 in class. 36 References 38

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) .

Date: November 21, 2018.

1

2 T. PRZEBINDA

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.

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.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 3

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.

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 θ2k

(2k)! + i

∞∑ k=0

(−1)k θ2k+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.

4 T. PRZEBINDA

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.

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)

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 5

In the particular case of this problem

z0 = 3, z1 = 3e i(2π/3), z2 = 3e

i(4π/3).

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.

6 T. PRZEBINDA

Problem 13. Prove that for z ∈ C\{1},

1 + z + z2 + · · · + zn = 1 −zn+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) = cos α cos β − sin α sin β (3) sin(α + β) = sin α cos β + cos α 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 = (cos α cos β − sin α sin β) + i(sin α cos β + cos α 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 4 χk,N

−1 4 χl) =

{ 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).

8 T. PRZEBINDA

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 −i2π

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 F4 = 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)| .

10 T. PRZEBINDA

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 = supp Tnv, n ∈ Z(N), cover the whole group Z(N):

Z(N) = ⋃

n∈Z(N)

supp Tnv . (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 = dim V 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

supp Tnv ⊆ supp Tn1v ∪ supp Tn2v ∪ ...∪ supp Tnmv . 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)

supp Tnv ⊆ supp Tn1v ∪ supp Tn2v ∪ ...∪ supp Tnmv .

By combining this with (11) we see that

Z(N) ⊆ supp Tn1v ∪ supp Tn2v ∪ ...∪ supp Tnmv . Hence the cardinality

|Z(N)| ≤ |supp Tn1v| + |supp Tn2v| + ... + |supp Tnmv| . But

|supp v| = |supp Tn1v| = |supp Tn2v| = ... = |supp Tnmv| . 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

dim V = 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

12 T. PRZEBINDA

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.

14 T. PRZEBINDA

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) dxe−2πiyn dy

=

∫ 1 0

∫ 1 0

u(y −x)v(x) dxe−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).

3.2. Exam 1, due Wednesday 10/3/2018 in class.

Problem 43. Find an error in the proof of Theorem 2.1, page 39, in the text and correct it

The assumption is that the function f is integrable, but not necessarily bounded. On page 50, the authors write “since f is integrable, hence bounded.” This is not true. (There are unbounded integrable functions, like |x|−1/2.) They use this statement in their proof of the last inequality on that page. The correct estimate looks as follows∣∣∣∣ ∫ δ≤|x|≤π

f(x)pk(x) dx

∣∣∣∣ ≤ 2π ∣∣∣∣ ∫ δ≤|x| |f(x)|dx

∣∣∣∣ (1 − �/2)k ≤ 2π ∫ ≤|x|≤π

|f(x)|dx(1 − �/2)k .

18 T. PRZEBINDA

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) .

By assumption, m = ax + by and n = cx + dy for some integers a, b, c and d. Hence

m + n = (ax + by) + (cx + dy) = (a + c)x + (b + d)y ∈ I(x,y) . Also,

mk = (ak)x + (bk)y ∈ I(x,y) . 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| < |d|. 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.

Suppose m ∈ I(x,y) and d does not divide m. Then (27) holds with r 6= 0, and r = m− qd ∈ I(x,y)

by Problem 44. Since |r| < d we get a contradiction. Thus d divides every element of I(x,y).

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 .

Let d be as in Problem 45. Then d divides x and d divides y. Since x and y are rel- atively prime, d = 1. Since d ∈ I(x,y), the claim follows.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 19

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.)

Since x ∈ {1, 2, ...,p − 1} and since p is prime, the greatest common divisor of x and p is 1. Hence, by Problem 46, there are integers a and b such that

ax + bp = 1 .

Now we take this equation modulo p and get

cx = 1 ,

where c ∈{1, 2, ...,p− 1} is equal to a 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).

Since the set Z(p) is finite, it is enough to see that this map is injective. Let y,z ∈ Z(p). Suppose

xy = xz .

Then

x(y −z) = 0 .

We multiply both sides by the inverse of x (which exists thanks to Problem 47) and get

y −z = 0 ,

i.e. y = z.

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 .

20 T. PRZEBINDA

We compute

|Ff(1)|2 = Ff(1)Ff(1) = 1 √ p

∑ x∈Z(p)

f(x)e −2πi

p x 1√

p

∑ y∈Z(p)

f(y)e 2πi p y

= 1

p

∑ x,y∈Z(p)

f(x)f(y−1)e −2πi

p (x−y)

= 1

p

∑ x,y∈Z(p)

f(xy−1)e −2πi

p (x−y)

= 1

p

∑ z,y∈Z(p)

f(z)e −2πi

p ((z−1)y)

= ∑ z∈Z(p)

f(z) 1

p

∑ y∈Z(p)

e −2πi

p ((z−1)y)

= ∑ z∈Z(p)

f(z)δ(z − 1)

= f(1) = 1 .

Here we used Problem 48 in the fifth equality.

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|.

For each x2 ∈ H there are exactly two elements ±x ∈ G whose squares are x2. Fur- thermore, if x2 6= y2 then

{x,−x}∩{y,−y} = ∅ . Hence the claim follows.

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}.

Since yH has the same number of elements as H and since |G| = 2|H|, by Problem 50, it’ll suffice to show that

H ∩yH = ∅ . If not then there are x,z ∈ G such that

x2 = yz2 .

Hence

y = (xz−1)2 ∈ H , which is a contradiction.

Problem 52. With the notation of Problem 50 show that product of two non-squares is a square. In other words

yH ·yH ⊆ H .

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 21

This is clear, because

yx2yz2 = (yxz)2 .

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 , −1 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.

Let x,y ∈ G. If both are squares then xy is a square, so f(xy) = 1, f(x) = 1 and f(y) = 1, so f(xy) = f(x)f(y).

If both are non-squares then by Problem 52, xy is a square, so f(xy) = 1, f(x) = −1 and f(y) = −1, so f(xy) = f(x)f(y).

If x is a square and y is a non-square, then xy is a non-square, so f(x) = 1, f(y) = −1 and f(xy) = −1, so f(xy) = f(x)f(y).

Also, f(x) = f(x) and f(x) = f(x−1) because x is a square if and only if x−1 is a square.

If x = 0, then xy = 0, so f(xy) = f(x)f(y).

Problem 54. Show that if p is an odd prime then∣∣∣∣∣ p−1∑ x=0

e −2πi

p x2

∣∣∣∣∣ = √p. Notice that with the notation of Problem 53

p−1∑ x=0

e −2πi

p x2

=

p−1∑ x=0

(1 + f(x))e −2πi

p x

=

p−1∑ x=0

e −2πi

p x

+

p−1∑ x=0

f(x)e −2πi

p x

=

p−1∑ x=0

f(x)e −2πi

p x

= √ pFf(1) .

Hence the claim follows from Problem 49.

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.

The function defined in Problem 53 is called the Legendre symbol. It is usually denotes as

f(x) =

( x

p

) .

22 T. PRZEBINDA

The expression p−1∑ x=0

e −2πi

p x2

is known as a Gauss sum. These are basic objects in Number Theory.

3.3. The L1 Theory continued.

Problem 56. For N = 2, 3, 6, plot the functions

FN (x) = 1

N

( sin(Nπx)

sin(πx)

)2 , −

1

2 < x <

1

2

using wolframalpha.com .

See the attached pdfs: F2explicit.pdf, F3explicit.pdf and F6explicit.pdf on crystal.ou.edu.

As we checked in class, the convolution of a continuous function f with the Fejer kernel FN is equal to

f ∗FN (x) = 1

N

N−1∑ n=0

n∑ k=−n

f̂(k)e2πikx . (28)

Problem 57. Suppose f(x) = f(−x). Show that f̂(k) = f̂(−k) and consequently,

f ∗FN (x) = f̂(0) + N−1∑ n=1

2 N −n N

f̂(n) cos(n2πx) .

The first part is straightforward

f̂(−k) = ∫ 1

2

−1 2

f(x)e2πikx dx =

∫ 1 2

−1 2

f(−x)e−2πikx dx

=

∫ 1 2

−1 2

f(x)e−2πikx dx = f̂(k) .

Hence, from (28),

f ∗FN (x) = 1

N

( f̂(0) +

N−1∑ n=1

n∑ k=0

f̂(k)2 cos(2πikx)

)

= 1

N (f̂(0)

+ f̂(0) + f̂(1)2 cos(2πx)

+ f̂(0) + f̂(1)2 cos(2πx) + f̂(2)2 cos(2 · 2πx) + f̂(0) + f̂(1)2 cos(2πx) + f̂(2)2 cos(2 · 2πx) + ... + f̂(N − 1)2 cos((N − 1) · 2πx)) ,

which verifies our formula.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 23

Problem 58. Let

f(x) =

{ x + 1

2 −1

2 ≤ x ≤ 0 ,

1 2 −x 0 ≤ x ≤ 1

2 .

Show that the Fourier transform is given by

f̂(0) = 1

4 ,

f̂(k) = 1

π2k2 , k = 1, 3, 5, 7, ...

f̂(k) = 0 , k = 2, 4, 6, ...

We use integration by parts and coffee. (Partially done in class.)

Problem 59. With the notation of Problem 58 show that

f ∗F2(x) = 1

4 +

1

π2 cos(2πx) ,

f ∗F4(x) = 1

4 +

3

2

1

π2 cos(2πx) +

1

2

1

π29 cos(6πx) ,

f ∗F6(x) = 1

4 +

5

3

1

π2 cos(2πx) +

3

3

1

π29 cos(6πx) +

1

3

1

π225 cos(10πx) .

This is obvious from Problem 58.

Problem 60. Use wolframalpha.com to plot the functions f ∗F2, f ∗F4 and f ∗F6 to see that they approximate f.

See the separate files F2.pdf, F4.pdf and F6.pdf on crystal.ou.edu.

3.4. The L2 Theory. Set

en(x) = e 2πinx (n ∈ Z,x ∈ R) .

Recall the inclusion

L2(R/Z) ⊆ L1(R/Z) . For f ∈ L2(R/Z) define the partial Fourier sums

SN (f) = N∑

n=−N

f̂(n)en (N = 0, 1, 2, ...) .

The crown jewel of the L2 theory is the following theorem.

24 T. PRZEBINDA

Theorem 3. For any f ∈ L2(R/Z) lim N→∞

‖ f −SN (f) ‖2= 0 (29)

and ‖ f ‖22=

∑ n∈Z

|f̂(n)|2 . (30)

Recall that the support of a function is the closure of the set where the function is non-zero. If the function is defined on discrete set like Z, the set of the integers, then the measure of the support is the same as the cardinality of the support.

Based on the Theorem 3 we checked that the following (essentially empty) Uncertainty Principle holds.

Theorem 4. For any non-zero f ∈ L2(R/Z) we have either

measure of supp(f) · cardinality of supp f̂ = ∞ (31) or

measure of supp(f) · cardinality of supp f̂ = cardinality of supp f̂ ≥ 1 . (32) In the case (32), f is a trigonometric polynomial.

In the following problems use wolframalpha or any other helpful computer program. In wolframalpha f(x + 2π) = f(x),

f̂(n) = 1

2π

∫ 2π 0

f(x)e−inx dx (n ∈ Z) ,

and

SN (f) = N∑

n=−N

f̂(n)einx (−π ≤ x ≤ π) .

Problem 61. In order to see that the Fourier transform of an odd function is odd, com- pute f̂(n) for f(x) = x, restricted to the interval [−π,π]. Similarly, in order to see that the Fourier transform of an even function is even, computef̂(n) for f(x) = x2, restricted to the interval [−π,π].

For f(x) = x,

f̂(n) =

{ i(−1)n n

if n 6= 0 , 0 if n = 0 .

For f(x) = x2,

f̂(n) =

{ 2(−1)n n2

if n 6= 0 , π3

3 if n = 0 .

Problem 62. For f(x) = x plot S0(f), S1(f), S2(f) and S3(f).

Type: fourier series of x .

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 25

Problem 63. For f(x) = x2 plot S0(f), S1(f), S2(f) and S3(f).

Type: fourier series of x squared .

Problem 64. Plot f(x) = |x| x

for −π < x < π.

Type: plot |x| x ,−π < x < π

Problem 65. For f(x) = |x| x

plot S0(f), S1(f), S2(f) and S9(f) and notice how these sums approximate the original function.

For the last term type FourierSeries[|x|/x,x, 9] .

Problem 66. For f(x) = cos(2πx) − sin(3πx) plot S0(f), S1(f), S2(f), S3(f), S4(f), S5(f). This is to notice that for a trigonometric polynomial, the SN does not depend on N if N is large enough.

As in Problem 65.

Problem 67. Compute ∞∑ n=1

1

n2 .

From Problem 61 and Theorem 3, ∞∑ n=1

1

n2 =

1

2

∑ n∈Z\{0}

1

n2 =

1

2

1

2π

∫ π −π x2 dx =

1

2

π2

3 = π2

6 .

Problem 68. Show that if f is differentiable and periodic (f(x + 1) = f(x)), then

Ff ′(n) = 2πinFf(n) (n ∈ Z) .

We integrate by parts,

Ff ′(n) = ∫ 2π 0

f ′(x)e−2πinx dx = f(x)e−2πinx|2π0 − ∫ 2π 0

f(x)(−2πin)e−2πinx dx

=

∫ 2π 0

f(x)2πine−2πinx dx = 2πinFf(n) .

Problem 69. Let V be a vector space over the complex numbers with a positive definite hermitian product ( , ) and the corresponding norm ‖ v ‖2= (v,v). Verify the following identity

4(u,v) =‖ u + v ‖2 −‖ u−v ‖2 +i ‖ u + iv ‖2 −i ‖ u− iv ‖2 .

26 T. PRZEBINDA

This is completely straightforward.

Problem 70. Let V and V′ be two vector spaces over the field of complex numbers. Let ( , ) be a positive definite hermitian product on V and let ( , )′ be a positive definite hermitian product on V′. Denote by ‖ v ‖2= (v,v) and ‖ v ‖′ 2 = (v′,v′)′ the corresponding norms. Suppose T : V → V′ is a linear map such that

‖ Tv ‖′=‖ v ‖ (v ∈ V).

Show that

(Tu,Tv)′ = (u,v) (u,v ∈ V).

We use the identity of the previous problem:

4(Tu,Tv)′ = ‖ Tu + Tv ‖′ 2−‖ Tu−Tv ‖′ 2 + i ‖ Tu + iTv ‖′ 2 − i ‖ Tu− iTv ‖′ 2

= ‖ T(u + v) ‖′ 2−‖ T(u−v) ‖′ 2 + i ‖ T(u + iv) ‖′ 2 − i ‖ T(u− iv) ‖′ 2

= ‖ u + v ‖2 −‖ u−v ‖2 +i ‖ u + iv ‖2 −i ‖ u− iv ‖2

= 4(u,v).

Problem 71. Deduce from Problem 70 and Parseval’s identity (Theorem 3, formula 30) that for any 1-periodic, square integrable functions f,g : R → C∫ 1

0

f(x)g(x) dx = ∑ n∈Z

Ff(n)Fg(n).

Take V to be the space of all the 1-periodic, square integrable functions f : R → C with the hermitian product ∫ 1

0

f(x)g(x) dx

and V′ to be the space of all the sequences an ∈ C such that ∑

n∈Z |an| 2 < ∞, with the

hermitian product ∑ n∈Z

anbn.

Since the linear map F : V → V′ preserves the norm, the formula follows from Problem 70.

Problem 72. Read the proof of the isoperimetric inequality in the text, all the way to the end with understanding.

Congratulations and thank you.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 27

4. The Fourier Transform on the Schwartz space S(R)

4.1. Basic properites of The Fourier Transform. Here we follow [Hör83, section 7.1]. Recall that the Schwartz space S(R) consists of all infinitely many times differentiable functions f : R → C such that for any two integers n,k ≥ 0

sup x∈R |xn∂kxf(x)| < ∞ .

In particular S(R) ⊆ L1(R) and we have the well defined Fourier transform

Ff(y) = ∫ R e−2πixyf(x) dx (y ∈ R, f ∈S(R)) .

As we checked in class F maps S(R) itno S(R) F : S(R) →S(R) .

Problem 73. Let f ∈ S(R) be such that f(0) = 0. Set g(x) = ∫ 1 0 f ′(tx) dt. Show that

f(x) = xg(x).

This is immediate from the Fundamental Theorem of Calculus, via a change of variables y = tx.

Problem 74. With the notation of Problem 73 show that g ∈S(R).

Fix two non-negative integers n and k. Suppose |x| ≤ 1. Then∣∣xng(k)(x)∣∣ = ∣∣∣∣xn ∫ 1 0

tkf(k+1)(tx) dt

∣∣∣∣ ≤ ∫ 1 0

|f(k+1)(tx)|dt ≤ max y∈R |f(y)| < ∞.

Notice that

xng(k)(x) = xn ( d

dx

)k (x−1f(x))

= xn k∑ p=0

k!

p!(k −p)! (−1)(−2) . . . (−p)x−p−1f(k−p)(x)

and that max |x|≥1

∣∣xn−p−1f(k−p)(x)∣∣ < ∞. Hence

max x∈R |xng(k)(x)| < ∞.

Problem 75. Fix y ∈ R. Let φ ∈S(R) be such that φ(y) = 0. Show that there is ψ ∈S(R) such that

φ(x) = (x−y)ψ(x).

28 T. PRZEBINDA

Let f(x) = φ(x + y). Then f ∈ S(R) and f(0) = 0. Let g be as in Problem 74. Set ψ(x) = g(x−y). Then ψ ∈S(R) and

φ(x) = f(x−y) = (x−y)g(x−y) = (x−y)ψ(x) .

Problem 76. Let T : S(R) →S(R) be a linear map with the property that if φ(y) = 0 for some y ∈ R then Tφ(y) = 0 for the same y. Show that there is a function c(x) such that

Tφ(x) = c(x)φ(x) (φ ∈S(R).

(In other words, T is the multiplication by the function c.)

Let φ1(x) = e −x2 . As we know this function belongs to S(R). Fix x ∈ R. Then for

any φ2 ∈S(R)

(φ2(x)φ1 −φ1(x)φ2) (x) = φ2(x)φ1(x) −φ1(x)φ2(x) = 0.

Hence, by the assumption on T ,

0 = T (φ2(x)φ1 −φ1(x)φ2) (x). (33)

Since T is linear

T (φ2(x)φ1) = φ2(x)T(φ1) and T(φ1(x)φ2) = φ1(x)T(φ2).

Thus evaluation at x and using (33) we see that

0 = φ2(x)T(φ1)(x) −φ1(x)T(φ2)(x).

Therefore

T(φ2)(x) = T(φ1)(x)

φ1(x) φ2(x).

Thus the claim holds with

c(x) = T(φ1)(x)

φ1(x) .

Problem 77. Suppose T : S(R) →S(R) is a linear map which commutes with the multi- plication by x:

T(xf(x)) = xT(f)(x) (f ∈S(R)). Show that T has the property of Problem 76.

Fix y ∈ R. Let φ ∈ S(R) be such that φ(y) = 0. Then, by Problem 3 there is ψ ∈ S(R) such that

φ(x) = (x−y)ψ(x). Hence

T(φ)(x) = T((x−y)ψ(x)) = (x−y)T(φ)(x), which is zero if x = y.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 29

Problem 78. Let T be as in Problem 77. Show that there is a function c(x) such that

Tφ(x) = c(x)φ(x) (φ ∈S(R)).

This is immediate from Problems 77 and 78.

Problem 79. Suppose T : S(R) →S(R) is a linear map which commutes with the multi- plication by x:

T(xf(x)) = xT(f)(x) (f ∈S(R)) and with the derivative

T(f ′) = T(f)′ (f ∈S(R)). Show that there is a constant c such that

Tφ(x) = cφ(x) (φ ∈S(R)).

We know from Problem 78. that T coincides with the multiplication by a function c(x). Since T commutes with the derivative we see that for any f ∈S(R)

c(x)f ′(x) = (c(x)f(x))′.

But

(c(x)f(x))′ = c′(x)f(x) + c(x)f ′(x).

Hence c′(x) = 0. therefore c(x) is a constant.

Problem 80. Let F : S(R) →S(R) denote the Fourier transform. Show that

F2(xf(x)) = −xF2(f)(x) and F2(f ′) = − ( F2(f)

)′ .

(Here F2(f) = F(F(f)) is the Fourier transform of the Fourier transform of f.)

We compute using the formulas of Proposition 1.2 on page 136 in the text.

F2(−2πixf(x))(ξ) = F(F(−2πixf(x)))(ξ) = F((F(f))′)(ξ) = 2πiξF(F(f))(ξ)

and

F2(f ′)(ξ) = F(F(f ′))(ξ) = F(2πixF(f))(ξ) = −F((F(f))′(ξ).

Problem 81. Let Rf(x) = f(−x). Show that the linear map T = RF2 : S(R) → S(R) commutes with the multiplication by x and with the derivative.

Since

R(xf(x)) = −xf(−x) = −xR(f)(x) and R(f ′) = −R(f) the claim follows from Problem 80.

30 T. PRZEBINDA

Problem 82. Show that the map RF2 : S(R) →S(R) is the identity, RF2f = f (f ∈S(R)).

We know from Problem 81 and 79 that the map RF2 is a constant multiple of the identity: cI = RF2. In order to find the constant c we evaluate both sides at f(x) = e−πx2 . Since Ff = f and since Rf = f,

cf = RF2f = f. Thus c = 1.

Problem 83. Deduce from Problem 82 the Fourier inversion formula:

F−1f(x) = ∫ R f(ξ)e2πixξ dξ (f ∈S(R)) .

The equation RF2 = I

implies F−1 = RF

which, in terms of the integrals, is exactly what we want.

Problem 84. Deduce from Problem 82 the formula:

F4 = I .

We see from Problem 82 that F2 = R−1 = R.

Hence, F4 = R2 = I .

4.2. Eigenvectors of Fourier Transform and Quantum Oscillator. Here we follow [HT92, section 2.1]. Let

A = 2πx−∂ , B = 2πx + ∂ . We view A and B as linear maps from S(R) to S(R).

Problem 85. Check that FA = −iAF

This follows from the equalities

Ff ′(y) = 2πiyFf(y) , F(2πixf(x))(y) = −(Ff)′(y) ,

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 31

which were checked in class.

For any two maps R,T : S(R) →S(R)

define their commutator by [R,T] = RT −TR

Problem 86. Check that or any three maps R,S,T : S(R) →S(R) [R,TS] = T[R,S] + [R,T]S .

This is completely straightforward and was checked in class.

Problem 87. Check that or any two maps R,T : S(R) →S(R) and n = 1, 2, 3, ...,

[R,Tn] = n−1∑ j=0

Tj[R,T]Tn−j−1 .

This is done by induction on n and was checked in class.

Problem 88. Check that for n = 1, 2, 3, ...,

[B,An] = 4πnAn−1 .

Explicite computation shows that

[x,x] = 0, [∂,∂] = 0, [∂,x] = 1 .

Hence [B,A] = 4π.

Hence, Problem 87 applied to R = B and T = A implies the formula. was checked in class.

Recall the Gaussian γ(x) = e−πx

2

(x ∈ R) . Let

γj = A jγ (j = 0, 1, 2, ...)

Problem 89. Check that Bγ = 0

and for j = 1, 2, 3, ..., Bγj = 4πjγj−1 .

The first forlumla is easy and for the second one we use Problem 88.

32 T. PRZEBINDA

Recall the scalar product

(f,g) =

∫ R f(x)g(x) dx (f,g ∈S(R)) .

Problem 90. Show that

(Af,g) = (f,Bg) (f,g ∈S(R)) .

This follows the the formula∫ R xf(x)g(x) dx =

∫ R f(x)xg(x) dx

and ∫ R f ′(x)g(x) dx = −

∫ R f(x)g′(x) dx,

where the second one is verified via integration by parts.

Problem 91. Check that for j,k = 1, 2, 3, ..., with j > k,

(γj,γk) = 4πk(γj−1,γk−1) .

We use Problems 89 and 90 and compute

(γj,γk) = (Aγj−1,γk) = (γj−1,Bγk) = (γj−1, 4πkγk−1)

Problem 92. Check that for j,k = 0, 1, 2, 3, ..., with j > k,

(γj,γk) = 0 .

We use Problem 91 and compute

(γj,γk) = (γj−1, 4πkγk−1) = ... = (4π) kk!(γj−k,γ0) = (4π)

kk!(γj−k−1,Bγ0) = 0 .

Problem 93. Check that for j,k = 0, 1, 2, 3, ..., with j 6= k, (γj,γk) = 0 .

Since (γj,γk) = (γk,γj), we may assume that j > k. Then the conclusion follows from Problem.

Problem 94. Check that for j,k = 0, 1, 2, 3, ...,

(γj,γj) = (4π)jj! √

2 .

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 33

From the solution to Problem 93,

(γj,γj) = (4π) jj!(γ0,γ0) .

Also

(γ0,γ0) =

∫ R e−2πx

2

dx = 1 √

2

∫ R e−πx

2

dx = 1 √

2 .

Problem 95. For n = 0, 1, 2, 3, ..., define

vn =

√ √ 2

(4π)nn! γn .

Show that

(vn,vk) =

{ 1 if n = k , 0 if n 6= k ,

and Fvn = (−i)nvn .

This is immediate from previous Problems.

Problem 96. Let H = 1 2 (AB + BA). Check that

H = 4π2x2 −∂2 .

Notice that

BA = (2πx + ∂)(2πx−∂) = 4π2x2 − 2πx∂ + ∂2πx−∂2 = 4π2x2 + 2π −∂2

and AB + BA = 2AB + BA−AB = 2(4π2x2 + 2π −∂2) + 4π = 2H .

Problem 97. Show that

Hvn = 4π(n + 1

2 )vn (n = 0, 1, 3, ...) .

Notoce that ABvn = A4πnvn−1 = 4πnvn

and BAvn = Bvn+1 = 4π(n + 1)vn .

Hence (AB + BA)vn = (4πn + 4π(n + 1))vn = 4π(2n + 1)vn .

34 T. PRZEBINDA

In Quantum Mechanics, B is known as the creation operator, A as the annihilation operator and H as the energy operator (or Hamiltonian) of the quantum oscillator. The functions vn are the states of this oscillator and the numbers 4π(n+

1 2 ) as the corresponding

energies. Also, the following theorem holds, but we shall not go through the proof in this class.

It would take too long, but the statement is easy to remember.

Theorem 5. The functions vn, n = 0, 1, 2, ..., form an orthonormal basis of the space S(R) in the sense that each f ∈S(R) may be written as the following convergent series

f = ∞∑ n=0

(f,vn)vn .

Problem 98. Is there a non-zero function f ∈S(R) such that Ff = 3f?

No, because the only eigenvalues of the Fourier transform are ±1 and ±i.

Problem 99. Are there a non-zero function f,g ∈S(R) such that∫ R f(x)g(x) dx = 4 and

∫ R Ff(x)Fg(x) dx = 6?

No, because the Fourier transform preserves the scalar product given by the above integral.

Problem 100. Is there a function f ∈S(R) such that f(x) > 0 for all x ∈ R and Ff(x) > 0 for all x ∈ R?

Yes, the Gaussian

f(x) = e−πx 2

.

Problem 101. Is there a function f ∈S(R) such that f(x) > 0 for all x ∈ R and Ff(0) = 0?

No, because

Ff(0) = ∫ R f(x) dx > 0.

Problem 102. Recall the functions vn, defined in Problem 95. (They are called Hermite functions.) Compute the integral∫

R (v1(x) + 2v2(x) + 3v3(x))

2 dx.

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 35

We recognize that the above integral is equal to

(v1 + 2v2 + 3v3,v1 + 2v2 + 3v3) = (v1,v1) + (v1, 2v2) + (v1, 3v3)

+ (2v2,v1) + (2v2, 2v2) + (2v2, 3v3)

+ (3v3,v1) + (3v3, 2v2) + (3v3, 3v3)

= (v1,v1) + (2v2, 2v2) + (3v3, 3v3)

= (v1,v1) + 4(v2,v2) + 9(v3,v3) = 1 + 4 + 9 = 14 .

Problem 103. Compute the Fourier transform

F(v1 + 2v2 + 3v3).

Since

Fvj = (−i)vj We see that

F(v1 + 2v2 + 3v3) = −iv1 − 2v2 + i3v3 .

4.3. The Heisenberg Uncertainty Principle. The following theorem, proven by Herman Weyl in 1931, was verified in class.

Theorem 6. Let f ∈S(R) be such that∫ R |f(x)|2 dx = 1 . (34)

Define

µ(f) =

∫ R x|f(x)|2 dx, σ2(f) =

∫ R (x−µ(f))2|f(x)|2 dx, (35)

µ̃(f) =

∫ R x|f̂(x)|2 dx, σ̃2(f) =

∫ R (x− µ̃(f))2|f̂(x)|2 dx.

Then

σ(f)σ̃(f) ≥ 1

4π . (36)

The equality occurs in (36) if and only if f is a translation or a modulation of a normalized

Gaussian e−ax 2 , a > 0.

Problem 104. Define a dilation of a function f : R → C by

ft(x) = t −1

2 f(t−1x) (t > 0) .

Show that

‖ ft ‖2=‖ f ‖2 (t > 0) .

36 T. PRZEBINDA

We compute

‖ ft ‖22= ∫ R |ft(x)|2 dx =

∫ R t−1|f(t−1x)|2 dx =

∫ R |f(x)|2 dx.

Problem 105. Show that

F(ft) = (Ff)t−1 (t > 0) .

We compute

F(ft)(y) = ∫ R e−2πixyt−

1 2 f(t−1x) dx =

∫ R e−2πixtyt

1 2 f(x) dx = (Ff)t−1 (y) .

Problem 106. Show that

σ(ft)σ̃(ft) = σ(f)σ̃(f) (t > 0) .

We compute

µ(ft) =

∫ R x|t−

1 2 f(t−1x)|2 dx =

∫ R tx|f(x)|2 dx = tµ(f)

σ2(ft) =

∫ R (x−µ(ft))2|ft(x)|2 dx =

∫ R (tx− tµ(f))2|f(x)|2 dx = t2σ2(f) .

Similarly, we see from Problem 105 that

σ̃2(ft) = t −2σ̃2(f) .

Hence the formula follows.

5. Exam 2, due Monday 11/26/2018 in class.

Problem 107. Let k = 0, 1, 2, .... Suppose f ∈ L1(R/Z) is k - times differentiable (i.e. the derivatives f ′,f ′′, ...,f(k) exist and belong to f ∈ L1(R/Z). Show that there is a constant C such that

|Ff(n)| ≤ C(1 + |n|)k (n ∈ Z) .

Problem 108. Let k = 0, 1, 2, .... Suppose f ∈ L1(R) is k - times differentiable (i.e. the derivatives f ′,f ′′, ...,f(k) exist and are integrable, i.e. belong to f ∈ L1(R). Show that there is a constant C such that

|Ff(y)| ≤ C(1 + |y|)k (y ∈ R) .

MATH 4123, HOMEWORK, EXAMS AND SOLUTIONS, FALL 2018 37

Problem 109. Show that ∞∑ n=1

1

n4 = π4

90 .

Hint: use Parseval’s formula and Problem 61.

Problem 110. Compute the integral∫ R

( 1

π

1

x2 + 1

)2 dx.

Hint: use Parseval’s formula and Lemma 2.4 on page 150 in the text.

Problem 111. Give an example of a non-zero function f ∈ L1(R/Z) is such that f∗f = f.

Problem 112. Suppose f ∈ L1(R) is such that f ∗f = f. Show that f = 0.

Problem 113. Let p(x) be a polynomial and let

f(x) =

{ p(x−1)e−x

−1 if x > 0 ,

0 if x ≤ 0 .

Show that the derivative f ′(0) = lim x→0

f(x)−f(0) x

exists and is equal to 0.

Problem 114. Let

φ(x) =

{ e−x

−1 if x > 0 ,

0 if x ≤ 0 . Show that the derivative φ(k)(0) exists and is equal to 0, and therefore φ(k)(x) exists for any x ∈ R.

Problem 115. Show that the space C∞c (R) is non-trivial in the sense that it contains a non-zero function.

38 T. PRZEBINDA

Problem 116. Let f,g ∈ S(R) and let a0,a1, ...,aN ∈ C be constants. Show that the functions f and g satisfy the differential equation

a0f(x) + a1∂xf(x) + a2∂ 2 xf(x) + ... + a2∂

N x f(x) = g(x) (x ∈ R)

if and only if their Fourier transforms f̂ and ĝ satisfy the the following algebraic equation,

(a0 + a1(2πiy) + a2(2πiy) 2 + ... + a2(2πiy)

N )f̂(y) = ĝ(y) (y ∈ R) .

Problem 117. With the notation of Problem 116, show that the only solution f ∈ S(R) of the differential equation

a0f(x) + a1∂xf(x) + a2∂ 2 xf(x) + ... + a2∂

N x f(x) = 0 (x ∈ R) ,

where not all an are zero, is f = 0.

References

[GGI05] D. Goldstein, R. Guralnick, and I. M. Isaacs. Inequalities for Finite Group Permutation Modules. Trans. Amer. Math. Soc., 357:4017–4042, 2005.

[Hör83] L. Hörmander. The Analysis of Linear Partial Differential Operators I. Springer Verlag, 1983. [HT92] R. Howe and E. C. Tan. Non-Abelian Harmonic Analysis, Applications of SL2(R). Springer

Verlag, 1992. [MS73] Matolcsi and J. Szücs. Intersections des mesures spectrales conjugees. C. R. Acad. Sci. Paris,

277:841–843, 1973.