Honors 1: Undergraduate Math Lab^{∗}

Peter Sarnak^{†}, Steven Miller^{‡}, Alex Barnett^{§}

October 16th, 2002

Courant Institute of Mathematical Sciences New York University New York, NY

Abstract

More on Graphs; Arithmetic maps mod p. Lecture by Peter Sarnak; notes by Steven Miller. Appendix: Introduction to Fourier Series and {nα}, by Steven Miller.

1 Kesten’s Measure

For a k-regular graph, define

dμ_{k}(t) = \begin{cases}^{\frac{c_{k}\sqrt{(k−1)− \frac{t^{2}}{4}}}{1−( \frac{t}{k})^{2}}} dt & if |t| ≤ 2\sqrt{k − 1} \\ 0 & otherwise \end{cases}  (1)

Fix a and b, and consider the N eigenvalues λ_{j}. Count

\frac{#\{j : λ_{j} ∈ [a, b]\}}{N}.  (2)

Then Claim 1.1.

∗Homepage: http://www.math.nyu.edu/∼millerj/ †E-mail: sarnak@math.princeton.edu ‡E-mail: millerj@cims.nyu.edu or sjmiller@math.princeton.edu §E-mail: barnett@nmr.mgh.harvard.edu

lim_{N→∞} \frac{#\{j : λ_{j} ∈ [a, b]\}}{N} = μ_{k}([a, b]).  (3)

We have

(ρ_{k}(2n))^{\frac{1}{2n}} → 2\sqrt{k − 1}.  (4)

2 Generating Functions on k-Regular Trees

2.1 R(z)

Fix k. The generating function R(z) is

R(z) = ∑_{n=0}^{∞} r_{n}z^{n},  (5)

where

r_{n} = \frac{ρ(n)}{k^{n}} = Probability that we return to v in n steps.  (6)

For |z| < 1, the series expansion for R(z) converges.

Let q_{n} be the probability of starting at v and ending at v for the first time (after n steps).

2.2 Q(z)

Define

Q(z) = ∑_{n=0}^{∞} q_{n}z^{n}.  (7)

Exercise 2.1. Prove

R(z) = \frac{1}{1 − Q(z)}.  (8)

2.3 T (z)

Define

T (z) = ∑_{n=0}^{∞} t_{n}z^{n},  (9)

where for w adjacent to v, t_{n} is the probability of going from w to v in n-steps for the first time.

Further, let t_{w,v}(n) be the probability of going from w to v in n-steps first time and d(w, v) = m ≥ 1. Remember that d(w, v) is the distance from w to v.

Exercise 2.2. Prove

1. Q(z) = zT (z).

2. ∑_{n=0}^{∞} t_{w,v}(n)z^{n} = (T (z))^{n}.

Exercise 2.3. Prove

T (z) = \frac{z}{k} + \frac{k − 1}{k}zT^{2}(z).  (10)

Note that this explicitly gives us T (z) by application of the quadratic formula:

T (z) = \frac{1 ± \sqrt{1 − 4(\frac{k−1}{k}z)^{2} \frac{z}{k}}}{2\frac{k−1}{k}z} .  (11)

Now that we have T (z) we have Q(z), from which we get R(z). T (z) will have a square-root – it will be an algebraic function of z.

3 Recovering the Measure f(x) from R(z)

We have

R(z) = ∑_{n=0}^{∞} r_{n}z^{n}.  (12)

As we know T (z), we know R(z), hence we know the numbers r_{n}. Now,

r_{n} = ∫_{∞}^{∞} x^{n}f(x)dx = ∫_{−k}^{k} x^{n}f(x)dx.  (13)

How do we recover f(x) given the numbers r_{n}? We’ve now normalized the eigenvalues to lie in [−1, 1].

R(z) = ∑_{n=0}^{∞} ( ∫_{−1}^{1} x^{n}f(x)dx)

= ∫_{−1}^{1} ( ∑_{n=0}^{∞}(xz)^{n})f(x)dx

= ∫_{−1}^{1} \frac{f(x)}{1 − xz} dx
B(z) = \frac{1}{z}R(\frac{1}{z}) = ∫_{−1}^{1} \frac{f(x)}{z − x}dx.  (14)

Suppose we know the LHS. Can we recover f(x)? If z ∈ [−1, 1], the function will have a singularity. Thus, if f(x) is a nice function, we do not expect to be able to make sense of the above relation if z ∈ [−1, 1]. We will, however, consider z close to the interval [−1, 1].

Let z = ξ + iy, ξ ∈ [−1, 1], y > 0. Later we will take z = ξ − iy. Look at

B(ξ + iy) − B(ξ − iy) = ∫_{−1}^{1} f(x)[\frac{1}{ξ + iy − x} − \frac{1}{ξ − iy − x}]dx

= 2i ∫_{−1}^{1} \frac{yf(x)}{(ξ − x)^{2} + y^{2}} dx.  (15)

We will study the above as y → 0.

3.1 Poisson Kernel

Recall ξ ∈ [−1, 1], f(x) fixed, we are integrating f(x) against the Poisson Kernel

\frac{y}{(ξ − x)^{2} + y^{2}} .  (16)

As y → 0, the above looks singular at x = ξ. At x = ξ, the kernel has height \frac{1}{y}, which is quite large. If x = ξ + ϵ, then as y → 0, the kernel goes to 0 very rapidly. Basically, as y → 0, the kernel becomes a higher, thinner spike centered

at ξ.

Now

∫_{∞}^{∞} \frac{y}{(x − ξ)^{2} + y^{2}} dx = ∫_{−∞}^{∞} \frac{y}{t^{2} + y^{2}} dt
= ∫_{−∞}^{∞} \frac{y^{2}}{y^{2}η^{2} + y^{2}} dη, from \frac{t}{y} = η
= ∫_{−∞}^{∞} \frac{1}{1 + η^{2}} dη
= π.  (17)

This is an approximation to the identity. Thus,

B(ξ + iy) − B(ξ − iy) → 2πif(ξ).  (18)

3.2 Cauchy Integral Formula

If you have an analytic function f(z) and γ is a curve enclosing z then

\frac{1}{2πi} ∫_{γ} \frac{f(ζ)}{z − ζ} dζ  (19)

In our case above, we cannot apply Cauchy’s Integral Formula, as our function f(x) is not analytic. It is compactly supported, and no non-zero analytic function is compactly supported.

Call this permutation ϕ:

ϕ : 𝔽_{p}^{∗} → 𝔽_{p}^{∗},  (20)

where ϕ^{2} is the identity.

Question 3.1. Does ϕ behave like a random permutation?

4 Third Problem

4.1 Introduction

Let p be a large prime, and consider the map

x ↦ x^{−1} mod p, x ≠ 0.  (21)

This is a map from 𝔽_{p} → 𝔽_{p}. The map is not completely random, as

1 ↦ 1
2 ↦ \frac{p + 1}{2}
p − 1 ↦ p − 1.  (22)

The map which sends x → x^{−1} is a permutation of 𝔽_{p}^{∗}. It is not a completely arbitrary permutation, as it pairs x with x^{−1} (away from a few very special x’s, such as p − 1 and 1).

Thus, this permutation is a product of transpositions.

4.2 Character Sums

Let

1 ≤ A ≤ B ≤ p, B − A large.  (23)

Let

\overline{m} be the inverse of m mod p  (24)

IE, m\overline{m} ≡ 1 mod p. Let

e(z) = 2^{2πiz}.  (25)

For ν ∈ ℤ/pℤ, consider

S = ∑_{A≤m≤B} e(\frac{\overline{m}ν}{p} ).  (26)

These sums will measure how equidistributed or random the map m → \overline{m} is.

Exercise 4.1. Prove the trivial bound for |S|:

|S| ≤ B − A.  (27)

Let N = B − A + 1. By the Central Limit Theorem, with high probability if we add N random numbers of modulus one we expect square-root cancellation. Thus, we expect (if the inverse map is random) that |S| ≈ \sqrt{N}.

4.3 Completing the Square

Let a, b ∈ ℤ, and consider the Kloosterman Sum

Kl(a, b, p) = ∑_{{x mod p \atop x≠0}} e(\frac{ax + b\overline{x}}{p}).  (28)

How large can the Kloosterman Sum be? If a = b = 0, then trivially Kl(0, 0, p) = p − 1.

If b = 0 and a ≠ 0 (or, by symmetry, the other way around) then

Kl(a, b, p) = ∑_{{x mod p \atop x≠0}} e(\frac{ax}{p})

= ∑_{{y mod p \atop y≠0}} e(\frac{y}{p})

= ∑_{y=0}^{p−1} e(\frac{y}{p}) − 1 = −1.  (29)

Exercise 4.2. Prove ∑_{y=0}^{p−1} e(\frac{y}{p}) = 0. Hint: Let T be this sum. Then show

e(\frac{1}{p})T = T; thus T = 0.

4.4 Weil’s Bound

Let a ≢ 0 mod p. Then

|Kl(a, b, p)| ≤ 2\sqrt{p}.  (30)

This is a very deep result.

How big is

∑_{a mod p} |Kl(a, 1, p)|^{2}  (31)

If we believe Weil’s bound, each term is of size at most 2\sqrt{p}, we square, then sum p terms. Thus, we expect a size of at most 4p^{2}. We will show on average that Weil’s bound is correct.

∑_{a mod p} |Kl(a, 1, p)|^{2} = ∑_{a(p)} | ∑_{{x mod p \atop x≠0}} e(\frac{\overline{x} + ax}{p})|

= ∑_{a(p)} ∑_{{x_{1},x_{2} mod p \atop x_{1},x_{2}≠0}} e(\frac{\overline{x}_{1} − \overline{x}_{2} + a(x_{1} − x_{2})}{p})

= ∑_{{x_{1},x_{2} mod p \atop x_{1},x_{2}≠0}} e(\frac{\overline{x}_{1} − \overline{x}_{2}}{p} ) ∑_{a(p)} e(\frac{a(x_{1} − x_{2})}{p})
= (p − 1)p,  (32)

where the last line follows from the a-sum vanishes unless x_{1} = x_{2}, which then collapses the sums. There are p−1 ways x_{1} = x_{2}, and when this occurs, the a-sum gives p.

Exercise 4.3. Consider

∑_{a(p)} |Kl(a, 1, p)|^{4}.  (33)

Above there are p-terms, each term of size (2\sqrt{p})^{4} = 16p^{2}. Thus, show the sum is at most 16p^{3}. You will find cp^{3} for some c independent of p.

By looking at one term, as every summand is positive, we find

|Kl(a, 1, p)|^{4} ≤ cp^{3}.  (34)

Thus, taking the fourth-root yields

|Kl(a, 1, p)| ≤ c^{\frac{1}{4}} p^{\frac{3}{4}} .  (35)

4.5 Fourier Expansion of Sums

Define the indicator function

I(y) = \begin{cases}1 & A ≤ y ≤ A + N \\ 0 & otherwise \end{cases}  (36)

Consider

S = ∑_{A≤x≤A+N} e(\frac{ν\overline{x}}{p})
= ∑_{x(p)} e(\frac{ν\overline{x}}{p})I(x).  (37)

We want to write I(x) in terms of its Fourier Coefficients

Î(m) = ∫_{0}^{1} e(−mt)I(t)dt.  (38)

Then

I(y) = ∑_{m=−∞}^{∞} Î(m)e(\frac{my}{p} ).  (39)

4.6 Brief Review of Fourier Series

Consider the unit interval [0, 1]. Define

ϕ_{m}(x) = e(mx).  (40)

Then (if our function is sufficiently nice)

f(x) = ∑_{m∈ℤ} f̂(m)e(mx),  (41)

where

f̂(m) = ∫_{0}^{1} f(x)e(−mx)dx.  (42)

5 Fourier Analysis and the Equi-Distribution of {nα}

5.1 Inner Product of Functions

We define the exponential function by means of the series

e^{x} = ∑_{n=0}^{∞} \frac{x^{n}}{n!} ,  (43)

which converges everywhere. Given the Taylor series expansion of sin x and cos x, we can verify the identity

e^{ix} = cos x + i sin x.  (44)

Exercise 5.1. Prove e^{x} converges for all x ∈ ℝ (even better, for all x ∈ ℂ. Show the series for e^{x} also equals

lim_{n→∞} (1 + \frac{x}{n})^{n},  (45)

which you may remember from compound interest problems.

Exercise 5.2. Prove, using the series definition, that e^{x+y} = e^{x}e^{y}. Use this fact to calculate the derivative of e^{x}. If instead you try to differentiate the series directly, you must justify the derivative of the infinite sum is the infinite sum of the derivatives.

Remember the definition of inner or dot product: for two vectors v⃗ = (v_{1}, ⋯, v_{n}), w⃗ = (w_{1}, ⋯, w_{n}), we take the inner product v⃗ · w⃗ (also denoted ⟨v, w⟩) to mean

v⃗ · w⃗ = ⟨v, w⟩ = ∑_{i} v_{i}w̄_{i}.  (46)

Further, the length of a vector v is

|v| = ⟨v, v⟩.  (47)

We generalize this for functions. For definiteness, assume f and g are functions from [0, 1] to ℂ. Divide the interval [0, 1] into n equal pieces. Then we can represent the functions by

f(x) ⟷ (f(0), f(\frac{1}{n}), …, f(\frac{n − 1}{n})),  (48)

and similarly for g. Call these vectors f_{n} and g_{n}. As before, we consider

⟨f_{n}, g_{n}⟩ = ∑_{i=0}^{n−1} f(\frac{i}{n}) · ḡ(\frac{i}{n}).  (49)

In general, as we continue to divide the interval (n → ∞), the above sum diverges. For example, if f and g are identically 1, the above sum is n.

There is a natural rescaling: we multiply each term in the sum by \frac{1}{n}, the size of the sub-interval. Note for the constant function, the sum is now independent of n.

Thus, for good f and g we are led to

⟨f, g⟩ = lim_{n→∞} ∑_{i=0}^{n−1} f(\frac{i}{n}) · ḡ(\frac{i}{n})\frac{1}{n} = ∫_{0}^{1} f(x)\overline{g(x)}dx.  (50)

The last result follows by Riemann Integration.

Definition 5.3. We say two continuous functions on [0, 1] are orthogonal (or perpendicular) if their dot product equals zero.

Exercise 5.4. Prove x^{n} and x^{m} are not perpendicular on [0, 1] for n ≠ m.

We will see that the exponential function behaves very nicely under the inner product. Define

e_{n}(x) = e^{2πinx} for n ∈ ℤ.  (51)

Then a straightforward calculation shows

⟨e_{n}(x), e_{m}(x)⟩ = \begin{cases}1 & if n = m \\ 0 & otherwise. \end{cases}  (52)

Thus e_{0}(x), e_{1}(x), e_{2}(x), ⋯ are an orthogonal set of functions, which means they are pairwise perpendicular. As each function has length 1, we say the functions e_{n}(x) are an orthonormal set of functions.

Exercise 5.5. Prove ⟨e_{n}(x), e_{m}(x)⟩ is 1 if n = m and 0 otherwise.

5.2 Fourier Series and {nα}

5.2.1 Fourier Series

Let f be continuous and periodic on ℝ with period one. Define the nth Fourier coefficient f̂(n) of f to be

f̂(n) = a_{n} = ⟨f(x), e_{n}(x)⟩ = ∫_{0}^{1} f(x)e^{−2πinx}dx.  (53)

Returning to the intuition of ℝ^{m}, we can think of the e_{n}(x)’s as an infinite set of perpendicular directions. The above is simply the projection of f in the direction of e_{n}(x).

Exercise 5.6. Show

⟨f(x) − f̂(n)e_{n}(x), e_{n}(x)⟩ = 0.  (54)

This agrees with our intuition, namely, that if you remove the projection in a certain direction, what is left is perpendicular to that direction.

The N^{th} partial Fourier series of f is

s_{N}(x) = ∑_{n=−N}^{N} f̂(n)e_{n}(x).  (55)

Exercise 5.7. Prove

1. ⟨f(x) − s_{N}(x), e_{n}(x)⟩ = 0 if |n| ≤ N.

2. |f̂(n)| ≤ ∫_{0}^{1} |f(x)|dx.

3. If ⟨f, f⟩ < ∞, then ∑_{n=−∞}^{∞} |f̂(n)|^{2} ≤ ⟨f, f⟩.

4. If ⟨f, f⟩ < ∞, then lim_{|n|→∞} f̂(n) = 0.

As ⟨f(x)−s_{N}(x), e_{n}(x)⟩ = 0 if |n| ≤ N, we might think that we just have to let N go to infinity to obtain a series s_{∞} such that

⟨f(x) − s_{∞}(x), e_{n}(x)⟩ = 0.  (56)

Assume that for a periodic function g(x) to be orthogonal to e_{n}(x) for every n it must be zero for every x. Then f(x) − s_{∞}(x) = 0, and hence f = s_{∞}. Voilá – an expression for f as a sum of exponentials! Be careful, however. We have just glossed over the two central issues – completeness and, even worse, convergence. We will now see a way of avoiding some of our problems.

5.2.2 Weighted partial sums

Define

D_{N}(x) = ∑_{n=−N}^{N} e_{n}(x) = \frac{sin((2N + 1)πx)}{sin πx},

F_{N}(x) = \frac{sin^{2}(Nπx)}{N sin^{2} πx} = \frac{1}{N} ∑_{n=0}^{N−1} D_{n}(x).

(57)

Here F stands for Féjer, D for Dirichlet. In general, functions which we are interested in taking their inner product against f are called kernels; thus, the Dirichlet kernel, the Féjer kernel, etc.

Note that, no matter what N is, F_{N}(x) is positive for all x.

We say that a sequence f_{1}(x), f_{2}(x), f_{3}(x), … of functions is an approximation to the identity if

1. f_{N}(x) ≥ 0 for all x and every N;

2. ∫_{0}^{1} f_{N}(x)dx = 1;

3. lim_{N→∞} ∫_{δ}^{1−δ} f_{N}(x)dx = 0 if 0 < δ < \frac{1}{2}.

Theorem 5.8. The Féjer kernels F_{1}(x), F_{2}(x), F_{3}(x), … are an approximation to the identity.

Proof: The first property is immediate. The second follows from the observation that F_{N}(x) can be written as

F_{N}(x) = e_{0}(x) + \frac{N − 1}{N}(e_{−1}(x) + e_{1}(x)) + ⋯,  (58)

and all integrals are zero but the first, which is 1.

To prove the third property, note that F_{N}(x) ≤ \frac{1}{N sin^{2} πδ} for δ ≤ x ≤ 1−δ. □

Let f be a continuous, periodic function on ℝ with period one. Thus, we can consider f as a function on just [0, 1], with f(0) = f(1). Define

T_{N}(x) = ∫_{0}^{1} f(y)F_{N}(x − y)dy.  (59)

Recall the following definition and theorem:

Definition 5.9 (Uniform Continuity). A continuous function is uniformly continuous if given an ϵ > 0, there exists a δ > 0 such that |x−y| < δ implies |f(x) − f(y)| < ϵ. Note that the same δ works for all points.

Theorem 5.10. Any continuous function on a closed, compact interval is uniformly continuous.

Exercise 5.11. Show x^{n} is uniformly continuous on [a, b] for −∞ < a < b < ∞.

Theorem 5.12. Given ϵ > 0, there is an N such that

|f(x) − T_{N}(x)| ≤ ϵ  (60)

for every x ∈ [0, 1].

Proof. For any positive N,

T_{N}(x) − f(x) = ∫_{0}^{1} f(x − y)F_{N}(y)dy − f(x) · 1

= ∫_{0}^{1} f(x − y)F_{N}(y)dy − ∫_{0}^{1} f(x)F_{n}(y)dy (property 2 of F_{N})

= ∫_{0}^{δ} (f(x − y) − f(x))F_{N}(y)dy
+ ∫_{δ}^{1−δ} (f(x − y) − f(x))F_{N}(y)dy
+ ∫_{1−δ}^{1} (f(x − y) − f(x))F_{N}(y)dy.

(61)

Let δ ∈ (0, 1/2). Then, using the fact that the F_{N}(x)’s are an approximation to the identity, we find

|∫_{δ}^{1−δ} (f(x − y) − f(x))F_{N}(y)dy| ≤ 2 max |f(x)| · ∫_{δ}^{1−δ} F_{N}(y)dy. (62)

Since

lim_{N→∞} ∫_{δ}^{1−δ} F_{N}(y)dy = 0,  (63)

we obtain

lim_{N→∞} ∫_{δ}^{1−δ}(f(x − y) − f(x))F_{N}(y)dy = 0.  (64)

Thus, by choosing N large enough (where large depends on δ), we can insure that this piece is at most \frac{ϵ}{3}.

It remains to estimate what happens near zero. Since f is continuous and [0, 1] is compact, f is uniformly continuous. Thus, we can choose δ small enough that |f(x − y) − f(x)| < \frac{ϵ}{3} for any x and any positive y < δ. Then

|∫_{0}^{δ} (f(x − y) − f(x))F_{N}(y)dy| ≤ ∫_{0}^{δ} \frac{ϵ}{3}F_{N}(y)dy ≤ \frac{ϵ}{3} ∫_{0}^{1} F_{N}(y)dy ≤ \frac{ϵ}{3}.

(65)

Similarly

|∫_{1−δ}^{1} (f(x − y) − f(x))F_{N}(y)dy| ≤ \frac{ϵ}{3}.  (66)

Therefore

|T_{N}(x) − f(x)| ≤ ϵ  (67)

for all N sufficiently large.

Definition 5.13 (Trigonometric Polynomials). Any finite linear combination of the functions e_{n}(x) is called a trigonometric polynomial.

From Theorem 5.12 we immediately get the Stone-Weierstrass theorem:

Theorem 5.14 (Stone-Weierstrass). Any continuous period function can be uniformly approximated by trigonometric polynomials.

5.3 Equidistribution

We say that a sequence {x_{n}}, x_{n} ∈ [0, 1] is equidistributed if

lim_{N→∞} \frac{1}{2N + 1}#\{n : |n| ≤ N, x_{n} ∈ (a, b)\} = b − a  (68)

for all (a, b) ⊂ [0, 1].

Theorem 5.15 (Weyl). Let α be an irrational number in [0, 1]. Let x_{n} = {nα}, where {y} denotes the fractional part of y. Then the sequence {x_{n}} is equidistributed.

Proof. We will estimate \frac{1}{2N+1} ∑_{−N}^{N} χ_{(a,b)}(x_{n}) as N → ∞, where χ_{(a,b)} is the function taking the value 0 outside (a, b) and 1 inside (a, b). We call χ_{(a,b)} the characteristic function of the interval (a, b).

Thus, we must show

lim_{N→∞} \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) = b − a.  (69)

Consider e_{k}(x) = e^{2πikx}. Since x_{n} = {nα} = nα − [nα] and e_{k}(x) = e_{k}(x + m) for every integer m,

e_{k}(x_{n}) = e^{2πiknα}.  (70)

Hence

\begin{aligned}\frac{1}{2N + 1} ∑_{n=−N}^{N} e_{k}(x_{n}) & = \frac{1}{2N + 1} ∑_{n=−N}^{N} e_{k}(nα) \\ & = \frac{1}{2N + 1} ∑_{n=−N}^{N}(e^{2πikα})^{n} \\ & = \begin{cases}1 & if k = 0 \\ \frac{1}{2N+1} \frac{e_{k}(−Nα)−e_{k}((N+1)α)}{1−e_{k}(α)} & if k > 0. \end{cases} \end{aligned}  (71)

Now for a fixed irrational α, |1 − e_{k}(α)| > 0. Therefore if k ≠ 0:

lim_{N→∞} \frac{1}{2N + 1} \frac{e_{k}(−Nα) − e_{k}((N + 1)α)}{1 − e_{k}(α)} = 0.  (72)

Let P (x) = ∑_{k} a_{k}e_{k}(x) be a finite sum (ie, P (x) is a trigonometric polynomial). By possibly adding some zero coefficients, we can write P (x) as a sum over a symmetric range: P (x) = ∑_{k=−K}^{K} a_{k}a_{k}(x).

Exercise 5.16. Show ∫_{0}^{1} P (x)dx = a_{0}.

By the above arguments, we have shown that for any (finite) trigonometric polynomial P (x):

lim_{N→∞} \frac{1}{2N + 1} ∑_{n=−N}^{N} P (x_{n}) → a_{0} = ∫_{0}^{1} P (x)dx.  (73)

Consider two approximations to the characteristic function χ_{(a,b)}:

1. f_{1m}: f_{1m}(x) = 1 if a + \frac{1}{m} ≤ x ≤ b − \frac{1}{m}, drops linearly to 0 at a and b, and is zero elsewhere.

2. f_{2m}: f_{1m}(x) = 1 if a ≤ x ≤ b, drops linearly to 0 at a − \frac{1}{m} and b + \frac{1}{m}, and is zero elsewhere.

Note there are trivial modifications if a = 0 or b = 1. Clearly

f_{1m}(x) ≤ χ_{(a,b)}(x) ≤ f_{2m}(x).  (74)

Therefore

\frac{1}{2N + 1} ∑_{n=−N}^{N} f_{1m}(x_{n}) ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} f_{2m}(x_{n}).

(75)

By Theorem 5.12, for each m, given ϵ > 0 we can find trigonometric polynomials P_{1m}(x) and P_{2m}(x) such that |P_{1m}(x) − f_{1m}(x)| < ϵ and |P_{2m}(x) − f_{2m}(x)| < ϵ.

As f_{1m} and f_{2m} are continuous functions, we can replace

\frac{1}{2N + 1} ∑_{n=−N}^{N} f_{im}(x_{n}) with \frac{1}{2N + 1} ∑_{n=−N}^{N} P_{im}(x_{n})  (76)

at a cost of at most ϵ. As N → ∞,

\frac{1}{2N + 1} ∑_{n=−N}^{N} P_{im}(x_{n}) → ∫_{0}^{1} P_{im}(x)dx.  (77)

But ∫_{0}^{1} P_{1m}(x)dx = (b−a)− \frac{1}{m} and ∫_{0}^{1} P_{2m}(x)dx = (b−a)+ \frac{1}{m}. Therefore, given m and ϵ, we can choose N large enough so that

(b − a) − \frac{1}{m} − ϵ ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) ≤ (b − a) + \frac{1}{m} + ϵ. (78)

Letting m tend to ∞ and ϵ tend to 0, we see \frac{1}{2N+1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) → b − a. Exercise 5.17. Rigorously do the necessary book-keeping to prove the previous theorem. Exercise 5.18. Prove

1. If α ∈ ℚ, then {nα} is periodic.

2. If α ∉ ℚ, then no two {nα} are equal.