Fourier Analysis, Stein and Shakarchi
Chapter 4 Some Applications of Fourier Series
Yung-Hsiang Huang∗
2018.05.31
Notation: T := [−π, π].
Exercises
1. Let γ : [a, b] → R2 be a parametrization for the closed curve Γ.
(a) Prove that γ is a parametrization by arc-length if and only if the length of the
curve from γ(a) to γ(s) is precisely s − a for each s ∈ [a, b], that is,
Z s
|γ 0 (t)| dt = s − a.
a
(b) Prove that any curve Γ admits a parametrization by arc-length. [Hint: If η is
Rs
any parametrization, let h(s) = a |η 0 (t)| dt and consider γ = η ◦ h−1 .]
Proof. (a) (⇒) is trivial. (⇐) This is the definition of arc-length.
(b) is proved as hint and (a).
2. Suppose γ : [a, b] → R2 is a parametrization for a closed curve Γ, with γ(t) = (x(t), y(t)).
(a) Use integration by parts, one can show that
Z
Z b
Z b
1 b
0
0
0
(x(s)y (s) − y(s)x (s)) ds =
x(s)y (s) ds = −
y(s)x0 (s) ds.
2 a
a
a
(b) Define the reverse parametrization of γ by γ − : [a, b] → R2 with γ − (t) = γ(b + a − t). The
image of γ − is precisely Γ, except that the points γ − (t) and γ(t) travel in opposite directions.
Thus γ − ”reverses” the orientation of the curve. It’s easy to prove that
Z
Z
(xdy − ydx) = −
(xdy − ydx).
γ−
γ
∗
Department of Math., National Taiwan University. Email: d04221001@ntu.edu.tw
1
In particular, we may assume (after a possible change in orientation) that
Z b
Z
1 b
0
0
x(s)y 0 (s) ds.
A=
(x(s)y (s) − y(s)x (s)) ds =
2 a
a
3. Freshman’s Calculus should teach how to calculate the area for
Ω = {(x, y) : 0 ≤ x ≤ 1 and g(x) ≤ y ≤ f (x)}.
4. Observe that with the definition of ` and A given in the text, the isoperimetric
inequality continues to hold (with the same proof ) even when Γ is not simple.
Show that this stronger version of the isoperimetric inequality is equivalent to
Wirtinger’s inequality, which says that if f is 2π-periodic, of class C 1 , and satisfies
R 2π
f (t) dt = 0, then
0
Z 2π
Z 2π
2
|f (t)| dt ≤
|f 0 (t)|2 dt
0
0
with equality if and only if f (t) = A sin t + B cos t (Exercise 11, Chapter 3).
Remark 1. There are some ways to relax the smoothness assumption of Γ, one way is through
the geometric measure theory (GMT), see for example Book III’s Section 3.4 or others textbook
for GMT (e.g. Evans-Graiepy, Simon or Federer).
In 2013, Cabre, Ros-Oton and Serra found that the ABP method for elliptic PDEs (treated in
the classical textbooks like Section 9.1 of Gilbarg-Trudinger, Chapter 5 of Lin-Han, or Chapter
3 of Cabre-Caffarelli) can be applied to derive some sharp isoperimetric inequalities, their paper
is published in J. Eur. Math. Soc., 18, 2016, 2971 - 2998.
Proof. (Isoperimetric ⇒ Wirtinger)
If we re-parametrize t = ks with k = T /(2π), then the change of variable shows that it is
sufficient to prove the claim for the period being 2π.
Rt
Given f with mean 0. we found F (t) := 0 f (s) ds is a 2π-periodic function. So the isoperimetric
and Hölder inequalities imply
Z 2π
Z 2π
Z
2
1 2π p 0
2
0
(f (t))2 + (F 0 (t))2 dt
f (t) dt =
f (t)F (t) dt ≤
4π 0
0
0
Z 2π
1
≤
(f 0 (t))2 + f (t)2 dt,
2 0
which is equivalent to the Wirtinger’s inequality.
(Isoperimetric ⇐ Wirtinger)
2
Assume the closed curve Γ(s) = (x(s), y(s)) is parametrized by the arc-length from [0, L] → R2 .
Then x, y are L- periodic and
dx 2
ds
+
dy 2
ds
≡ 1.
L
L
Note that f (θ) = x( 2π
θ) and g(θ) = y( 2π
θ) are 2π-periodic and
df 2
dθ
+
dg 2
dθ
≡
L2
.
4π 2
Let f¯ denotes the mean of f , then the Wirtinger’s inequality and
R 2π
0
g0 = 0 =
R 2π
0
(f − f¯)
imply
Z 2π
0
¯
f g dθ = 2
(f − f )g dθ =
(f − f¯)2 + (g 0 )2 − (f − f¯ − g)2 dθ
2A = 2
0
0
0
Z 2π
Z 2π
L2
(f − f¯)2 + (g 0 )2 dθ ≤
(f 0 )2 + (g 0 )2 dθ = 2π 2 ,
≤
4π
0
0
Z
2π
Z
0
2π
which is the isoperimetric inequality stated here.
Remark 2. [1, 5, 9, 3, 12] are classical textbooks and surveys on isoperimetric inequality and
Brunn-Minkowski inequality.
5. Prove that the sequence {γn }∞
n=1 , where γn is the fractional part of
1 + √ 5 n
,
Wn :=
2
is not equidistributed in [0, 1].
Proof. According to the hint, one observes that Un = Wn + Wn satisfies Ur = Ur−1 + Ur−2 for
r ≥ 2 and U0 = 2, U1 = 1. So Un ∈ N for all n ∈ Z≥0 . Note that Wn → 0 as n → ∞ since
√
| 1−2 5 | < 1. So we complete the proof since γn 6∈ ( 14 , 43 ) for all large n.
6. Let θ = p/q be a rational number where p and q are relatively prime integers (that
is, θ is in lowest form). We assume without loss of generality that q > 0. Define
a sequence of numbers in [0, 1) by ξn = hnθi where h·i denotes the fractional part.
Show that the sequence {ξ1 , ξ2 , · · · } is equidistributed on the points of the form
0, 1/q, 2/q, · · · , (q − 1)/q.
In fact, prove that for any 0 ≤ a < q, one has
#{n : 1 ≤ n ≤ N, hnθi = a/q}
1
1
= + O( ).
N
q
N
3
Proof. By Euclidean algorithm, we know that there is x, y ∈ Z such that xp + yq = 1, that is
(ax + tq)p + (ay − tp)q = a for every t ∈ Z. For each k ∈ Z≥0 , by adjusting t, we can find
n = ax + tq ∈ [kq, (k + 1)q) with hn pq i = aq . On the other hand, one can easily show such n is
unique. For N = lq + r where 0 ≤ l and 0 ≤ r < q, one then have the inequalities
p
a
l ≤ #{n : 1 ≤ n ≤ N, hn i = } ≤ l + 1.
q
q
Then the desired conclusion follows easily.
7. Prove the second part of Weyl’s criterion: if a sequence of numbers ξ1 , ξ2 , · · · in
[0, 1) is equidistributed, then for all k ∈ Z \ {0}
N
1 X 2πikξn
e
→ 0 as N → ∞.
N n=1
Proof. First, the equidistributed property of {ξn } implies that for each (a, b) ⊂ [0, 1)
N
1 X
χ(a,b) (ξn ) → b − a as N → ∞.
N n=1
By the linearity, we know for each step function S(x) =
PM
k=1
ak χIk (x).
Z 1
N
1 X
S(ξn ) →
S(x) dx as N → ∞.
N n=1
0
Finally, for each continuous function f (x) on [0, 1], its uniform continuity implies that for each
> 0 there is a step function S such that kf − S k∞ < . Then for each N ∈ N
Z 1
Z 1
N
N
N
N
1 X
1 X
1 X
1 X
|
f (ξn ) −
f (x) dx| ≤ |
f (ξn ) −
S (ξn )| + |
S (ξn ) −
S (x) dx|
N n=1
N n=1
N n=1
N n=1
0
0
Z 1
Z 1
+|
S (x) dx −
f (x) dx|
0
1
≤ 2 + |
N
Hence | N1
PN
n=1 f (ξn ) −
R1
0
0
N
X
Z
S (ξn ) −
1
S (x) dx|.
0
n=1
f (x) dx| ≤ 3 for large N .
8. Show that for any a 6= 0, and σ with 0 < σ < 1, the sequence hanσ i is equidistributed
in [0, 1). For arbitrary non-integer σ > 0, see Problem 3.
4
Proof. We use mean value theorem in the following calculation. Given k ∈ Z\{0}, one observes
N
X
σ
cos(2πkhan i) =
n=1
N
X
cos(2πkanσ )
n=1
N
Z
σ
=
cos(2πkax ) dx + cos(2πka) +
1
N
X
σ
sin 2πkaxσ
=
2πkaσxσ−1
N
−
x=1
1
N
N
cos(2πkaxσ ) dx
cos(2πkan ) −
1
n=2
Z
Z
N
X
sin 2πkaxσ
−σ
(1 − σ)x dx + cos(2πka) +
cos(2πkanσ ) − cos(2πkaxσn )
2πkaσ
n=2
= O(N 1−σ ) + cos(2πka) −
N
X
2πkaσ sin(2πka(yn )σ )ynσ−1 (n − xn ),
n=2
where xn ∈ [n − 1, n] and yn ∈ [xn , n] for each n.
P
P
σ
1−σ
σ−1
So | N
) + O( N
) = O(N 1−σ ) + O(N σ ).
n=1 cos(2πkhan i)| ≤ O(N
n=1 n
Similar for the sine (imaginary) part, so {hanσ i}n is equidistributed by Weyl’s criterion.
Remark 3. Such estimate can not be applied to Problem 3 directly, that is, σ > 1. For
σ ∈ (1, 2), One can modify the integral ”piecewisely” like van der Corput’s lemma, see my
solution to Exercise 3.16(d) and the second part of Problem 3.
9. In contrast with the result in Exercise 8, prove that ha log ni for any a.
Proof. We use mean value theorem in the following calculation. One observes that
N
X
cos(2πha log ni) =
N
X
cos(2πa log n)
n=1
n=1
Z
=
N
cos(2πa log x) dx +
1
Z
log N
=
0
cos(2πaz)ez dz +
N
X
Z
N
cos(2πa log n) −
n=2
N
X
cos(2πa log x) dx
1
cos(2πa log n) − cos(2πa log xn )
n=2
N
N [cos(2πa log N ) + 2πa sin(2πa log N )] − (1 + 2πa) X
2πa sin(2πa log yn )yn−1 (n − xn ),
=
−
1 + (2πa)2
n=2
where xn ∈ [n − 1, n] and yn ∈ [xn , n] for each n. So
lim sup
N →∞
N
N
1 X
1
1 X −1
1
cos(2πkha log ni) ≥ p
− lim sup C
n = p
.
N n=1
N n=1
N →∞
2 1 + (2πa)2
2 1 + (2πa)2
and {ha log ni}n is not equidistributed by Weyl’s criterion.
10. Suppose that f is a periodic function on R of period 1, and {ξn } is a sequence which
is equidistributed in [0, 1). Prove that:
5
(a) If f is continuous and satisfies
R1
0
f (x) dx = 0, then
N
1 X
f (x + ξn ) = 0 uniformly in x.
N →∞ N
n=1
lim
(b) If f is merely in L2 (0, 1) and satisfies
Z
1
lim
N →∞
0
Proof. (a) In general, we have limN →∞
R1
0
f (x) dx = 0, then
N
2
1 X
f (x + ξn ) dx = 0.
N n=1
1
N
PN
n=1
f (x + ξn ) =
R1
0
f (t) dt uniform in x.
First, if f (x) = e2πikx , k 6= 0, then
lim |
N →∞
N
N
1 X
1 X 2πikξn
f (x + ξn )| = lim |e2πikx ||
e
| = 0 uniformly in x.
N →∞
N n=1
N n=1
By linearity, this is true for any trigonometric polynomials without zero order term. We then
complete the proof by Weierstrauss approximation theorem.
(b) This is accomplished by (a) and approximate f by continuous functions in L2 sense. (In
the estimations, one may need to use Minkowski’s inequality somewhere). The approximation
steps are the same as my answer to Problem 3.2(d). Note that this is also true for Lp case
where p ∈ [1, ∞). (In this problem p = 2.)
Remark 4. For more results in ergodic theory, check [2], [6], [10] and [13].
11. Show that if u(x, t) = (f ∗ Ht )(x) where Ht is the heat kernel, and f is Riemann
integrable, then
Z
1
|u(x, t) − f (x)|2 dx → 0 as t → 0.
0
Proof. Let f (x) ∼
Z
0
P
n∈Z
an e2πinx . Given > 0, we have, by Parseval’s identity
1
|u(x, t) − f (x)|2 dx =
X
|an |2 (1 − e−4π
2 n2 t
n∈Z
)2 ≤
X
0<|n|≤N
|an |2 (1 − e−4π
2 n2 t
)2 +
X
|an |2 ,
|n|>N
where N = N () is chosen to be large enough such that the second term is less than . Then
we see that there is a t > 0 such that the first term is less than whenever 0 < t < t .
6
12. A change of variables in (8) leads to the solution
u(θ, τ ) =
X
2
an e−n τ einθ = (f ∗ hτ )(θ)
of the equation
∂ 2u
∂u
= 2 with 0 ≤ θ ≤ 2π and τ > 0,
∂τ
∂θ
P
P
−n2 τ inθ
with boundary condition u(θ, 0) = f (θ) ∼
an einθ . Here hτ (θ) = ∞
e . This
n=−∞ e
version of the heat kernel on [0, 2π] is the analogue of the Poisson kernel, which can be written
P
−|n|τ inθ
as Pr (θ) = ∞
e with r = e−τ (and so 0 < r < 1 corresponds to τ > 0).
n=−∞ e
13. The fact that the kernel Ht (x) is a good kernel, hence u(x, t) → f (x) at each point
of continuity of f , is not easy to prove. This will be shown in the next chapter.
However, one can prove directly that Ht (x) is ”peaked” at x = 0 as t → 0 in the
following sense:
R 1/2
(a) Show that −1/2 |Ht (x)|2 dx is of the order of magnitude of t−1/2 as t → 0. More
R 1/2
precisely, prove that t1/2 −1/2 |Ht (x)|2 dx converges to a non-zero limit as t → 0.
R 1/2
(b) Prove that −1/2 x2 |Ht (x)|2 dx = O(t1/2 ) as t → 0.
Proof. (a) One note that from Parseval’s identity
R 12
− 21
|Ht (x)|2 dx =
P
n∈Z
e−4π
2 n2 t
. Then we
have the following upper estimates
X
−4π 2 n2 t
e
=
=
e
−4π 2 n2 t
=
e
√
+
=
R 1/2
−1/2
1
4
−1≥
−4π 2 x2 t
e
0
Z
e−4π
dx +
2 x2 t
dx − 1
−∞
0
n≤0
e
−4π 2 n2 t
+
X
e
−4π 2 n2 t
Z
+1≤
−4π 2 x2 t
e
0
n≤−1
∞
Z
0
dx +
e−4π
2 x2 t
dx + 1
−∞
√
π(2π t)−1 + 1
|Ht (x)|2 dx →
(b) As hint, x2 ≤
e
∞
Z
−4π 2 n2 t
π(2π t)−1 − 1
X
√
X
√
n≥1
n∈Z
So t1/2
−4π 2 n2 t
n≥0
n∈Z
X
X
1
√
2 π
as t → 0+ .
sin2 πx for |x| ≤
1
,
2
and hence by mean-value theorem and Parseval’s
7
1
identity (note that {e2πi(n+ 2 )x }n∈Z also form an orthonormal basis for L2 ([− 21 , 12 ]), we have
Z 1
Z 1 X
2
2
1 2 eiπx − e−iπx 2
1
2 2
2
x |Ht (x)| dx ≤
|Ht (x)| =
(eiπx − e−iπx )e−4π n t e2πinx dx
4 − 12
2i
16 − 12 n∈Z
− 12
Z 1 X
2
2
1
1 X −4π2 n2 t
1
2 2
2
2
2
2
=
e−4π n t − e−4π (n−1) t e2πi(n+ 2 )x dx =
|e
− e−4π (n−1) t |2
1
16 − 2 n∈Z
16 n∈Z
1 X −4π2 (n−δnt )2 t 2
| 16π 4 (2n − 1)2 t2 , where δnt ∈ (0, 1) ∀ n ∈ Z
=
|e
16 n∈Z
hX
i
X
2 2
2
2
≤ π 4 t2
|e−4π n t |2 (2n − 1)2 +
|e−4π (n−1) t |2 (2n − 1)2
Z
1
2
2
2
n≤0
n≥1
Z ∞
Z 0
i
h
−8π 2 x2 t
2
−8π 2 x2 t
2
4 2
e
(2x + 3) dx
e
(2x − 3) dx + 1 +
≤π t 1+
0
−∞
Z ∞
h
i
1
1
−8π 2 x2 t
2
4 2
e
(4x + 12x + 9) dx = Ct 2 + o(t 2 ) as t → 0+ .
= 2π t 1 +
0
Similarly, we use symmetry of the integrand and π 2 x2 ≥ sin2 πx for x ∈ [0, 12 ] to obtain
R 21 2
1
1
x |Ht (x)|2 dx ≥ Dt 2 + o(t 2 ) as t → 0+ .
−1
2
Problems
Problem 2 and 3 are taken from [4, Section 2,3 of Chapter 1].
1. This problem explores another relationship between the geometry of a curve and
Fourier series. The diameter of a closed curve Γ parametrized by γ(t) = (x(t), y(t))
on [−π, π] is defined by
d = sup |P − Q| =
P,Q∈Γ
sup
|γ(t1 ) − γ(t2 )|.
t1 ,t2 ∈[−π,π]
If an is the n-th Fourier coecient of γ(t) = x(t) + iy(t) and l denotes the length of Γ,
then (a) 2|an | ≤ d for all n 6= 0. (b) l ≤ πd, whenever Γ is convex.
Rπ
1
Property (a) follows from the fact 2an = 2π
[γ(t) − γ(t + π/n)]e−int dt.
−π
The equality l = πd is satisfied when Γ is a circle, but surprisingly, this is not the
only case. In fact, one finds that l = πd is equivalent to 2|a1 | = d. We re-parametrize
γ so that for each t in [−π, π] the tangent to the curve makes an angle t with the
y-axis. Then, if a1 = 1 we have
γ 0 (t) = ieit (1 + r(t)),
where r is a real-valued function which satisfies r(t) + r(t + π) = 0,
Rπ
0
eix r(x) dx = 0
and |r(t)| ≤ 1. Figure 7 (a) shows the curve obtained by setting r(t) = cos 5t. Also,
8
Figure 7 (b) consists of the curve where r(t) = h(3t), with h(s) = −1 if −π ≤ s ≤ 0
and h(s) = 1 if 0 < s < π. This curve (which is only piecewise of class C 1 ) is known
as the Reuleaux triangle and is the classical example of a convex curve of constant
width which is not a circle.
Remark 5. This is from [7, 8]. Also see Exercise 2.7 of Book II. After checking the details in
Rπ
those papers, I think the condition 0 eix r(x) dx = 0 omitted in [8] is actually needed.
We summarize [7, 8, Theorem 1] and its proof here. In the end, we give the proof for (b).
Let R = Rγ be the radius of the smallest closed disc containing Γ. Then it’s easy to see d ≤ 2R
and a weaker inequality |an | ≤ R for all n 6= 0. As stated and proved in (a), one have a stronger
inequality |an | ≤ d2 . Using Fourier series and the Parseval’s identity, one can easily see that
the equality in weaker inequality holds precisely when γ(t) = a0 + an eint , |an | = R. The papers
cited here aim to determine when the equality holds in the above stronger inequality |an | ≤ d2 .
Because the case d = 0 is trivial, one can normalize the function γ by the condition an = 1, i.e.
to confine ourselves to functions of the type
Γ(t) = eint +
X
ak eikt ,
k6=n
and so to the class
Fn = {Γ : an = 1, dΓ = 2}.
From the identity used in (a), |Γ(t) − Γ(t + πn )| ≡ 2. Moreover, application of Parseval’s identity
P
πik
πik
to Γ(t) − Γ(t + πn ) ∼ k ak (1 − e n )eikt implies that ak (1 − e n ) = 0 for all k 6= n and hence
ak = 0 except k = 2nν for some ν ∈ Z. Therefore
Γ(t) = eint +
X
c2nν e2nνit .
(1)
ν∈Z
With γ(t) := Γ( nt ) and g(t) :=
P
ν∈Z c2nν e
2νit
equation (1) reduces to
γ(t) = eit + g(t), g(t + π) = g(t)
9
(2)
Because the curves γ, Γ have the same diameter, the problem of determing the class Fn (for
any n 6= 0) is thus reduced to the particular case of n = 1.
One of the main results in those papers are
Theorem 6. Let the function γ be absolutely continuous on R and 2π-periodic, and let γ have
its first Fourier coefficient normalized by a1 = 1. Then the curve corresponding to this function
has a diameter dγ ≥ 2 and the equality holds if and only if γ 0 admits the representation
γ 0 (t) = ieit (1 + r(t)),
where r is a real-valued function which satisfies r(t) + r(t + π) = 0,
Rπ
0
eix r(x) dx = 0 and
|r(t)| ≤ 1 for all t.
Proof. For the if part, one assume that θ ∈ (0, π). There is α = αt,θ such that
Z
t+θ
ieix (1 + r(x)) dx = −Reiα
γ(t + θ) − γ(t) =
t
and R ≥ 0. This implies
Z
t+θ
(1 + r(x)) sin(x − α) dx = R
t
We are going to show R = Rt,θ ≤ 2 for all t, θ. Let J denote the subset of [t, t + θ] on which
sin(x − α) is ≥ 0. Then
Z α+π
Z
(1 + r(x)) sin(x − α) dx
R ≤ (1 + r(x)) sin(x − α) dx ≤
α
J
Z α+π
=2+
r(x) sin(x − α) dx = 2
α
where the last equality is due to the periodicity of r. Then we completes the proof for dγ = 2
Rπ
since |γ(π) − γ(0)| = | 0 ieix + ieix r(x) dx| = 2π.
For the only if part, we assume first γ is twice differentiable. Let t be arbitrary but fixed,
and let z(x) = γ(t) − γ(t + x). Since γ is of the form (2) by the analysis before (1), we have
z(π) = γ(t) − γ(t + π) = 2eit , and by the assumption, |z(x)| ≤ dγ = 2 for any x, then function
|z(x)|2 attains its maximum at x = π. Hence
d
dz |z(x)|2 = 2 Re z(x) (x) = 0 at x = π.
dx
dx
That is, Re {2e−it (ieit − g 0 (t + π))} = 0 or Re {e−it g 0 (t))} = 0 for all t. Letting
e−it g 0 (t) = ir(t),
10
we conclude that r(t) is real-valued and satisfies r(t + π) = −r(t) for all t. Furthermore,
R π ix
Rπ
e r(x) dx = −i 0 g 0 (x) dx = −i[g(π) − g(0)] = 0.
0
Finally, as the function |z(x)|2 attains its maximum at x = π,
dz
2
d2 z d2
2
(x)
+
z(x)
|z(x)|
=
2
Re
(x) ≤ 0 at x = π.
dx2
dx
dx2
That is, −2(1 − r2 (t)) ≤ 0 for all t and is equivalent to |r(t)| ≤ 1.
Proof for (b) l ≤ πd. The proof is taken from here.
Let γ(t) = x(y) + iy(t), using complex notation. For every θ ∈ [0, π] the function fθ (t) =
Re(eiθ γ(t)) takes values within some interval of length at most d. By convexity, this function has
two intervals of monotonicity on the circle (think of [−π, π] as the circle by gluing the endpoints
together)(one can use the notion of support line from the proof of Jensen’s inequality to justify
this sentence rigorously). Therefore, its total variation is at most 2d. The total variation
is the integral of absolute value of derivative, provided that we chose a sensible (absolutely
continuous) parametrization γ: the arclength parametrization would do, for example. Hence,
Z π
|Re (eiθ γ 0 (t))| dt ≤ 2d
−π
Integrate both sides with respect to θ and exchange the order of integration (by Fubini-Tonelli
theorem), we have
π
Z
Z
−π
Note that for any ζ ∈ C we have
π
|Re (eiθ γ 0 (t))| dθdt ≤ 2πd.
0
Rπ
|Re(eiθ ζ)| dθ = 2|ζ|. So the above inequality becomes
Z π
2|γ 0 (t)| dt ≤ 2πd.
2l =
0
−π
2. Here we present an estimate of Weyl which leads to some interesting results.
P
2πif (n)
(a) Let SN = N
. Show that for H ≤ N , one has
n=1 e
H
N −h
N X X 2πi(f (n+h)−f (n))
|SN | ≤ c
e
,
H h=0 n=1
2
for some constant c > 0 independent of N, H, and f .
(b) Use this estimate to show that the sequence hn2 γi is equidistributed in [0, 1)
whenever γ is irrational.
11
(c) More generally, show that if {ξn } is a sequence of real numbers so that for all
positive integers h the difference hξn+h − ξn i is equidistributed in [0, 1), then hξn i is
also equidistributed in [0, 1).
(d) Suppose that P (x) = ck xk + · · · + c0 is a polynomial with real coefficients, where
at least one of c1 , · · · , ck is irrational. Then the sequence hP (n)i is equidistributed
in [0, 1).
Remark 7. (a) is a special form of van der Corput’s Inequality. (c) is called the Van der
Corput’s difference theorem.
Proof. (a) As hint, we consider an = e2πif (n) if 1 ≤ n ≤ N and zero for other indices. Then by
Cauchy-Schwarz, we have a (general) form of van der Corput’s Inequality as follows
H 2|
X
an |2 =
H X
X
N
−1
X
2
an+k
=
= (N + H − 1)
H
H X
XX
n
= (N + H − 1)
2
≤ (N + H − 1)
an+k
n
an+k an+j = (N + H − 1)
k=1 j=1
X
1≤j≤k≤H
n
X
X
1≤k<j≤H
n
an+k an+j +
n X
|an |2 +
= (N + H − 1) H
X
X
1≤j<k≤H
n
n X
|an |2 + 2 Re
= (N + H − 1) H
k=1
n
X
1≤j<k≤H
n
H−1
n X
X
|an |2 + 2 Re
= (N + H − 1) H
n
X
n
X
1≤k<j≤H
n
H−1
X
X
h=1
n
(H − h)
an an+j−k
o
o
an+h an
o
n
j=1 1≤h≤H−j
n X
|an |2 + 2 Re
= (N + H − 1) H
o
X
an+k−j an
X
an+k an+j
n
an+k an+j
an+k−j an +
X
2
an+k
H X
H X
X
k=1 j=1
X
n
H
X X
n
n=1−H k=1
k=1 n∈Z
n∈Z
H
X
an+h an
o
In particular,
|
N
X
n=1
e
N −h
H−1
H−1 N −h
N +H −1 X
2N X X 2πi(f (n+h)−f (n))
h X 2πi(f (n+h)−f (n))
≤
| ≤
2
1−
e
2
e
.
H
H n=1
H h=0 n=1
h=0
2πif (n) 2
(b) Let k ∈ Z \ {0}. Note that for f (n) = kn2 γ in (a), f (n + h) − f (n) = k(2nhγ + h2 γ). Since
hnγi is equidistributed in [0, 1), we have, for each H ∈ N
N
H
N −h
1 X 2πikn2 γ 2
c X 1 X 2πi2nkhγ
lim sup |
e
| ≤ lim sup
e
N n=1
N n=1
N →∞
N →∞ H
h=0
H
N
−h
X
X
c(N − h) 1
c
c
=
+ lim sup
e2πi(2kh)nγ = .
H
N
N − h n=1
H
N →∞
h=1
12
So limN →∞
1
N
PN
2 γi
2πikhn
n=1 e
= limN →∞
1
N
PN
2γ
n=1
e2πikn
= 0. Since k is arbitrary, hn2 γi is
equidistributed by Weyl’s criterion.
(c) Modify the proof for (b) slightly.
(d) First we consider the case c1 ∈ Qc and c2 , · · · ck ∈ Q. Write P (x) = Q(x)+c1 x+c0 . Let D be
the least common factor of the denominators of c2 , · · · ck . Then we have hQ(Dk + d)i = hQ(d)i
for k ≥ 0 and d ≥ 1. Therefore, for every s ∈ Z \ {0},
N
1 X 2πisP (n)
1
e
=
N n=1
N
1
=
N
N
N
X
2πisP (n)
e
n=[ N
]D+1
D
N
X
D [ D ]−1
1 X X
+
e2πis(Q(Dk+d)+c1 (Dk+d)+c0 )
N d=1 k=0
e2πisP (n) +
D
X
n=[ N
]D+1
D
N
e2πis(Q(d)+c1 d+c0 )
d=1
]−1
D
1 [ X
N
e2πisDc1 k
k=0
Since c1 6∈ Q, hc1 ki is equidistributed and hence the second term of the above inequality tends
to zero as N → ∞. Note that the first term is bounded by
D
N
which also tends to zero as
N → ∞. So hP (n)i is equidistributed in [0, 1).
For general case, we use induction argument. Let k ≥ q ≥ 1 be the largest index such that
cq ∈ Qc . Now we see the claim is true for k = 1. Now we assume the claim is true for k = m
and consider case k = m + 1, for each h ∈ N, P (n + h) − P (n) = Qh (n) for some polynomial
Qh whose indices for irrational coefficients are all ≤ m. So the Weyl’s criterion implies for each
s ∈ Z \ {0}
N −h
1 X 2πisQh (n)
e
→0
N − h n=1
and hence, by (a), for each H ∈ N
lim sup |
N →∞
H
N
N
−h
X
X
1 X 2πisP (n) 2
c
c(N − h) 1
c
e
| ≤
+ lim sup
e2πisQh (n) = .
N n=1
H
N
N − h n=1
H
N →∞
h=1
Therefore limN →∞
1
N
PN
n=1
e2πishP (n)i = limN →∞
1
N
PN
n=1
e2πisP (n) = 0. Since s is arbitrary,
hP (n)i is equidistributed in [0, 1) by Weyl’s criterion.
3. If σ > 0 is not an integer and a 6= 0, then hanσ i is equidistributed in [0, 1).
Before the proof taken from [4, Theorem 1.3.5], we present a proof for σ ∈ (1, 2) inspired
by [4, Exercise 1.2.23] which uses similar idea as Exercise 8. See Remark 3. This reflects the
importance of Weyl’s estimate (Van der Corput’s trick) in Problem 2 for the theory of sequences
of uniform distribution.
13
A proof for the special case σ ∈ (1, 2). By Weyl’s criterion, it suffices to show for all k ∈ Z\{0}
N
1 X 2πikanσ
e
→ 0 as N → ∞.
N n=1
We recall the Van der Corput’s lemmas proved in Exercise 3.16 (d).
Given a real-valued function f (u) and numbers a < b, we set F (u) = e2πif (u) ,
Z b
X
I(F ; a, b) =
F (u) du, S(F ; a, b) =
F (n), D(F ; a, b) = I(F ; a, b) − S(F ; a, b).
a
a<n≤b
Lemma 8. (i) If f has a monotone derivative f 0 , and if there is a λ > 0 such that f 0 ≥ λ or
f 0 ≤ −λ in (a, b), then |I(F ; a, b)| < λ−1 .
1
(ii) If f 00 ≥ ρ > 0 or f 00 ≤ −ρ < 0, then |I(F ; a, b)| ≤ 4ρ− 2 .
Lemma 9. If f 0 is monotone and |f 0 | ≤
1
2
in (a, b), then
|D(F ; a, b)| ≤ A
where A is an absolute constant independent of a, b. In fact, we have A =
P∞
2
n=1 πn(n− 1 )
2
+ 2.
Lemma 10. If f 00 ≥ ρ > 0 or f 00 ≤ −ρ < 0, then
1
|S(F ; a, b)| ≤ |f 0 (b) − f 0 (a)| + 2 (4ρ− 2 + A).
WLOG, we assume a > 0 and σ ∈ (1, ∞) \ Z. The function f (u) = kauσ has an increasing
derivative f 0 (u) = kaσuσ−1 . In the following estimates, C stands for different generic constants
depending on k, a, σ only.
Now we deal the case σ ∈ (1, 2) first, f 00 (u) = kaσ(σ − 1)uσ−2 ≥ C2j(σ−2) on [2j , 2j+1 ] for each
j ∈ Z≥0 . A simple application of Lemma 10 shows that for each j ≥ 0
|S(F ; 2j , 2j+1 )| ≤ {C2j(σ−1) + 2}{C2j(2−σ)/2 + A} ≤ C2jσ/2 .
Similarly, |S(F ; 2n , N )| ≤ C2nσ/2 if 2n < N ≤ 2n+1 and hence
|
N
X
σ
e2πikan | ≤ 1 + |S(F ; 1, 2)| + |S(F ; 2, 4)| + · · · + |S(F ; 2n , N )|
k=1
≤ 1 + C(1 + 2σ/2 + · · · + 2nσ/2 ) ≤ C2nσ/2 < CN σ/2 .
Since σ ∈ (1, 2), | N1
PN
k=1
σ
e2πikan | ≤ CN σ/2−1 → 0 as N → ∞.
A proof for all σ ∈ R+ \ Z. This is a special case of the following theorem.
14
Theorem 11. Let k ∈ N, and let f (x) be a function defined for x ≥ 1, which is k times differentiable for x ≥ x0 . If f (k) (x) tends monotonically to zero as x → ∞ and if limx→∞ x|f (k) (x)| =
∞, then hf (n)i is equidistributed in [0, 1).
In our cases, f (x) = axσ and k = [σ].
Proof for Theorem 11. We use induction argument and assuming it’s true for k = 1 for a while.
Now assume it’s true for k = m and consider the case k = m + 1. For each h ∈ N, we set
R x+h
(m)
gh (x) = f (x + h) − f (x). Then gh (x) = f (m) (x + h) − f (m) (x) = x f (m+1) (t) dt which is
(m)
monotone since f (m+1) is. Also, gh
decays to 0 since f (m+1) does.
(m)
Note that x|gh (x)| = x|f (m+1) (x + θx h)| for some θx ∈ [0, 1] for each x. So
(m)
lim inf x|gh (x)| = lim inf (x + θx h)|f (m+1) (x + θx h)|
x→∞
x→∞
x
= ∞.
x + θx h
By the induction hypothesis, hgh (n)i is equidistributed in [0, 1) for each h ∈ N. Hence, hf (n)i
is equidistributed in [0, 1) by Problem 2 (c).
Now we recall the following discrete analogy of k = 1 due to Fejér.
Theorem 12 (Fejér’s Theorem). Let {f (n)}∞
n=1 ⊂ R with ∆f (n) := f (n+1)−f (n) is monotone
in n. If limn→∞ ∆f (n) = 0 and limn→∞ n|∆f (n)| = ∞, then hf (n)i is equidistributed in [0, 1).
Now, for the case k = 1, ∆f (n) satisfies the conditions of Fejér’s Theorem by mean-value
theorem, at least for sufficiently large n. The finitely many exceptional terms do not influence
the uniform distribution of the sequence.
Proof for Fejér’s Theorem. First we recall a fundamental inequality (think about Taylor expansion), for each u, v ∈ R,
2πiu
|e
−e
2πiv
− 2πi(u − v)e
2πiv
| = |e
2πi(u−v)
− 1 − 2πi(u − v)| = 4π
2
Z
u−v
(u − v − w)e2πiw dw
0
≤ 4π 2
Z
u−v
(u − v − w) dw = 2π 2 (u − v)2
0
Now set u = sf (n + 1) and v = sf (n), where s ∈ Z \ {0}, in the above inequality, we have
e2πisf (n+1) e2πisf (n)
−
− 2πise2πisf (n) ≤ 2π 2 s2 |∆f (n)| ∀ n ≥ 1.
∆f (n)
∆f (n)
Hence
e2πisf (n+1)
e2πisf (n)
1
1
−
− 2πise2πisf (n) ≤
−
+ 2π 2 s2 |∆f (n)| ∀ n ≥ 1.
∆f (n + 1)
∆f (n)
∆f (n) ∆f (n + 1)
15
Then
2πis
N
−1
X
e
2πisf (n)
=
n=1
≤
N
−1 X
n=1
N
−1
X
n=1
2πise2πisf (n) −
e2πisf (n+1)
e2πisf (n) e2πisf (N ) e2πisf (1)
+
+
−
∆f (n + 1)
∆f (n)
∆f (N )
∆f (1)
N −1
X
1
1
1
1
−
+
+
2π 2 s2 |∆f (n)| +
∆f (n) ∆f (n + 1)
|∆f (N )| |∆f (1)|
n=1
By the monotonicity of ∆f (n), we get
N −1
−1
π|s| N
X
1 X 2πisf (n)
1 1
1
e
≤
+
+
|∆f (n)|,
N n=1
π|s| N |∆f (N )| N |∆f (1)|
N n=1
which tends to zero as N → ∞. Since s is arbitrary, hf (n)i is equidistributed in [0, 1).
Remark 13. Any direct proof for this problem without using the abstract Theorem 11?
Remark 14. This theorem was first proved by Paul Csillag ”Über die gleichmässige Verteilung
nichtganzer positiver Potenzen mod. 1.” Acta Szeged 5, 13-18 (1930).
4. An elementary construction of a continuous but nowhere differentiable function is
obtained by ”piling up singularities,” as follows.
On [−1, 1] consider the function
ϕ(x) = |x|
and extend ϕ to R by requiring it to be periodic of period 2. Clearly, ϕ is continuous
on R and |ϕ(x)| ≤ 1 for all x so the function f defined by
f (x) =
∞ X
3 n
n=0
4
ϕ(4n x)
is continuous on R.
(a) Fix x0 ∈ R. For every positive integer m, let δm = ± 12 4−m where the sign is chosen
so that no integer lies in between 4m x0 and 4m (x0 + δm ). Consider the quotient
γn =
ϕ(4n (x0 + δm )) − ϕ(4n x0 )
.
δm
Prove that if n > m, then γn = 0, and for 0 ≤ n ≤ m one has |γn | ≤ 4n with |γm | = 4m .
(b) From the above observations prove the estimate
f (x0 + δm ) − f (x0 )
1
≥ (3m + 1),
δm
2
and conclude that f is not differentiable at x0 .
16
Proof. See [11, Theorem 7.18]. By using Baire Category Theorem, one can show nonwhere
continuous functions are many in a topological sense, that is, is of second category in the
sup-norm topology, see Section 1.2 of Chapter 4 in Book IV.
5. Let f be a Riemann integrable function on the interval [−π, π]. We define the
generalized delayed means of the Fourier series of f by
σN,K =
SN + · · · + SN +K−1
.
K
Note that in particular
σ0,N = σN , σN,1 = SN and σN,N = ∆N ,
where ∆N are the specific delayed means used in Section 3.
(a) Show that
σN,K =
1
((N + K)σN +K − N σN ),
K
and
X
σN,K = SN +
1−
N +1≤|ν|≤N +K−1
|ν| − N b
f (ν)eiνθ .
K
From this last expression for σN,K conclude that
X
|σN,K − SM | ≤
|fb(ν)|
N +1≤|ν|≤N +K−1
for all N ≤ M < N + K.
(b) Use one of the above formulas and Fej́er’s theorem to show that with N = kn
and K = n, then
σkn,n (f )(θ) → f (θ) as n → ∞
whenever f is continuous at θ, and also
σkn,n (f )(θ) →
f (θ+ ) + f (θ− )
as n → ∞
2
at a jump discontinuity (refer to the preceding chapters and their exercises for the
appropriate definitions and results). In the case when f is continuous on [−π, π],
show that σkn,n (f ) → f uniformly as n → ∞.
(c) Using part (a), show that if fb(ν) = O(1/|ν|) and kn ≤ m < (k + 1)n, we get
|σkn,n − Sm | ≤
17
C
k
for some constant C > 0.
(d) Suppose that fb(ν) = O(1/|ν|). Prove that if f is continuous at θ then
SN (f )(θ) → f (θ) as N → ∞,
and if f has a jump discontinuity at θ then
f (θ+ ) + f (θ− )
as N → ∞.
SN (f )(θ) →
2
Also, show that if f is continuous on [−π, π], then SN (f ) → f uniformly.
P
(e) The above arguments show that if cn is Cesáro summable to s and cn = O(1/n),
P
then
cn converges to s. This is a weak version of Littlewood’s theorem (Problem
3, Chapter 2).
Remark 15. Another proof for (e) can be found in my Exercise 2.14(d).
Proof. (a) is easy. (For the SM part, one rewrite the second identity by moving some terms
from the second series to SN , this is why there is a 1 there).
(b) is proved by Fejér’s theorem and the first identity in (a).
(c)(d) are consequences of (a)(b).
(e) All the arguments still works when fb(ν)eiνθ is replaced by cν .
6. Dirichlet’s theorem states that the Fourier series of a real continuous periodic
function f which has only a finite number of relative maxima and minima converges
everywhere to f (and uniformly).
Prove this theorem by showing that such a function satisfies fb(n) = O(1/|n|).
[Hint: Argue as in Exercise 17, Chapter 3; then use conclusion (d) in Problem 5
above, or Problem 2.3, the harder Big-O theorem of Littlewood.]
Proof. The key observation is the following:
When {x1 , x2 , · · · xk } are relative extreme points inside (−π, π). If x1 is a relative maximum,
then f is increasing on (−π, x1 )∪(x2 , x3 )∪· · · and decreasing on (x1 , x2 )∪(x3 , x4 )∪· · · . Similar
for x1 is a relative minimum.
By using the integration by parts as Exercise 3.17, we have fb(n) = O(1/|n|).
18
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19