Katznelson Problems
Prakash Balachandran
Duke University
June 19, 2009
Chapter 1
1. Compute the Fourier coefficients of the following functions (defined by their values on [−π, π)):
(√
2π |t| < 12
f (t) =
1
0
≤ |t| ≤ π
2
(
1 − |t| |t| < 1
∆(t) =
0
1 ≤ |t| ≤ π

−1 < t ≤ 0

1
g(t) = −1 0 < t < 1

0
1 ≤ |t|
−π <t<π
h(t) = t
What relation do you see between f and ∆? What about g and ∆?
Solution:
1
fˆ(n) =
2π
1
ˆ
∆(n)
=
2π
Z
π
−int
∆(t)e
−π
π
Z
1
f (t)e−int dt =
2π
−π
1
dt =
2π
Z
1
(1 − |t|)e
−1
r
=
1
ĝ(n) =
2π
Z
π
−int
g(t)e
−π
1
ĥ(n) =
2π
1
dt =
2π
Z
2 sin n2
π n
Z
h(t)e
√
2πe
dt =
2 sin n2
.
π n
!2
−int
= fˆ(n)2 .
Z
dt −
1
dt =
2π
Z
1
−int
e
0
1
−int
Z 1
1 2 sin n
2
n
−int
dt =
−
|t|e
dt =
sin2
2
2π
n
πn
2
−1
−1
−int
r
− 21
0
e
π
−π
−int
1
2
Z
2i sin2
dt =
πn
π
−π
te−int dt =
n
2
i(−1)n
. n
ˆ
= in∆(n).
2. Show that the Fourier series of a function f ∈ L1 (T ) is formally equal to
∞
A0 X
+
(An cos nt + Bn sin nt)
2
n=1
where An = fˆ(n) + fˆ(−n) and Bn = i(fˆ(n) − fˆ(−n)). Equivalently:
Z
1
An =
f (t) cos ntdt
π
Z
1
f (t) sin ntdt.
Bn =
π
Show also that if f is real valued, then An and Bn are all real; if f is even, then Bn = 0 for all n; and if f
is odd, then An = 0 for all n.
Solution:
∞
X
S[f ] ∼
fˆ(n)eint = fˆ(0) +
∞
X
(fˆ(n)eint + fˆ(−n)e−int )
n=−∞
=
n=1
∞
X
eint + e−int
eint − e−int
2fˆ(0)
(fˆ(n) + fˆ(−n))
+
+ i(fˆ(n) − fˆ(−n))
2
2
2i
n=1
∞
=
A0 X
(An cos nt + Bn sin nt).
+
2
n=1
Z
1
f (t)(e
+ e )dt =
f (t) cos ntdt
π
Z
Z
Z
i
i
1
−int
int
Bn =
f (t)(e
− e )dt =
f (t) · −2i sin ntdt =
f (t) sin ntdt.
2π
2π
π
1
An =
2π
Z
−int
int
Obviously if f is real, then An and Bn are all real.
If f is even, then f sin nt is odd for all n. Taking the fundamental domain to be [−π, π], this implies that
Bn = 0 for all n.
If f is odd, then f cos nt is odd for all n, so that again taking the fundamental domain to be [−π, π] implies
that An = 0 for all n .
P
aj cos jt then S̃ ∼ aj sin jt
ijt −ijt P
P
P∞
e +e
ijt
Proof: Given S ∼ ∞
a
cos
jt
=
a
= ∞
where bj =
j=1 j
j=1 j
n=−∞ bj e
2
j > 0 b0 = 0.
3. Show that if S ∼
P
aj
,
2
b−j =
aj
2
for
By definition:
S̃ ∼
∞
X
n=−∞
−isgn(n)bn e
int
=
∞
X
−ibn e
int
+ ib−n e
n=1
−int
=
∞
X
j=1
2
aj ·
eint − e−int
2i
=
∞
X
j=1
aj sin jt. 4. Let f ∈ L1 (T ) and let P (t) =
PN
an eint . Compute the Fourier coefficients for f P .
n=−N
Solution: Let g(t) = f (t)P (t). Then,
1
ĝ(n) =
2π
Z
−int
g(t)e
1
dt =
2π
Z
Z
N
N
X
1 X
−i(n−m)t
an f (t)e
dt =
an fˆ(n − m). f (t)P (t)dt =
2π m=−N
m=−N
5. Let f ∈ L1 (T ) and let m be a positive integer. Write
f(m) (t) = f (mt).
Show that
(
fˆ
fd
(n)
=
(m)
0
n
m
if m|n
otherwise
Proof: First, assume that m|n. Notice that the function f(m) (t) has period 2π
. Using this fact and the
m
substitution u = mt:
Z
Z
Z 2π
Z
m
n
1
1
m
1
−int
−int
−int
d
f(m) (n) =
f(m) (t)e
dt =
f (mt)e
dt =
f (mt)e
dt =
f (u)e−i m u du
2π T
2π T
2π 0
2π T
n
= fˆ
.
m
P
int
be a trigonometric polynomial. Then:
Now suppose that m - n. Let p(t) = N
n=−N an e
Z 2π
N
1 X
pd
ak
ei(mk−n)t dt = 0
(m) (n) =
2π k=−N
0
since m - n.
Now, since f ∈ L1 (T ), choose a trigonometric polynomial p(t) =
(proven in the next chapter). Then:
PN
j=−N
aj eijt such that ||f − p||1 < Z
Z
1
d 1
−int −int
f(m) (t)e
dt = f (mt)e
dt
f(m) (n) = 2π T
2π T
Z
Z
1 −int −int
p(mt)e
≤
(f
(mt)
−
p(mt))e
dt
+
dt
2π T
1
≤
2π
Z
T
1
|f (mt) − p(mt)|dt =
m
2π
T
Z
0
2π
m
1
|f (mt) − p(mt)|dt =
2π
Since was arbitrary, we have that fd
(m) (n) = 0. 3
Z
2π
|f (t) − p(t)|dt = ||f − p||1 < .
0
6. Show that no trigonometric polynomial of degree n > 0 can have more than 2n zeros on T . Note: The
trigonometric polynomial cos nt = 21 (eint + e−int ) is of degree n and has exactly 2n zeros on T .
P
Proof: Let P (t) = nj=−n aj eijt be a trigonometric polynomial of degree n. Then, we can identify this
P
P
polynomial with P (z) = z −n nj=−n aj z n+j = z −n p(z) on |z| = 1, where p(z) = nj=−n aj z n+j . Since
p(z) is a polynomial of degree 2n the fundamental theorem of algebra implies that p has at most 2n zeros in
C, and hence at most 2n zeros on T . Since z −n is always nonzero on |z| = 1, we have that P (z) can have
no more than 2n zeros on T . 7. Denote by C ∗ the multiplicative group of complex numbers different from zero. Denote by T ∗ the subgroup
of all z ∈ C ∗ such that |z| = 1. Prove that if G is a subgroup of C ∗ which is compact, then G ⊆ T ∗ .
Proof: Let z ∈ G ≤ C ∗ . Then, z = reiθ for some r > 0 and θ ∈ [0, 2π). Consider the sequence
∗
j
{z j }∞
j=1 . Since G is a subgroup of C , {z } ⊆ G. Furthermore, since G is compact, this sequence contains
a convergent sequence, so that z jm → z0 as m → ∞. Since G is a compact subset of C ∗ it is closed, so that
z0 ∈ G.
Now, z jm = rjm eijm θ . If r < 1 then z jm → 0 which is certainly not in G. If r > 1 then this sequence
diverges, so that we must have r = 1. Thus, z = eiθ ∈ T ∗ . Since z was arbitrary, we must have G ⊆ T ∗ . 8. Let G be a compact proper subgroup of T . Prove that G is finite and determine its structure.
Proof:
Suppose that G is infinite. Then, since G is compact, there exists a convergent sequence {gn0 } in G. Since
G is closed, gn0 → g0 ∈ G. Thus, gn = g0−1 gn0 → e in G. Identifying T with [0, 2π) we may WLOG assume
gn ↓ 0 as n ↑ ∞.
Now, let t ∈ T . Given > 0 there exists N such that gN < . Consider the grid of [0, 2π) formed by
{m · gN }M
m=1 where M is the largest integer such that M · gN < 2π. If (m − 1)gN ≤ t < mgN for
m ∈ {1, . . . , M }, then |t − (m − 1)gN | < so that since G a subgroup implies (m − 1)gN ∈ G, we can get
close to t by elements in G.
If M · gN ≤ t < 2π ≤ (M + 1) · gN then |t − M · gN | < . Thus, we can always get within of t ∈ T by
elements of G, so that there exists {fn } in G such that fn → t. Since G is closed, this implies t ∈ G. Since
t ∈ T was arbitrary, G = T , contradicting that G was proper.
2πp
q
for p, q ∈ N.
D E
n
Remark: This does not hold for T , as can be exhibited by the compact proper subgroup 2π
× [0, 2π].
p
Any such group G must obviously generated by finitely many elements of the form
9. Let G be an infinite subgroup of T . Prove that G is dense in T .
Proof: G ≤ T so that one can easily verify that Ḡ ≤ T . Thus, Ḡ is a compact subgroup of T with |Ḡ| = ∞.
If it were proper, then (8) would imply that G is finite, which is a contradiction. Thus, Ḡ = T so that G is
dense in T . 10. Let α be an irrational multiple of 2π. Prove that {nα(mod2π)}n∈Z is dense in T .
Proof: Let
G = hαi = {nα(mod2π)}n∈Z
Then, G is an infinite subgroup of T , so that the density statement follows by (9). 4
11. Prove that a continuous homomorphism of T into C ∗ is necessarily given by an exponential function.
Proof: Let φ be the continuous
SinceT is compact φ(T ) ⊆ C ∗ is compact, sothat by (7)
homomorphism.
q
2πi
2π
2π
2π
q .
φ(T ) ⊆ T ∗ . Now, φ q · 2π
=
φ
is
a
q-th
root
of
unity;
that
is
φ
=
e
=
1
so
that
φ
q
q
q
q
Now, let G be the subgroup generated by rational multiples of 2π. Then, G is an infinite subgroup of T so
that (9) implies that it is dense in T . If r is a an element in T , then there exists qn → r such that qn are
rational multiples of 2π. Continuity of φ implies that φ(r) = limn→∞ φ(qn ) = limn→∞ eiqn = eir so that
φ(t) = eit for all t. 12. If E is a subset of T and τ0 ∈ T , define E + τ0 = {t + τ0 : t ∈ E}. Say that E is invariant under translation
by τ if E = E + τ . Show that, given a set E, the set of τ ∈ T such that E is invariant under translation by
τ is a subgroup of T . Hence, prove that if E is a measurable set on T and E is invariant under translation
by infinitely many τ ∈ T then E or its complement has measure zero.
Proof: Given E measurable, let ΓE = {τ ∈ T : E + τ = E}. The subgroup statement is trivial.
Let t ∈ ΓE . Then,
⇒
{E ∩ (−, )} + t = E ∩ (t − , t + )
µ{E ∩ (−, )}
µ{{E + (−, )} + t}
µ{E ∩ (t − , t + )}
=
=
. (∗)
2
2
2
Now, if |ΓE | = ∞ then ΓE is an infinite subgroup of T so that it is dense in T by (9).
So, given t ∈ T there exists tn ∈ ΓE such that tn → t. The bounded convergence theorem guarantees that
µ{E ∩ (tn − , tn + )} → µ{E ∩ (t − , t + )} so that (∗) implies
µ{E ∩ (−, )}
µ{E ∩ (t − , t + )}
=
2
2
for all t ∈ T . Thus,
lim
→0
µ{E ∩ (t − , t + )}
µ{E ∩ (−, )}
= lim
→0
2
2
for all t ∈ T .
So, if t is a point of density of E, then every point in T is a point of density of E. The relation
µ{E ∩ (t − , t + )} + µ{E c ∩ (t − , t + )}
µ{(t − , t + )}
=
=1
2
2
then implies that no point of T is a point of density of E c so that µ{E c } = 0.
Switching the roles of E and E c , the result follows. 13. If E and F are subsets of T , write
E + F = {t + τ : t ∈ E, τ ∈ F }
and call E + F the algebraic sum of E and F . We define similarly the sum of any finite number of sets. A
set E is called a basis for T if there exists an integer N such that E + E + · · · + E (N times) is T . Prove
that every set E of positive measure on T is a basis.
Proof: If E contains an interval, then the assertion follows. To see this, suppose I ⊂ E is an interval, say
I = (a, b). Then, E + E contains the interval (2a, 2b), and so E + E + · · · + E (N times) contains the
interval (N a, N b) which has length N (b − a). So, for N large enough, 2π < N (b − a), so that the interval
5
(N a, N b) contains a (closed) interval of form [2nπ, 2(n + 1)π]. Since this is a translated interval of [0, 2π],
we have that (2N −1 a, 2N −1 b) (modulo 2π) covers T . Thus, for this N , E + E + · · · + E = T .
Now, let E be a set of positive measure. We claim that E + E contains an interval.
Since almost every point in E is a point of density of E, by translating E if necessary (which preserves
Lebesgue measure), we may assume that 0 ∈ E is a point of density.
Suppose ∀ > 0, ∃ y ∈ E c ∩ B(0, ) such that y − x ∈ E c , ∀ x ∈ E.
Let 0 < < 1. Then, there exists y ∈ E c ∩ B(0, 2 ) such that y − x ∈ E c for all x ∈ E. So:
(1−)
− 2 µ(E ∩ B(0, − 2 ))
µ(E ∩ B(0, − 2 ))
µ(E ∩ B(0, − |y|))
µ(E ∩ B(0, − 2 ))
=
=
≤
2
2
2( − )
2( − )
2
2
µ(y − E ∩ B(0, − |y|))
µ(E c ∩ B(0, ))
≤
≤ 1.
2
2
Since 0 < < 1 was arbitrary, and 0 is a point of density of E, taking → 0 we have that
=
1 = lim(1 − )
→0
µ(E ∩ B(0, − 2 ))
µ(E c ∩ B(0, ))
≤
lim
≤1
→0
2( − 2 )
2
so that
µ(E c ∩ B(0, ))
lim
=1
→0
2
implying that 0 is a point of density of E c contradicting that 0 is a point of density of E.
So, ∃ > 0 such that ∀y ∈ E c ∩ B(0, ) y − x ∈ E for some x ∈ E.
For this , we thus have E c ∩B(0, ) ⊆ E +E; since clearly E ∩B(0, ) ⊆ E +E we have B(0, ) ⊆ E +E.
By repeating the first part of the proof, we thus have that E + E + · · · + E = T N -times. 14. Show that a measurable proper subgroups of T have measure zero.
Proof: Suppose otherwise. Then, there exists G < T that is a measurable proper subgroup with positive
measure. (13) implies that G is a basis of T so that there exists an integer N such that G + G + · · · + G (N
times) is T . But, since G is a subgroup, G + G + · · · + G (N times) is exactly G. Thus, G = T contradicting
that G is proper. 15. Show that measurable homomorphisms of T into C ∗ map it into T ∗ .
16. Let f be a measurable homomorphism from T into T ∗ . Show that for all values of n except possibly one
value, fˆ(n) = 0.
Proof: Fix t ∈ T . Using the substitution x = y+t and the hypothesis that f is a measurable homomorphism:
Z
Z
Z
1
1
f (t)e−int
−inx
−in(y+t)
ˆ
f (n) =
f (x)e
dx =
f (y + t)e
dy =
f (y)e−iny dy = f (t)e−int fˆ(n)
2π T
2π T
2π
T
⇒ fˆ(n)(1 − f (t)e−int ) = 0.
Since t was arbitrary, if there exists t ∈ T for which 1 − f (t)e−int 6= 0 then since n was arbitrary, we must
have fˆ(n) = 0 for all n.
If every t ∈ T satisfies 1 − f (t)e−int = 0 ⇔ f (t) = eint , then obviously fˆ(m) = 0 for all values of m
except m = n in which case fˆ(n) = 1. 6