MATH 238: HOMEWORK #6 DUE MONDAY, 10/31/16
ALEX IOSEVICH
P
d
Problem #1: Let νf,g (t) =
x·y=t f (x)g(y), where f, g : Zp → R, d ≥ 2.
Suppose that f, g ≥ 0. For 1 ≤ p < ∞, define
 p1

X
||f ||p = 
|f (x)|p  .
x∈Zdp
Prove that
ν(t) = ||f ||1 ||g||1 p−1 + Rf,g (t),
where
|Rf,g (t)| ≤ ||f ||2 ||g||2 p
d−1
2
.
P 2
(t)
Problem #2: With νf,g (t) as above, formulate and prove a bound for t νf,g
in analogy with the bound we obtained in class in the case when f (x) = g(x) =
E(x), where E ⊂ Z2p . Note that I am asking for this bound in Zdp , d ≥ 2.
d
Use this bound to show that if A ⊂ Zp such that #A ≥ cp 2d−1 , then
#dA2 ≡ # {A · A + A · A + · · · + A · A} ≥ C(c)p.
Problem #3: Let p ≡ 3 mod 4. Let O2 (Zp ) denote the group of two by two
matrices M , with entries in Zp , such that M t M = I2 and det(M ) = 1. Suppose
that x, y ∈ Z2p , x 6= (0, 0), y 6= (0, 0), such that ||x|| = ||y||. Then there exists
M ∈ O2 (Zp ) such that y = M x.
Problem #4: Prove that if t 6= 0 and f : Z2p → R, p ≡ 3 mod 4, then
 21


 34
X
4
2
1 X

|fb(m)|  ≤ C  2
|f (x)| 3  .
p
2
||m||=t
x∈Zp
Problem #5: Let E ⊂ Z2p , p ≡ 3 mod 4. Define an equivalence relation ∼ on
Z2p , z ∼ w, z, w ∈ Z2p , z 6= (0, 0), w 6= (0, 0), if z = tw for some t ∈ Zp . Let D(E)
(the direction set of E) denote the set of equivalence relations of elements of
E − E = {x − y : x, y ∈ E}.
Prove that if #E > p, then D(E) = D(Zp ).
1
2
ALEX IOSEVICH
Hint: Prove that if D(E) 6= D(Zp ), then there exist v, w ∈ Z2p such that v·w = 0
and
E = {tv + f (t)w : t ∈ Zp },
where f : Zp → Zp is some function.
In particular, this will imply that #E = p, instantly giving you what you want.
Heuristically, what the hint says is that if a direction is missing, you can express
your set as a graph with respect to some coordinate system. This is why a semicircle is a graph, but the full circle is not.