Baire Category, Probabilistic Constructions and Convolution Squares

Exercise 9.8. Using the same kind of methods as we used to establish Theorem
7.3, establish the following result.
If q is an integer with q ≥ 1, then, given any α > 1/(2q), there exists a
probability measure µ such that
|ˆµ(r)| ≤ |r| α
for all r = 0, but, given distinct points x1, x2, ..., xq ∈ supp µ, the only
solution to the equation
q
mjxj = 0
with mj ∈ Z is the trivial solution m1 = m2 = · · · = mq = 0.
j=1
Notice that there is a very big gap between the result of Exercise 9.8 and
the result of Lemma 7.1.
Exercise 9.9. Consider the independent random variables Yu and the random
measure
σ = n −1
n
δYu
u=1
introduced in Lemma 8.4.
Show that, provided n is large enough, the probability that more than
n1/2 (log n) 1/2 nq of different q-tuples j(1), j(2), ..., j(q) satisfy
q
k=1
mkYj(k) ∈ [n −q+1/2 ,n −q+1/2 ] (⋆)
is very small indeed. By removing one element Yj(1) corresponding to every qtuple
which satisfies ⋆, show that with high probability, the set Y1, Y2, ...,Yn
contains a subset {W1, W2, ...,Wv} with v ≥ n − n1/2 (log n) −1/2 with the
following property. If j(1), j(2), ..., j(q) are distinct integers with 1 ≤
j(k) ≤ v then
q
k=1
mkYj(k) /∈ [n −q+1/2 ,n −q+1/2 ].
Let τ = v −1 v
u=1 δYu. By comparing ˆ
τ(r) and ˆσ(r) show that there exists
an A depending only on q such that, if n is large enough, then, with high
probability
|ˆτ(r)| ≤ An −1/2 (log n) 1/2
for all 1 ≤ |r| ≤ n 4q .
Hence show that we can replace the condition α > 1/(2q) in Exercise 9.8
by the condition α > 1/(2q + 1
2 ).
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