1.2. Roth’s theorem 31
Remark 1.2.5. The above argument relied heavily on the fact that there
was only a one-parameter family of linear relations between n, n + r, n + 2r.
The same argument does not work directly for finding arithmetic progres-
sions of length longer than three; we return to this point in later sections.
The second step converts correlation with a linear character into a den-
sity increment on a subprogression:
Proposition 1.2.6 (Fragmenting a linear character into progressions). Let
N 1, let ε 0, and let φ(n) := e(αn) be a linear phase. Then there exists
N = N (N, ε) which goes to infinity as N for fixed ε, and a partition
[N] =
Pj E
of [N] into arithmetic progressions Pj of length at least N , together with an
error term E of cardinality at most O(εN), such that φ fluctuates by at most
O(ε) on each progression Pj (i.e., |φ(x) φ(y)| ε whenever x, y Pj).
Proof. We may assume that N is sufficiently large depending on ε, as the
claim is trivial otherwise (just set N = 1).
Fix ε, and let N be a slowly growing function of N to be chosen later.
By using recurrence for the linear phase n αn, we can find a shift h 1
of size h = ON ,ε(1) such that αh
ε/N . We then partition [N]
into h arithmetic progressions of spacing h, and then partition each of those
progressions in turn into subprogressions Pj of spacing h and length N , plus
an error of cardinality at most N , leading to an error set E of cardinality
at most hN = ON ,ε(1). On each of the Pj, αn fluctuates by at most ε.
The claim then follows by choosing N to be a sufficiently slowly growing
function of N.
Now we can prove Proposition 1.2.3 (and thus Roth’s theorem). Let
N, δ, δ , A be as in Proposition 1.2.3. By Proposition 1.2.4 (if N is large
enough), we can find α for which
|En∈[N](1A(n) δ )e(−αn)|
We now let ε 0 be a small quantity depending on δ to be chosen later
(actually it turns out that we can take ε to be a small multiple of
apply Proposition 1.2.6 to decompose [N] into progressions P1,...,PJ and
an error term E with the stated properties. Then we have
En∈[N](1A(n) δ )e(−αn) =
j=1 n∈Pj
(1A(n) δ )e(−αn)) + O(ε).
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