Page 51 -

Substituting (4) in (3) gives

The Basic Approximation Lemma 45
m m
q m (ω) 2α f − C P +(f (n−m)−C P (n−m)). (5)
α α
Now we find it useful to replace the function fàx) − C P x by its
“monotonemajorant”Fàx) max t≤x (f (t)−C P t)andnotethatthis
Fàx) is nondecreasing and satisfies Fàx) o(x) since fàx) − C P x
satisfies the same. So (5) can be replaced by

m n
q m (ω) 2αF + Fàn − m) ≤ 2αF + F(n) (6)
α α
(a bound independent of m).
So choose n 0 so that x ≥ n 0 implies Fàx) ≤ 1x, and then choose
1
n 1 so that x ≥ n 1 implies Fàx) ≤ x. From now on we will pick
n 0
n
n ≥ n 1 and also will fix N [ ].
n 0
1
Dirichlet’s theorem on approximation by rationals now tells us
that the totality of arcs surrounding these ω with length 2 2π
αàN +1)
covers the whole circle. Thus using (2) for q(z), ζ ω and 7
2 2π ≤ 2 2πè 0 gives
αàN +1) nα

n n 0
q(z) [2αF + F(n)] 1 + 2π . à 7)
α α
We separate two cases:
n
Case I: α ≤ n 0 . Here we use Fà n ) ≤ F(n) and obtain [2αF( ) +
α α
2πè 0 2πè 0 2πè 0
F(n)](1+ ) ≤ (2α+1)(1+ )F(n) ≤ 3αà 1+ )F(n)
α α α
1
(3α + 6πè 0 )F(n) ≤ (6π + 3)n 0 F(n) ≤ (6π + 3)n 0 n ≤ 221n.
n 0
n 2πè 0 n
Case II: α> n 0 . Here [2αF( ) + F(n)](1 + ) ≤ [2αF( ) +
α α α
n n n n
F(n)](1 + 2πð . But still α ≤ ,or ≥ n 0 .So Fà ) ≤ 1 , and
n 0 α α α
the above is ≤ (21n + 1n)(1 + 2πð (3 + 6πð1n < 221n.
In either case Dirichlet’s theorem yields our lemma.
So let P be any affine property, and denote by A A(n;Pð the
number of arithmetic progressions from S(n;Pð (where order counts
N
1 Dirichlet’s theorem can be proved by considering the powers 1, z, z , ···, z for z
2
any point on the unit circle. Since these are N + 1 points on the circle, two of them
j
i
z , j must be within arc length 2π of one another. This means | arg z i−j |≤ 2π
N+1 N+1
and calling |i − j| α gives the result.