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We apply (A) to S 1 . To do so, we note that
V. The Waring Problem 55

m
a a
k k
e n − α 0 + e n − α

b b
n 1 b[m/b]<è ≤m

≤ (1 +|α|)b ≤ 2b.
a k a k
Also, the total variation of e[(x − )t ] is equal to 2π|x − |N ≤
√ b b
2π N , whereas M 1. The result is

b
√
|S 1 |≤ 4π N + 2b ≤ 5πN 2/3 . à 4)
Next we apply (B) to S 2 and obtain
√

N a 2π N
k
e x − t dt + . à 5)
|S 2 |≤
0 b b
∞
k
Since eàu )du converges we get
0

N J 1/k
a
e x − t dt eàu )du ≤ .
k N k NC 3
1/k 1/k
0 0
b J J
Combining this with (5) gives
C 4 N|α| √
|αS 2 |≤ + 2π N. à 6)
(1 + jð 1/k
C 1
Now if we apply Lemma 3 to the case N b, we obtain |α|≤ δ ,
b
δ 2 1−k , and by (3) the addition of (4) and (6) gives

N
C 5 N
k 2/3
eàxn ) ≤ + 7πN

δ
b (1 + jð 1/k

n 1
C 5 N C 6 N
≤ + .
δ
b (1 + jð 1/k (b + jð 1/2
√
Sincej ≤ N andb ≤ N 2/3 ,thechoiceC 2 C 5 +C 6 +C 1 +2π,
1 1
1 min δ, , completes the proof.
k 4