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it suffices to prove that there exists g and C for which
N g V. The Waring Problem 53
1
g−k
eàxn ) dx ≤ CN for all n> 0. à 1)
k

0 n 1
First some parenthetical remarks about this inequality. Suppose it is
N k

known to hold for some C 0 and g 0 . Then, since | eàxn )|≤ N,
n 1
it persists for C 0 and any g ≥ g 0 . Thus (1) is a property of large g’s,
in other words, it is purely a “magnitude property.” Again, (1) is a
best possible inequality in that, for each g, there exists a c> 0 such
that

N g
1
g−k
k
eàxn ) dx > cN for all n> 0. à 2)

0 n 1
N
k
To see this, note that eàxn ) has a derivative bounded by
n 1
2πN k+1 . Hence, in the interval (0, 1 k ),
4πN
N

k+1 1 N
eàxn ) ≥ N − 2πN ,
k
4πN k 2
n 1
1
and so (2) follows with c g .
4π2
The remainder of our paper, then, will be devoted to the derivation
of (1) from Lemma 3. Henceforth k is fixed. Denote by I a,b,N the
a 1 k a
x-interval |x − |≤ k−1/2 , and call J N |x − |, j [J],
b bN b
where a, b, N, j are integers satisfying N> 0, b> 0, 0 ≤ a< b,
1
(a, b) 1, b ≤ N k− 2 .
By Dirichlet’s theorem, these intervals cover (0, 1). Our main tool
is the following lemma:

Lemma 4. There exists 1> 0 and C 2 such that, throughout any
interval I a,b,N ,

N
C 2 N
k
eàxn ) ≤ .
1
(b + jð

n 1