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III. The Erd˝ os–Fuchs Theorem
34
our case of A(re ) (but it even holds in much greater “miraculous”
generalities). iθ
At any rate, Parseval’s identity gives us
π

2
iθ
2
|A(re )| dθ 2π r 2a 2πA(r ). (5)
−π a∈A
π
2 2iθ
The last integral we must cope with is |A(r e )|dθ, and,
−π
2
unlike integrals of |f | , there is no formula for integrals of |f |. But

2 1/2
there is always the Schwarz inequality |f |≤ ( 1 · |f | ) , and
so at least we can get an upper bound for such integrals, again by
Parseval. The conclusion is that
π

2 2iθ
|A(r e )|dθ ≤ 2π A(r ). (6)
4
−π
All four of the integrals in (2) have been spoken for and so, by (2)
through (6), we obtain

M C 1 + r
2 4
A(r ) ≤ A(r ) + + C + log . à 7)
2π π 1 − r
It is a nuisance that our function A is evaluated at two different
points, but we can alleviate that by the obvious monotonicity of A,
4
2
A(r ) ≤ A(r ), and obtain

C 1 + r
2 2
A(r ) ≤ A(r ) + M + log . à 8)
π 1 − r
Is something bounded in terms of its own square root? But if x ≤
√ √ √
1
1
1 2
x + a, we obtain ( x − ) ≤ a + , x ≤ a + 1 + ,x ≤
2 4 4 2

1 1
a + + a + . This yields a pure bound on x. Then
2 4

C 1 + r C 1 + r
2
A(r ) ≤ M + log + M + log .à 9)
π 1 − r π 1 − r
2
But, so what? This says that A(r ) grows only at the order of
2
log 1 as r → 1 , but it doesn’t say that A(r ) remains bounded,
−
1−r
does it? Wherein is the hoped contradiction? We must revisit (1)
2 2 4 2
for this. Thereby we obtain, in turn A (r ) − A(r ) Pàr ) +