Master Dissertation

Now we want to use this to show it for arbitrary a, b ∈ R+. Let
Then for a fixed,
g(x) := f((b − a)x).
f(x + k) − f(x) = f(x + a + k − a) − f(x + a) + f(x + a) − f(x)
x + a k − a
x + a
= g + − g + f(x + a) − f(x)
b − a b − a b − a
= g(y + h) − g(y) + f(x + a) − f(x), (6.6)
where y and h are defined in the obvious way. Now since k ∈ [a, b] we see
that h ∈ [0, 1]. Further y → ∞ is equivalent to x → ∞. Now by (6.4),
f(x − a) − f(x) → 0 for x → ∞. And by the proof above for I = [0, 1]
Hence by (6.6)
as claimed.
sup |g(y + h) − g(y)| → 0 for y → ∞.
h∈I
sup
k∈[a,b]
|f(x + k) − f(x)| → 0 for x → ∞.
Lemma 6.4. Let f(x) = ln r(e −x ), where r is the slowly varying function
defined by (6.2). Then f satisfies (6.4) and there exists a constant x0 such
that f is bounded on every interval [a, b] where x0 ≤ a.
FIXME: x0 comes from lemma 6.3 and tells us what the domain of f is.
Proof.
lim (f(x + k) − f(x)) = lim
x→∞ x→∞ (ln r(e−(x+k) ) − ln r(e x ))
r(e
= ln lim
x→∞
−(x+k) )
= ln lim
t→0
= ln 1 = 0,
r(e −x )
since (6.3) holds for r(t). Therefore f satisfies (6.4).
By Lemma 6.3 there exists a such that
for all x ≥ a and for all k ∈ [0, 1].
r(at)
r(t) , substituting t = e−x and a = e −k
|f(x + k) − f(x)| < 1 (6.7)
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