Master Dissertation

substituting t := s − x. We may write
f(t + x + 1) − f(t + x) = [f(x + t + 1) − f(x)] − [f(x + t) − f(x)].
Both terms on the right hand side converge uniformly in t towards 0 on
[0, k] as x → ∞ by Lemma 6.3. Hence
f ∗ (x + k) − f ∗ k
(x) =
0
f(x + t + 1) − f(x + t)
≤ k sup |f(x + t + 1) − f(x + t)| → 0
t∈[0,k]
for x → ∞. The same way it can be shown to hold for k < 0 as well.
Hence f ∗ satisfies (6.4).
Note that f ∗ is continuous in any interval [a, b], because given ɛ > 0,
choose δ = ɛ/ sup [a,b] |f(s + 1) − f(s)|. Then
x0
|f(x0) − f(x)| = |
≤
x x0
x
f(s + 1) − f(s)ds|
|f(s + 1) − f(s)|ds
≤ |x0 − x| sup
[a,b]
|f(s + 1) − f(s)|
< δ sup |f(s + 1) − f(s)| < ɛ.
[a,b]
Now we want to construct a representation of f ∗ .
First note that since f ∗ is continuous it is measurable.
Thus, by Lemma 6.3 we may conclude that
sup |f
[c,d]
∗ (x + k) − f ∗ (x)| → 0 for x → ∞.
Further, since by Lemma 6.4 f is bounded on every closed interval so is f ∗ ,
because
sup
[c,d]
|f ∗ (x)| = sup |
x
[c,d] a
x
≤ sup
[c,d]
a
f(s + 1) − f(s)ds|
|f(s + 1) − f(s)|ds
≤ (d − a) sup |f(t + 1) − f(t)| < ∞.
[a,d]
Since f ∗ is measurable and bounded on closed intervals, it is integrable.
By Lemma 6.6 f ∗ has the following representation for x ≥ a ∗ ,
FIXME: what is a ∗ ?.
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