270 CHAPTER 11. FUNDAMENTAL TRANSFORMS
with this last integral, do an integration by parts. Since h vanishes off some interval,∫R
sin(ru)h(u)dmp =−1r
∫R
cos(ru)h′ (u)dmp
Thus this last integral is dominated by Cr so it converges to 0. For r large enough, it follows
that |∫R sin(ru) f (u)du| ≤ 2ε and since ε is arbitrary, this establishes the claim. ■
Definition 11.3.4 The following notation will be used assuming the limits exist.
limr→0+
g(x+ r)≡ g(x+) , limr→0+
g(x− r)≡ g(x−)
Theorem 11.3.5 Suppose that g ∈ L1 (R) and that at some x, g is locally Holdercontinuous from the right and from the left. This means there exist constants K,δ > 0 andr ∈ (0,1] such that for |x− y|< δ ,
|g(x+)−g(y)|< K |x− y|r (11.2)
for y > x and|g(x−)−g(y)|< K |x− y|r (11.3)
for y < x. Then
limr→∞
2π
∫∞
0
sin(ur)u
(g(x−u)+g(x+u)
2
)du
=g(x+)+g(x−)
2.
Proof: As in the proof of Theorem 11.3.3, changing variables shows that for largepositive r,
2π
∫∞
0
sin(ru)u
du = 1.
Therefore,
2π
∫∞
0
sin(ur)u
(g(x−u)+g(x+u)
2
)du− g(x+)+g(x−)
2
=2π
∫∞
0
sin(ur)u
(g(x−u)−g(x−)+g(x+u)−g(x+)
2
)du
=2π
∫δ
0sin(ur)
(g(x−u)−g(x−)
2u+
g(x+u)−g(x+)
2u
)du
+2π
∫∞
δ
sin(ur)u
(g(x−u)−g(x−)
2+
g(x+u)−g(x+)
2
)du (11.4)
Second Integral: It equals
2π
∫∞
δ
sin(ur)u
(g(x−u)+g(x+u)
2− g(x−)+g(x+)
2
)du