1.7 Proof of Convergence 99
Under these conditions, the functionφ(y)of Eq. (16) has a jump disconti-
nuity aty=0 and again is sectionally continuous.
In either case, we see that the second integral in Eq. (14) is the Fourier
sine coefficient of a sectionally continuous function. By Lemma 3, then, it
too has limit 0 asNincreases, and the proof is complete for everyxwhere
fis continuous.
Part 4.Iffis not continuous atx.
Now let us suppose thatfhas a jump discontinuity atx.Inthiscase,we
must return to Part 2 and express the proposed sum of the series as
1
2(
f(x+)+f(x−))
=π^1∫π0f(x+)(
1
2 +
∑N
n= 1cos(ny))
dy+^1
π∫ 0
−πf(x−)(
1
2
+
∑N
n= 1cos(ny))
dy. (20)Here, we have used the evenness of the integrand in Lemma 1 to write
1
π∫π0(
1
2 +
∑N
n= 1cos(ny))
dy=1
π∫ 0
−π(
1
2 +
∑N
n= 1cos(ny))
dy=1
2. (21)
Next, we have a convenient way to write the quantity to be limited:
SN(x)−^12(
f(x+)+f(x−))
=^1
π∫π0(
f(x+y)−f(x+))
(
1
2
+
∑N
n= 1cos(ny))
dy+
1
π∫ 0
−π(
f(x+y)−f(x−))
(
1
2 +
∑N
n= 1cos(ny))
dy. (22)The interval of integration forSN(x)asshowninEq.(10)hasbeensplitin
half to conform to the integrals in Eq. (20).
The last step is to show that each of the integrals in Eq. (22) approaches 0
asNincreases. Since the technique is the same as in Part 3, this is left as an
exercise.
Let us emphasize that the crux of the proof is to show that the function
from Eq. (16),
φ(y)=f(x+y)−f(x)
2sin(^12 y)cos( 1
2
y)
(23)
(orasimilarfunctionthatarisesfromtheintegrandsinEq.(22)),doesnot
have a bad discontinuity aty=0.