286 CHAPTER 12. SERIES AND TRANSFORMS
Lemma 12.5.2 The Fejer kernel has the following properties.
1. Fn+1 (t) = Fn+1 (t +2π)
2.∫
π
−πFn+1 (t)dt = 1
3.∫
π
−πFn+1 (t) f (x− t)dt = ∑
nk=−n bkeikθ for a suitable choice of bk.
4. Fn+1 (t) =1−cos((n+1)t)
4π(n+1)sin2( t2 ), Fn+1 (t)≥ 0,Fn (t) = Fn (−t) .
5. For every δ > 0, limn→∞ sup{Fn+1 (t) : π ≥ |t| ≥ δ}= 0. In fact, for
|t| ≥ δ ,Fn+1 (t)≤2
(n+1)sin2(
δ
2
)4π
.
Proof: Part 1.) is obvious because Fn+1 is the average of functions for which this istrue.
Part 2.) is also obvious for the same reason as Part 1.). Part 3.) is obvious because it istrue for Dn in place of Fn+1 and then taking the average yields the same sort of sum.
The last statements in 4.) are obvious from the formula which is the only hard part of4.).
Fn+1 (t) =1
(n+1)sin( t
2
)2π
n
∑k=0
sin((
k+12
)t)
=1
(n+1)sin2 ( t2
)2π
n
∑k=0
sin((
k+12
)t)
sin( t
2
)Using the identity sin(a)sin(b) = cos(a−b)−cos(a+b) with a =
(k+ 1
2
)t and b = t
2 , itfollows
Fn+1 (t) =1
(n+1)sin2 ( t2
)4π
n
∑k=0
(cos(kt)− cos(k+1) t) =1− cos((n+1) t)(n+1)sin2 ( t
2
)4π
which completes the demonstration of 4.).Next consider 5.). Since Fn+1 is even it suffices to show
limn→∞
sup{Fn+1 (t) : π ≥ t ≥ δ}= 0.
For the given t,
Fn+1 (t)≤1− cos((n+1) t)
(n+1)sin2(
δ
2
)4π
≤ 2
(n+1)sin2(
δ
2
)4π
which shows 5.). This proves the lemma.Here is a picture of the Fejer kernels Fn+1(t) for n = 1,2,3,4.
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0.2
0.4
0.6
0.8