Parseval’s Theorem (complex version): Let 𝑓 ∈ 𝐿2 [−𝜋, 𝜋]. Then 𝑓 𝑡
𝜋
−𝜋
𝑓(𝑡) 2 𝑑𝑡 = 2𝜋
∞
𝑛=−∞
𝑎𝑛 2 , where 𝑎𝑛 =
Proof: We will instead prove that 𝑓, 𝑔 = 2𝜋
𝜋
−𝜋
2𝜋
1
∞
𝑛=−∞ 𝑎𝑛 𝑏𝑚
2
=
𝑓(𝑡)𝑒 −𝑖𝑛𝑡 𝑑𝑡.
, where 𝑓 =
∞
𝑖𝑛𝑡
,
𝑛=−∞ 𝑎𝑛 𝑒
𝑖𝑚𝑡
𝑖𝑛𝑡
𝑖𝑚𝑡
𝑔= ∞
, 𝑓𝑁 = 𝑁
, and 𝑔𝑁 = 𝑁
. We will examine the case where
𝑚 =−∞ 𝑏𝑚 𝑒
𝑛=−𝑁 𝑎𝑛 𝑒
𝑚 =−𝑁 𝑎𝑚 𝑒
𝑁
𝑓 = 𝑔. First, we need to show that 𝑓𝑁 , 𝑔𝑁 = 2𝜋 𝑛=−𝑁 𝑎𝑛 𝑏𝑚 . Now, 𝑓𝑁 , 𝑔𝑁
𝑖𝑛𝑡
𝑖𝑚𝑡
𝑖𝑛𝑡
𝑖𝑚𝑡
= 𝑁
, 𝑁
= 𝑁
, 𝑁
by linearity on the first term. By
𝑛=−𝑁 𝑎𝑛 𝑒
𝑚 =−𝑁 𝑏𝑚 𝑒
𝑛=−𝑁 𝑎𝑛 𝑒
𝑚 =−𝑁 𝑏𝑚 𝑒
𝑁
𝑖𝑛𝑡
additivity on the second term and conjugate symmetry, we obtain 𝑛=−𝑁 𝑎𝑛 𝑁
, 𝑒 𝑖𝑚𝑡 .
𝑚 =−𝑁 𝑏𝑚 𝑒
0, 𝑛 ≠ 𝑚
𝑒 𝑖𝑛𝑡 𝑒 𝑖𝑚𝑡
Now, since 2𝜋 , 2𝜋 is an orthonormal set, then 𝑒 𝑖𝑛𝑡 , 𝑒 𝑖𝑚𝑡 =
. Since 𝑚 = 𝑛, then we
2𝜋, 𝑛 = 𝑚
𝑁
𝑁
𝑁
obtain 𝑁
𝑛=−𝑁 𝑎𝑛 𝑚 =−𝑁 𝑏𝑚 2𝜋 = 𝑛=−𝑁 𝑎𝑛 𝑏𝑚 2𝜋. Hence, 𝑓𝑁 , 𝑔𝑁 = 2𝜋 𝑛=−𝑁 𝑎𝑛 𝑏𝑚 .
Now we must show that 𝑓𝑁 , 𝑔𝑁 → 𝑓, 𝑔 as 𝑁 → ∞. We will do so by showing that 𝑓𝑁 , 𝑔𝑁 −
𝑓, 𝑔 → 0 as 𝑁 → ∞. Now, 𝑓𝑁 , 𝑔𝑁 − 𝑓, 𝑔 = 𝑓𝑁 , 𝑔𝑁 − 𝑓, 𝑔𝑁 + 𝑓, 𝑔𝑁 − 𝑓, 𝑔 =
𝑓𝑁 − 𝑓, 𝑔𝑁 + 𝑓, 𝑔𝑁 − 𝑔 by additivity of the first and second terms. By the Schwarz inequality, we
obtain 𝑓𝑁 − 𝑓, 𝑔𝑁 + 𝑓, 𝑔𝑁 − 𝑔 ≤ 𝑓𝑁 − 𝑓 𝑔𝑁 + 𝑓 𝑔𝑁 − 𝑔 . By convergence in the norm, we
obtain 𝑓𝑁 − 𝑓, 𝑔𝑁 + 𝑓, 𝑔𝑁 − 𝑔 ≤ 0 𝑔𝑁 + 𝑓 0 = 0. Hence, 𝑓𝑁 , 𝑔𝑁 − 𝑓, 𝑔 → 0 as 𝑁 → ∞.
This completes the proof. ∎
Define the convolution of 𝑓 and 𝑔, denoted 𝑓 ∗ 𝑔 𝑡 , as
𝑓(𝑥) =
∞
𝑛=−∞
𝑎𝑛 𝑒 𝑖𝑛𝑥 and 𝑔(𝑥) =
∞
𝑛=−∞
1 π
2𝜋 −π
𝑓 𝑡 𝑔(𝑥 − 𝑡) 𝑑𝑡. If
𝑏𝑛 𝑒 𝑖𝑛𝑥 , then 𝑓 ∗ 𝑔 𝑡 =
Proof: From the definition of complex Fourier series, we know that 𝑓 ∗ 𝑔 𝑡 =
need to show that 𝑐𝑛 = 𝑎𝑛 𝑏𝑛 . Now, 𝑐𝑛 =
1 π
2π −π
∞
𝑖𝑛𝑥
.
𝑛=−∞ 𝑎𝑛 𝑏𝑛 𝑒
∞
𝑖𝑛𝑥
,
𝑛=−∞ 𝑐𝑛 𝑒
so we
𝑓 ∗ 𝑔 𝑡 𝑒 −𝑖𝑛𝑥 𝑑𝑥 =
1 π
1 π
1 π 1 π
𝑓 𝑡 𝑔(𝑥 − 𝑡) 𝑑𝑡 𝑒 −𝑖𝑛𝑥 𝑑𝑥 = 2π −π 2𝜋 −π 𝑓 𝑡 𝑔(𝑥 − 𝑡) 𝑒 −𝑖𝑛𝑥 𝑑𝑡 𝑑𝑥 =
2π −π 2𝜋 −π
π
1 π 1
𝑓 𝑡 −π 𝑔(𝑥 − 𝑡) 𝑒 −𝑖𝑛𝑥 𝑑𝑥 𝑑𝑡 by Fubini’s Theorem. Now, let 𝑢 = 𝑥 − 𝑡. Then 𝑑𝑢
2π −π 2𝜋
= 𝑑𝑡,
𝑥 = 𝑢 + 𝑡, and the limits of integration do not change. Hence, we obtain
π
1 π 1
𝑓 𝑡 −π 𝑔(𝑢) 𝑒 −𝑖𝑛 (𝑡+𝑢) 𝑑𝑢 𝑑𝑡
2π −π 2𝜋
1 π
1 π
𝑓 𝑡 𝑒 −𝑖𝑛𝑡 2𝜋 −π 𝑔(𝑢) 𝑒 −𝑖𝑛𝑢 𝑑𝑢
2π −π
This completes the proof. ∎
1
= 2π
𝑑𝑡 =
π 1
π
𝑓 𝑡 −π 𝑔(𝑢) 𝑒 −𝑖𝑛𝑡 𝑒 −𝑖𝑛𝑢 𝑑𝑢 𝑑𝑡 =
−π 2𝜋
1 π
1 π
𝑓 𝑡 𝑒 −𝑖𝑛𝑡 𝑏𝑛 𝑑𝑡 = 2π −π 𝑓 𝑡 𝑒 −𝑖𝑛𝑡
2π −π
𝑑𝑡 𝑏𝑛 = 𝑎𝑛 𝑏𝑛 .
Riemann-Lebesque Theorem: Let 𝑓(𝑥) be a piecewise-continuous function on
𝑏
𝑎
[𝑎, 𝑏]. Then lim𝑘→∞
𝑓 𝑡 cos 𝑘𝑡 𝑑𝑡 = lim𝑘→∞
𝑏
𝑎
𝑓(𝑡) sin 𝑘𝑡 𝑑𝑡 = 0.
𝑵
𝒏=𝟏
Lemma 1: Let 𝒇(𝒙) be a 𝟐𝝅-periodic function and 𝑺𝑵 𝒙 = 𝒂𝟎 +
𝝅
𝟏
Then 𝑺𝑵 𝒙 = 𝝅
𝒇(𝒙 + 𝒖)
−𝝅
𝟏
+
𝟐
𝑵
𝒌=𝟏 𝐜𝐨𝐬 𝒌𝒖
𝒅𝒖.
𝑁
𝑘=1
Proof: By the definitions of Fourier coefficients, we have 𝑆𝑁 𝑥 = 𝑎0 +
1 𝜋
2𝜋 −𝜋
𝑁
1 𝜋
−𝜋
𝜋
𝑘=1
𝑓(𝑡) 𝑑𝑡 +
𝜋
1
𝑓(𝑡) 𝑑𝑡 + 𝜋
−𝜋
𝑁
𝑓
𝑘=1
Now, let 𝑢 = 𝑡 − 𝑥. Then we obtain
we obtain
𝜋
−𝜋
𝑓(𝑡) sin 𝑘𝑡 𝑑𝑡 sin 𝑘𝑥 = 2𝜋
1
𝜋
−𝜋
𝑓(𝑡) 𝑑𝑡 +
𝑓 𝑡 cos 𝑘𝑡 cos 𝑘𝑥 + sin 𝑘𝑡 sin 𝑘𝑥 𝑑𝑡 = 2𝜋
𝜋
−𝜋
1
𝑥) 𝑑𝑡 = 2𝜋
𝜋
1
𝜋
1
2
𝑓 𝑥+𝑢
−𝜋
+
1
𝑓 𝑥+𝑢
−𝜋−𝑥
𝑁
𝑘=1 cos 𝑘𝑢
1
2
1
2
+
cos 𝑘𝑢 sin
𝑢
2
𝑁
then
sin 𝑘
𝑘=1
1
−2 𝑢
1
cos 𝑘𝑢 sin
1
sin
2
1
1
1
𝑢
2
𝑁 + 2 𝑢 ⇒ sin
1
2
= sin 𝑁 +
𝑢
2
1
2
+
𝟏 𝟏
+
𝟐
Lemma 3: Let 𝑷𝑵 𝒖 = 𝝅
1
2
−𝜋
𝑁
𝑘=1 cos 0
1
=2+
𝑢
𝑁
𝑘=1 1
− 𝑥) 𝑑𝑡.
𝑑𝑢. Since 𝑓(𝑥) is 2𝜋-periodic,
1
1
2
1
𝑢
2
𝑢 − sin
1
2
1
𝑢
2
1
= 2 sin 𝑁 + 2 𝑢 ⇒ 2 +
𝑵
𝒌=𝟏 𝐜𝐨𝐬 𝒌𝒖
. Then
𝝅
𝑷
−𝝅 𝑵
1
=2+
𝐬𝐢𝐧 𝑵+𝟏 𝟐 𝒖
.
,𝒖 ≠ 𝟎
𝟐 𝐬𝐢𝐧 𝒖 𝟐
1
𝑁 = 𝑁 + 2. Now, if 𝑢
1
𝑘=1 2
=
1
, we have sin
1
𝑵 +𝟏 𝟐,𝒖 = 𝟎
=
𝑁
𝑢
= 2 sin 𝑁 + 2 𝑢 − 2 sin
𝑁
𝑘=1 cos 𝑘𝑢
𝑓 𝑡 cos 𝑘(𝑡 −
𝑁
cos 𝑘(𝑡
𝑘=1
+
𝑵
𝒌=𝟏 𝐜𝐨𝐬 𝒌𝒖
sin 𝑘𝑢 + 2 − sin 𝑘𝑢 − 2
𝑢
2
𝑓(𝑡) 𝑑𝑡 +
𝑑𝑢 as desired. ∎
=2+
= 2 sin 𝑁 + 2 𝑢 − sin
𝑁
𝑘=1
𝑁
𝑘=1 cos 𝑘𝑢
𝑁
1
=
𝑘=1 2
1
2
𝑓 𝑡
𝑁
𝑘=1 cos 𝑘𝑢
+
Lemma 2: For all 𝒌 ∈ ℝ+ and for all 𝒖 ∈ [−𝝅, 𝝅], 𝟏 𝟐 +
Proof: If 𝑢 = 0, then
1 𝜋
−𝜋
𝜋
𝑘=1
𝜋
1
𝜋−𝑥
𝜋
−𝜋
𝑁
𝑡 cos 𝑘(𝑡 − 𝑥) 𝑑𝑡 = 𝜋
1
𝜋
𝑎𝑘 cos 𝑘𝑥 + 𝑏𝑘 sin 𝑘𝑥 =
1
𝑓(𝑡) cos 𝑘𝑡 𝑑𝑡 cos 𝑘𝑥 + 𝜋
𝑁
1 𝜋
−𝜋
𝜋
𝑘=1
𝒂𝒏 𝐜𝐨𝐬 𝒏𝝅𝒙 + 𝒃𝒏 𝐬𝐢𝐧 𝒏𝝅𝒙 .
𝑢
2
≠ 0,
1
sin 𝑘 + 2 𝑢 −
. Since
𝑁
+
cos 𝑘𝑢 sin
𝑘=1
𝑁
𝑘=1 cos 𝑘𝑢
=
𝑢
2
=
sin 𝑁+1 2 𝑢
.
2 sin 𝑢 2
∎
𝒖 𝒅𝒖 = 𝟏 .
Pointwise Convergence Theorem: If 𝑓(𝑥) is continuous and 2𝜋-periodic, then for
every point 𝑥 where 𝑓’(𝑥) exists, the trigonometric Fourier series, 𝐹(𝑥), converges to
𝑓(𝑥).
Proof: We need to show that lim 𝑆𝑁 𝑥 = 𝑓(𝑥). We will instead prove that lim 𝑆𝑁 𝑥 − 𝑓 𝑥 = 0. By
𝑁→∞
𝑁→∞
𝜋
Lemma 1, lim 𝑆𝑁 𝑥 − 𝑓 𝑥 = lim −𝜋 𝑃𝑁 𝑢 𝑓(𝑥 + 𝑢)𝑑𝑢 − 𝑓 𝑥 ⋅ 1. By Lemma 3, we obtain
𝑁→∞
𝑁→∞
𝜋
𝜋
𝜋
lim −𝜋 𝑃𝑁 𝑢 𝑓(𝑥 + 𝑢)𝑑𝑢 − 𝑓 𝑥 −𝜋 𝑃𝑁 𝑢 𝑑𝑢 = lim −𝜋 𝑃𝑁 𝑢 𝑓(𝑥 + 𝑢) − 𝑓 𝑥 𝑑𝑢. By Lemma 2,
𝑁→∞
𝑁→∞
𝜋
𝜋
sin 𝑁+1 2 𝑢
sin 𝑁𝑢 cos 𝑢 2 +sin 𝑢 2 cos 𝑁𝑢
we obtain lim
𝑓(𝑥
+
𝑢)
−
𝑓
𝑥
𝑑𝑢
=
lim
𝑓(𝑥 +
2𝜋
sin
𝑢
2
2𝜋 sin 𝑢 2
𝑁→∞ −𝜋
𝑁→∞ −𝜋
𝜋
sin 𝑁𝑢 cos 𝑢 2
cos 𝑁𝑢
𝑢) − 𝑓 𝑥 𝑑𝑢 = lim
+ 2𝜋
𝑓(𝑥 + 𝑢) − 𝑓 𝑥 𝑑𝑢 =
2𝜋 sin 𝑢 2
𝑁→∞ −𝜋
𝜋
lim
𝑁→∞ −𝜋
𝑓(𝑥+𝑢)−𝑓 𝑥 sin 𝑁𝑢 cos 𝑢 2
2𝜋 sin 𝑢 2
and 𝑕(𝑢) =
𝜋
lim
𝑁→∞ −𝜋
𝑓(𝑥+𝑢)−𝑓 𝑥
2𝜋
+
𝑓(𝑥+𝑢)−𝑓 𝑥 cos 𝑁𝑢
2𝜋
𝑑𝑢. Now, let 𝑔(𝑢) =
𝑓(𝑥+𝑢)−𝑓 𝑥 cos 𝑢 2
2𝜋 sin 𝑢 2
. By substitution and the Riemann-Lebesque Theorem, we obtain
𝜋
𝑁→∞ −𝜋
𝑔(𝑢) sin 𝑁𝑢 + 𝑕(𝑢) cos 𝑁𝑢 𝑑𝑢 = lim
𝜋
𝑁→∞ −𝜋
𝑔(𝑢) sin 𝑁𝑢 𝑑𝑢 + lim
𝑕(𝑢) cos 𝑁𝑢 𝑑𝑢 = 0. ∎