Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes

July 30, 2020 APM346 – Week 12 Justin Ko


1 Fourier Transform
We introduce the concept of Fourier transforms. This extends the Fourier method for finite intervals
to infinite domains. In this section, we will derive the Fourier transform and its basic properties.

1.1 Heuristic Derivation of Fourier Transforms
1.1.1 Complex Full Fourier Series
Recall that DeMoivre formula implies that

eiθ − e−iθ eiθ + e−iθ
sin(θ) = and cos(θ) = .
2i 2
This implies that the set of eigenfunctions for the full Fourier series on [−L, L]
n  πx   2πx   πx   2πx  o
1, cos , cos , . . . , sin , sin ,...
L L L L
inπx
is generated by the set of complex exponentials {e L }n∈Z . We define the inner product for complex
valued functions Z L
hf, gi = f (x)g(x) dx.
−L
inπx
− inπx
Since the complex conjugate e L =e L , it is also easy to check for n 6= m, that
Z L
inπx −imπx
e L e L dx = 0.
−L

so this is an orthogonal set. We have the following reformulation of the full Fourier series using complex
variables.
Definition 1. The complex form of the full Fourier series is given by
∞
inπx
X
f (x) = cn e L (1)
n=−∞

where the (complex valued) Fourier coefficients are given by
RL inπx
hf (x), e
inπx
i f (x)e− dx L
Z
L L
1 inπx
cn = inπx inπx = R −L
L inπx
= f (x)e− L dx. (2)
he L ,e L i e L e − inπx
L dx 2L −L
−L

The proof of Parseval’s equality also implies that
∞ Z L
X 1 2
|cn | = |f (x)|2 dx. (3)
n=−∞
2L −L


1.1.2 Fourier Transform
We now formally extend the Fourier series to the entire line by taking L → ∞. If we substitute (2)
into (1), then
∞ Z L 
1 X − inπx inπx
f (x) = f (x)e L dx e L .
2L n=−∞ −L




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, July 30, 2020 APM346 – Week 12 Justin Ko


nπ π
We define kn = L and ∆k = kn − kn−1 = L then this simplifies to
∞ Z L  ∞
1 X −kn x 1 X
f (x) = f (x)e dx ekn x ∆k = √ C(k)ekn x ∆k.
2π n=−∞ −L 2π n=−∞

where Z L
1
C(k) = √ f (x)e−ikx dx.
2π −L

If we take L → ∞, then
Z L Z ∞
1 1
C(k) = √ f (x)e−ikx dx → √ f (x)e−ikx dx (4)
2π −L 2π −∞

and interpreting the sum as a right Riemann sum,
∞ Z ∞
1 X 1
f (x) = lim √ C(k)ekn x ∆k → √ C(k)ekx dk. (5)
L→∞ 2π n=−∞ 2π −∞

Similarly, Parseval’s equality (3) becomes
Z L ∞ ∞ Z L 2 ∞
X 2π
X 1 inπx
X
2
|f (x)| dx = 2L 2
|cn | = 2L 2
√ f (x)e− L dx = |C(kn )|2 ∆k.
−L n=−∞ n=−∞
4L 2π −L n=−∞

so taking L → ∞ implies Z ∞ Z ∞
2 1
|f (x)| dx = √ |C(k)|2 dk. (6)
−∞ 2π −∞

Remark 1. The step where we took L → ∞ was not rigorous, because the bounds of integration and
the function depend on L. A rigorous proof of this extension is much trickier.

1.2 Definition of the Fourier Transform
The Fourier transform F is an operator on the space of complex valued functions to complex valued
functions. The coefficient C(k) defined in (4) is called the Fourier transform.

Definition 2. Let f : R → C. The Fourier transform of f is denoted by F[f ] = fˆ where
Z ∞
1
fˆ(k) = √ f (x)e−ikx dx (7)
2π −∞

Similarly, the inverse Fourier transform of f is denoted by F −1 [f ] = fˇ where
Z ∞
ˇ 1
f (x) = √ f (k)eikx dk. (8)
2π −∞

The follows from (5) that F and F −1 are indeed inverse operations.
Theorem 1 (Fourier Inversion Formula)
If f and fˆ are integrable, then F −1 [Ff ] = f and F[F −1 f ] = f . In particular,
Z ∞ Z ∞
1 1
f (x) = √ fˆ(k)eikx dk and f (k) = √ fˇ(x)e−ikx dx.
2π −∞ 2π −∞

The Fourier transforms of integrable and square integrable functions are also square integrable (6).


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