Distribution Theory and Fourier Analysis

Special Notation

Relatively open: no one in V is infinitely close to any member of U

U \subset^{\circ} V

Compactly Contained: any possible limit points of U lie within V

U \subset\subset V

Definition

We introduce the following important notations to be used for the preceding discussions
• for $f \in L_{loc}^{1} (U)$ we identify the function $f$ with the distribution $\tilde{f}$ from 28
• for $f \in L_{loc}^{1} (U)$ and $φ \in C_{c}^{\infty} (U)$ we define the pairing

(f, φ) = \int_{U} f (x) φ (x) d x

• for $u \in D^{'} (U)$ and $φ \in C_{c}^{\infty} (U)$ we define the pairing

$(u, φ) = u (φ)$

Remark

It is common to write

(u, φ) = \int_{U} u (x) φ (x) d x

as well but take caution that this only makes sense when you write it like this. If you try to isolate $u (x)$ and it as a classical function the result will not make sense

Example

Consider isolating the dirac delta function $δ (x - x^{'}) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{i k (x - x^{'})} d k$ from the fourier decomposition of the wave function

Ψ (x) = \frac{1}{\sqrt{2 π}} \int [\frac{1}{\sqrt{2 π}} \int Ψ (x^{'}) e^{- i k x^{'}} d x^{'}] e^{i k x} d k,

Ψ (x) = \int_{- \infty}^{\infty} Ψ (x^{'}) \frac{1}{2 π} \int e^{i k (x - x^{'})} d k d x^{'} .

if you took this in a classical sense after attempting to isolate the dirac delta "function" you get for $x = x^{'}$ (sifting property)

Ψ (x) = Ψ (x) ∙ \infty

which is nonsense

fourier transforms

derivation of fourier transform from fourier series. first consider the complex form of the fourier series

f (t) = \sum_{n = - \infty}^{\infty} c_{n} e^{i 2 π n t / T},

where the coefficients $c_{n}$ are given by

c_{n} = \frac{1}{T} \int_{- T / 2}^{T / 2} f (t) e^{- i 2 π n t / T} d t .

Question

what happens when $T \to \infty$

First recall that fourier are derived for periodic functions with period $T$ . So with $T \to \infty$ , we effectively are considering aperiodic functions. Next recalling the formula for the coefficients of fourier series as given above we see that by rewrite the coefficient $c_{n}$ in terms of angular frequency $ω_{n} = n ω_{0} = 2 π n / T$ :

c_{n} = \frac{1}{T} \int_{- T / 2}^{T / 2} f (t) e^{- i ω_{n} t} d t = \frac{ω_{0}}{2 π} \int_{- T / 2}^{T / 2} f (t) e^{- i ω_{n} t} d t,

since $ω_{0} = 2 π / T$ .
Now substitute back into the series:

f (t) = \sum_{n = - \infty}^{\infty} c_{n} e^{i ω_{n} t} = \sum_{n = - \infty}^{\infty} (\frac{ω_{0}}{2 π} \int_{- T / 2}^{T / 2} f (s) e^{- i ω_{n} s} d s) e^{i ω_{n} t},

where we've used s as the dummy integration variable to avoid confusion with t.
This can be rearranged as:

f (t) = \frac{1}{2 π} \sum_{n = - \infty}^{\infty} (\int_{- T / 2}^{T / 2} f (s) e^{- i ω_{n} s} d s) e^{i ω_{n} t} \cdot ω_{0} .

As $T \to \infty$ :
• The integration limits $- T / 2$ to $T / 2$ extend to $- \infty$ to $\infty$ .
• The sum $\sum_{n = - \infty}^{\infty} g (ω_{n}) \cdot Δ ω$ (where $Δ ω = lim_{T \to \infty} ((n + 1) w_{0} - n w_{0}) = ω_{0}$ ) becomes the integral $\int_{- \infty}^{\infty} g (ω) d ω$ , because the discrete points $ω_{n}$ fill the real line densely.

Let $g (ω) = \int_{- \infty}^{\infty} f (s) e^{- i ω s} d s \cdot e^{i ω t}$ . Then the expression becomes:

f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} f (s) e^{- i ω s} d s) e^{i ω t} d ω .

Remark

notice this kind of looks like the continuous version of the fourier series now and analagously we see that its "coeffcients" can be obtained by the expression in the brackets

we give this a proper definition specifically

Definition

Define the Fourier transform $F (ω)$ (often denoted $\hat{f} (ω)$ or $F (ω)$ ) as:

F (ω) = \int_{- \infty}^{\infty} f (t) e^{- i ω t} d t .

The inverse transform is:

f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} F (ω) e^{i ω t} d ω .

just like how we did for parsevel theorem previously we aim to find the equivalent for this "continuous version". Specifically we call this the planceral theorem. We will return this after the following important interlude.

Note that we essentially derived fourier transform from fourier series which exists on the space of periodic functions or the circle group( $T$ ).

Remark

basically we are approximating with fourier series first then taking the "period to infinity" to get fourier transform

In reality fourier transforms can be applied to functions on $R^{n}$ not just the circle group. A full approach is one that starts from schwartz functions(which are just certain rapidly decaying functions) to prove the fourier transform pair via fourier inversion formula(see ur 18.155 notes)

Theorem

(Fourier Transform on $L^{2}$ ) Assume that $f \in L^{2} (R^{n})$ then the fourier transform $\hat{f}$ (as defined above) also lies in $L^{2} (R^{n})$ and we have

‖ \hat{f} ‖_{L^{2} (R^{n})} = (2 π)^{n / 2} ‖ f ‖_{L^{2} (R^{n})}

Remark

this means in the case of 1D and the "unitary" fourier transform

F (ω) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} f (t) e^{- i ω t} d t .

f (t) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} F (ω) e^{i ω t} d ω .

we get

\int_{- \infty}^{\infty} | f (x) |^{2} d x = \int_{- \infty}^{\infty} | \hat{f} (ξ) |^{2} d ξ