📂Fourier Analysis

Fourier Transform

Definition

Fourier Transform as a Function

The Fourier transform of function $f \in$ $L^{1}$ is defined as follows.

$$ \hat{f}(\xi) := \int_{-\infty}^{\infty} f(t) e^{-i \xi t}dt $$

Fourier Transform as an Operator

The operator $\mathcal{F} : L^{1} \to$ $C_{0}$ defined as follows is called the Fourier transform.

$$ \mathcal{F}[f] (\xi) = \int_{-\infty}^{\infty} f(t) e^{-i \xi t}dt $$

Explanation

As seen in the definition, the term Fourier transform refers to both the operator $\mathcal{F}$ itself and the functional value $\hat{f} = \mathcal{F}f = \mathcal{F}[f]$ of $\mathcal{F}$. It’s guaranteed by the Riemann-Lebesgue lemma that the codomain of $\mathcal{F}$ is $C_{0}$. Moreover, it can be easily shown that the following holds for $f \in L^{1}$,

$$ \left\| \mathcal{F}f \right\|_{\infty} \le \left\| f \right\|_{1} $$

Proof

$$ \begin{align*} \left\| \mathcal{F}f \right\|_{\infty} = \max\limits_{\xi \in \mathbb{R}} \left| \mathcal{F}f(\xi) \right| &= \max\limits_{\xi \in \mathbb{R}} \left| \int_{-\infty}^{\infty} f(t) e^{-i \xi t}dt \right| \\ &\le \max\limits_{\xi \in \mathbb{R}} \int_{-\infty}^{\infty} \left| f(t) e^{-i \xi t} \right| dt \\ &= \int_{-\infty}^{\infty} \left| f(t) \right| dt = \left\| f \right\|_{1} \end{align*} $$

The Fourier transform is a type of integral transform and its inverse transform is as follows.

$$ f(t) = \mathcal{F}^{-1}\hat{f}(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\xi) e^{i t \xi} d \xi $$

The preceding constant $\dfrac{1}{2\pi}$ can be attached either at the front of the inverse transform or at the front of the transform, or $\frac{1}{\sqrt{2\pi}}$ can be attached on both sides. This depends on the author’s convenience and does not fundamentally make any difference. Also, from the definition, one can see that $f$ must satisfy being integrable, that is, satisfy condition $f\in L^{1}$, for the Fourier transform to be well-defined. If $\hat{f}$ is also integrable, the Fourier inverse transform is well-defined as well.

Fourier Transform of Multivariable Functions

The Fourier transform of multivariable functions is defined as follows. The Fourier transform of multivariable function $f \in L^{1}(\mathbb{R}^{n})$ is,

$$ \mathcal{F}f(\boldsymbol{\xi}):=\int f(x)e^{-i \boldsymbol{\xi} \cdot \mathbf{x} }d\mathbf{x} $$

$$ \mathcal{F} f(\xi_{1},\ \cdots ,\ \xi_{n}) := \int_{-\infty}^{\infty} \dots \int_{-\infty}^{\infty} f(x_{1},\ \cdots,\ x_{n})e^{-i(\xi_{1} x_{1}+\cdots+\xi_{n} x_{n})}dx_{1}\cdots dx_{n} $$

Notation

There are two common notations for the Fourier transform of $f$.

$$ \mathcal{F}(f),\quad \hat{f} $$

In textbooks, it depends on which symbol the author prefers to use, but both are widely used. Although the right-hand hat symbol might seem convenient, it can be confusing, so when precision is required, it’s better to use the left-hand expression. For example, if the input function’s symbol itself gets lengthy, using the hat symbol can be confusing or not aesthetically pleasing. In such cases, using $\mathcal{F}$ makes the meaning of the formula clear and neat. For instance, the Fourier transform of $W_{c}f$ is better represented as shown below by using $\mathcal{F}$.

$$ \mathcal{F}(\mathcal{W}_{c}f),\quad \hat{\mathcal{W}_{c}f},\quad \widehat{\mathcal{W}_{c}f} $$

However, when there’s no room for confusion, the hat symbol is more convenient. Just like that, having various notations for the same concept also applies to differentiation.

$$ f^{\prime}, \quad \dfrac{df}{dx} $$

The advantages and disadvantages of the two notations $\hat{f}$ and $\mathcal{F}$ are similar to how in differentiation, the left-hand Newton’s notation has better economy and convenience, while the right-hand Leibniz’s notation has superiority in terms of rigor and precision when calculating things like the chain rule.

Derivation¹

Functions defined over a finite interval can be approximated using Fourier series. While useful, this only applies to periodic functions, so a similar tool for non-periodic functions is needed. This is where the idea of the Fourier transform comes from. The key concept in deriving the Fourier transform is thinking of non-periodic functions as if they have the entire real number line as their period, with the period repeating once across the entire line.

Let’s say $f$ is a function defined in the interval $[-L,L)$. Then the Fourier series and complex Fourier coefficients of $f$ are as follows.

$$ \begin{equation} f(t)=\sum \limits_{n=-\infty}^{\infty} c_{n} e^{i\frac{n\pi t}{L}} \end{equation} $$

$$ c_{n} = \dfrac{1}{2L}\int_{-L}^{L}f(t)e^{-i\frac{n \pi t}{L} }dt $$

Let’s perform the following variable substitution.

$$ \Delta \xi = \dfrac{\pi}{L},\quad \xi_{n}=n\Delta\xi=\dfrac{n\pi}{L} $$

Then $(1)$ becomes as follows.

$$ f(t) = \sum \limits_{n=-\infty}^{\infty} c_{n} e^{i\xi_{n} t}, \quad c_{n} = \dfrac{1}{2L}\int_{-L}^{L}f(t)e^{-i \xi_{n} t }dt $$

Multiply $f(t)$ by an appropriate constant and let the integration term of $c_{n}$ be $\hat{f}(\xi_{n})$

$$ f(t)=\dfrac{L}{\pi}\sum \limits_{n=-\infty}^{\infty} c_{n} e^{i\xi_{n} t}\Delta \xi , \quad c_{n} = \dfrac{1}{2L}\hat{f}(\xi_{n}) $$

Assume that $f(t)$ converges rapidly to $0$ when $t \rightarrow \pm \infty$. Then extending the integration interval for $c_{n}$ from $[-L,L)$ to $(-\infty,\infty)$ should not significantly alter the original $c_{n}$.

$$ c_{n} \approx \dfrac{1}{2L} \int_{-\infty}^{\infty} f(t) e^{-i\xi_{n} t}dt $$

This is a function of $\xi_{n}$ alone, so let’s call it $c_{n} = \frac{1}{2L}\hat{f}(\xi_{n})$. Inserting into $f(t)$ yields

$$ f(t) \approx \dfrac{1}{2 \pi}\sum \limits_{n=-\infty}^{\infty} \hat{f}(\xi_{n}) e^{i\xi_{n} t}\Delta \xi $$

This looks very similar to a Riemann sum. Now, if we take the limit as $L\rightarrow \infty$, we get $\Delta\xi \rightarrow 0$, and the equation becomes equal while the sum turns into integration.

$$ f(t) = \dfrac{1}{2 \pi}\int_{-\infty}^{\infty} \hat{f}(\xi) e^{i\xi t} d\xi \quad \text{and} \quad \hat{f}(\xi)=\int_{-\infty}^{\infty} f(t) e^{-i\xi t}dt $$

At this point, $\hat{f}$ is called the Fourier transform of $f$, and $f$ is called the Fourier inverse transform of $\hat{f}$.