Definition

Fourier Transform as a Function

We define the Fourier transform of function $f \in$ $L^{1}$ as follows:

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

Fourier Transform as an Operator

An operator defined as $\mathcal{F} : L^{1} \to$ $C_{0}$ 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 can refer to the operator $\mathcal{F}$ itself as well as to the function value $\hat{f} = \mathcal{F}f = \mathcal{F}[f]$ of $\mathcal{F}$ . The codomain of $\mathcal{F}$ being $C_{0}$ is guaranteed by the Riemann-Lebesgue Lemma. Additionally, it can be easily shown that the following holds true. 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 placed either before the inverse transform or before the transform since it doesn’t matter where it’s placed. Alternatively, $\frac{1}{\sqrt{2\pi}}$ can be placed on both sides. The difference in notation is merely a matter of the author’s preference and does not fundamentally affect the concept. From the definition, it is clear that the Fourier transform is well-defined if $f$ is integrable, i.e., satisfies the condition $f\in L^{1}$ . Also, if $\hat{f}$ is integrable, the Fourier inverse transform is well-defined as well.

Fourier Transform of Multivariable Functions

The Fourier transform of a multivariable function is defined as follows. The Fourier transform of a 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 commonly used notations for the Fourier transform of $f$ .

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

Textbooks often use different symbols depending on the author’s preference. Though the hat notation on the right seems convenient, it can cause confusion, so it is preferable to use the notation on the left for clarity. For example, if the input function symbol is lengthy, using the hat notation can be confusing or less visually appealing. In such cases, using $\mathcal{F}$ clearly and precisely conveys the meaning of the equation. For instance, the Fourier transform of $W_{c}f$ is better represented as $\mathcal{F}$ , as shown below.

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

However, when there is no potential for confusion, the hat notation is more convenient. Similarly, having multiple notations for the same concept also occurs with differentiation.

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

The advantages and disadvantages of the notations $\hat{f}$ and $\mathcal{F}$ resemble those of the Newton notation and the Leibniz notation in differentiation. The Newton notation is advantageous in terms of efficiency and convenience, whereas the Leibniz notation excels in precision and clarity when calculating chain rules, for instance.

Derivation¹

A function with a finite interval (i.e., defined on a finite interval) can be approximated using the Fourier series. While this is useful, it can only be applied to periodic functions, thus necessitating a tool for non-periodic functions. This led to the development of the Fourier transform. The key idea in deriving the Fourier transform is to consider a non-periodic function as if it were a periodic function with a period spanning the entire real line, repeating once over the entire number line.

Let $f$ be a function defined on 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 us perform the following variable substitution.

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

Then, $(1)$ is as follows.

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

Multiplying $f(t)$ by an appropriate constant and denoting the integral of $c_{n}$ as $\hat{f}(\xi_{n})$ ,

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

Assuming that $t \rightarrow \pm \infty$ converges quickly to $0$ as $f(t)$ approaches $t \rightarrow \pm \infty$ , it would not significantly alter the original $c_{n}$ if we extend the integration interval from $[-L,L)$ to $(-\infty,\infty)$ .

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

This is a function of $\xi_{n}$ only, so denoting it as $c_{n} = \frac{1}{2L}\hat{f}(\xi_{n})$ , substituting it into $f(t)$ ,

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

This closely resembles a Riemann sum. Taking the limit $L\rightarrow \infty$ and the resulting $\Delta\xi \rightarrow 0$ , the sum turns into an integral and the expression becomes $\approx$ , which is an identity.

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

Here, denoting $\hat{f}$ as the Fourier transform of $f$ , we refer to $f$ as the Fourier inverse transform of $\hat{f}$ .