Question

Let \(f \in C\left(\mathbb{R}^{n}\right)\) be absolutely integrable on \(\mathbb{R}^{n} .\) For real \(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\) we define \(\mathcal{F}[f]\left(\omega_{1}, \ldots, \omega_{n}\right)=\) $$ \frac{1}{(2 \pi)^{n}} \underbrace{\int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi}}_{n \text { times }} f\left(x_{1}, \ldots, x_{n}\right) e^{-i\left(\omega_{1} x_{1}+\cdots+\omega_{n} x_{n}\right)} d x_{1} \cdots d x_{n} . $$ The function \(\mathcal{F}[f]\) (defined on \(\mathbb{R}^{n}\) ) is said to be the multivariate Fourier transform of \(f\). Show that if \(f_{1}, f_{2}, \ldots, f_{n}\) are univariate continuous functions in \(G(\mathbb{R})\), and if $$ f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right) \cdots f_{n}\left(x_{n}\right) $$ then $$ \mathcal{F}[f]\left(\omega_{1}, \ldots, \omega_{n}\right)=\mathcal{F}\left[f_{1}\right]\left(\omega_{1}\right) \mathcal{F}\left[f_{2}\right]\left(\omega_{2}\right) \cdots \mathcal{F}\left[f_{n}\right]\left(\omega_{n}\right) $$

Short answer

The multivariate Fourier transform of the product is the product of the individual transforms.

Step-by-step solution

Write the Multivariate Fourier Transform

The multivariate Fourier transform of a function \( f(x_1, x_2, \ldots, x_n) \) is defined as:\[ \mathcal{F}[f](\omega_1, \ldots, \omega_n) = \frac{1}{(2 \pi)^n} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} f(x_1, \ldots, x_n) e^{-i(\omega_1 x_1 + \cdots + \omega_n x_n)} \, dx_1 \cdots dx_n.\]Our goal is to show that if \( f(x_1, x_2, \ldots, x_n) = f_1(x_1) f_2(x_2) \cdots f_n(x_n) \), then this reduces to the product of the Fourier transforms of the individual univariate functions.

Substitute the Product Expression for f

Substitute the expression \( f(x_1, x_2, \ldots, x_n) = f_1(x_1) f_2(x_2) \cdots f_n(x_n) \) into the Fourier transform:\[ \mathcal{F}[f](\omega_1, \ldots, \omega_n) = \frac{1}{(2 \pi)^n} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} \prod_{j=1}^{n} f_j(x_j) e^{-i \omega_j x_j} \, dx_1 \cdots dx_n.\]The product inside the integral can be separated into \( n \) separate integrals because each \( f_j(x_j) e^{-i \omega_j x_j} \) depends only on \( x_j \).

Separate Each Integral

Separate the multiple integral into \( n \) distinct integrals:\[ \prod_{j=1}^{n} \left( \frac{1}{2 \pi} \int_{-\pi}^{\pi} f_j(x_j) e^{-i \omega_j x_j} \, dx_j \right).\]Each integral under product corresponds to the Fourier transform of the function \( f_j(x_j) \).

Recognize the Univariate Fourier Transforms

Recognize that each integral \( \frac{1}{2 \pi} \int_{-\pi}^{\pi} f_j(x_j) e^{-i \omega_j x_j} \, dx_j \) is precisely the Fourier transform \( \mathcal{F}[f_j](\omega_j) \) of the univariate function \( f_j \).

Conclude the Product of the Transforms

Conclude that the Fourier transform of the product \( f(x_1, x_2, \ldots, x_n) = f_1(x_1) f_2(x_2) \cdots f_n(x_n) \) is:\[ \mathcal{F}[f](\omega_1, \ldots, \omega_n) = \prod_{j=1}^{n} \mathcal{F}[f_j](\omega_j)\]This shows that the multivariate Fourier transform of \( f \) is the product of the individual Fourier transforms \( \mathcal{F}[f_j](\omega_j) \).

Key concepts

Univariate Functions

Univariate functions are mathematical expressions that involve only one variable. These functions are represented commonly as \( f(x) \), where \( x \) is the sole variable. Univariate functions are foundational in calculus and algebra because they allow us to investigate the behavior of relationships along a single axis.
Recognizing a univariate function is straightforward: if the function only contains one variable, it is univariate. A simple example is \( f(x) = x^2 \), where the relationship between \( x \) and \( f(x) \) can be plotted on a two-dimensional graph. In the context of Fourier transforms, when you examine the transform of a univariate function, it involves integrating over a single variable, simplifying the transformation process.
Understanding univariate functions is critical for breaking down more complex, multivariate functions, especially when dealing with transforms in multiple dimensions. Each univariate segment might represent a piece of a larger puzzle, illustrating how isolated changes affect the whole.

Continuous Functions

Continuous functions are a crucial concept in mathematics, especially when dealing with calculus and transforms. A function \( f(x) \) is termed continuous if there are no “jumps” or interruptions in its graph. This means that small changes in \( x \) correspond to small changes in \( f(x) \).
Mathematically, a function \( f(x) \) is continuous at a point if \( \lim_{{x \to a}} f(x) = f(a) \), where the limit exists and is equal to the function’s value at that point. This property of smoothness allows for various calculus operations, such as differentiation and integration, to be performed more readily.
Continuous functions ensure that Fourier transforms, which rely heavily on integration over a domain, provide meaningful results. In the context of multivariate functions, each component function must be continuous to enable the integral over the product of functions, ensuring a valid and predictable multivariate Fourier transform.

Product of Transforms

The product of transforms is a concept especially relevant in the study of Fourier transforms. When dealing with multivariate functions, this property showcases that the Fourier transform of a product function can be split into the product of individual Fourier transforms of its factors.
This principle is elegantly illustrated through the process of transforming a multivariate function such as \( f(x_1, x_2, ..., x_n) = f_1(x_1) f_2(x_2) ... f_n(x_n) \). When performing a multivariate Fourier transform on this composition, each function part \( f_j(x_j) \) is transformed separately.
The final result reveals that \( \mathcal{F}[f](\omega_1, ..., \omega_n) = \mathcal{F}[f_1](\omega_1) \mathcal{F}[f_2](\omega_2) ... \mathcal{F}[f_n](\omega_n) \). This break-down simplifies complex integrals into manageable parts, making the analysis of signal processing or frequency analysis more efficient by reducing the complexity of operations.

Integral Calculus

Integral calculus is one of the two fundamental branches of calculus, the other being differential calculus. It concerns itself with the concept of integration, which is essentially the inverse operation of differentiation.
A core aspect of integral calculus is finding the total size or value, such as areas under a curve or total accumulation from a given rate. This process involves evaluating the integral of a function, which can be indefinite or definite.
In the context of Fourier transforms, integral calculus is employed to convert a function from its original domain into the frequency domain. This involves calculating integrals of the form \( \int_{-\pi}^{\pi} f(x) e^{-ix} \, dx \), which are part of the transformation process. Being comfortable with integral calculus is vital to understand Fourier transforms, as it not only governs the conversion but also underpins the smooth operation and extraction of useful information from the transformed data.