# Question Fourier Transform Smoothness/Compactness

• mnb96
In summary, the author is discussing the concept of smoothness and compactness in the context of the Fourier transform. They present a formula for F(\omega) and state that as \omega approaches infinity, \omega F(\omega) tends to 0. The question arises as to why this is the case and the author clarifies that it is because |\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx| tends to 0. This is generally true unless f'(x) is a pathological function.

#### mnb96

Hello,
my question arises from reading the section on Smoothness/Compactness from Bracewell's "The Fourier Transform and Its Applications" page 162.

I don't quite understand the following reasoning:

$$F(\omega) = \ldots = \frac{1}{i\omega}\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx$$

and at this point the author says that when $\omega\to\infty$ then $\omega F(\omega) \to 0$.

But why $\omega F(\omega)$ is supposed to tend to zero, and not just $F(\omega) \to 0$ ?

Thanks.

ωF(ω) -> 0 is a stronger statement than F(ω) -> 0 for ω -> ∞.

Ok, but I thought that if you multiply F by $\omega$ that quantity does not tend to zero anymore.

In a similar way 1/x -> 0 but x(1/x) clearly tends to 1.

*EDIT:
On the other hand, I was thinking that if F(w) itself didnt tend to zero, its inverse FT, that is f(x), would not exist.

Last edited:
mnb96 said:
Hello,
my question arises from reading the section on Smoothness/Compactness from Bracewell's "The Fourier Transform and Its Applications" page 162.

I don't quite understand the following reasoning:

$$F(\omega) = \ldots = \frac{1}{i\omega}\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx$$

and at this point the author says that when $\omega\to\infty$ then $\omega F(\omega) \to 0$.

But why $\omega F(\omega)$ is supposed to tend to zero, and not just $F(\omega) \to 0$ ?

Thanks.

The author is simply saying that $|\int_{-\infty}^{+\infty}f'(x)e^{-i\omega x}dx|$ tends to 0. This makes sense unless f'(x) is completely pathological.

## 1. What is the Fourier Transform and why is it important in science?

The Fourier Transform is a mathematical tool used to decompose a complex function into simpler components. It is important in science because it allows us to analyze and understand the behavior of complex systems, such as signals and waves, in terms of their frequency components.

## 2. What is the relationship between the Fourier Transform and smoothness?

The Fourier Transform can be used to determine the smoothness of a function. A smooth function will have a Fourier Transform with a high concentration of its energy towards lower frequencies, whereas a less smooth function will have a more evenly distributed Fourier Transform.

## 3. How is compactness related to the Fourier Transform?

The compactness of a function refers to its ability to be represented by a finite number of parameters. The Fourier Transform can be used to analyze the compactness of a function by looking at the decay of its Fourier coefficients. A more compact function will have faster decaying coefficients, while a less compact function will have slower decaying coefficients.

## 4. Can the Fourier Transform be applied to any function?

In theory, the Fourier Transform can be applied to any function. However, in practice, the function must meet certain conditions, such as being integrable, in order for the Fourier Transform to be well-defined. Additionally, the accuracy of the Fourier Transform depends on the regularity and smoothness of the function.

## 5. How is the Fourier Transform used in practical applications?

The Fourier Transform has a wide range of applications in science and engineering. It is commonly used in fields such as signal processing, image processing, and data analysis. It has also been applied in various areas of physics, such as quantum mechanics and electromagnetic theory. Additionally, the Fourier Transform is used in practical applications such as sound and image compression, and in the design of filters and other signal processing systems.