Problem 7 - Basic Analysis: Introduction to Real Analysis

Question

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a Riemann integrable function on every interval \([a, b],\) and such that \(\int_{-\infty}^{\infty}|f(x)| d x<\infty .\) Show that the Fourier sine and cosine transforms exist. That is, for every \(\omega \geq 0\) the following integrals converge $$ F^{s}(\omega):=\frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \sin (\omega t) d t, \quad F^{c}(\omega):=\frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \cos (\omega t) d t. $$ Furthermore, show that \(F^{s}\) and \(F^{c}\) are bounded functions.

Step-by-Step Solution

Answer

The Fourier sine and cosine transforms exist and are bounded for \(\omega \geq 0\).

Step 1: Verify Conditions for Existence of Integrals

To show the existence of the Fourier sine and cosine transforms, we start by verifying the condition \( \int_{-\infty}^{\infty} |f(x)| \, dx < \infty \). This implies that the function \( f(x) \) is absolutely integrable on the real line.

Step 2: Define Fourier Sine and Cosine Transforms

The Fourier sine transform \( F^s(\omega) \) is defined as \( \frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \sin(\omega t) \, dt \) and the Fourier cosine transform \( F^c(\omega) \) as \( \frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \cos(\omega t) \, dt \). We need to show that these integrals converge for every \( \omega \geq 0 \).

Step 3: Convergence of Fourier Sine Transform

To prove convergence, consider the integral \( \int_{-\infty}^{\infty} |f(t) \sin(\omega t)| \, dt \). This is bounded by \( \int_{-\infty}^{\infty} |f(t)| \, dt \) since \( |\sin(\omega t)| \leq 1 \). As \( \int_{-\infty}^{\infty} |f(t)| \, dt < \infty \), the integral for \( F^s(\omega) \) converges.

Step 4: Convergence of Fourier Cosine Transform

Similarly, consider \( \int_{-\infty}^{\infty} |f(t) \cos(\omega t)| \, dt \). This is also bounded by \( \int_{-\infty}^{\infty} |f(t)| \, dt \) because \( |\cos(\omega t)| \leq 1 \). Therefore, this integral also converges, ensuring that \( F^c(\omega) \) exists.

Step 5: Show Boundedness of Fourier Transforms

Next, let's demonstrate that \( F^s(\omega) \) and \( F^c(\omega) \) are both bounded. Since \( F^s(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \sin(\omega t) \, dt \), it follows that \( |F^s(\omega)| \leq \frac{1}{\pi} \int_{-\infty}^{\infty} |f(t)| \, dt \). Similarly, \( |F^c(\omega)| \leq \frac{1}{\pi} \int_{-\infty}^{\infty} |f(t)| \, dt \). Since \( \int_{-\infty}^{\infty} |f(t)| \, dt < \infty \), \( F^s(\omega) \) and \( F^c(\omega) \) are bounded.

Key Concepts

Riemann integrable functions

Riemann integrable functions are a foundational concept in calculus, used to describe functions that can be integrated over an interval. This concept is essential for understanding how integrals behave. To check whether a function is Riemann integrable, it must satisfy two main conditions:

• The function must be bounded on the interval \([a, b]\), which means there is a limit to how large the function values can be.
• Its set of discontinuities must have measure zero, which implies the function is continuous almost everywhere with very few discontinuities.

These conditions ensure we can sum the areas of rectangles under the curve, approaching the area under the curve as the rectangles become thinner.
To further apply Riemann integrability to functions defined on the entire real line, the function must also be absolutely integrable, meaning \( \int_{-\infty}^{\infty} |f(x)| \, dx < \infty \). This absolute integrability guarantees that the function's total area under the curve is finite, which is crucial for the existence of Fourier transforms.

Fourier sine transform

The Fourier sine transform is a specific integral transform, useful in breaking down functions into sinusoidal components. It is used particularly for functions defined on the half real line, emphasizing the odd symmetry property of sine functions. The transform of a function \ f(t) \ is defined as follows:
\[ F^{s}(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \sin(\omega t) \, dt \]
This transform decomposes a function into its sine components, and it is particularly useful in solving boundary value problems involving odd functions.
To find the convergence of this integral, we check that the function is multiplied by \ \sin(\omega t) \, which remains bounded between -1 and 1. This ensures that:

• The absolute value of the integrand, \ |f(t) \sin(\omega t)| \, does not exceed \ |f(t)| \.
• Therefore, the convergence of the Fourier sine integral follows from the convergence of \ \int_{-\infty}^{\infty} |f(t)| \, dt \.

Thus, the Fourier sine transform \ F^s(\omega) \ exists for every \ \omega \geq 0 \, assuming \ f(t) \ is absolutely integrable.

Fourier cosine transform

The Fourier cosine transform is another integral transform, similar in nature to the Fourier sine transform but dealing with even functions. The Fourier cosine transform is particularly effective for functions extending over the full real line, focusing on the even symmetry property of cosine functions. It's represented mathematically as:
\[ F^{c}(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(t) \cos(\omega t) \, dt \]
This transform helps in expressing a function in terms of its cosine components, which is crucial for applications involving heat and wave equations.
Like the sine transform, the cosine transform's convergence is also dependent on the property that:

• The function \ \cos(\omega t) \ is bounded between -1 and 1.
• This implies \ |f(t) \cos(\omega t)| \leq |f(t)| \, ensuring that the integral \ \int_{-\infty}^{\infty} |f(t) \cos(\omega t)| \, dt \ is bounded by \ \int_{-\infty}^{\infty} |f(t)| \, dt \.

Hence, if \ f(t) \ is absolutely integrable, the Fourier cosine transform \ F^c(\omega) \ exists for all \ \omega \geq 0 \.

Convergence of integrals

The convergence of integrals is a critical concept in advanced calculus, especially in the context of Fourier transforms. Convergence, in this framework, refers to whether an integral reaches a finite value as the upper and lower limits extend to infinity. With Fourier sine and cosine transforms, convergence is assured under the condition that the given function is absolutely integrable on the entire real line, represented by:
\( \int_{-\infty}^{\infty} |f(x)| \, dx < \infty \)
For practical purposes, this implies:

• The total area between the function \ f(x) \ and the x-axis is finite, ensuring no divergence.
• This property impacts both the sine and cosine transforms as their respective equivalences in bounds are derived from this initial condition.

This condition is necessary as it guarantees the functions under the integral remain controlled and do not lead to infinite oscillations or discontinuities, which could cause divergence. Hence, the absolute integrability of \ f(t) \, ensuring convergence, is pivotal for the existence of valid and bounded Fourier sine and cosine transforms.