Question

(a) Suppose \(g^{\prime}\) is continuous on \([a, b]\) and \(\omega \neq 0 .\) Use integration by parts to show that there's a constant \(M\) such that $$ \left|\int_{a}^{b} g(x) \cos \omega x d x\right| \leq \frac{M}{\omega} \quad \text { and } \quad\left|\int_{a}^{b} g(x) \sin \omega x d x\right| \leq \frac{M}{\omega}, \quad \omega>0 $$ (b) Show that the conclusion of (a) also holds if \(g\) is piecewise smooth on \([a, b] .\) (This is a special case of Riemann's Lemma. (c) We say that a sequence \(\left\\{\alpha_{n}\right\\}_{n=1}^{\infty}\) is of order \(n^{-k}\) and write \(\alpha_{n}=O\left(1 / n^{k}\right)\) if there's a constant \(M\) such that $$ \left|\alpha_{n}\right|<\frac{M}{n^{k}}, \quad n=1,2,3, \ldots $$ Let \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) and \(\left\\{b_{n}\right\\}_{n=1}^{\infty}\) be the Fourier coefficients of a piecewise smooth function. Conclude from (b) that \(a_{n}=O(1 / n)\) and \(b_{n}=O(1 / n)\)

Short answer

Question: Prove that for a piecewise smooth function \(g(x)\), the Fourier coefficients satisfy \(|a_n| = O(1/n)\) and \(|b_n| = O(1/n)\). Answer: Since \(g(x)\) is a piecewise smooth function, we have that the inequality derived in part (a) holds on each subinterval of \([0, \pi]\) where \(g(x)\) is continuously differentiable. By adding up the inequalities for all the subintervals, we can show that the inequalities hold for the whole interval \([0, \pi]\). Hence, the Fourier coefficients are given by: $$ |a_n| \le \frac{2}{\pi} \cdot \frac{M}{n}, \qquad |b_n| \le \frac{2}{\pi} \cdot \frac{M}{n} $$ This implies that \(|a_n| = O(1/n)\) and \(|b_n| = O(1/n)\) for piecewise smooth functions.

Step-by-step solution

Part (a): Proving the Inequalities using Integration by Parts

To show the inequalities, we will use integration by parts with the following formula: $$ \int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du $$ First, let's prove the inequality for the cosine integral: Let \(u = g(x)\) and \(dv = \cos{\omega x} \, dx\). Then, we have: - \(du = g'(x) \, dx\) - \(v = \frac{1}{\omega} \sin{\omega x}\) Applying integration by parts, we obtain: $$ \int_a^b g(x) \cos{\omega x} \, dx = \left[\frac{1}{\omega}g(x)\sin{\omega x}\right]_a^b - \int_a^b \frac{1}{\omega}g'(x)\sin{\omega x} \, dx $$ Taking the absolute value of both sides and applying the triangle inequality, we have: $$ \left|\int_{a}^{b} g(x) \cos \omega x \, dx\right| \le \frac{1}{\omega}\left|\left[g(x)\sin{\omega x}\right]_{a}^{b}\right| + \frac{1}{\omega}\left|\int_{a}^{b} g'(x) \sin{\omega x} \, dx\right| $$ Now, note that \(|\sin{\omega x}| \le 1\) and \(|g(a) - g(b)| \le M\) for some constant \(M\). So, we have: $$ \left|\int_{a}^{b} g(x) \cos \omega x \, dx\right| \le \frac{1}{\omega}M $$ The proof for the sine integral is analogous.

Part (b): Generalizing the Result for Piecewise Smooth Functions

Suppose \(g(x)\) is a piecewise smooth function on \([a, b]\). Then, \(g(x)\) is continuous and has a continuous derivative except for a finite number of points. Let's divide the interval \([a, b]\) into subintervals \([x_{i-1}, x_i]\) such that \(g(x)\) is continuously differentiable within each subinterval. Since the result from part (a) holds for each of these subintervals, we have: $$ \left|\int_{x_{i-1}}^{x_i} g(x) \cos \omega x \, dx\right| \le \frac{M_i}{\omega} \quad \text{ and } \quad\left|\int_{x_{i-1}}^{x_i} g(x) \sin \omega x \, dx\right| \le \frac{M_i}{\omega} $$ for some constants \(M_i\). By adding up the inequalities for all the subintervals, we can show that the inequalities hold for the whole interval \([a, b]\).

Part (c): Analyzing Fourier Coefficients for Piecewise Smooth Functions

For a piecewise smooth function, the Fourier coefficients \(a_n\) and \(b_n\) are given by: $$ a_n = \frac{2}{\pi} \int_{0}^{\pi} g(x) \cos(nx) \, dx, \qquad b_n = \frac{2}{\pi} \int_{0}^{\pi} g(x) \sin(nx) \, dx $$ Using the results from part (b), we have: $$ |a_n| \le \frac{2}{\pi} \cdot \frac{M}{n}, \qquad |b_n| \le \frac{2}{\pi} \cdot \frac{M}{n} $$ Thus, we can conclude that \(a_n = O(1/n)\) and \(b_n = O(1/n)\) for piecewise smooth functions.

Key concepts

Riemann's Lemma

Riemann's Lemma provides an insightful look into the behavior of integrals involving trigonometric functions. When dealing with functions that are piecewise smooth, it guarantees certain diminishing effects for integration results as the frequency parameter increases.
This lemma is especially useful in studying Fourier series, where it aids in establishing that Fourier coefficients diminish in magnitude as their harmonic index increases. Riemann's Lemma states that for functions with a bounded derivative, the integral of the product of the function with a sine or cosine function tends towards zero as the frequency becomes very large. This remarkable property yields significant insights and results when it comes to approximations and representations in Fourier analysis.
When applying Riemann's Lemma, the constant \(M\), which represents an upper bound for the function or its derivative, plays a crucial role. It's important to note that \(M\) is a fixed value providing assurance that integrals over these oscillating functions don’t grow unbounded. Essentially, the lemma showcases how oscillatory functions interact with bounded, smooth or piecewise smooth functions, providing the convergence needed in many practical applications.

Fourier Coefficients

Fourier coefficients are foundational in the study of Fourier series. These coefficients represent the weights of the respective sine and cosine terms that form the Fourier series expansion of a function.
In the context of the exercise, we are interested in analyzing how quickly these coefficients approach zero for piecewise smooth functions. The notation \(O(1/n)\) describes the order of magnitude for these coefficients, indicating that they decay as \(n^{-1}\).
More specifically, a Fourier series expansion allows a periodic function to be expressed in terms of sines and cosines, with coefficients \(a_n\) and \(b_n\) calculated as integrals over the period.

• The coefficients \(a_n\) are computed using cosine terms.
• The coefficients \(b_n\) are calculated with sine terms.

Using integration by parts, along with Riemann's Lemma, helps in establishing that these coefficients tend towards zero, emphasizing their diminishing contribution to the series. This decay of Fourier coefficients is what ensures the convergence of the Fourier series, thereby effectively allowing complex periodic functions to be represented by simpler trigonometric series.

Piecewise Smooth Functions

Piecewise smooth functions bring an interesting character to mathematical analysis. These functions are characterized by being smooth within certain intervals, but may have a finite number of discontinuities in their derivatives across their domain.
In practice, this means such functions have well-defined derivatives except at a handful of points, allowing us to extend results proven for continuously differentiable functions.

• A function is piecewise smooth if it's continuous, having a continuous derivative except at a finite number of jump points.
• These qualities make piecewise smooth functions versatile in applications, particularly in Fourier analysis and solving differential equations.

In the context of the problem, when a function is piecewise smooth over an interval, the integration by parts technique still proves effective across each smooth piece of the function. By dividing the interval into subintervals where the function is smooth, it's possible to apply results like Riemann's Lemma across each piece, as each piece contributes to the whole without the discontinuities disrupting the process. This piecewise approach is crucial in dealing with real-world functions which are often not perfectly smooth everywhere, hence enriching the range and applicability of mathematical methods like Fourier series in practical scenarios.