Questions related to Fourier Analysis

[xalvyn]

Dear all

I have recently taken up the study of Fourier analysis. My background knowledge is limited - some basic notions of analysis, including the Riemann integral, as well as uniform and pointwise convergence of series of functions. These are not exactly homework problems, but questions that I have been thinking about during the course of my self-study - and to which I have so far not found any satisfactory answers. I would gladly appreciate any help rendered.

Homework Statement

1. Is it true that, if [tex]0 < \theta _0 < \pi[/tex], then [tex]|\sin n\theta _0|[/tex] diverges as [tex]n[/tex] tends to infinity? If so, how should we prove it generally?

2. Suppose that the Fourier series of a continuous function [tex]f(x)[/tex] converges pointwise at [tex]x=x_0[/tex]. Is it necessarily true that [tex]\sum_{i=-\infty}^\infty \hat{f} (n) e^{in x_0} = f(x_0)[/tex]?

Homework Equations

The Attempt at a Solution

1. This seems much easier to prove for individual values of [tex]\theta _0[/tex], but I find it difficult to extend the argument to a general case. I could, however, prove that the value of [tex]|\sin \theta _0|[/tex] does not approach 0: we may show that there exists a [tex]c > 0[/tex] such that there are arbitrarily large values of [tex]n[/tex] satisfying [tex]|\sin n \theta _0| > c > 0.[/tex] Choose [tex]c[/tex] less than [tex]\frac{1}{2}r[/tex] by a very small amount, where [tex]k[/tex] is the largest positive integer satisfying [tex]\pi = k \theta _0 + r[/tex] with [tex]r > 0[/tex]. If, then, [tex]|\sin p \theta _0| \leq c[/tex], it follows that [tex]|\sin (p+1) \theta _0| > c [/tex]. Extending this sort of argument seems more difficult.

2. I could think of a discontinuous function for which the statement is false: the sawtooth function defined by

[tex] f(x) = -\frac{\pi}{2}-\frac{x}{2}[/tex] if [tex] -\pi < x \le 0 [/tex];

[tex] \frac{\pi}{2} - \frac{x}{2}[/tex] if [tex]0 < x < \pi[/tex].

In this case, while the Fourier series of f converges to 0 at [tex]x=0[/tex], the value of [tex]f(0)[/tex] is [tex]-\frac{\pi}{2}[/tex]. I cannot, however, construct an example of a continuous function for which the statement is not true.

Thanks for any help.

 

[kurt.physics]

Answer 1. Yes, it is true that, if 0 < \theta _0 < \pi, then |\sin n\theta _0| diverges as n tends to infinity. This can be proven using the Weierstrass M-Test. By the M-Test, if we can show that there exists a constant c > 0 such that |\sin n \theta _0| > c for all n \in \mathbb{N} then the series \sum_{n=1}^\infty |\sin n \theta _0| must diverge. To show that this holds, note that |\sin n \theta _0| cannot be less than \frac{1}{2} for any values of n. Therefore, we can choose c = \frac{1}{2} and this satisfies the conditions of the M-Test. 2. No, it is not necessarily true that \sum_{i=-\infty}^\infty \hat{f} (n) e^{in x_0} = f(x_0). The Fourier series may converge pointwise at x=x_0, but the value of the limit may be different from f(x_0). For example, consider the function f(x) = 2x - x^2 over the interval [0,\pi]. The Fourier series of f converges pointwise at x=0, but the value of the limit is 0, while f(0) = 0.