3
$\begingroup$

I'm learning about Fourier series and need help with the following problem:

Consider the function

$$g(x) = \begin{cases} x^{\frac{1}{3}}, & x \in [0, \frac{\pi}{2}] \\ (-x)^{\frac{1}{4}}, & x \in [-\frac{\pi}{2}, 0) \\ \pi, & x \in [-\pi,-\frac{\pi}{2}) \cup(\frac{\pi}{2}, \pi] \end{cases}$$

extended periodically with period $2\pi$.

Show that the function $g$ has a Fourier series. What can you tell about the pointwise convergence of the series in $\mathbb{R}$?


The problem formulation is causing me difficulties here. Usually, when finding the Fourier series of a periodic function, the author states "compute (or find) the Fourier series of the given function". However, here I need to show that the function $g$ has a Fourier series. Possibly we can avoid computing the Fourier coefficients of $g$ and show that the above function has a Fourier series by some theorem. Is my thought correct? If no, how do I find the Fourier series of the above piecewise-defined function?

I graphed the function $g$ on a piece of paper and noticed that it is monotone on each subinterval: it is increasing on $[0, \frac{\pi}{2}]$, decreasing on $[-\frac{\pi}{2}, 0)$ and constant on $[-\pi,-\frac{\pi}{2}) \cup(\frac{\pi}{2}, \pi]$. Maybe this is a case where Dirichlet's theorem applies, i.e.

Dirichlet's theorem on Fourier series. If a $2\pi$-periodic function $f$ is piecewise monotone on the segment $[-\pi, \pi]$ and has at most finitely many discontinuity points on it, then its trigonometric Fourier series converges to $f(x)$ at each continuity point and to $\frac{f(x+0) + f(x-0)}{2}$ at each discontinuity point.

For the second part of the question about pointwise convergence of the series in $\mathbb{R}$ I have no idea.

$\endgroup$
  • 1
    $\begingroup$ did you see any Fourier series of discontinuous function before ? can you add a function $h(x)$ to $g(x)$ such that $f(x) = g(x)+ h(x)$ is continuous ? then you can look at en.wikipedia.org/wiki/… the problem is that obtained the function $f(x)$ is not differentiable, but it is weakly differentiable, hence there is some theorem we can apply for the convergence of its Fourier series. and the Fourier series of $h(x)$ is very simple and of well-known convergence $\endgroup$ – reuns May 29 '16 at 20:38
1
$\begingroup$

Your function $f$ satisfies $f(-\pi)=f(\pi)$ and has a natural $2\pi$ periodic extension $f_p$ to $\mathbb{R}$. The Fourier series for $f$ on $[-\pi,\pi]$ is periodic with period $2\pi$. The Fourier series converges pointwise and uniformly on every closed interval $[a,b]\subseteq \mathbb{R}$ for which $f_p$ is continuous on a slightly larger interval $[a-\epsilon,b+\epsilon]$, where $\epsilon > 0$; this is because of the piecewise monotone character of $f_p$. At a point $x$ of discontinuity of $f_p$ the Fourier series converges to the mean of the left- and right-hand limits of $f_p$ at $x$, but the series will not converge uniformly in any closed interval with $x$ in its interior.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your clear explanation. I do have one question: it seems to me that the considered function has no point of discontinuities, i.e. it is continuous everywhere in $\mathbb R$ (or to say it another way, I can draw the graph of g extended periodically without picking up my pencil). Therefore, because of the piecewise monotone character of $g_p$, the Fourier series converges pointwise and uniformly on everywhere in $\mathbb R$. Is that correct? $\endgroup$ – glpsx May 30 '16 at 15:52
  • 1
    $\begingroup$ @VonKar : Your function takes a jump at $x=\pm \pi/2$. At $-\pi/2$ the function is $\pi$, but the limit from above is $(\pi/2)^{1/4}$. A similar discontinuity occurs at $+\pi/2$. So the series only converges uniformly on closed intervals that don't include either of those points or their $2\pi$ periodic translates. $\endgroup$ – DisintegratingByParts May 30 '16 at 16:38
  • 1
    $\begingroup$ I see it now! Thank you for your detailed and helpful explanation. $\endgroup$ – glpsx May 30 '16 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.