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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.

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    $\begingroup$ 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, 2016 at 20:38

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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.

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  • $\begingroup$ 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, 2016 at 15:52
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    $\begingroup$ @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. $\endgroup$ May 30, 2016 at 16:38
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    $\begingroup$ I see it now! Thank you for your detailed and helpful explanation. $\endgroup$
    – glpsx
    May 30, 2016 at 16:48

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