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Prove that $f(x) = \sum_{n=1}^{\infty} \frac{\sin^3(nx)}{n}$ is discontinuous at point $0$.
I rewrote this function using $\sin^3(x) = \frac{1}{4} (3 \sin(x) - \sin(3 x))$. $f(x) = \frac{3}{4}(\sum \frac{\sin(nx)}{n} - \sum \frac{\sin(3nx)}{3n})$ So now my problem is that I have to find some cool epsilon that approaches zero so that $|f(\epsilon) - f(2\epsilon)|$ would be GREATER than some constant (or function that is greater than some constant). $f(\epsilon) - f(2\epsilon) = \sum \frac{\sin(n\epsilon)(1 - 2\sin(\pi/2 - n\epsilon))}{n}$ if sum contains only n that are not divisible by $3$. I checked this series in the desmos and indeed it can be quite large but for some particular epsilon. I will be very grateful.

Gary
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tryinghard
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  • I mean for every given distance there some points near zero such that (x_1 - x_2 < distance) and |f(x_1) - f(x_2)| > some constant – tryinghard Oct 23 '24 at 12:39
  • But what does "near $0$" mean ? We can say that a function is uniformly continuous on a domain, but not at a point (it is a global notion, not a local one). – TheSilverDoe Oct 23 '24 at 12:40
  • I have to prove that this function is not uniformly continuous on [0; pi/4]. But I noticed that it is discontinuous at zero, so I'm trying to prove it. I just misused terms, sorry – tryinghard Oct 23 '24 at 12:46
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    I don't think taking the three out of the second term does much good. The expression is, if $g(x)=\sum\frac{\sin(nx)}{n},$ then the function is $\frac14(3g(x)-g(3x)).$ You can Find $g(x)$ in closed form by realizing it is the imaginary part of $\sum\frac{e^{inx}}{n}=-\ln(1-e^{ix}).$ – Thomas Andrews Oct 23 '24 at 13:12
  • That doesn't seem to me as a more easy task. Actually I cannot understant how I could prove that the imaginary part of $-ln(1-e^{ix}$ is discontinuous at zero. – tryinghard Oct 23 '24 at 13:24
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    This post may help you https://math.stackexchange.com/questions/3241994/showing-that-sum-n-1-infty-left-frac-sin22n7n-right3-frac12?noredirect=1 – Riemann Oct 23 '24 at 13:32

2 Answers2

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Let $$g(x) :\equiv \sum_{n=1}^\infty\frac{\sin nx}n$$

Then $g(x) = \dfrac{\pi-x}2$ for $x\in(0,2\pi)$ (notice this is a Fourier series; you can find a proof here).

As you have cleverly noticed, we have $4f(x) \equiv 3g(x) - g(3x)$. Then, for $x\in\left(0, \dfrac{2\pi}3\right)$, we have $$4f(x) = 3g(x)-g(3x) = \frac{3\pi-3x}2-\frac{\pi-3x}2 = \pi$$ This means $\displaystyle\lim_{x\to0^+} f(x) = \frac\pi4\ne0=f(0)$, therefore, $f$ is discontinuous at $0$.

Alma Arjuna
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  • Oh, this is cool. But I cannot understant last line. Is there a rigorous prove that if it does not equal to f(0) so it is discontinous? Like, $g(x)$ is defined on $(0; 2\pi)$ and not on [0; 2\pi$ why can we conclude such thing? – tryinghard Oct 23 '24 at 16:48
  • @tryinghard in fact, $g$ is defined for all $x\in\mathbb R$. However, it equals $0$ for integral multiples of $2\pi$ (because this nullifies the sine). – Alma Arjuna Oct 23 '24 at 16:54
  • Great! Can you send a link why is it equals to $\frac{\pi - x}{2}$. Cause I still do not understand why is $\frac{\pi - x}{2}$ at point x = 0 $\not = f(0)$ is equivalent to discontinuity :-( – tryinghard Oct 23 '24 at 16:58
  • @tryinghard I tried to make the conclusion clearer. and added a link to the proof of the fact you requested. – Alma Arjuna Oct 23 '24 at 17:06
  • Thank you so much! – tryinghard Oct 23 '24 at 17:33
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This is a sketch for a different approach that works with similar trigonometric series $f(x)=\sum a_n \sin nx, a_n$ decreasing to zero $na_n$ uniformly bounded but $na_n$ not converging to zero, which are not so easily computable like here, or more generally $f(x)=\sum a_n \sin^{2k+1} nx$ ($k \ge 0$ fixed integer) for which still $\sum _{n=1}^N \sin^{2k+1} nx$ are uniformly bounded away from numbers of the form $2m\pi$ by simple geometric series manipulations (while they also become $\sum b_n \sin nx$ by expanding $\sin ^{2k+1}nx$ in terms of $\sin nx, \sin 3nx, ..\sin (2k+1)nx$).

First one notes that if $f_n$ are the partial series then $f_n(\pi/(2n)$ doesn't converge to zero by simple inequalities ($\sin x \ge cx, 0 \le x \le \pi/2$) so the partial sums do not converge uniformly - unfortunately that is not enough to conclude discontinuity at zero, but gives us a clue to use the Caesaro means $\sigma_n$ of the partial series (and a similar argument as for $f_n$) to show that $\sigma_n(\pi/(2n)$ doesn't converge to zero so $\sigma_n$ do not converge uniformly.

Feijer theorem tells us that $f$ cannot be continuous on $[-\pi, \pi]$ since it is periodic, bounded (by partial summation) so $\sum a_n \sin nx$ (or more generally $\sum a_n \sin^{2k+1} nx=\sum b_n \sin nx$) is its Fourier series. But by partial summation it also immediately follows that the series is uniformly continuous on any compact interval not containing a number of the form $2m\pi$ for some integer $m$, so the only possible discontinuity point is zero.

Conrad
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