Given $f(x)$ is the characteristic function of the interval $[a,b]\subset [-\pi,\pi]$ ($a\neq b$), so $f(x) = 1$ for $x\in [a,b]$ and $f(x)=0$ otherwise.
From this definition, I obtained the Fourier series of $f$ is $\frac{b-a}{2\pi} + \sum_{n\neq 0, n\in Z} \frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}$.
Question If $a\neq -\pi$ or $b\neq \pi$, show that the Fourier series above does not converge absolutely.
My attempt By some algebraic manipulations, I got $|e^{-ina}-e^{-inb}| = 2|\sin(n\theta)|$ where $\theta = \frac{b-a}{2}$. Now, I want to find a constant lower bound $c > 0$ such that $|sin(n\theta)|\geq c$ for many positive integers $n$, but I couldn't get further from this (I don't see how to use the given condition to find such $c$). Can anyone please help with this step?
It seems to me that if $a=\frac{-\pi}{2}$ and $b=\frac{\pi}{2}$, which satisfies the given condition, we have: $\sum_{n\neq 0, n\,\in Z}\ \frac{2|\sin(n\theta)|}{|n|} =\sum_{n\neq 0, n\,\text{is odd}} \frac{2}{|n|}$, which diverges. The problem statement is wrong?
Best Answer
We have the following theorem:
$\textbf{Theorem}$. Let $|a_n| \downarrow 0$. Then, if $\sum_{n=1}^{\infty}a_n\sin nx$ converges absolutely at some point $x_0$ which is not a multiple of $\pi$, then $\sum_{n=1}^{\infty}|a_n|<\infty$.
This theorem can be found in N. K. Bary's book: A Treatise on Trigonometric Series, Vol. II, page 334. Next, your series can be rewritten as $$ \sum_{n\in \mathbb{Z} \backslash \{0 \}} 2\frac{|\sin nx|}{|n|}=4\sum_{n=1}^{\infty} \frac{|\sin nx|}{n}.$$ Since $1/n \downarrow 0$, if the series converges absolutely at some $x_0$ which is not multiple of $\pi$, then we would obtain the convergence of the harmonic series, which is obviously false.
I recommend you having a look at the proof of the cited theorem so you can see the difference between it and your approach.
EDIT: Proof of theorem:
If $\sum |a_n \sin nx_0|<\infty$, then $\sum |a_n \sin^2 nx_0|<\infty$ and $\sum |a_{n-1} \sin^2 (n-1)x_0|<\infty$. But \begin{align} |a_n|\sin^2 nx_0 +|a_{n-1}|\sin^2(n-1)x_0 & \geq |a_n|(\sin^2 nx_0 +\sin^2 (n-1)x_0)\\ & = |a_n|((1-\cos 2nx_0)/2+(1-\cos 2(n-1)x_0)/2) \\ &= |a_n| (1-\cos x_0 \cos(2n-1)x_0) \geq |a_n|(1-|\cos x_0|). \end{align}
Hence, $$ \infty> \sum |a_n|(1-|\cos x_0|) = (1-|\cos x_0|)\sum |a_n|, $$ which gives the desired result, since $x_0$ is not multiple of $\pi$.