1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $\pi$.

$\endgroup$
  • $\begingroup$ Many thanks for your great help, Alberto! Can you show me the proof for that theorem please? I cannot find it online somehow:P $\endgroup$ – user177196 Apr 29 '15 at 14:20
  • 1
    $\begingroup$ There you got it! $\endgroup$ – Alberto Debernardi Apr 29 '15 at 14:31
1
$\begingroup$

If the series converges uniformly, the sum is a continuous function. Since the characteristic function of $[a,b]$ is not continuous on $[-\pi,\pi]$ if $a\ne-\pi$ or $b\ne\pi$, the convergence cannot be uniform.

$\endgroup$
  • $\begingroup$ how did you relate from the continuity of a function to the continuity of characteristic function? And can you please show me the proof for uniform convergent series -> continuous function? $\endgroup$ – user177196 Apr 29 '15 at 14:30
  • $\begingroup$ It is a standard fact that the limit of a uniformly convergent sequence of continuous functions is continuous. An it is clear that the characteristic functions has a discontinuity with jump $1$ at $a$ or $b$. $\endgroup$ – Julián Aguirre Apr 29 '15 at 14:52

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.