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Does the series $$ \sum^{\infty}_{k=1} \frac{\sin(kx)}{k^{\alpha}}, $$ converge for all $\alpha > \frac{1}{2}$ and for all $x \in [0,2 \pi]$?

It is obvious that it does when $\alpha > 1$, but I have no idea how to deal with the case $$ \frac{1}{2} < \alpha \le 1. $$

I already appreciate your hints/ideas.

Marcelo
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the8thone
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    Did you try replacing $\sin(kx)$ by $e^{ikx}$? This is probably some Fourier fact or something, since for $\alpha > \frac 12$ you have $\sum_{k \ge 1} \left( \frac 1{k^{\alpha}} \right)^2 < \infty$. – Patrick Da Silva Dec 28 '13 at 20:51
  • @PatrickDaSilva Thanks, but after replacing with $\frac{e^{ikx}-e^{-ikx}}{2i}$ how should I go on .... ? – the8thone Dec 28 '13 at 20:56
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    Dirichlet's test plus the fact the partial sums $\sum_{k=1}^n sin(kx)$ are bounded is enough to show convergence for $\alpha > 0$. – achille hui Dec 28 '13 at 20:58
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    @achille hui : How do you show the partial sums are bounded? I don't believe this statement. – Patrick Da Silva Dec 28 '13 at 21:00
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    @Roozbeh-unity : You're gonna have $$ \sum_{k=-\infty}^{\infty}^* \frac{k/|k|}{2i |k|^{\alpha}} e^{ikx} $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier. – Patrick Da Silva Dec 28 '13 at 21:01
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    @Roozbeh-unity: this will help you(believe):http://math.stackexchange.com/questions/108486/sx-sum-n-1-inftya-n-sinnx-a-n-is-monotonic-decreasing-a-n-to-0/108586#108586 – Salech Alhasov Dec 28 '13 at 21:04
  • @PatrickDaSilva see the proof in Salech's link. – achille hui Dec 28 '13 at 21:07

2 Answers2

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Actually, this sum converges for every $\alpha>0$.

Step I. For every $x\in\mathbb R$, the sequence $s_n=\sum_{k=1}^n\sin kx$ is bounded.

Indeed, if $x=m\pi$, then $s_n=0$. If $x\ne m\pi$, then $\sin(x/2)\ne 0$, and $$ s_n=\sum_{k=1}^n\sin kx=\mathrm{Im}\left(\mathrm{e}^{xi}+\mathrm{e}^{2xi}+\cdots\mathrm{e}^{nxi}\right)= \mathrm{Im}\left(\mathrm{e}^{xi}\frac{\mathrm{e}^{nxi}-1}{\mathrm{e}^{xi}-1}\right) $$ But $$ \left|\mathrm{e}^{xi}\frac{\mathrm{e}^{nxi}-1}{\mathrm{e}^{xi}-1}\right|\le \frac{2}{|\mathrm{e}^{xi}-1|}=\frac{2}{|\mathrm{e}^{xi/2}-\mathrm{e}^{-xi/2}|}=\frac{1}{|\sin (x/2)|}. $$ and hence $\lvert s_n\rvert\le \lvert\sin(x/2)\rvert^{-1}$.

Step II. Use Abel's summation method. \begin{align} \sigma_n &=\sum_{k=1}^n\frac{\sin kx}{k^\alpha}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k^\alpha} =\sum_{k=1}^n\frac{s_k}{k^\alpha}-\sum_{k=1}^n\frac{s_{k-1}}{k^\alpha} \\ &=\sum_{k=1}^n\frac{s_k}{k^\alpha}-\sum_{k=0}^{n-1}\frac{s_{k}}{{(k+1)}^\alpha}= \frac{s_n}{n^a}+\sum_{k=1}^{n-1}s_k\left(\frac{1}{k^a}-\frac{1}{(k+1)^a}\right). \end{align} But $$ \frac{1}{k^\alpha}-\frac{1}{(k+1)^\alpha}\le\frac{a}{k^{1+\alpha}}, $$ and hence the series $$ \sum_{n=1}^{\infty}s_n\left(\frac{1}{n^a}-\frac{1}{(n+1)^a}\right), $$ converges (indeed absolutely) due to the comparison test.

Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval $[\varepsilon,2\pi-\varepsilon]$.

Yiorgos S. Smyrlis
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The set $\{ e^{inx} \, | \, n \in \mathbb Z \}$ forms an Hilbert basis of $L^2([0,2\pi])$ which means that for a function $f \in L^2([0,2\pi])$, we have $$ f = \sum_{n \in \mathbb Z} \langle f, e_n \rangle e_n $$ with $$ \langle f,e_n\rangle = \frac{n/|n|}{2i n^{\alpha}} $$ if $n \neq 0$ and $0$ otherwise, if and only if $$ \sum_{n \in \mathbb Z} |\langle f,e_n \rangle|^2 < \infty. $$ Since $\alpha > \frac 12$, in our situation this is the case, therefore $\sum_{n \in \mathbb Z} \frac{\sin(nx)}{n^{\alpha}}$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2\pi])$.

Hope that helps,

Patrick Da Silva
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    $L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember. – Daniel Fischer Dec 28 '13 at 21:15
  • @Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need. – Patrick Da Silva Dec 28 '13 at 21:16
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    No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_\delta$. Since $C(T) \subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows. – Daniel Fischer Dec 28 '13 at 21:24
  • @Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better. – Patrick Da Silva Dec 28 '13 at 21:27
  • It's also because I thought "this was the proof" because of the $\alpha > 1/2$ condition. – Patrick Da Silva Dec 28 '13 at 21:28