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 Post subject: Uniform convergence of series Posted: Sun Jan 17, 2016 3:53 pm
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
Let $\{a_n\}_{n \in \mathbb{N}}$ be a decreasing sequence such that $na_n \xrightarrow{n \rightarrow +\infty}0$. Prove that the series $\sum a_n \sin n x$ converges uniformly if and only if $n a_n \rightarrow 0$.

Edit:I added the red bold letters after a solution was given proving both directions. ${\color{red} {05/14/2016}}$

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Imagination is much more important than knowledge. Top   Post subject: Re: Uniform convergence of series Posted: Tue May 10, 2016 4:02 pm

Joined: Tue May 10, 2016 3:56 pm
Posts: 33
Let $a_n$ be a nonnegative and decreasing sequence. Show that $\sum_{n=1}^\infty\,a_n\sin nx$ converges uniformly on $\mathbb{R}$ if and only if $\lim_{n \to \infty}\,na_n = 0$.

$(\Rightarrow)$ Assume the series $\sum_{n=1}^\infty\,a_n\sin nx$ converges uniformly on $\mathbb{R}$. Then for every $\epsilon > 0$, there is an $N \in \mathbb{N}$ such that
$$\left|\sum_{k=m}^n\,a_k\sin kx\right| < \epsilon \hspace{0.2in} \mbox{for all x \in \mathbb{R} whenever n > m> N}.$$
Let $n > 2N, m =[n/2 +1]$. Then $m > n/2> N$. Choosing $x = \pi/2n$, we have
$$\left|\sum_{k=m}^n\,a_k\sin(k\pi/2n)\right| < \epsilon. \tag{1}$$
When $m \leq k \leq n$, we have
$$\frac{\pi}{4} < \frac{m\pi}{2n} \leq \frac{k\pi}{2n} \leq \frac{\pi}{2}.$$
Now (1) implies that
$$\epsilon > \sum_{k=m}^n\,a_k\sin(k\pi/2n) \geq \sin\frac{\pi}{4}\sum_{k=m}^n\,a_k \geq \frac{\sqrt{2}}{2}\,(n-m+1)a_n \geq \frac{\sqrt{2}}{4}\,na_n.$$
This proves that $\lim_{n \to \infty}\,na_n = 0$.

$(\Leftarrow)$ Assume that $\lim_{n \to \infty}\,na_n = 0$. Let
$$\alpha_n = \sup_{k \geq n}\,\{ka_k\}.$$
Then $\alpha_n$ is decreasing and converges to zero as $n \to \infty$. Let
$$S_{m, n}(x) = \sum_{k=m}^n\,a_k\sin kx.$$
we want to show that
$$|S_{m, n}(x)| \leq (\pi + 3)\alpha_{m}\hspace{0.2in} \mbox{for every x \in \mathbb{R}}. \tag{2}$$
Notice that $S_{m, n}(x)$ is an odd and periodic function with period $2\pi$. It suffices to show that (2) holds on $[0, \pi]$. To do so, we divide $[0, \pi]$ into $[0, \pi/n], [\pi/n, \pi/m], [\pi/m, \pi]$ and show that (2) holds on each such subinterval.

(a). If $0 \leq x \leq \pi/n$, then $kx \in [0, \pi]$ for all $m < k \leq n$. Using $\sin\theta \leq \theta$ yields
$$|S_{m, n}(x)| = \sum_{k=m}^n\,a_k\sin kx \leq x\,\sum_{k=m}^n\,ka_k$$
$$\leq x\,\sum_{k=m}^n\,\alpha_k \leq \frac{\pi}{n}(n-m+1)\alpha_{m} \leq \pi\alpha_{m}.$$

(b). If $\pi/n \leq x \leq \pi$, in view of that $\sin\theta \geq \frac{2}{\pi}\,\theta$ for $\theta \in [0, \pi/2]$, we have
$$\left|\sum_{k=m}^n\,\sin kx\right| = \frac{|\cos(n-1/2)x -\cos(m+1/2)x|}{2\sin(x/2)} \leq \frac{1}{\sin(x/2)} \leq \frac{\pi}{x} \leq m.$$
By Abel's lemma, we have
$$|S_{m, n}(x)| = m(a_m + 2a_n) \leq 3ma_m \leq 3\alpha_m.$$

(c). If $\pi/n \leq x \leq \pi/n$, then $m \leq \pi/x \leq n$. Let $L = [\pi/x]$. We have
\begin{eqnarray*}
S_{m, n}(x) & = & \sum_{k=m+1}^L\,a_k\sin kx + \sum_{k=L+1}^n\,a_k\sin kx = S_{m, L}(x) + S_{L+1, n}(x).
\end{eqnarray*}
Since $L \leq \frac{\pi}{x} < L +1$, it follows that $x \leq \frac{\pi}{L}$. The result of (a) implies that
$$|S_{m, L}(x)| \leq \pi\alpha_m.$$
For $\frac{\pi}{L+1} < x \leq \frac{\pi}{m}$ and $L +1 > \frac{\pi}{x} \geq m$, by (b) and the decreasing of $\alpha_n$, we have
$$\left|S_{L+1, n}(x)\right| \leq 3\alpha_{L +1}$$
and so
$$\left|S_{m, n}(x)\right| \leq (3 + \pi)\alpha_m.$$
In summary, we have proved that (1) holds for every $x \in [0, \pi]$. Therefore, the series converges uniformly on $\mathbb{R}$.

Remark. This result does not apply to $\sum_{n=1}^\infty\,a_n\cos nx$. Can you give a counterexample?

Top   Post subject: Re: Uniform convergence of series Posted: Thu May 04, 2017 1:26 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 157
Location: Melbourne, Australia
mathofusva wrote:
Remark. This result does not apply to $\sum_{n=1}^\infty\,a_n\cos nx$. Can you give a counterexample?

Up!!

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