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Uniform convergence

Let \{a_n\}_{n \in \mathbb{N}} be a decreasing sequence. Prove that \sum \limits_{n=1}^{\infty} a_n \sin n x converges uniformly if-f n a_n \xrightarrow{n \rightarrow +\infty} 0.


Since n a_n \rightarrow 0 then for a given \epsilon>0 there exists positive integer n_0 such that if n \geq n_0 to hold

(1)   \begin{equation*} \left | na_n \right |-na_n < \frac{\epsilon}{2(\pi+1)}   \end{equation*}

It suffices to prove the result for x \in [0, \pi] due to symmetry and periodicity. So, it suffices to prove that for n \geq n_0 it holds \displaystyle \left | \sum_{k=n+1}^{n+p} a_k \sin kx \right |< \epsilon for all x \in [0, \pi] and all p \in \mathbb{N}.

We distinguish cases:

  • x \in \left[ 0, \frac{\pi}{n+p} \right];

        \begin{align*} \left | \sum_{k=n+1}^{n+p}a_k \sin k x \right | &=\sum_{k=n+1}^{n+p}a_k \sin kx \\ &\leq \sum_{k=n+1}^{n+p}a_k kx \quad \left ( \sin kx \leq kx , \;\; x\geq 0 \right ) \\ &= \sum_{k=n+1}^{n+p}(ka_k)x\\ &\overset{(1)}{\leq } \frac{\epsilon}{2(\pi+1)}\frac{\pi p}{n+p}\\ &< \epsilon \end{align*}

  • x \in \left[ \frac{\pi}{n+p}, \pi \right];

        \begin{align*} \left | \sum_{k=n+1}^{n+p}a_k \sin kx \right | &\leq \sum_{k=n+1}^{\left \lfloor \frac{\pi}{x} \right \rfloor} a_k k x+ \left | \sum_{k=\left \lfloor \frac{\pi}{x} \right \rfloor+1}^{n+p} a_k \sin kx \right | \\ &\leq \sum_{k=n+1}^{\left \lfloor \frac{\pi}{x} \right \rfloor} a_k kx + \frac{2a_{m+1}}{\sin \frac{x}{2}} \quad \text{(summation by parts)} \\ &\leq \frac{\epsilon}{2 (\pi+1)} \left ( \left \lfloor \frac{\pi}{x} \right \rfloor -n\right ) x +\frac{2a_{m+1}}{\sin \frac{x}{2}} \\ &\leq \frac{\epsilon}{2 (\pi+1)}mx + 2a_{m+1} \frac{\pi}{x} \;\;\; \left( \text{since} \;\; \frac{2x}{\pi}\leq \sin x  \right) \\ &\leq \frac{\epsilon}{2(\pi+1)}\pi + 2a_{m+1} \left ( m+1 \right)\\ &< \frac{\epsilon}{2}+ 2\cdot \frac{\epsilon}{2 (\pi+1)}\\ &< \epsilon \end{align*}

    where \displaystyle{m=\left \lfloor \frac{\pi}{x} \right \rfloor}.

This completes the proof of the first part. Now the converse is much easier. For each n such that k\leq n \leq 2k-1 we have \frac {n\pi}{6k} \in \left( \frac {\pi}{6}, \frac {\pi}{2} \right). Hence \frac {1}{2} \le \sin \frac {n\pi }{6k}. Therefore, picking x=\frac{\pi}{6k} we have:

(2)   \begin{align*} \sum _{n=k}^{2k-1} a_n\sin \frac {n\pi}{6k} &\geq \frac {1}{2} \sum _{n=k}^{2k-1} a_n\\ &\geq \frac {1}{2} \sum _{n=k}^{2k-1} a_{2k-1} \\ &= \frac {1}{2} k a_{2k-1}\\ &\geq \frac {1}{4}(2 k-1) a_{2k-1} \\ &\geq 0   \end{align*}

Since the series converges uniformly we have that \sum \limits_{n=k}^{2k-1} a_n \sin nx \rightarrow 0 uniformly. More specifically, \sum \limits_{n=k}^{2k-1} a_n\sin \frac {n\pi}{6k} \rightarrow 0 as k \rightarrow +\infty. Using the sandwich theorem it follows that (2k-1)a_{2k-1} \rightarrow 0. It only remains to prove that 2ka_{2k} \rightarrow 0. This follows from

    \[0\leq 2ka_{2k} \leq 2ka_{2k-1}\leq 4(2k-1)a_{2k-1} \rightarrow 0\]

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