Uniform convergence

Let $\{a_n \}$ be a decreasing sequence of positive real numbers such that $na_n \xrightarrow{n \rightarrow +\infty}0$. Prove that the series $\sum a_n \sin nx $ converges uniformly.

Solution:



 \( (\Rightarrow) \) Suppose that the series converges uniformly on \( \mathbb{R} \). Then for a given \( \epsilon >0 \) there exists a positive integer \( N \) such that \( n \geq N \) and :

$$\left | \sum_{k=n}^{2n-1}a_k \sin kx \right |< \epsilon \tag{1}$$

Choose \( x =\frac{1}{n} \) hence \( \sin \frac{1}{2} \leq \sin kx \leq \sin 1 \). Therefore for \( n \geq N \) we always have by \( (1) \) that:

$$\begin{aligned}
\epsilon > \left | \sum_{k=n}^{2n-1} a_k \sin kx\right | &= \sum_{k=n}^{2n-1}a_k \sin kx\\
 &\geq \sum_{k=n}^{2n-1}a_{2k}\sin \frac{1}{2}  \quad \left ( a_k >0  \; {\rm and \; decreasing} \right ) \\
 &= \left(\frac{1}{2}\sin \frac{1}{2} \right)(2n a_{2n})
\end{aligned}$$

So, we proved that \( 2na_{2n} \xrightarrow{n \rightarrow +\infty} 0 \). Similarly , we can prove that \( (2n-1)a_{2n-1}  \xrightarrow{n \rightarrow +\infty} 0 \) and the first part is complete.

\( (\Leftarrow) \) Suppose that \( na_n \xrightarrow{n \rightarrow +\infty} 0 \). Thus, for a given \( \epsilon >0 \) there exists a positive integer \(n_0 \) such that \( n \geq n_0 \) and:

$$\left | na_n \right |-na_n < \frac{\epsilon}{2(\pi+1)} \tag{2}$$

In order to prove that the series converges uniformly on \( \mathbb{R} \) it is sufficient to prove that it converges uniformly on \( [0, \pi] \). So, it is sufficient to prove that if \( n \geq n_0 \) ,

$$\left | \sum_{k=n+1}^{n+p} a_k \sin kx \right |< \epsilon$$

holds for all \( x \in [0, \pi] \) and for all \( p \in \mathbb{N} \).

We distinguish the following cases:

\( \color{gray} \bullet \quad    x \in \left[ 0, \frac{\pi}{n+p} \right] \) In this case we have:

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

\( \color{gray} \bullet \quad    x \in \left[ \frac{\pi}{n+p}, \pi \right] \)  In this case we have:

$$\begin{aligned}
\left | \sum_{k=n+1}^{n+p}a_k \sin kx \right | &\leq \sum_{k=n+1}^{\left \lfloor \frac{\pi}{x} \right \rfloor} a_k k x+ \left | \sum_{k=\left \lfloor \frac{\pi}{x} \right \rfloor+1}^{n+p} a_k \sin kx \right | \\
 &\leq \sum_{k=n+1}^{\left \lfloor \frac{\pi}{x} \right \rfloor} a_k kx + \frac{2a_{m+1}}{\sin \frac{x}{2}}  \quad \left ( {\rm summation \; by \; parts} \right )\\
 &\leq  \frac{\epsilon}{2 (\pi+1)} \left ( \left \lfloor \frac{\pi}{x} \right \rfloor -n\right ) x +\frac{2a_{m+1}}{\sin \frac{x}{2}} \\
 &\leq \frac{\epsilon}{2 (\pi+1)}mx + 2a_{m+1} \frac{\pi}{x} \quad \left ( {\rm since}  \;\; \frac{2x}{\pi}\leq \sin x , \; x \in \left [ 0, \frac{\pi}{2} \right ] \right )\\
 &\leq \frac{\epsilon}{2(\pi+1)}\pi + 2a_{m+1} \left ( m+1 \right)\\
 &< \frac{\epsilon}{2}+ 2\cdot \frac{\epsilon}{2 (\pi+1)}\\
 &< \epsilon 
\end{aligned}$$

whereas $m= \left \lfloor \frac{\pi}{x} \right \rfloor$.
This leads us to extract the conclusion that the series converges uniformly on \( \mathbb{R} \) as wanted.


Source: mathimatikoi.org