Uniform convergence

Question

How would you show that

$$f(x)=\sum_{n=2}^\infty \frac{\sin(2\pi n x)}{n\log n}$$ converges uniformly on $x\in[0,1]$.

The pointwise convergence can be proved by Drichlet test

Answer 1

With $a_n = 1/(n \log n)$ and $b_n(x)=\sin(2\pi n x)$, we have $a_n$ decreasing monotonically and converging to $0$ uniformly with respect to $x \in [\delta,1]$ with $\delta > 0$.

Also $b_n(x)$ has uniformly bounded partial sums.

If $2x \neq m\in \mathbf{N},$

$$\left|\sum_{k=2}^n \sin(2\pi k x)\right|\leq 1+ \left|\frac{\cos(\pi x)-\cos[2(n+1/2)\pi x]}{2\sin(\pi x)}\right|\leq 1+\frac{2}{|\sin(\pi \delta)|},$$

and if $2x = m \in\mathbf{N},$

$$\left|\sum_{k=2}^n \sin(2\pi k x)\right|= 0$$

Therefore the series converges uniformly by the extension of the Dirichlet test for uniform convergence. This can be proved as follows

We show that the sequence of partial sums $S_n(x) = \sum_{k=1}^{n} a_kb_k(x)$ satifies the Cauchy criterion uniformly.

Using summation by parts,

$$|S_m(x) - S_n(x)|= \left|\sum_{k=n+1}^{m-1} a_kb_k(x)\right|= \left|a_mB_m(x)- a_{n+1}B_n(x) + \sum_{k=n+1}^{m-1} (a_k-a_{k+1})B_k(x)\right|,$$

where

$$B_n(x) = \sum_{k=1}^{n}b_n(x).$$

Note that $B_n$ is uniformly bounded, and there exists $B > 0$ such that $|B_n(x)| \leq B$ for all $n$ and all $x \in [\delta,1]$.

Furthermore, $(a_n)$ is a non-increasing sequence and converges uniformly to $0$. For any $\epsilon > 0$, there exists $N$ such that if $n > N$, then for every $x \in [\delta,1]$ we have $a_n < \epsilon/(2B)$.

Hence,

$$|S_m(x) - S_n(x)| \leq |a_m||B_m|+ |a_{n+1}||B_n| + \sum_{k=n+1}^{m-1} |a_k-a_{k+1}||B_k|\\ \leq B\left(a_m+ a_{n+1} + \sum_{k=n+1}^{m-1} (a_k-a_{k+1})\right)=2Ba_{n+1}.$$

Thus, for any $\epsilon > 0$ and for every $x \in [\delta,1]$, if $n > N$ we have

$$|S_m(x) - S_n(x)| \leq 2Ba_{n+1}< 2B\frac{\epsilon}{2B}= \epsilon,$$

and the sequence of partial sums is uniformly convergent.