Q:

Let $\{a_n\}$ be decreasing and $\geq 0$. Let $0<\varepsilon<1/2$. Prove $\sum a_n\sin(2\pi nx)$ is unif. converg. in $[\epsilon, 1-\epsilon]$.

I'm doing the following exercise:

Let $\{a_n\}$ be a decreasing and $\geq 0$ sequence. Let $0<\epsilon<1/2$. Prove that $\sum a_n \sin(2\pi nx)$ is uniformly convergent in $[\epsilon, 1-\epsilon]$, and deduce that it's continuous in there.

I can't use Abel's criterion, nor Dirichlet's, to see that for every $x\in [\epsilon,1-\epsilon]$ the series converges pointwise. But I don't know how to see that the convergence is uniform. What's the best way to see that it converges uniformly in there?

A:

Hint.
If you suppose that $(a_n)$ is decreasing and converging to zero, then the result can be proven.
The main lines of the proof are following ones:

For $\varepsilon > 0$ and $n \ge m \ge 0$, $\displaystyle \sum_{k=m}^n \sin(2\pi kx)$ is uniformly bounded by a constant $A$ on the interval $[\varepsilon, 1-\varepsilon]$. This can be proven using the classical formulae for the sum of sinus.
You can then use uniform Cauchy criteria.
To apply uniform Cauchy criteria, do an Abel transformation of $\displaystyle \sum_{k=m}^n a_k \sin(2\pi nx)$ to get $\displaystyle \vert \sum_{k=m}^n a_k \sin(2\pi nx) \vert \le a_m A$

This last inequality can probably be reworked to prove the second result, analyzing the way $A$ depends on $\varepsilon$.

