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 $[\epsilon, 1-\epsilon]$.

Question

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?

Answer 1

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:

1. 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.
2. You can then use uniform Cauchy criteria.
3. 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 $\varepsilon$.