Let $f \colon [0, 1] \to \mathbb{R}$ be defined by $$ f(x) = \sum_{n=1}^\infty \frac{a_n}{n} \sin(n\pi x), $$ for some square-summable real sequence $a_n$.
Is it true that $f$ is absolutely continuous? I tried a basic estimate, $$ \sum_{k=1}^K |f(u_k) - f(v_k)| \leq \sum_{n \geq 1} \sum_{k \leq K} \frac{|a_n|}{n} |\sin(n \pi u_k) - \sin(n\pi v_k)| \lesssim \delta \sum_{n \geq 1} |a_n|, $$ provided that $\sum_k |u_k - v_k| \leq \delta$, however this does not seem to be strong enough.
Is there another argument? Or is this not true?
Let $g(x)=\sum a_n \cos (n\pi x)$. In view of the orthogonality of $(\cos (n \pi x))$ it follows that $g$ is a well defined $L^{2}$ function. Hence $g \in L^{1}$ and $\int_0^{x} g(t)dt$ is absolutely continuous. But $\int_0^{x} g(t)dt=\frac 1 {\pi} f(x)$, so $f$ is absolutely continuous.