Let $f(x)$ be a $2\pi$-periodic function on $(0,2\pi)$ (so we extend $f$ to the reals by $f(x)=f(x+2\pi)$. ) $f$ is also absolutely improperly integrable, i.e., $$ \int_0^{2 \pi } |f| \, dx := \lim_{x \rightarrow 0} \int_x^c |f| \, dx + \lim_{y \rightarrow 2 \pi} \int_c^y |f| \, dx .$$ exists, where the terms in limits are Riemann Integrals exists. (So $|f|$ is Riemann Integrable in any interval $(\alpha, \beta) \subset [0, 2\pi] $.) $$ G_m(\zeta):= \int_{\delta} ^{2 \pi - \delta} f(x+\zeta) \Big(\sum_{k=m+1}^{\infty} \frac{ \sin k x } {k} \Big) \, dx $$
Then there exists $m$ such that $|G_m(\zeta)| < \varepsilon$ for all $\zeta \in \mathbb{R}$.
I know that $\sum_{k=1}^\infty \frac{ e^{ikx} } {k} $ is uniformly convergent on $[\delta, 2 \pi - \delta]$, and the sum is uniformly bounded on $\mathbb{R}$, say by $M$. How does one prove this?
This is my attempt, is it right?: As $|f|$ is absolutely integrable, by change of variables $u=x+\zeta$, $$ |G_m(\zeta)| \le M \Big[ \int_{\delta + \zeta}^{2 \pi -\delta + \zeta} |f(u)| \Big| \sum_{m+1}^\infty \frac{ \sin ku }{k} \Big| + \int_{\delta + \zeta}^{2 \pi -\delta + \zeta} |f(u)| \Big| \sum_{m+1}^\infty \frac{ \cos ku }{k} \Big| \, du \Big] $$ It only suffice to consider first integral. Wlog, suppose $2 \pi \in (\delta+ \zeta, 2 \pi - \delta + \zeta)$. By Absolute Integrability, exists $\delta'>0$ such that we may split the integral into three parts,
$$ \int_{\delta+\zeta}^{ 2 \pi - \delta'} + \int_{2 \pi - \delta'}^{ 2 \pi + \delta'} + \int_{2 \pi + \delta'}^{2 \pi - \delta + \zeta} |f(u)| \Big| \sum_{m+1}^\infty \frac{ \sin ku }{k} \Big| \, du \le M \varepsilon + \int_{\delta'}^{2 \pi - \delta'} |f(u)| \Big| \sum_{m+1}^\infty \frac{ \sin ku }{k} \Big| \, du $$ where the last summation limits to $0$ due to uniform convergence of zero, and $|f|$ being bounded.
First, convince yourself that if $(a,b) \subseteq \mathbb{R}$ is any interval of length smaller than $2\pi$, i.e. $b-a \leq 2 \pi$, then we have $$\int_{a}^{b} |f| ~\mathrm{d}x \leq \int_{0}^{2\pi} |f| ~ \mathrm{d}x =: I$$ due to the periodicity of $f$. Because of the uniform convergence, for any $\epsilon > 0$ you can find a $M \in \mathbb{N}$ such that $$\left| \sum_{k=m+1}^{\infty} \frac{\sin(kx)}{x}\right| < \frac{\epsilon}{I}, \quad x \in [\delta, 2\pi - \delta], \quad m \geq M.$$ We find that $$|G_m (\zeta)| \leq \frac{\epsilon}{I}\int_{\delta}^{2\pi - \delta}|f(x+\zeta)| ~ \mathrm{d}x = \frac{\epsilon}{I}\int_{\zeta + \delta}^{\zeta + 2\pi - \delta} |f(y)|~\mathrm{d}y \leq \frac{\epsilon}{I}I = \epsilon $$ The last inequality holds because the length of the integration interval is smaller than $2\pi$.