Let $f$ be an absolutely integrable function on $(0,2 \pi)$. We have $F(x) := \int_0^x f(t) \, dt$, $$F(x)= \frac{\alpha_0}{2} x + \sum_{k=1}^\infty \frac{\alpha_k \sin k x + \beta_k ( 1 - \cos k x ) }{k}. $$ The last series is uniformly convergent throughout. (Rogosinski, Fourier Series p29)
Proof:
The text first shows the equality: \begin{align*} \sum_{k=1}^\infty \frac{\beta_k(\zeta)}{k} &= \frac{1}{2 \pi } \int_0^{2 \pi} f(x+\zeta) ( \pi - x ) \, dx \\ &= \frac{1}{2 \pi} \int_\zeta^{2 \pi + \zeta} f(x) (\pi -x + \zeta ) \, dx \\ &= \frac{\zeta \alpha_0}{2} + \sum_{k=1}^\infty \frac{\beta_k}{k} - F(\zeta) \end{align*}
Form uniform continuity it seems like we have to use the equality (the integrand being the same):
\begin{align*} \sum_{k=1}^m \frac{\beta_k(\zeta)}{k} - \int_{0}^{2 \pi} f(x+ \zeta) (\pi -x ) \, dx &= \Big(\frac{1}{\pi} \int_\delta^{2\pi -\delta} \Big) + \Big[ \frac{1}{\pi} \int_{0}^\delta + \frac{1}{\pi} \int_{2 \pi - \delta}^{ 2 \pi } f(x+\zeta ) \Big( \sum_{k=m+1}^\infty \frac{\sin kx}{k} \Big) \, dx \Big] \\ &= I_1 + I_2 \end{align*}
What I don't understand is how they get uniform continuity in $\zeta$
I already know the series $\sum_{1}^\infty \frac{\sin kx }{ k }$ is uniformly convergent everywhere. How do I bound $I_1$ and $I_2$?
