I have a query about Fourier series.
Let $f(x)\sim \sum_{n=-\infty}^\infty c_n e^{inx}$ where $x\in[-\pi,\pi]$ and assume that this series converges absolutely. Define the tail sum by $E_n = \sum_{\vert k \vert > n} \vert c_k \vert$ for $n\in\mathbb{Z}_0^+$. Then we have the following
$$ \vert f(x+h) - f(x) \vert \leq 2E_{N_h} - hN_hE_{N_h} + h\sum_{n=1}^{N_h} E_{n-1} \leq 2E_{N_h} + \frac{1}{N_h}\sum_{n=0}^{N_h-1}E_n, $$
where $N_h=[1/h]$, $h>0$, and $[x]$ denotes the whole part of $x$. I am having difficulty proving the following.
If $E_n=\mathcal{O}(n^{-\alpha})$ for some $0<\alpha \leq 1$, then $\omega_h(f)=\mathcal{O}(h^\alpha)$ for $0<\alpha<1$ and $\omega_h(f)=\mathcal{O}(h\log(1/h))$ for $\alpha=1$. Showing the latter one is
$$ \omega_h(f) \leq 2E_{N_h} + \frac{1}{N_h}\sum_{n=0}^{N_h-1}E_n \leq C_0h + hE_0 + \int_1^{1/h} \frac{h}{x} dx = \mathcal{O}(h\log{\frac{1}{h}}) $$
for a sufficiently small $h\leq 1/e$ and $E_0$ is considered as $\mathcal{O}(1)$. But when I tried to figure out the former case $0<\alpha<1$, a term like $C(h^\alpha-h)/(1-\alpha)$ appears. How can I remove $h$?
I post an answer to this own question, because I made a simple mistake. If $h<1$, then $h^\alpha \geq h$ for $0<\alpha<1$. Therefore, we have
$$ \omega_h(f) \leq 2E_{N_h}+h\sum_{n=0}^{N_h-1}E_n \leq C_0h^\alpha + C_1h + \int_1^{1/h} \frac{h}{x^\alpha}dx = \mathcal{O}(h^\alpha) $$
for a sufficiently small $h<1$.