Recently, I have tried to resolve the following problem.
Let periodic $f$ be defined by the following Fourier series, \begin{align} f(x) = \sum_{k=1}^\infty \frac{1}{k^{3/2}} e^{ikx}. \end{align} Prove that $f$ belongs to Nikol'skii space $H^1_2(-\pi,\pi)$ but \begin{align} \frac{1}{2\pi}\int_{-\pi}^\pi \vert f(x+h) - f(x) \vert^2 dx \geq \frac{4h^2}{\pi^2}\log \frac{\pi}{\vert h \vert} \end{align} for $0<\vert h \vert <1$. It is not that hard to prove the first claim, since $f$ is absolutely convergent, $f\in L^1(-\pi,\pi)$; hence, we can integrate the series term by term. This yields that \begin{align} c_n(f)=\frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=1}^\infty \frac{e^{-ix(n-k)}}{k^{3/2}}dx = \frac{1}{2\pi} \sum_{k=1}^\infty \frac{1}{k^{3/2}} \int_{-\pi}^\pi e^{-ix(n-k)} dx = \frac{1}{n^{3/2}} \end{align} if $n=k$. That is, $c_n(f)=0$ for $n\leq0$. Thus, we have for all $0\leq j \in \mathbb{Z}$, \begin{align} \sum_{2^j\leq \vert n \vert < 2^{j+1}} \vert n \vert^2 \vert c_n(f) \vert^2 = \sum_{2^j\leq n < 2^{j+1}} \frac{n^2}{n^3} \leq 2^j2^{-j}< \infty, \end{align} which implies that $f\in H^1_2(-\pi,\pi)$. For any $0<\alpha<1$, $f\in H^\alpha_2(-\pi,\pi)$ is equivalent to $L^2$-Holder condition with $\alpha$. But $\alpha=1$ in this case, so I have tried to get a certain lower bound of the following with sufficiently small $\vert h \vert$. Using Parseval equality, \begin{align} \frac{1}{2\pi}\int_{-\pi}^\pi \vert f(x+h) - f(x) \vert^2 dx = \sum_{n=1}^\infty \frac{\vert e^{inh}-1 \vert^2}{n^3} = 4\sum_{n=1}^\infty \frac{\sin^2(nh/2)}{n^3}\geq \frac{4h^2}{\pi^2}\sum_{n=1}^N \frac{1}{n^3} \end{align} where $N$ satisfying $0<\vert h \vert < N\vert h \vert < \pi$ since $\vert \sin(nh/2) \vert \geq \vert h \vert / \pi$ for all $1\leq n\leq N$. After getting this far, I stopped for a bit. How can we derive the logarithmic term?
Now I have fully resolved the OP. It was my simple mistake.
Since $\vert \sin(nh/2) \vert \geq (2/\pi)(n\vert h\vert /2)$ for all $n$ satisfying $n\vert h \vert \leq \pi$ due to the concavity of $\sin(z)$ on $z\in[0,\pi/2]$, then we have \begin{align} \frac{1}{2\pi}\int_{-\pi}^\pi \vert f(x+h) - f(x) \vert^2 dx \geq \frac{4h^2}{\pi^2}\sum_{n\vert h \vert \leq \pi} \frac{1}{n}. \end{align}
There is $0<N\in\mathbb{Z}$ such that $N \vert h \vert \leq \pi < (N+1)\vert h \vert$ by Archimedean property, and using Riemann-sum implies that \begin{align} \sum_{n=1}^N \frac{1}{n} \geq \log(N+1) \geq \log\frac{\pi}{\vert h \vert}, \end{align} for all $N>0$ since $1/n$ is monotonic decreasing. This concludes the proof.