When I read some paper, it says
$u\in H^{3/2}$ if and only if $u$ is absolutely continuous with $u'\in H^{1/2}$, where $$H^s=\left\{ u\in L^1(\mathbb{S}^1):\|u\|_{H^{s}} \doteq\left(\sum_{n=-\infty}^{+\infty}|n|^{2 s}\left|a_{n}(u)\right|^{2}\right)^{\frac{1}{2}} \right\}, a_{n}(u)=\frac{1}{2 \pi} \int_{0}^{2 \pi} {u}(e^{i\theta}) e^{-i n \theta} d \theta.$$
"If part" is easy to understand, but how to prove if $u\in H^{3/2}$, $u$ is absolutely continuous?
Any hint will be appreciated!
The absolute continuous part of the assertion is somewhat useless, it is just here to say that u' makes sense for people who do not know distribution theory.
Nevertheless your question boils down to the fact that $H^{3/2}$ is imbedded into absolutely continuous functions. The proof depends on how you define $H^{3/2}$. If for example you follow the definition with Fourier coefficients, then you can check that $v=\sum ina_ne^{inx}$ is an $L^2$ function (this is true even in $H^1$), and that actually $u-\int_0^x v$ is a constant. This last point is due to the fact that for any smooth periodic $\phi$, $\int_0^{2\pi}(u-\int_0^x v)\phi'dx=0$. (corrected after the comment)
Now obviously $\int_0^x v$ is absolutely continuous, as it is the integral of an $L^2$ (hence $L^1$) function. Note that this argument works for any $H^s$ function, $s\geq 1$.
There is maybe a simpler argument, but this is not trivial as you can only embed $H^{3/2}$ into $C^{0,\alpha},\ \alpha<1$, and unfortunately these spaces are not comparable to absolutely continuous functions.