For any $s\in \mathbb{R}$ define the Hilbert space $H^s(\mathbb{T})$ by means of norm $$\|f\|^2_{H^s}=|\widehat{f}(0)|^2+\sum_{n\in\mathbb{Z}}|n|^{2s}|\widehat{f}(n)|^2.$$
Show that for any $0\leq s\leq 1$ one has $\|f(\cdot +\theta)-f\|_2\leq 2\pi \|f\|_{H^s}|\theta|^s$. How would you prove this estiamte, I tried Prseval formula but it is not working. How can I form the constant $2\pi$ frpom my original equation.
Your third comment is on the right track, but you miss $n$ in the exponent: you can reduce the left hand side of the inequality to $$\|(e^{-2\pi i n\theta }-1)\widehat{f}(n)\|^2_{l^2}$$ To be more explicit, the desired inequality can be stated as $$ \sum_{n\in\mathbb Z} |e^{-2\pi i n\theta }-1|^2 |\hat f(n)|^2 \le (2\pi)^2 |\theta|^{2s} \sum_{n\in\mathbb Z} \max(|n|^{2s},1) |\hat f(n)|^2 \tag{1}$$
After looking at (1) for a while, one realizes that it either holds elementwise, or fails. Namely, if $$|e^{-2\pi i n\theta }-1|^2 \le (2\pi)^2 |\theta|^{2s} \max(|n|^{2s},1),\quad \forall n\in\mathbb Z \tag{2}$$ then (1) is true. And if (2) fails for some $n$, then (1) fails, because the function $t\mapsto e^{2\pi i nt}$ is a counterexample.
So... is (2) true? For $n=0$ it is, so we can assume $n\ne 0$ and simplify (2) to $$|e^{-2\pi i n\theta }-1| \le 2\pi |n\theta|^{ s} \quad \forall n\in\mathbb Z\setminus\{0\} \tag{3}$$
The function $t\mapsto e^{it}$ is $1$-Lipschitz, since its derivative has unit modulus. Hence, $|e^{-2\pi i n\theta}-1|\le 2\pi |n \theta|$. This implies (3) in the case $|n\theta|\le 1$. And if $|n\theta|>1$, then the right hand side of (3) is greater than $2\pi$, while the left hand side is at most $2$.