Let $f\in C_c^\infty((0,1)^n)$, i.e., $f$ is smooth and supported in the unit cube $(0,1)^n\subset\mathbb{R}^n$. Then $f$ can be regarded as a function on $\mathbb{R}^n$ which vanishes outside $(0,1)^n$, with Sobolev norm $$\|f\|_{H^s(\mathbb{R}^n)}:=\bigg(\int(1+|\xi|^2)^s|\hat{f}(\xi)|^2\,\mathrm{d}\xi\bigg)^{1/2}.$$ Similarly, it is also a function on $\mathbb{T}^n$, with Sobolev norm $$\|f\|_{H^s(\mathbb{T}^n)}:=\bigg(\sum_{\xi\in\mathbb{Z}^n}(1+|\xi|^2)^s|\hat{f}(\xi)|^2\bigg)^{1/2}.$$
If $s\in\mathbb{N}$, then obviously $\|f\|_{H^s(\mathbb{R}^n)}=\|f\|_{H^s(\mathbb{T}^n)}$.
Question: If $s\in\mathbb{R}$, is there a constant $C=C(s)>0$ such that $C^{-1}\|f\|_{H^s(\mathbb{T}^n)}\leq\|f\|_{H^s(\mathbb{R}^n)}\leq C\|f\|_{H^s(\mathbb{T}^n)}$ for all $f\in C_c^\infty((0,1)^n)$?
I think it is true. An argument which I could not make rigorous is the following: By "duality" $\|f\|_{H^s(\mathbb{R}^n)}=\|f\|_{H^s(\mathbb{T}^n)}$ for $s\in-\mathbb{N}$. By "interpolation" these norms are equivalent. But what justifies this argument? And is there any direct proof of this fact?