Let $H^1(\mathbb{T})=\{f\in\mathcal{D}'(\mathbb{T}): \sum_{-\infty}^{\infty}n^{2}|c_n|^2\}<\infty$. If $f\in H^1,$ do we have $f\in L^2(-\pi, \pi)$? I know some books explicitly define $H^k$ to be a subspace of $L^2(\mathbb{T})$, like in this Wikipedia article(https://en.wikipedia.org/wiki/Sobolev_space), $H^k({\mathbb T}) = \left \{ f\in L^2({\mathbb T}) : \sum_{n=-\infty}^\infty \left (1+n^2 + n^4 + \dots + n^{2k} \right ) \left |\widehat{f}(n) \right |^2 < \infty \right \}$. But the book I'm reading uses the definition as above, and still treats it as a subset of $L^2(\mathbb{T})$. How can I prove this?
I would greatly appreciate any comments or suggestions.
You may consider $L^2(\mathbb{T})$ to be the set of functions $f(x)=\sum_{n=-\infty}^{\infty}c_ne^{inx}$ for which $\sum_{n=-\infty}^{\infty}|c_n|^2 < \infty$. This is because $\{ e^{inx}\}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^2(\mathbb{T})$. So, in that case, it is possible to consider $H^1(\mathbb{R})$ to be a linear subspace of $L^2(\mathbb{T})$ because $f(x)=\sum_{n}c_ne^{inx}$ converges in $L^2(\mathbb{R})$ if $\sum_{n}n^2|c_n|^2 < \infty$. For higher order derivatives you don't need all the powers of $n$ in the sum, only the highest power because convergence with respect to the higher powers always forces the convergence with respect to lower powers.