Let $\mathbb T=\mathbb R/(2\pi\mathbb Z)$ and define the positive <self-adjoint operator>
$$
B=(1-\partial_x^2)^{1/2}
$$
on $L^2(\mathbb T)$. On the Fourier mode $e^{inx}$ it acts by multiplication by $\sqrt{1+n^2}$. Consequently
$$
D(B)=H^1(\mathbb T),
\qquad
D(B^2)=H^2(\mathbb T).
$$
The energy space is the <periodic Sobolev space>
$$
\boxed{\mathcal H=H^1(\mathbb T)\times L^2(\mathbb T)},
$$
with inner product
$$
((u,v),(p,q))_{\mathcal H}
=(Bu,Bp)_{L^2}+(v,q)_{L^2}.
$$
If $u=\sum u_ne^{inx}$ and $v=\sum v_ne^{inx}$, then
$$
\|(u,v)\|_{\mathcal H}^2
=2\pi\sum_{n\in\mathbb Z}
[(1+n^2)|u_n|^2+|v_n|^2],
$$
which is precisely the stated energy norm.

With $v=u_t$, the periodic <Klein-Gordon equation> becomes
$$
\dot Z=AZ,
\qquad
A=\begin{pmatrix}0&I\\-B^2&0\end{pmatrix},
\qquad
A(u,v)=(v,-B^2u).
$$
For $AZ$ to belong to $H^1\times L^2$, one needs $v\in H^1$ and $u\in H^2$. Thus
$$
\boxed{D(A)=H^2(\mathbb T)\times H^1(\mathbb T)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 319 / 1 / c / Solution

{title2=$H^s(\mathbb T^d)$}

The periodic Sobolev space $H^s(\mathbb T^d)$ consists of periodic distributions whose Fourier coefficients obey
$$
\sum_{n\in\mathbb Z^d}(1+|n|^2)^s|\widehat u_n|^2<\infty.
$$
In one dimension, $H^1(\mathbb T)$ embeds continuously into $L^\infty(\mathbb T)$ and is a Banach algebra.

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 Incoming links: Periodic Sobolev space