Double integral definition of periodic Sobolev spaces Preliminaries: For $s \geq 0$ define the periodic Sobolev spaces $H^s(\mathbb T)$ with norm
$$
\| f \|_{H^s(\mathbb T)}^2 = \sum_{k \in \mathbb Z} \big(1 + |k|^2 \big)^s \, \big|\widehat{f}_k \big|^2,
$$
where
$$
\widehat{f}_k = \frac{1}{\sqrt{2\pi}} \int_{\mathbb T} f(x) \, e^{-ikx} \, dx
$$
are the Fourier coefficients.
Question: In "For which $s\in\mathbb R$, is $H^s(\mathbb T)$ a Banach algebra?"  one answer @Normal Human stated that

the norm of  $H^s(\mathbb T)$ has an equivalent expression
  $$ \|f\|_{H^s(\mathbb T)}^2 \approx \iint_{\mathbb T\times\mathbb
 T}\frac{|f(x)-f(y)|^2}{|x-y|^{1+2s}}\,dx\,dy
$$

I would like to see references or some argument for the validity of this expression.
 A: As pointed out in my earlier comment, we need to assume $s\in (0,1)$. Out of habit, I'm going using the convention $\mathbb{T}\cong[0,1)$. You can modified everything to your own convention.
The expression
$$[f]_{W^{2,s}(\mathbb{T})}=\left(\iint_{\mathbb{T}\times\mathbb{T}}\dfrac{\left|f(x)-f(y)\right|^{2}}{\left|x-y\right|^{1+2s}}dxdy\right)^{1/2}$$
sometimes called the Gagliardo seminorm, isn't a norm as its name would indicate--$[f]_{W^{2,s}}=0$ implies $f$ is constant. A norm would be
$$\left\|f\right\|_{W^{2,s}(\mathbb{T})}:=\left(\left\|f\right\|_{L^{2}(\mathbb{T})}^{2}+[f]_{W^{2,s}(\mathbb{T})}^{2}\right)^{1/2}$$
and what you want to show is that $\left\|f\right\|_{W^{2,s}(\mathbb{T})}$ and $\left\|f\right\|_{H^{s}(\mathbb{T})}$ are equivalent. More generally, one approach to fractional Sobolev spaces $W^{s,p}(\Omega)$ is to define the seminorm
$$[f]_{W^{s,p}(\Omega)}:=\left(\iint_{\Omega\times\Omega}\dfrac{\left|f(x)-f(y)\right|^{p}}{\left|x-y\right|^{n+sp}}dxdy\right)^{1/p}$$
where $s\in (0,1)$, $1\leq p<\infty$, and $\Omega\subset\mathbb{R}^{n}$. One defines $W^{s,p}(\Omega)$ by
$$W^{s,p}(\Omega):=\left\{f\in L^{p}(\Omega):[f]_{W^{s,p}(\Omega)}<\infty\right\}$$
and equips this space with the norm
$$\left\|f\right\|_{W^{s,p}(\Omega)}:=\left(\left\|f\right\|_{L^{p}}^{p}+[f]_{W^{s,p}}^{p}\right)^{1/p}$$
For $s=m+\sigma\geq 1$, where $m=[s]$ and $\sigma=s-m$, we define $W^{s,p}(\Omega)$ by
$$\left\{f\in W^{m,p}(\Omega) : D^{\alpha}f\in W^{\sigma,p}(\Omega) \forall\left|\alpha\right|=m\right\}$$
and equip with norm
$$\left\|f\right\|_{W^{s,p}(\Omega)}:=\left(\left\|f\right\|_{W^{m,p}}^{p}+\sum_{\left|\alpha\right|=m}\left\|D^{\alpha}f\right\|_{W^{\sigma,p}}^{p}\right)^{1/p}$$
See the paper "A Hitchhiker's Guide to Fractional Sobolev Spaces" for more details.

Write
$$\iint_{\mathbb{T}\times\mathbb{T}}\dfrac{\left|f(x)-f(y)\right|^{2}}{\left|x-y\right|^{1+2s}}dxdy=\iint_{\mathbb{T}\times\mathbb{T}}\dfrac{\left|f(x)-f(x-h)\right|^{2}}{\left|h\right|^{1+2s}}dxdh \tag{1}$$
By Parseval's identity, the RHS in (1) equals
$$\int_{\mathbb{T}}\left|h\right|^{-1-2s}\sum_{k\in\mathbb{Z}}\left|\widehat{f}(k)\right|^{2}\left|e^{2\pi i hk}-1\right|^{2}dh\tag{2}$$
Observe that by standard trigonometric identities,
$$\left|e^{2\pi i hk}-1\right|^{2}=(\cos(2\pi hk)-1)^{2}+\sin^{2}(2\pi hk)=2\left[1-\cos(2\pi hk)\right]$$
By Taylor's theorem,
$$\dfrac{1-\cos y}{\left|y\right|^{1+2s}}\leq C\dfrac{1}{\left|y\right|^{-1+2s}},\qquad\forall y\in (0,1]$$
where $C>0$ is some constant. Since $0<2s<2$, $\left|y\right|^{1-2s}$ is integrable in a neighborhood of the origin; whence, $\left|y\right|^{1-2s}(1-\cos y)$ is integrable in a neighborhood of the origin. By the monotone convergence, (2) equals
$$\sum_{k\in\mathbb{Z}}\left|\widehat{f}(k)\right|^{2}\int_{0}^{1}\dfrac{1-\cos(2\pi hk)}{\left|h\right|^{1+2s}}dh=\sum_{k\in\mathbb{Z}}\left|\widehat{f}(k)\right|^{2}\left|k\right|^{2s}c_{k}\tag{3}$$
Making the change of variable $hk\mapsto h$, we see that
$$c_{k}=\int_{0}^{|k|}\dfrac{1-\cos(2\pi h)}{\left|h\right|^{1+2s}}dh\leq\int_{0}^{1}\dfrac{1-\cos(2\pi h)}{\left|h\right|^{1+2s}}dh+2\int_{\mathbb{R}}\dfrac{1}{(1+\left|h\right|)^{1+2s}}dh=:C_{1}$$
Since $1-\cos(2\pi h)\geq 0$ for all $h$ and strictly positive on an open set, we see that $c_{k}\geq c_{1}>0$ for all $k \neq 0$.
I think you can fill in the rest of the details.
