The Bessel Potential Space is defined for $s\in\mathbb{R}$ as,


$H^s(\mathbb{R}^d) = \{f\in L\_2(\mathbb{R}^n) : (1+|\cdot|)^{s/2}\hat{f}(\cdot)\in L\_2(\mathbb{R}^n)\}.
$


This defines a Hilbert space such that for any $f,g\in H^s(\mathbb{R}^n)$, 


$
\langle f, g\rangle = \int\_{\mathbb{R}^n} \hat{f}(\omega)\overline{\hat{g}(\omega)} (1+|\omega|)^{s}d\omega.
$


For any open set $\Omega\subset\mathbb{R}^n$ we have $H^s(\Omega)$ being the set of restrictions with norm,


$
\left\|f\right\|\_{H^s(\Omega)} = 
$


$
\inf\_{g\in H^s(\mathbb{R}^n)}\{\left\|g\right\|\_{H^s(\mathbb{R}^n)} : g|\Omega=f \}
$


Does this definition of the norm ensure we have the following: Given an open set 
$\Omega\subset \mathbb{R}^n$ and open sets $\Omega\_1, \Omega\_2\subset \mathbb{R}^n$ such that $\Omega = \Omega\_1\cup \Omega\_2$ and $f\in H^s(\Omega)$,


$
\left\|f\right\|^2\_{H^s(\Omega)}\leq \left\|f\right\|^2\_{H^s(\Omega\_1)} + \left\|f\right\|^2\_{H^s(\Omega\_2)}.
$



Bessel Potential Space inequality
The answer is negative anyway. Take $\mathbb R=(-\infty,a)\cup(-a,+\infty)$ with small $a>0$. Take $f=e^{-|x|}$. Then $\widehat f(y)\approx \frac 1{1+y^2}$. Now take $s=3-\delta$. $\|f\|\_{H^s(\mathbb R)}$ is huge if $\delta$ is small. On the other hand, we can expand $f$ from $(-\infty,0)$ to a Schwartz function $g$. To make it into an extension from $(-\infty,a)$, we can just create a small "triangular dip" in $g$ with base and height of order $a$. This dip tends to $0$ in $H^s(\mathbb R)$ as $a\to 0$ (its Fourier transform is bounded by $C\min(a^2,y^{-2})$). Thus, we can get an extension of the norm dominated by the norm of $g$ in $H^3$ for each half.


Side note: most people I know will denote your $H^s$ by $H^{s/2}$ (having in mind that $s$ is the "number of derivatives"). :)


