Bessel Potential Space inequality

2011-07-25T13:02:19Z

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)}. $

Answer by fedja for Bessel Potential Space inequality

2011-07-25T13:45:12Z

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"). :)

Bessel Potential Space inequality - MathOverflow

Bessel Potential Space inequality

Answer by fedja for Bessel Potential Space inequality