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Let $\Omega \subset \mathbb R^n$ be a domain with a smooth boundary $\partial \Omega$. I know that for $s > 1/2$, one can define a (zeroth order) trace operator $$ \gamma \colon H^s(\Omega) \to H^{s-1/2}(\partial \Omega) \subset L^2(\partial \Omega)\text.$$ Here, I wrote $H^s$ to denote the fractional Sobolev space.

Q: Is it possible to extend this definition to the case $s < 1/2$? The image would then consist of distributions.

In particular, let's choose $s = 0$. Is there a trace operator $\gamma \colon L^2(\Omega) \to H^{-1/2}(\partial \Omega)$? That is to say, is it possible to make sense of the expression $$ \int_{\partial \Omega} v w \quad \text{with $v \in L^2(\Omega)$, $w \in H^1(\Omega)$?}$$

anonymous
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    I doubt that this is possible. The easiest thing would be to show that $v \in C(\bar \Omega) \mapsto \int_{\partial\Omega} v , w$ is not continuous w.r.t. the $L^2$-norm of $v$. – gerw May 03 '13 at 19:46

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I agree with gerw's comment: this does not look possible. For example, the constant function $w\equiv 1$ on the unit ball $D\subset \mathbb R^n$ is in $H^1(D)$ (and of course in all $H^m(D)$). So, for the trace operator to be defined, we would need to make sense out of $\int_{\partial D} v$ for $v\in L^2(D)$. But for functions such as $v(x)=\sin \left(1/(1-|x|)\right)$ there is no usable notion of $\int_{\partial D} v$.

75064
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    For $s<\frac{1}{2}$, $C_c^\infty(\Omega)$ is dense in $H^s(\Omega)$ even when $\Omega$ is smooth bounded domain. So, constant functions are approximated by test functions. Hence trace operator cannot exist. – Kaushik Dec 29 '19 at 19:15