Why is that not possible (if it is not) to define the trace of a function in a very weak regularity space $H^{-s}(\partial \Omega)$? We usually encounter trace theorem as

$$Tr: H^{s(\Omega)} \rightarrow H^{s- 1/2}(\partial \Omega)$$

$$ || Tr \, (u) ||_{s - 1/2, \partial \Omega} \leq C || u ||_{s,\Omega} \,,$$

which is good for s>1/2 with suitable regularity of $\partial \Omega$, e.g. s< 3/2 if Liptchitz.)

However, I always wonder, if I only aim for a very weak functional, and not expect a well-defined "function", can we just extend the definition of trace of $g$ in $H^{-s}(\partial \Omega)$ as follows?

$$\langle \frac{\partial f }{\partial \nu} , Tr (g) \rangle_{\partial \Omega} := \langle \nabla f, \nabla g \rangle_{\Omega} + \langle \Delta f, g \rangle_{\Omega}$$

where now the regularity of test functions $f$ shall be relative high, e.g. $H^{s+2}_0(\Omega)$.

Is there some issue that makes this definition pathological?

Why is that not possible (if it is not) to define the trace of a function in a very weak regularity space $H^{-s}(\partial \Omega)$? We usually encounter trace theorem as

$$Tr: H^{s}(\Omega) \rightarrow H^{s- 1/2}(\partial \Omega)$$

$$ \| Tr \, (u) \|_{s - 1/2, \partial \Omega} \leq C \| u \|_{s,\Omega} \,,$$

which is good for $s>1/2$ with suitable regularity of $\partial \Omega$, e.g. $s< 3/2$ if Lipschitz.)

However, I always wonder, if I only aim for a very weak functional, and not expect a well-defined "function", can we just extend the definition of trace of $g$ in $H^{-s}(\partial \Omega)$ as follows?

$$\bigl\langle \frac{\partial f }{\partial \nu} , Tr (g) \bigr\rangle_{\partial \Omega} := \langle \nabla f, \nabla g \rangle_{\Omega} + \langle \Delta f, g \rangle_{\Omega}$$

where now the regularity of test functions $f$ shall be relative high, e.g. $H^{s+2}_0(\Omega)$.

Is there some issue that makes this definition pathological?

I have a stupid question:

Why is that not possible (if it is not) to define the trace of a function in a very weak regularity space $H^{-s}(\partial \Omega)$? We usually encounter trace theorem as

$$Tr: H^{s(\Omega)} \rightarrow H^{s- 1/2}(\partial \Omega)$$

$$ || Tr \, (u) ||_{s - 1/2, \partial \Omega} \leq C || u ||_{s,\Omega} \,,$$

which is good for s>1/2 with suitable regularity of $\partial \Omega$, e.g. s< 3/2 if Liptchitz.)

However, I always wonder, if I only aim for a very weak functional, and not expect a well-defined "function", can we just extend the definition of trace of $g$ in $H^{-s}(\partial \Omega)$ as follows?

$$\langle \frac{\partial f }{\partial \nu} , Tr (g) \rangle_{\partial \Omega} := \langle \nabla f, \nabla g \rangle_{\Omega} + \langle \Delta f, g \rangle_{\Omega}$$

where now the regularity of test functions $f$ shall be relative high, e.g. $H^{s+2}_0(\Omega)$.

Is there some issue that makes this definition pathological?

Why is that not possible (if it is not) to define the trace of a function in a very weak regularity space $H^{-s}(\partial \Omega)$? We usually encounter trace theorem as

$$Tr: H^{s(\Omega)} \rightarrow H^{s- 1/2}(\partial \Omega)$$

$$ || Tr \, (u) ||_{s - 1/2, \partial \Omega} \leq C || u ||_{s,\Omega} \,,$$

which is good for s>1/2 with suitable regularity of $\partial \Omega$, e.g. s< 3/2 if Liptchitz.)

However, I always wonder, if I only aim for a very weak functional, and not expect a well-defined "function", can we just extend the definition of trace of $g$ in $H^{-s}(\partial \Omega)$ as follows?

$$\langle \frac{\partial f }{\partial \nu} , Tr (g) \rangle_{\partial \Omega} := \langle \nabla f, \nabla g \rangle_{\Omega} + \langle \Delta f, g \rangle_{\Omega}$$

where now the regularity of test functions $f$ shall be relative high, e.g. $H^{s+2}_0(\Omega)$.

Is there some issue that makes this definition pathological?