Excellent answer. By the way, counterexample is a one word...

This comment is very useful to me. When I numerically-tested singular integrals, I used to isolate the singularity manually, then manually decrease the size of the isolated neighborhood. Thank you.

Thanks a lot. I appreciate the time and effort you put into making the answer so clear.

Thank you, but I am afraid there is a significant factor of $|\vec{r}|^2$ missing in your computation.

@Jeff. Okay. Let us now connect the two half balls by the small ball $B(0; \epsilon )$ as you suggested. Will $x\mapsto sing (x_{1})$ still be a $W^{1,p}$ function on the connected domain ?

@Jeff. I am confused. On Jan 16 at 22:40, you commented "Either half has a Lipschitz boundary and the trace is well-defined for either half considered separately". Did you mean to say neither "Neither half has a Lipschitz boundary" ?

@Jeff. So, we still do not have an example where $f\in W^{1,p}(\Omega_{i})$, $i=1,2$, $\Omega_{1}$ is a domain not Lipschitz and $f$ does not have trace on $\partial \Omega_{1}$, while $\Omega_{2}$ is a Lipschitz domain and $f$ does have trace on $\Omega_{2}$. This will show that Lipschitz condition on the boundary is necessary for a $W^{1,p}$ function to have trace. Notice that the domain in your previous examples are Lipschitz. Thanks