Let $\Omega:=B(0,1)\subset \mathbb R^N$ the unite ball and $N\geq 2$. Given $u\in H^1(\Omega)$. Then the trace $T[u]$ is well defined over $\partial \Omega$. (by $H^1$ I mean $W^{1,2}$ space)
Now let's consider the domain $\Omega':=B(0,1)\setminus \overline{B(0,1/2)}$. Let us use polar coordinate for $x=(\theta,r)$. Then I define a new function $v$ over $\Omega'$ by $$ v(x):=T[u](x') $$ where $x'=(\theta,1)$ for $x=(\theta,r)\in \Omega'$.That is, $v(x)=T[u](x')$ for $x'=x/|x|$, $x\in \Omega'$.
My question: what kind of function $v$ is over domain $\Omega'$? Is it a $H_1(\Omega')$ function?
Given a general $H^1$ function, it would be at best a $H^{\frac{1}{2}}$ function, consider the case where $T[u]\in H^{\frac{1}{2}}(\partial\Omega)\setminus H^1(\partial\Omega)$, and $N=2$. The function you have defined can be represented in polar coordinates by a function $w(r,\theta)=v(x)$, where $r=|x|$ and $\theta=\arctan\frac{x_2}{x_1}$. Notice that it is constant in $r$, and thus we see that $$\|v\|_{H^1(\Omega')}^2=\int_0^{2\pi}\int_{1/2}^1w^2r+w_\theta^2r\,dr\,d\theta$$ $$=\int_{1/2}^1r\,dr\int_0^{2\pi}w^2+w_\theta^2\,d\theta$$ $$=\frac{1}{4}\int_0^{2\pi}w^2+w_\theta^2\,d\theta.$$ We know that $w$ coincides with $T[u]$, and also the derivative with respect to theta is the tangential derivative on $\partial\Omega'$, thus we obtain $$\|v\|_{H^1(\Omega')}^2=\frac{1}{4}\|T[u]\|_{H^1(\partial\Omega)}=\infty.$$