double a Sobolev function along the boundary of the domain Let $u$ be a $W^{1,p}$-Sobolev function defined on the half disk $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$.
If I define a function $\tilde u$ by
$$ \tilde u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}, $$
is this $\tilde u$ a Sobolev $W^{1,p}$-function on the whole unit disk?
It seems obvious to me that if $\tilde u$ does have generalized derivatives, these have to be simply $\partial_x \tilde u (x,y) = \partial_x u (x,|y|)$ and $\partial_y \tilde u(x,y) = \partial_y u(x,y)$ if $y >0$ and $\partial_y \tilde u(x,y) = - \partial_y u(x,-y)$ if $y <0$.  These functions are all $L^p$, so the only question that is left is whether these really are the generalized derivatives of $\tilde u$ and how to check this.
Thank you for any help.
 A: Since no one answers, I write some ideas that can be useful.
I prove the analogous problem in $\mathbb{R}$ ( the same idea can be use in your problem ).
I want use that 
\begin{equation}u\in W^{1,1}((a,b)) \text{ iff } u\in AC((a,b))\\ \text{
(In the sense that there is an absolutely continuous representative function)}\end{equation}

Thm1:
  Let $u\in W^{1,1}((0,1))$ and $\bar{u}(x):=\begin{cases}u(x)\quad
> x\in(0,1)\\ u(-x)\quad x\in(-1,0)\end{cases}$
Then $\bar{u}\in W^{1,1}((-1,1))$.

Proof: \begin{equation}u\in W^{1,1}((0,1)),\ \text{ then }\ u\in AC((0,1))\end{equation}
For how $\bar{u}$ is defined, we have that
\begin{equation}\bar{u}\in AC((-1,1))\end{equation}
Then $\bar{u}\in W^{1,1}((-1,1))$
Indeed, since $\bar{u}$ is absolutly continuous, we can use integration by parts and we obtain
\begin{equation}\bar{u}'(x)=\begin{cases}u'(x)\quad x\in(0,1)\\ -u'(-x)\quad x\in(-1,0).\end{cases}\end{equation}
Now, using Thm1, it's easy prove that

Thm2:
  Let $u\in W^{1,p}((0,1))$ and $\bar{u}(x):=\begin{cases}u(x)\quad
> x\in(0,1)\\ u(-x)\quad x\in(-1,0)\end{cases}$
Then $\bar{u}\in W^{1,p}((-1,1))$.

Check that there are not any mistakes because I'm not used to working with absolutly continuos function. 
A: Thank you Revzora for your answer.  I'll just add in a few more details to your solution.
Indeed by a theorem due to Nykodim  a function $f$ is Sobolev $W^{1,p}(\Omega)$ if and only if its restriction to almost every line parallel to the coordinate directions is absolutely continuous, and if $f$ and $\nabla f$ are both in $L^p(\Omega)$.
If $u\in W^{1,p}(D_+^2)$ with $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$ the half-disk, we define $\bar u$ by
$$ \bar u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}. $$
Since $u$ is Sobolev, clearly the restriction of $u$ to almost all lines parallel to the coordinate axis will be absolutely continuous.  From this we directly see that the restriction of $\bar u$ is absolutely continuous along almost all lines that are parallel to the $x$-direction, and it only remains to verify the lines in $y$-direction.
The defintion of absolute continuity means for a function $f\colon [a,b] \to \mathbb{R}$ that for every $\epsilon > 0$ there exists a $\delta >0$ with the following property:  choose any finite sequence $a \le a_1 < b_1 \le a_2 < b_2 \le a_3 < \dotsb \le a_N < b_N \le b$ with $\sum_{j=1}^N (b_j-a_j) < \delta$, then
$$
  \sum_{j=1}^N |f(b_j) - f(a_j)| < \epsilon.
$$
If $x_0\in (-1,1)$ such that $y\mapsto u(x_0,y)$ is absolutely continuous then $y\mapsto \bar u(x_0,y)$ will inherit this property.  In fact, it trivially holds if we only use intervals that do not include the point $0$ at which we have mirrored the half-line, and if one of the intervals $(a_j,b_j)$ does contain that point, there is no danger either, because we can split $(a_j, b_j)$ into $(a_j,0)$ and $(0,b_j)$, and for such intervals the statement holds.
It still remains to see that the restriction to the lines is $L^p$, but this is obvious.  To see that the derivatives are $L^p$, we can use that the generalized partial derivatives agree almost everywhere with the classical partial derivatives.
