If $u \in W^{3,p}(\mathbb{R}^{+})$ how can we construct the catoptric extension $\overline{u}$ of $u$ in $\mathbb{R}$ (reflection) such that $\overline{u} \in W^{3,p}(\mathbb{R})$ ?
EDIT: By setting $v(x)=\left\{\begin{matrix} u''(x) &, x>0 \\ u''(-x) &, x<0 \end{matrix}\right.$ we get:
$$w(x)=u'(0)+\int_0^x v(t) dt=\left\{\begin{matrix} u'(x) &, x>0 \\ 2u'(0)-u'(-x) &, x<0 \end{matrix}\right.$$
How do we use this in order to get an $\overline{u}$ such that $\overline{u}=u$ in $(0,\infty)$ where $\overline{u} \in W^{3,p}(\mathbb{R})$?
EDIT 2: If we would want the extension to be in $W^{2,p}(\mathbb{R})$ we could pick the following:
$$\overline{u}=\left\{\begin{matrix} u(x) &, x \geq 0 \\ -3u(-x)+4{u\left(-\frac{x}{2} \right )} & , x \leq 0 \end{matrix}\right.$$
How can we get an extension that is in $W^{3,p}(\mathbb{R})$ ?
Here are some ideas to get you started:
If you were using the space $W^{1,p}(\newcommand{\R}{\mathbb R} \R^+)$ you would define the reflection $\bar u(x) = u(-x)$ for $x < 0$ and $\bar u(x) = u(x)$ for $x > 0$. This gives you continuity of $\bar u$ at $x = 0$ from which the absolute continuity of $\bar u$ follows.
Suppose instead you are using the space $W^{2,p}(\R^+)$. In this case both $u$ and its derivative $u'$ are absolutely continuous. If you use the same definition of $\bar u$ you run into the problem that $\bar u'(x) = - u'(-x)$ for negative $x$ and $\bar u'(x) = u'(x)$ for $x > 0$ so that $\bar u'$ can fail to be continuous at the origin. One way around this is to define the derivative of $\bar u$ first using reflection: define $$v(x) = \left\{ \begin{array}{cl} u'(x) & x > 0 \\ u'(-x) & x < 0 \end{array} \right.$$ so that $v$ is absolutely continuous and then define $$\bar u(x) = u(0) + \int_0^x v(t) \, dt.$$ This is almost the right extension--you can check that it satisfies $$\bar u(x) = \left\{ \begin{array}{cl} u(x) & x > 0 \\ 2u(0) - u(-x) & x < 0 \end{array} \right.$$ but it has the problem that $\bar u$ may fail to be integrable on the whole line. You can fix this if you multiply $\bar u$ by a suitable cut-off function that equals zero for negative numbers sufficiently far from the origin, and that equals $1$ for all positive numbers.
What about $W^{3,p}(\R^+)$? Start by reflecting the second derivative.