Proof of the Extension theorem of sobolev spaces

partial differential equations

I just go through the proof of the extension theorem in the book of Evan – Partial differential equations (First edition, section 5.4, pages 254-257).

I was thinking about point 4 of the proof, that is

  1. Using these calculations we readily check as well
    $$\|\bar{u}\|_{W^{1 , p}(B)} \leq C\|u\|_{W^{1, p}\left(B^{+}\right)}$$
    for some constant $C$ which does not depend on $u$..

Here, $\bar u$ is the higher order reflection of $u$ from $B^+ := B\cap \{x_n>0\}$ to $B^- := B\cap \{ x_n < 0 \}$ for some ball $B$, given by
$$
\bar{u}(x):=\left\{\begin{array}{ll}
u(x) & \text { if } x \in B^{+} \\
-3 u\left(x_{1}, \ldots, x_{n-1},-x_{n}\right)+4 u\left(x_{1}, \ldots, x_{n-1},-\frac{x_{n}}{2}\right) & \text { if } x \in B^{-}
\end{array}\right.
$$

this inequality should follow from the calculations before. But it looks like that it isn't that simple or at least it isn't that simple for me.

Could you help me with this inequality. Thank you!

Edit: If I have the inequality $\int_{B^-} |D^{\alpha} \bar{u} |^p \leq C \int_{B^+} |D^{\alpha} u |^p$, then I see how the upper inequality follows.

For $B^+$ I should get: $\int_{B^+} |D^{\alpha} \bar{u} |^p = \int_{B^+} |D^{\alpha} u |^p$

and

for $B^-$: $\int_{B^-} |D^{\alpha} \bar{u} |^p = \int_{B^-} |D^{\alpha} (-3u(x_1, …,x_{n-1}, -x_n) + 4u(x_1, …,x_{n-1}, – \frac{x_n}{2})) |^p \\
= \int_{B^-} |-3D^{\alpha} u(x_1, …,x_{n-1}, -x_n) + 4D^{\alpha} u(x_1, …,x_{n-1}, – \frac{x_n}{2}) |^p $

so here there have to exists a constant c_1 such that

$\int_{B^-} |-3D^{\alpha} u(x_1, …,x_{n-1}, -x_n) + 4D^{\alpha} u(x_1, …,x_{n-1}, – \frac{x_n}{2}) |^p \\ \leq c_1 \int_{B^-} |-D^{\alpha} u(x_1, …,x_{n-1}, -x_n) + D^{\alpha} u(x_1, …,x_{n-1}, – \frac{x_n}{2}) |^p$

If I now get the estimation
$c_1 \int_{B^-} |-D^{\alpha} u(x_1, …,x_{n-1}, -x_n) + D^{\alpha} u(x_1, …,x_{n-1}, – \frac{x_n}{2}) |^p \\ \leq c_2 \int_{B^-} |-D^{\alpha} u(x_1, …,x_{n-1}, -x_n)|^p + |D^{\alpha} u(x_1, …,x_{n-1}, – \frac{x_n}{2}) |^p$

I could maybe use a transformation to get to $B^+$? But it dosn't look that trivial for me or is that your trivial way how to get to the inequality?

Best Answer

From the definition of $\bar u$, its trivial that for any multiindex $\alpha$, $0\le |\alpha|\le 1$, $$ \int_{B^{\pm}} |D^\alpha\bar u|^p \le C \int_{B^+} |D^\alpha u|^p .$$ The point of the calculations leading to this claim is to check that $D^\alpha\bar u$ is actually defined on $B$, i.e. that $\bar u$ is differentiable on $B$.

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