I don't understand the proof that $ \| \overline{u} \|_{W^{1,p}(B)} \leq C \| u \|_{W^{1,p}(B^+)}$

Question

This is from Evan's PDE's Extension Theorem (Theorem 1 in Section 5.4.) One can find a proof like this here. I have an old edition of the book so I don't think my page is going to work for anyone here.

Here are the relevant information

Define

$$\overline{u}(x) = \left\{\begin{matrix} u(x) & x \in B^+ \\ -3u(x_1,\dots,x_{n-1},-x_n) + 4u(x_1,\dots,x_{n-1},-x_n/2)& x \in B^- \end{matrix}\right.$$

where $B$ is an open ball with centre $x^0$ and radius $r$ and

$$\left\{\begin{matrix} B^+ := B \cap \{x_n \geq 0\} \subset \overline{U} \\ B^- := B \cap \{x_n \leq 0\} \subset R^n -\overline{U} \end{matrix}\right.$$

where $U$ is a bounded (open) set.

$W^{1,p}(U)$ is the Sobolev space of $U$.

Q1: In the pdf I linked (Step 4), why is $\| \overline{u} \|_{W^{1,p} (B) } =\|-3u(...,-x_n) + 4u(...,-x_n/2) \|_{W^{1,p} (B) }$? Isn't the quantity $-3u(...,-x_n) + 4u(...,-x_n/2)$ only defined on the lower half of the ball?

Q2: What exactly is $\| \overline{u} \|_{W^{1,p} (B) }?$ Is it defined as

$$\int_B \| \overline{u} \|^p + \int_B \|D\overline{u} \|$$?

And is that further equal to

$$\int_B \| \overline{u} \|^p = \int_{x_n = 0} \| \overline{u} \|^p + \int_{B^{+}} \| u \|^p + \int_{B^{-}} \| -3u + 4u \|^p$$
and similarly for $\int_B \|D\overline{u} \|.$

Q3: Lastly I just don't quite understand the last estimation involving $B \cap \{x_n < r\}$

Answer 1

1. Yes. For example, $u(x_{1},\ldots,x_{n-1},-x_{n})$ is not defined if $x_{n}>0$.

$$\|\overline{u}\|_{W^{1,p}(B)}:=\left( \sum_{|\alpha| \leq 1} \int_{B}|D^{\alpha}\overline{u}|^{p}dx\right)^{\frac{1}{p}},$$ where $\alpha=(\alpha_{1},\ldots,\alpha_{n}) \in \mathbb{Z}_{\geq 0}^{n}$ is a multiindex and $|\alpha|=\sum_{i=1}^{n}\alpha_{i}$ is its norm. If you expand this, you obtain $$\|\overline{u}\|_{W^{1,p}(B)}=\left( \sum_{|\alpha| \leq 1} \int_{B}|D^{\alpha}\overline{u}|^{p}dx\right)^{\frac{1}{p}}=\left(\int_{B}|\overline{u}|^{p}dx+\sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx\right)^{\frac{1}{p}}.$$ Also $$\int_{B}|\overline{u}|^{p}dx=\int_{B^{+}}|u|^{p}dx+\int_{B^{-}}|-3u(x_{1},\ldots,x_{n-1},-x_{n})+4u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx. $$ The integral over $\{x_{n}=0\}$ vanishes because this set has Lebesgue measure zero.

3. Their stimations do not have sense because $u(\ldots,-x_{n}/2)$ is only defined for $B^{+}$, not for the whole ball.

Finally, let me prove that $\|\overline{u}\|_{W^{1,p}(B)} \leq C \|u\|_{W^{1,p}(B^{+})}$. Note that \begin{align*} \int_{B}|\overline{u}|^{p}dx &=\int_{B^{+}}|u|^{p}dx+\int_{B^{-}}|-3u(x_{1},\ldots,x_{n-1},-x_{n})+4u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx \\ &\leq \int_{B^{+}}|u|^{p}dx+3\int_{B^{-}}|u(x_{1},\ldots,x_{n-1},-x_{n})|^{p}dx+4\int_{B^{-}}u(x_{1},\ldots,x_{n-1},-x_{n}/2)|^{p}dx \\ &=\int_{B^{+}}|u|^{p}dx+3\int_{B^{+}}|u(y_{1},\ldots,y_{n-1},y_{n})|^{p}dy+2\int_{B^{+}}u(y_{1},\ldots,y_{n-1},y_{n})|^{p}dy, \\ &=6\int_{B^{+}}|u|^{p}dx. \end{align*} where in the penultimate inequality I used the substitutions $(y_{1},\ldots,y_{n})=(x_{1},\ldots,-x_{n})$ and $(y_{1},\ldots,y_{n})=(x_{1},\ldots,-x_{n}/2)$, respectively. Also, \begin{align*} \sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx=&\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+\int_{B^{-}}|(-3u(\ldots,-x_{n})+4u(\ldots,-x_{n}/2))_{x_{i}}|^{p}dx\right) \\ &+\int_{B^{+}}|u_{x_{n}}|^{p}dx+\int_{B^{-}}|3u_{x_{n}}(\ldots,-x_{n})-2u(\ldots,-x_{n}/2)|^{p}dx \\ \leq &\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+3\int_{B^{-}}|u_{x_{i}}(\ldots,-x_{n})|^{p}dx+4\int_{B^{-}}|u_{x_{i}}(\ldots,-x_{n}/2)|^{p}dx\right) \\ &+\int_{B^{+}}|u_{x_{n}}|^{p}dx+3\int_{B^{-}}|u_{x_{n}}(\ldots,-x_{n})|^{p}dx+2\int_{B^{-}}|u(\ldots,-x_{n}/2)|^{p}dx \\ =&\sum_{i=1}^{n-1}\left(\int_{B^{+}}|u_{x_{i}}|^{p}dx+3\int_{B^{+}}|u_{x_{i}}|^{p}dx+2\int_{B^{+}}|u_{x_{i}}|^{p}dx\right) \\ &+\int_{B^{+}}|u_{x_{n}}|^{p}dx+3\int_{B^{+}}|u_{x_{n}}|^{p}dx+\int_{B^{+}}|u|^{p}dx \\ =&6 \sum_{i=1}^{n} \int_{B^{+}}|u_{x_{i}}|^{p}dx, \end{align*} where I used the same substitutions as before. Finally \begin{align*} \|\overline{u}\|_{W^{1,p}(B)} &=\left(\int_{B}|\overline{u}|^{p}dx+\sum_{i=1}^{n}\int_{B}|\overline{u}_{x_{i}}|^{p}dx\right)^{\frac{1}{p}} \\ & \leq 6^{1/p}\left(\int_{B^{+}}|u|^{p}dx+\sum_{i=1}^{n}\int_{B^{+}}|u_{x_{i}}|^{p}dx\right)^{\frac{1}{p}} \\ &=6^{1/p} \|u\|_{W^{1,p}(B^{+})}. \end{align*}

Answer 2

re:Q2, Evans defines $$\|f\|_{W^{1,p}(B)}^p :=\sum_{0\le\alpha\le 1}\|D^\alpha f\|_{L^p(B)}^p= \sum_{0\le\alpha\le 1}\int |D^\alpha f(x)|^p \ dx.$$ Note $\|f\|_X$ is usually reserved for a function norm; the Euclidean norm / absolute value is written $|x|$.

Note that it is different from \begin{align} \| f\|_{L^p(B)}^p+\|D f\|_{L^p(B)}^p &= \int_B | f(x)|^p \ dx + \int_B | Df(x)|^p \ dx \\ &= \int_B | f(x)|^p \ dx +\int_B \left(\sum_{|\alpha|=1}| D^\alpha f(x)|^2\right)^{p/2} \ dx ,\end{align} but they define equivalent norms. What you wrote with $\int |Du| dx$ is NOT equivalent because its missing a $p$. Unless $p=1$.

So re:Q1, you have indeed found a typo; $u$ is only defined on the upper ball. In fact there are more typos, the notes write $\alpha<1$ but this means $\alpha=0$. And the calculation involves $|2^{-p}v|^p$ which is $2^{-p^2}|v|^p$? It may be better to read Evans (sorry Chee Han!). Perhaps what they meant to write was $$ \|\bar u\|_{W^{1,p}(B^-)} = \|{-3} u(\dots,-x_n)+4 u(\dots,-x_n/2)\|_{W^{1,p}(B^-)}.$$ In order to use that $\|\bar u\|_{W^{1,p}(B^-)} + \|\bar u\|_{W^{1,p}(B^+)} = \|\bar u\|_{W^{1,p}(B)} $, for which we first need weak differentiability, one uses some continuity arguments (which you have because of the way the function was extended to the lower ball; this is the earlier part of the proof in Evans.)

Back to Q2, specialising to $f=\overline u$:

$$\int_B | \overline{u} |^pdx = \int_{B^{+}} | u |^p dx+ \int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n.$$ We used that $B=B^+\cup B^-$ up to a $\mathbb R^n$-Lebesgue null set (a subset of $\{x:x_n=0\}$). Note that the last term $\int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n$ is different from $$ \int_{B^{-}} | {-3}u + 4u |^p dx= \int_{B^{-}} | {-3}u(x',x_n) + 4u(x',x_n) |^p dx' dx_n,$$ which is what you wrote. As you noted, $u$ is not defined for $x_n<0$, so this doesn't make sense. It is useful to write things out fully, instead of trying to use short notation.

As for Q3, using $$ |x+y|^p \le 2^{p-1}( |x|^p + |y|^p),$$ (this follows by the convexity of $t\mapsto t^p$ on $[0,\infty)$) you can split $\int_{B^{-}} | {-3}u(x',-x_n) + 4u(x',-x_n/2) |^p dx' dx_n$ into two types of integrals, times a constant depending on $p$: One is $$I_1=\int_{B^{-}} | u(x',-x_n) |^p dx' dx_n = \int_{B^{+}} | u(x',y_n) |^p dx' dy_n, $$ (by the obvious change of variables $B^- \to B^+$) and the other is $$ I_2=\int_{B^{-}} | u(x',-x_n/2) |^p dx' dx_n = 2\int_{V} | u(x',z_n) |^p dx' dz_n ,$$ where $z_n = -x_n/2, dx_n = 2dz_n$ (as unsigned measures), and $$V:=\{ (x_1,\dots,x_{n-1},x_n/2) : x\in B^+\}.$$ Note that $V\subset B^+$; this means we can bound $I_2$ by $$ I_2 \le 2\int_{B^+} | u(x) |^p dx.$$ In total you should get for the $L^p$ norm, $\| \overline u\|_{L^p(B)} \le C_p \|u\|_{L^p(B^+)}$ with $C_p=1+2^{p-1}(3+4\times 2).$ But what's important is that $C_p < \infty$, not the exact value.

This set $V$ is the correct one to use; note carefully it is different from $B^+\cap \{ x_n < r/2\}$ as in the screenshot.

For the derivatives, its the same calculation. The only thing is to be careful with chain rule if you want to compute the constant correctly: for $x\in B^-$,

$$ \partial_{x_n} \overline u(x) = \partial_{x_n}(-3u(x',-x_n)+4u(x',-x_n/2)) = 3 (\partial_{x_n} u)(x',-x_n)-2 (\partial_{x_n} u)(x',-x_n/2)$$

The amusing fact that $-3+4=1=3-2$, which you should have already seen when verifying the statements in the earlier part of the proof, is what makes the particular choice of "reflection" work.