% import coprime class manually because of an usage of tectonic
importfile* (@/coprime/coprime.cls)
importfile* (@/coprime/coprime-math.sty)
importfile* (@/coprime/coprime-reference.sty)
importfile* (@/coprime/fancythm.sty)
docclass coprime (geometry, fancy)
importpkg { xparse, fontspec }
\settitle{Partial Differential Equations Homework 1}{2022021350 Sungbae Jeong}
\setgeometry{a4paper, margin=2.5cm}
%-
\newfontfamily{\hebrewSans}{Noto Serif Hebrew}
\NewDocumentCommand{\cone}{o}{
\makeatletter
\def\c@ne{\hebrewSans\char"05E7}
\IfNoValueTF{#1}{\!\mathop{\text{\c@ne}}\nolimits}{\!\mathop{\text{\c@ne}}\nolimits_{#1}}
\makeatother
}
-%
defun cl() \overline enddef
defun supp() \mathop{\rm{supp}}\nolimits enddef
defun diff() \partial enddef
\fancyon
startdoc
\section{Some notations used in here}
In here, denote $B(x,r)\subset\R^n$ be an \it{open} ball centered at $x$ with a radius $r$.
To write a closed ball, it is used $B[x,r]$ instead.
If $x\in\R^n$, then write $x^j$ be a $j^"th"$ coordinate for $x$ for each $1<=j<=n$.
In other words,
$$
x = (x^1,\cdots,x^n).
$$
Also denote $x'\in\R^{n-1}$ by $x'=(x^1,\cdots,x^{n-1})$ for each $x\in\R^n$.
If $E,F\subset\R^n$, $x\in\R^n$ and $\lambda\in\R$, we define
$$
useenv aligned {
x + E &:= \{ x + y : y\in E \} \\
E + F &:= \{ x + y : x\in E,\ y\in F \} \\
\lambda E &:= \{ \lambda y : y \in E\}
}
$$
If there is a constant $C=C(p_1,\cdots,p_n)>0$ which depends on $p_1,\cdots,p_n$, then the notation
$f\lesssim_{p_1,\cdots,p_n} g$ means that $f<= Cg$. Also write $f\lesssim g$ to omit such dependencies.
We shall use a special open set so called a \it{cone} in $\R^n$ defined by
$$
\cone[\alpha] := \{x\in\R^n : x^n>0,\ |x'|<\alpha|x^n|\} \qquad (\alpha>0).
$$
The reason why I choose this special character is that the
first letter of word \it{cone} in Hebrew is $\cone$.
\section{Global approximation by smooth functions}
The homework which I should solve is proving the following theorem:
useenv thm [Global approximation by functions smooth up to the boundary] {
\label{thm:main}
Assume that $\Omega$ is bounded and $\partial\Omega\in C^{0,1}$.
Suppose that $u\in W^{k,p}(\Omega)$ for some $1<=p<\infty$.
Then there exist functions $u_m\in C^\infty(\overline\Omega)$ such that
$$
u_m -> u \quad "in"\quad W^{k,p}(\Omega).
$$
}
To prove this theorem, we need some steps to work with.
\section{The one and only one lemma}
useenv lemma { \label{lemma:1}
Let
$$
\Omega := \{x\in\R^n : x^n > h(x')\}
$$
where $h:\R^{n-1}->\R$ is Lipschitz continuous. Then there exists $\alpha>0$ such that
$x+\cone[\alpha]\subset\Omega$ for every $x\in\cl\Omega$.
\proof
First of all, take $M<\infty$ such that
$$
|h(u)-h(v)| <= M|u-v|
$$
for every $u,v\in\R^{n-1}$.
If $x\in\cl\Omega$, then $x_n>=h(x')$ holds, thus if $y\in x+\cone[1/M]$, we get
$$
h(y') <= h(x') + M|x'-y'|
< h(x') + |y^n-x^n|
<= x^n + (y^n-x^n) = y^n.
$$
Therefore, $y\in\Omega$ holds so $x-\cone[1/M]\subset\Omega$. \proved
}
\section{Special case for \cref{thm:main}}
Before proving \cref{thm:main}, we will prove the somewhat \it{special} case of \cref{thm:main}, which is the
following.
useenv prop { \label{prop:1}
\cref{thm:main} is valid for an open set $\Omega$ whose definition is
$$
\Omega := \{x\in\R^n : x^n > h(x')\}
$$
for Lipschitz function $h:\R^{n-1}->\R$.
\proof
Fix $\alpha$ as in \cref{lemma:1} and simply write $\cone:=\cone[\alpha]$.
and take $\zeta\in C_c^\infty(\R^n)$ such that $\zeta>=0$,
$\supp\zeta\subset -\cone$, and $\int\zeta=1$.
Then define $\zeta_\epsilon(x) := \epsilon^{-n}\zeta(x/\epsilon)$ and
$$
u_\epsilon(x) := \int_{-\cone} \zeta_\epsilon(y)u(x-y)dy \qquad (x\in\cl\Omega).
$$
Indeed, if $y\in-\cone$ and $x\in\cl\Omega$, we get $x-y\in x+\cone\subset\Omega$ by \cref{lemma:1}.
This shows that $u_\epsilon$ is well-defined in $\cl\Omega$.
Also, for fixed $x\in\cl\Omega$, if $y\in\Omega\setminus(x+\cone)$, then $x-y\notin-\cone$.
Since
$$
\supp\zeta_\epsilon = \epsilon\supp\zeta \subset \epsilon(-\cone) = -\cone,
$$
we get $\zeta_\epsilon(x-y)=0$. Hence, by the change of variables, we can write
$$
u_\epsilon(x) = \int_{x+\cone}\zeta_\epsilon(x-y)u(y)dy = \int_\Omega\zeta_\epsilon(x-y)u(y)dy.
$$
First, claim that $u_\epsilon\in C^\infty(\cl\Omega)$ for every $\epsilon>0$.
To prove this, first we need some fact to deal with.
Put $\delta=(1/2)d(\supp\zeta,\diff(-\cone))$. Then if $x\in\cl\Omega$, $y\in\supp\zeta_\epsilon$ and
$h\in\R^n$ for which $|h|<\delta\epsilon$.
Since $-\cone$ is open, we have
$$
y-h \in B(y,\delta\epsilon) = \epsilon B(y/\epsilon,\delta) \subset \epsilon (-\cone) = -\cone.
$$
This implies that $x-y+h\in x+\cone\subset\Omega$ by \cref{lemma:1}.
In other words, for fixed $x\in\cl\Omega$ and $h\in\R^n$ with $|h|$ suffciently near to zero,
we can extend the value of $u_\epsilon$ at $x+h$ by defining
$$
u_\epsilon(x+h) := \int_{\supp\zeta_\epsilon}\zeta_\epsilon(y)u(x-y+h)dy.
$$
This extension does not harm the case when $x+h\in\cl\Omega$ because $\supp\zeta_\epsilon\subset-\cone$.
Also, the change of variables gives that
$$
u_\epsilon(x+h) = \int_{\Omega}\zeta_\epsilon(x-y+h)u(y)dy.
$$
Now, let $\{e_j\}_1^n$ be a standard basis for $\R^n$.
Then fix $x\in\cl\Omega$, $1<=j<=n$, and $\epsilon>0$. If $h$ is sufficiently near zero, we get
useenv equation { \label{eq:1}
useenv aligned {
{u_\epsilon(x+he_j) - u_\epsilon(x)//h}
&= \int_\Omega [?{\zeta_\epsilon(x-y+he_j) - \zeta_\epsilon(x-y)//h}?]u(y)dy \\
&= {1//\epsilon^n}\int_\Omega {1//h}[?\zeta(?{x-y+he_j//\epsilon}?) \zeta(?{x-y//\epsilon}?)?]u(y)dy \\
&= {1//\epsilon^n}\int_V {1//h}[?\zeta(?{x-y+he_j//\epsilon}?) \zeta(?{x-y//\epsilon}?)?]u(y)dy
}
}
for some open $V\Subset\Omega$ because $\zeta$ has a compact support.
Since
$$
{1//h}[?\zeta(?{x-y+he_j//\epsilon}?) \zeta(?{x-y//\epsilon}?)?]\uniform
{1//\epsilon}\zeta_{x_j}(?{x-y//\epsilon}?)
$$
on $y\in V$ as $h->0$. This can be proved using the mean value theorem on $h$ and the fact that
$\zeta\in C_c^\infty(\R^n)$.
Using this fact on \cref{eq:1},
$$
useenv aligned {
u_{\epsilon,x_j} &= {1//\epsilon^n}\int_V {1//\epsilon}\zeta_{x_j}(?{x-y//\epsilon}?)u(y)dy \\
&= {1//\epsilon^n}\int_\Omega {1//\epsilon}\zeta_{x_j}(?{x-y//\epsilon}?)u(y)dy \\
&= \int_\Omega \zeta_{\epsilon,x_j}(x-y)u(y)dy \\
}
$$
because the support of $\zeta_{x_j}$ is contained in the one of $\zeta$.
Continuing this process, we get
$$
D^\alpha u_\epsilon(x) = \int_\Omega D_x^\alpha\zeta_\epsilon(x-y)u(y)dy
$$
for all $|\alpha|<=k$. This shows that $u_\epsilon\in C^\infty(\cl\Omega)$ as $x\in\cl\Omega$ is arbitrary.
Now as $\zeta_\epsilon\in C_c^\infty(\R^n)$ and $u\in W^{k,p}(\Omega)$,
we get
$$
D^\alpha u_\epsilon(x)
= \int_\Omega D_x^\alpha\zeta_\epsilon(x-y)u(y)dy
= (-1)^{|\alpha|}\int_\Omega D_y^\alpha\zeta_\epsilon(x-y)u(y)dy
= \int_\Omega \zeta_\epsilon(x-y)D^\alpha u(y)dy,
$$
thus $D^\alpha u_\epsilon(x) = [\zeta_\epsilon * D^\alpha u](x)$ for all $x\in\Omega$ and $|\alpha|<=k$.
This implies that $\|D^\alpha u_\epsilon-D^\alpha u\|_{W^{k,p}(\Omega)}->0$ as $\epsilon->0$, so it is
finished. \proved
}
Now we are ready to prove the main theorem \cref{thm:main} and the proof of which is in the next section.
\section{The proof of \cref{thm:main}}
Let $\Omega$ be a bounded Lipschitz domain and $u\in W^{k,p}(\Omega)$.
For each $x\in\diff\Omega$ corresponds a positive $r_x$ and a Lipschitz function $h_x:\R^{n-1}->\R$ such that
by relabling, translating and rotating coordinate axes if necessary, we have
$$
\Omega\cap B[x,r_x] = \{y\in B[x,r_x] : y^n > h_x(y')\}.
$$
Being compact, there are $x_1,x_2,\cdots,x_N\in\diff\Omega$ such that
$$
\diff\Omega \subset B(x_1,r_{x_1}/2) \cup B(x_2,r_{x_2}/2) \cup\cdots\cup B(x_N,r_{x_N}/2).
$$
Define $V_j=\Omega\cap B(x_j,r_{x_j}/2)$ for $1<=j<=N$. Finally, take $V_0\Subset\Omega$ such that
$$
\cl\Omega \subset V_0\cup B(x_1,r_{x_1}/2)\cup B(x_2,r_{x_2}/2)\cup\cdots\cup B(x_N,r_{x_N}/2).
$$
Take a smooth partition of unity $\{\zeta_j\}_0^N$ of $\cl\Omega$ subordinate to the open cover
$$
\{V_0,B(x_1,r_{x_1}/2),\cdots,B(x_N,r_{x_N}/2)\}.
$$
Fix $\epsilon>0$.
If $j!=0$, first of all, take $\eta_j\in C_c^\infty(B(x_j,r_{x_j}))$ such that $\eta_j\equiv 1$ on $\cl{V_j}$.
Then we can apply \cref{prop:1} to take $u_j \in C^\infty(\cl{\Omega_j})$
such that $\|u_j - u\eta_j\|_{W^{k,p}(\Omega_j)}<\epsilon$ where
by relabling, translating and rotating coordinate axes if necessary,
$$
\Omega_j := \{y\in\R^n : y^n > h_{x_j}(y') \}.
$$
For the case when $j=0$,
we can take $u_0\in C^\infty(\cl{V_0})$ such that $\|u_0-u\zeta_0\|_{W^{k,p}(V_0)}<\epsilon$
by the theorem in the textbook page 264.
Finally, define $\tilde u := \sum_{j=0}^N u_j\zeta_j$.
Then clearly $\tilde u\in C^\infty(\cl\Omega)$.
Note that $u=\sum_0^N u\zeta_j$ and $u\eta_j=u$ in $V_j$ for $1<=j<=N$.
Hence, for each $|\alpha|<=k$ and $1<=j<=N$, we get
$$
useenv aligned {
\|D^\alpha u_j\zeta_j - D^\alpha u\zeta_j\|_{L^p(\Omega)}
&= \|D^\alpha u_j\zeta_j - D^\alpha u\zeta_j\|_{L^p(V_j)} \\
&<= \sum_{\beta<=\alpha}\binom{\alpha @\beta}\sup_{x\in V_j}|D^{\alpha-\beta}\zeta_j(x)|
\|D^\beta u_j - D^\beta u\|_{L^p(V_j)} \\
&<= \sum_{\beta<=\alpha}\binom{\alpha @\beta}\sup_{x\in V_j}|D^{\alpha-\beta}\zeta_j(x)|
\|D^\beta u_j - D^\beta u\eta_j\|_{L^p(V_j)} \\
&\lesssim_{\alpha,V_j,\zeta_j} \|u_j-u\eta_j\|_{W^{k,p}(V_j)} \\
&<= \|u_j-u\eta_j\|_{W^{k,p}(\Omega_j)} \\
&< \epsilon.
}
$$
Also, for $j=0$, we get
$$
useenv aligned {
||D^\alpha u_0\zeta_0 - D^\alpha u\zeta_0||_{L^p(\Omega)}
&= \|D^\alpha u_0\zeta_0 - D^\alpha u\zeta_0\|_{L^p(V_0)} \\
&<= \sum_{\beta<=\alpha}\binom{\alpha @\beta}\sup_{x\in V_0}|D^{\alpha-\beta}\zeta_0(x)|
\|D^\beta u_0 - D^\beta u\|_{L^p(V_0)} \\
&<= \sum_{\beta<=\alpha}\binom{\alpha @\beta}\sup_{x\in V_0}|D^{\alpha-\beta}\zeta_0(x)|
\|D^\beta u_0 - D^\beta u\eta_0\|_{L^p(V_0)} \\
&\lesssim_{\alpha,V_0,\zeta_0} \|u_0-u\eta_0\|_{W^{k,p}(V_0)} \\
&< \epsilon.
}
$$
This finishes the proof because
$$
useenv aligned {
\|\tilde u - u\|_{W^{k,p}(\Omega)}
&<= \sum_{j=0}^N \|u_j\zeta_j - u\zeta_j\|_{W^{k,p}(\Omega)} \\
&<= \sum_{j=0}^N \sum_{|\alpha|<=k} \|D^\alpha(u_j\zeta_j) - D^\alpha(u\zeta_j)\|_{L^p(\Omega)} \\
&\lesssim \epsilon(N+1)\#\{\alpha : |\alpha|<=k\}. \\
}
$$
\hbox{}\proved