Machine-translated from Chinese.
The Mass of an Asymptotically Flat Manifold - Robert Bartnik
带权Sobolev空间
我们首先在$\mathbb{R}^n,$ $n\ge 3$中讨论. 记$r=|x|,$
$\sigma=\sqrt{1+r^2},$ 分别在$L_{\mathrm{loc} }^p(\mathbb{R}^n),$
$L_{\mathrm{loc} }^p(\mathbb{R}^n\setminus\{0\})$中定义带权Lebesgue空间:
$$
L_\delta^p:\quad \Vert u\Vert_{p,\delta}=
\left\{
\begin{aligned}
&\left(\int_{\mathbb{R}^n}|u|^p\sigma^{-\delta p-n}dx\right)^{\frac{1}{p} },&p<\infty,\\
&\operatorname{ess}\operatorname{sup}_{\mathbb{R}^n}(\sigma^{-\delta}|u|), &p=\infty,
\end{aligned}
\right.
$$
$$
{L'}_\delta^p:\quad \Vert u\Vert'_{p,\delta}=
\left\{
\begin{aligned}
&\left(\int_{\mathbb{R}^n\setminus\{0\} }|u|^pr^{-\delta p-n}dx\right)^{\frac{1}{p} },&p<\infty,\\
&\operatorname{ess}\operatorname{sup}_{\mathbb{R}^n\setminus\{0\} }(r^{-\delta}|u|), &p=\infty,
\end{aligned}
\right.
$$
接下来, 定义带权Sobolev空间:
$$
W_\delta^{k,p}:\quad \Vert u\Vert_{k,p,\delta}=\sum_{j=0}^k\Vert D^ju\Vert_{p,\delta-j},
$$
$$
{W'}_\delta^{k,p}:\quad \Vert u\Vert'_{k,p,\delta}=\sum_{j=0}^k\Vert D^ju\Vert'_{p,\delta-j},
$$
注意这里权重的不同使得相加项量纲基本一致. $p\in [1,\infty)$时,
$C_c^\infty(\mathbb{R}^n),$
$C_c^\infty(\mathbb{R}^n\setminus\{0\})$分别是在它们中稠密的.
权重的选取使得对于函数$u_R(x)=u(Rx),$
$$
\Vert u\Vert'_{k,p,\delta}=R^{-\delta}\Vert u_R\Vert'_{k,p,\delta},
$$
$$
\Vert u\Vert_{k,p,\delta;A_R}\approx R^{-\delta}\Vert u_R\Vert_{k,p,\delta;A_1}, \quad R\ge 1.
$$
这里$A_R:=B_{2R}\setminus B_R,$ $B_R$为原点处半径为$R$的闭球.
“$\approx$”即指两者可由与$R$无关的常数倍控制, 考虑控制$\sigma$即可.
接下来我们不加证明地列举如下性质:
定理 1. 对于$1\le p\le q\le \infty,$ $\delta_2<\delta_1,$ $u\in L_{\delta_2}^q,$ 有$\Vert u\Vert_{p,\delta_1}\le c\Vert u\Vert_{q,\delta_2},$ 因此$L_{\delta_2}^q\subset L_{\delta_1}^p.$ 即$p$越高, $\delta$越小, $L_\delta^p$性质越好, 对函数要求越高.
定理 2 (Hölder不等式). $u\in L_{\delta_1}^q,$ $v\in L_{\delta_2}^r,$ $\delta=\delta_1+\delta_2,$ $1\le p,q,r\le \infty,$ $\frac{1}{p}=\frac{1}{q}+\frac{1}{r},$ 则
$$
\Vert uv\Vert_{p,\delta}\le \Vert u\Vert_{q,\delta_1}\Vert v\Vert_{r,\delta_2}.
$$
定理 3 (插值不等式). $\,\forall\,\varepsilon>0,$ $\,\exists\,C(\varepsilon)>0,$ $\,\forall\,u\in W_{\delta}^{2,p},$ $p\in [1,\infty],$
$$
\Vert u\Vert_{1,p,\delta}\le \varepsilon\Vert u\Vert_{2,p,\delta}+C(\varepsilon)\Vert u\Vert_{0,p,\delta}.
$$
定理 4 (Sobolev不等式). 若$u\in W_{\delta}^{k,p},$ 则
$$
\Vert u\Vert_{q,\delta}\le C\Vert u\Vert_{k,p,\delta},\quad 0<\frac{n}{p}-k\le \frac{n}{q}\le \frac{n}{p};
$$
$$
\Vert u\Vert_{\infty,\delta}\le C\Vert u\Vert_{k,p,\delta},\quad \frac{n}{p}-k<0,\quad \text{且} |u(x)|=o(r^{\delta}).
$$
定义带权Hölder模:
$$
\Vert u\Vert_{C_\delta^{0,\alpha} }:=\sup_{x\in \mathbb{R}^n}\left(\sigma^{-\delta+\alpha}(x)\sup_{4|x-y|\le \sigma(x)}\frac{|u(x)-u(y)|}{|x-y|^{\alpha} }\right)+\sup_{x\in\mathbb{R}^n}\sigma^{-\delta}(x)|u(x)|.
$$
定理 5 (负指数Sobolev不等式). 若$u\in W_{\delta}^{k,p},$ $0<\alpha\le k-\frac{n}{p}\le 1,$ 则
$$
\Vert u\Vert_{C_{\delta}^{0,\alpha} }\le C\Vert u\Vert_{k,p,\delta},\quad \Vert u\Vert_{C_{\delta}^{0,\alpha}(A_R)}=o(1).
$$
定理 6 (Poincaré不等式). $\,\forall\,u\in {W'}_{\delta}^{1,p},$ $1\le p<\infty,$ $\delta\neq 0,$
$$
\Vert u\Vert_{p,\delta}'\le |\delta|^{-1}\Vert u_r\Vert'_{p,\delta-1}\le |\delta|^{-1}\Vert Du\Vert'_{p,\delta-1},\quad u_r=\partial_ru=r^{-1}x\cdot Du.
$$
$$
\Vert u\Vert_{p,\delta}\le C\Vert u_r\Vert_{p,\delta-1},\quad \delta<0,\quad u\in W_{\delta}^{1,p}.
$$
引理 7. $k>j,$ $\delta<\varepsilon,$ $1\le p<\infty,$ $W_\delta^{k,p}\Subset W_{\varepsilon}^{j,p}$为紧嵌入.
渐进椭圆算子
不等式估计
定义 8. 算子$Pu:=a^{ij}(x)\partial_{ij}^2u+b^i(x)\partial_iu+c(x)u$称为(以速率$\tau$)渐进(asymptotic)至$\Delta$的, 若$\,\exists\,n<q<\infty,$ $\tau\ge 0,$ $C_1,\lambda,$ 使得
$$
\lambda |\xi|^2\le a^{ij}(x)\xi_i\xi_j\le \lambda^{-1}|\xi|^2,\quad \,\forall\,x,\xi\in \mathbb{R}^n;
$$
$$
\Vert a^{ij}-\delta_{ij}\Vert_{1,q,-\tau}+\Vert b^i\Vert_{0,q-1,-1-\tau}+\Vert c\Vert_{0,\frac{q}{2},-2-\tau}\le C_1.
$$
由前面的定理, $a^{ij}$是Hölder连续的,
且$|a^{ij}-\delta_{ij}|=o(r^{-\tau}).$ 对于$1\le p\le q,$
$\delta\in \mathbb{R},$
$P:W_{\delta}^{2,p}\rightarrow W_{\delta-2}^{0,p}$是有界算子.
事实上有如下估计成立:
命题 9. 设$P$渐进至$\Delta,$ $1<p\le q,$ $\delta\in \mathbb{R},$ 则存在常数$C=C(n,p,q,\delta,C_1,\lambda),$ 使得若$u\in L_\delta^{0,p},$ $Pu\in L_{\delta-2}^{0,p},$ 则$u\in W_{\delta}^{2,p},$
$$
\Vert u\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta}).
$$
欲证该命题, 做二阶椭圆型方程中通常的$L^p$估计即可.
注意该估计对${W'}_{\delta}^{k,p}$模也成立.
接下来我们来讨论算子$P$的Fredholm性, 即考虑其像是否是闭的,
以及核与余核维数的有限性. 上述命题并不足以说明它具有该性质.
我们需要加强估计, 使得右侧误差项关于$W_\delta^{2,p}$是紧的.
权重参数$\delta$称为非例外的(nonexceptional),
若$\delta\in \mathbb{R}\setminus\{k\in \mathbb{Z},k\neq -1,-2,\cdots,3-n\}.$
例外值(exceptional
value)$\{k\in \mathbb{Z}, k\neq -1,-2,\cdots,3-n\}$对应了$\mathbb{R}^n\setminus B_1$上调和函数的增长阶数.
定义
$$
k^-(\delta)=\max\{k \text{ exceptional},k<\delta\}.
$$
定理 10. 设$\delta$非例外, $p\in (1,\infty),$ $s$为非负整数. 那么有同构
$$
\Delta:{W'}_\delta^{s+2,p}\rightarrow {W'}_{\delta-2}^{s,p},
$$
且存在常数$C=C(n,p,\delta,s),$
$$
\Vert u\Vert'_{s+2,p,\delta}\le C\Vert\Delta u\Vert'_{s,p,\delta-2}.
$$
证: 只需对$s=0$的情况证明.
首先我们说明$\Delta$的分布逆有卷积核$K(x,y):$
$$
c_nK(x,y)=
\left\{\begin{aligned}
&|x-y|^{2-n} &2-n<\delta<0,\\
&|x-y|^{2-n}-|y|^{2-n}\sum_{j=0}^k P_j^\lambda(\mu)\left(\frac{|x|}{|y|}\right)^j &k\ge 0,\\
&|x-y|^{2-n}-|x|^{2-n}\sum_{j=0}^k P_j^\lambda(\mu)\left(\frac{|y|}{|x|}\right)^j &k<2-n.
\end{aligned}\right.
$$
这里$k=k^-(\delta),$ $\mu=\frac{\left<{}x,y\right>}{|x||y|},$
$P_j^\lambda$是超球(ultraspherical)函数, 出现在泰勒展开式中:
$$
|x-y|^{2-n}=|x|^{2-n}\sum_0^\infty P_j^\lambda(\mu)\left(\frac{|y|}{|x|}\right)^j,\quad |y|<|x|.
$$
先来验证卷积核$K$的确定义了${W'}_{\delta-2}^{0,p}\rightarrow {W'}_{\delta}^{0,p}$的有界算子.
以$k\ge 0$的情况为例, 我们有估计
$$
|K(x,y)|\le c(n,k)|x-y|^{2-n}
\left\{\begin{aligned}
&\left(\frac{|x|}{|y|}\right)^{k+1} &\frac{|x|}{|y|}<\frac{1}{2},\\
&\left(\frac{|x|}{|y|}\right)^{n+k-2} & \frac{|x|}{|y|}\ge \frac{1}{2}.
\end{aligned}\right.
$$
我们有引理如下:
引理 11. *固定$p\in (1,\infty),$ $\frac{1}{p}+\frac{1}{p'}=1,$ $a,b\in \mathbb{R}$且$a+b>0.$ 设有核
$$
K'(x,y)=|x|^{-a}|x-y|^{2-n}|y|^{-b},\quad x\neq y.
$$
- 对$u\in L^p(\mathbb{R}^n),$ 定义$K'u(x)=\int K'(x,y)u(y)dy,$ 那么$K':L^p\rightarrow L^p$是有界算子当且仅当$a<\frac{n}{p},$ $b<\frac{n}{p'}.$
令$K_1,K_2$是核算子$|x-y|^{2-n}\left(\frac{|x|}{|y|}\right)^\alpha,$
分别对应$\alpha=k+1,$ $\alpha=n+k-2$.
上述引理说明当$3-n+k<\delta<k+1$时, 核
$$
K_1'(x,y)=|x|^{-\delta-\frac{n}{p} }K_1(x,y)|y|^{\delta-2+\frac{n}{p} }
$$
定义了$L^p\rightarrow L^p$的有界算子. 进而
$$
\Vert K_1u\Vert_{p,\delta}'\le c\Vert r^{2-\delta-\frac{n}{p} }u\Vert_{L^p}\le c\Vert u\Vert'_{p,\delta-2},
$$
即$K_1:{W'}_{\delta-2}^{0,p}\rightarrow {W'}_{\delta}^{0,p}$是有界算子.
类似地, 当$k<\delta<n+k-2$时, $K_2'$是$L^p\rightarrow L^p$上的有界算子,
进而$K_2$也是有界算子. 这两个估计就说明了当$k<\delta<k+1$时,
$K$也是有界算子, 而$k=k^-(\delta),$ 条件自动满足.
$K$定义中的另外两种情况类似处理.
我们有分布恒等式:
$$
\Delta_x K(x,y)=\Delta_y K(x,y)=\delta(x-y),
$$
因此$K(\Delta u)=u,$
$\,\forall\,u\in C_c^\infty(\mathbb{R}^n\setminus \{0\}).$
于是由$K$的有界性及$C_c^\infty$的稠密性给出
$$
\Vert u\Vert_{0,p,\delta}'=\Vert K(\Delta u)\Vert_{0,p,\delta}'\le C\Vert\Delta u\Vert'_{0,p,\delta-2},\quad \,\forall\,u\in {W'}_\delta^{k,p}.
$$
结合前面的$L^p$估计, 就给出了定理所述不等式的证明($s=0$).
由于$\Delta:{W'}_{\delta}^{2,p}\rightarrow {W'}_{\delta-2}^{0,p}$是有界算子,
该不等式立刻说明其像是闭的.
由分布恒等式, 我们看到,
$$
\Delta(Kf)(x)=f(x),\quad f\in C_c^\infty(\mathbb{R}^n\setminus \{0\}),\quad Kf\in {W'}_{\delta}^{0,p}.
$$
所以$\Delta$是同构, 定理证毕.
定理给出了如下熟知的Liouville定理:
推论 12. 设$\Delta u=0,$ $u\in L_\delta^p,$ $1<p<\infty.$ 令$k=k^-(\delta).$ 若$k<0,$ 那么$u\equiv 0;$ 若$k\ge 0,$ 那么$u$是$k$阶及以下的调和多项式.
证: 由于$0\in {W'}_\delta^{s,p},$ $\,\forall\,s\ge 0,$
推出$u\in {W'}_\delta^{s+2,p}.$ 由Sobolev不等式,
$u\in C^\infty(\mathbb{R}^n).$ 令$h_k(x)$为$u(x)$的$k$次 Taylor 展开式,
即
$$
u(x)=h_k(x)+O(|x|^{k+1}).
$$
由于$u$是调和的, 我们知道$h_k$为调和多项式.
此时$u-h_k\in {L'}_\delta^p,$ 由前面的命题与定理,
$u-h_k\in {W'}_\delta^{2,p},$
且有估计$\Vert u-h_k\Vert'_{2,p,\delta}\le C\Vert\Delta(u-h_k)\Vert_{0,p,\delta-2}'=0,$
从而$u=h_k.$
接下来, 我们给出与前面命题类似的估计. 乍一看不等式的量纲有些混乱,
因此称为scale-broken估计(我猜的).
定理 13. 设$P$渐进于$\Delta,$ $\delta\in \mathbb{R}$非例外, 渐进条件中的$q\ge p>1.$ 那么算子$P:W_\delta^{2,p}\rightarrow W_{\delta-2}^{0,p}$的核是有限维的, 具闭值域, 且对任意$u\in W_{\delta}^{2,p},$ 有常数$C,R$由$P,\delta,n,p,q$控制, 使得
$$
\Vert u\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{L^p(B_R)}).
$$
证: 取算子模
$$
\Vert P-\Delta\Vert_{op}:=\sup\{\Vert(P-\Delta)u\Vert_{0,p,\delta-2}:u\in W_{\delta}^{2,p},\Vert u\Vert_{2,p,\delta}=1\}.
$$
类似地定义$\Vert\cdot\Vert_{op,R},$
要求选取的函数$u$支集落在$E_R=\mathbb{R}^n\setminus B_R$中,
即在$B_R$内取零. 那么若$\operatorname{supp}(u)\subset E_R,$ 由$q>n,$
$$
\begin{aligned}
\Vert(P-\Delta)u\Vert_{0,p,\delta-2}\le& \sup_{|x|>R}\{|a^{ij}(x)-\delta_{ij}|\}\Vert D^2u\Vert_{0,p,\delta-2}\\
&+C\Vert b\Vert_{0,q,-1;E_R}\Vert Du\Vert_{1,p,\delta-1}+C\Vert c\Vert_{0,\frac{q}{2},-2;E_R}\Vert u\Vert_{2,p,\delta}
\end{aligned}
$$
由于$P$渐进于$\Delta,$
$$
\Vert P-\Delta\Vert_{op,R}=o(1),\quad R\rightarrow \infty.
$$
取$\chi\in C_c^\infty(B_2),$ $\chi\in [0,1],$ $\chi|_{B_1}=1.$
取待定常数$R,$ 令$\chi_R(x)=\chi(\frac{x}{R}),$ $u_0=\chi_Ru,$
$u_\infty=(1-\chi_R)u,$ $u=u_0+u_\infty.$ 由前面的定理, 我们有:
$$
\begin{aligned}
\Vert u_\infty\Vert_{2,p,\delta}&\le C\Vert\Delta u_{\infty}\Vert_{0,p,\delta-2}=C\Vert(\Delta-P+P) u_{\infty}\Vert_{0,p,\delta-2}\\
&\le C(\Vert Pu_{\infty}\Vert_{0,p,\delta-2}+\Vert P-\Delta\Vert_{op,R}\Vert u_{\infty}\Vert_{2,p,\delta})
\end{aligned}
$$
由于$\Vert P-\Delta\Vert_{op,R}=o(1),$ 对充分大的$R,$ 有
$$
\Vert u_{\infty}\Vert_{2,p,\delta}\le C\Vert Pu_\infty\Vert_{0,p,\delta-2}.
$$
进一步估计, 有
$$
\begin{aligned}
\Vert Pu_\infty\Vert_{0,p,\delta-2}&\le \Vert Pu\Vert_{0,p,\delta-2}+\Vert 2a^{ij}u_i\partial_j\chi_R+(a^{ij}\partial^2_{ij}\chi_R+b^i\partial_i \chi_R)u\Vert_{0,p,\delta-2;A_R}\\
&\le \Vert Pu\Vert_{0,p,\delta-2}+C\Vert u\Vert_{1,p,\delta;A_R}
\end{aligned}
$$
再对$\Vert\cdot\Vert_{1,p,\delta}$项利用插值不等式进一步改造,
以及$L^p$估计, 即有
$$
\Vert u_\infty\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta;A_R}).
$$
同时由$L^p$估计, 对$u_0$有
$$
\Vert u_0\Vert_{2,p,\delta}\le C(\Vert Pu_0\Vert_{0,p,\delta-2}+\Vert u_0\Vert_{0,p,\delta})\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert Pu_{\infty}\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta;{B_{2R} } }).
$$
再利用一遍前面关于$Pu_\infty$已有的估计, 就得到了定理所需的不等式:
$$
\Vert u\Vert_{2,p,\delta}\le \Vert u_0\Vert_{2,p,\delta}+\Vert u_\infty\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{L^p(B_R)}).
$$
接下来讨论核维数的有限性. 设$\{u_k\}$为$\ker P$中的函数列,
满足$\Vert u_k\Vert_{2,p,\delta}=1.$ 由紧嵌入定理,
不妨设$\{u_k\}$本身在$L^p(B_R)$中强收敛. 此时由于$Pu_k\equiv 0,$
刚刚证得的不等式就说明了$\{u_k\}$是$W_\delta^{2,p}$中的Cauchy列.
因此$\ker P$中单位球面是紧的, 从而其维数有限.
由于$\ker P$维数有限, 有分解$W_\delta^{2,p}=\ker P\oplus Z,$ $Z$为闭集.
断言存在常数$C$使得
$$
\Vert u\Vert_{2,p,\delta}\le C\Vert Pu\Vert_{0,p,\delta-2},\quad \,\forall\,u\in Z.
$$
不然, 存在$\{u_k\},$ $\Vert u_k\Vert_{2,p,\delta}\equiv 1,$
而$\Vert Pu_k\Vert_{0,p,\delta-2}\rightarrow 0.$ 此时由紧嵌入定理,
利用定理中的不等式再一次得到$\{u_k\}$是$W_{\delta}^{2,p}$中的Cauchy列.
由$Z$的闭性, 它收敛到$u_0\in Z.$ 然而又由$P$的连续性, $Pu_0=0,$
$u_0\in \ker P.$ 从而$u_0=0,$
但这又与$\Vert u_k\Vert_{2,p,\delta}\equiv 1$矛盾,
因为$u_k\rightarrow u_0.$ 于是断言成立, 立即说明$P$具闭值域.
半Fredholm性质
接下来我们进一步考虑$P$的核的维数. 记
$$
N(P,\delta)=\dim\ker (P:W_\delta^{2,p}\rightarrow W_{\delta-2}^{0,p}), \quad 1<p\le q.
$$
由$L^p$估计与Sobolev不等式, 可以说明$N(P,\delta)$与$p$选取无关.
下面为了简化记号, 我们记$X=W_\delta^{2,p},$ $Y=W_{\delta-2}^{0,p},$
$\delta$非例外, $1<p\le q.$ 那么由刚刚的定理,
我们知道$P:X\rightarrow Y$是半Fredholm的, 即核是有限维的,
且具闭值域.
命题 14. 设$P:X\rightarrow Y$是Banach空间间的半Fredholm算子, 那么存在常数$C>0,$ 有估计
$$
\Vert u-\ker P\Vert_{X}=\Vert u\Vert_{X/\ker P}\le C\Vert Pu\Vert_Y.
$$
同时, 有$\varepsilon>0,$
使得对于任意满足$\Vert P-P'\Vert_{op}<\varepsilon$的另一个半Fredholm算子$P',$
$$
\dim \ker P'\le \dim \ker P.
$$
证: 已知$\operatorname{Im}P$是闭的, 那么它是Banach空间,
因此第一个估计由逆算子定理立即得到. 现令$\varepsilon=\frac{1}{2C},$
设核维数不等式不成立, 那么存在$u\in \ker P',$ $\Vert u\Vert_X=1,$
$\Vert u-\ker P\Vert_X>\frac{1}{2}.$ 此时
$$
\frac{1}{2}<\Vert u-\ker P\Vert_{X}\le C\Vert Pu\Vert_{Y}\le C\Vert P-P'\Vert_{op}\Vert u\Vert_X<C\varepsilon,
$$
产生矛盾.
可见$\frac{1}{C}$衡量着$P$与其它核更大的半Fredholm算子的距离.
下面反方向的命题使这一观点更为深刻:
命题 15. $X,Y$同上, 记$\mathcal{F}$为$X$到$Y$上渐进于$\Delta$的算子全体. 设$\mathcal{U}\subset \mathcal{F}$为其中满足$\dim\ker P\equiv N$的算子全体, 那么给定$P\in \mathcal{U},$ $\,\exists\,\varepsilon,C>0,$ 使得
$$
\Vert u-\ker P'\Vert_{X}\le C\Vert P'u\Vert_Y,\quad \,\forall\,u\in X,P'\in \mathcal{U}, \Vert P-P'\Vert_{op}\le \varepsilon.
$$
证: 不然, 存在序列$P_k\rightarrow P\in \mathcal{U},$
$\{u_k\}\subset X,$ 满足$\Vert u_k\Vert_X=1,$
$\Vert u_k-\ker P_k\Vert_{X}>\frac{1}{2},$ 且
$$
\Vert u_k-\ker P_k\Vert_{X}\ge k\Vert P_ku_k\Vert_Y.
$$
这说明$P_ku_k\rightarrow 0.$
此时继续通过前面定理中的紧嵌入方法即可取子列$u_k\rightarrow u,$
满足$Pu=0,$ $\Vert u\Vert_X=1,$
且对充分大的$k$有$\Vert u-\ker P_k\Vert\ge \frac{1}{4}.$
类似地, 任取$v_k\in \ker P_k,$ $\Vert v_k\Vert_X=1,$
同理它也有子列收敛到$v\in \ker P.$
这就说明$N$维的$\ker P_k$收敛到$\ker P$中的$N$维子空间.
然而由上段的估计, $u$不在这样的子空间中, 但$u\in \ker P.$
这就说明$\dim \ker P\ge N+1,$ 矛盾.
最后我们给出如下的定理, 说明核更大的半Fredholm算子集更小.
定理 16. 设$P:X\rightarrow Y$是半Fredholm的, $A:X\rightarrow Y$是$P$-有界的, 即$\Vert Au\Vert_Y\le C(\Vert Pu\Vert_Y+\Vert u\Vert_X),$ $\,\forall\,u\in X.$ 那么存在$\lambda_0>0,$ 使得对于$|\lambda|<\lambda_0,$ $P+\lambda A$总是半Fredholm的, 且$\dim \ker (P+\lambda A)$在$0<\lambda<\lambda_0$时为常数(注意$\lambda=0$不在其中).
Fredholm指标
我们有伴随算子$P^\ast :W^{0,p'}_{2-n-\delta}\rightarrow W^{-2,p'}_{-n-\delta},$
$\frac{1}{p'}+\frac{1}{p}=1.$ 若其也是渐进的,
那么我们可以通过Fredholm指标得到更多信息. 由前面的命题,
当$1<p'\le \widetilde{q}$时, $\ker P^\ast \subset W^{2,p'}_{2-n-\delta}.$
当$\delta$非例外时, 伴随也是半Fredholm算子, 从而:
$$
\dim \operatorname{coker}P=\dim \ker P^\ast =N(P^\ast ,2-n-\delta).
$$
因此$P$成为Fredholm算子, 具Fredholm指标:
$$
\iota(P,\delta)=N(P,\delta)-N(P^\ast ,2-n-\delta).
$$
由于指标在Fredholm算子空间上是局部常数的, 我们有
$$
\iota(P,\delta)=\iota(\Delta,\delta)=:\iota_0(\delta).
$$
而$\iota_0(\delta)$是可以计算的.
记$H_k$为$\mathbb{R}^n$中$k$次齐次调和多项式全体,
我们有$n_k=\dim H_k=\frac{(n-2+2k)(n-3+k)!}{k!(n-2)!}.$
接下来定义$N_0(\delta):=N(\Delta,\delta),$ 我们有:
$$
N_0(\delta)=n_0+\cdots+n_k=\frac{(n-1+2k)(n-2+k)!}{k!(n-1)!},\quad k=k^-(\delta)>0.
$$
因此, $\iota_0(\delta)=\begin{cases}
N_0(\delta),&\delta>0;\\
-N_0(2-n-\delta),&\delta<0.
\end{cases}$ 上述推导给出了如下命题:
命题 17. 设$P,P^\ast $都是渐进于$\Delta$的, $\delta$非例外, $1<p\le q$时, $P:W^{2,p}_\delta\rightarrow W^{0,p}_{\delta-2}$是Fredholm算子, $N(P,\delta)=\dim\ker P$与$p$选取无关. 若$k^-(\delta)<\delta'\le \delta,$ 那么$N(P,\delta)=N(P,\delta'),$ 且若$u\in W_\delta^{2,p},$ $Pu\in W_{\delta'}^{0,p},$ 则$u\in W_{\delta'}^{2,p}.$
证: 由于$k^-(\delta)=k^-(\delta'),$
$\iota(P,\delta)-\iota(P,\delta')=\iota_0(\delta)-\iota_0(\delta')=0.$
展开即有
$$
N(P,\delta)-N(P,\delta')=N(P^\ast ,2-n-\delta)-N(P^\ast ,2-n-\delta').
$$
由于$L^p_{\delta'}\subset L^p_{\delta},$ 左侧非负右侧非正,
因此两侧皆为零, $N(P,\delta)=N(P,\delta').$
这说明$\ker(P,\delta)=\ker(P,\delta').$ 现若$Pu\in W_{\delta'}^{0,p},$
则$\left<{}Pu,v\right>=\left<{}u,P^\ast v\right>=0,$
$\,\forall\,v\in \ker(P^\ast ,2-n-\delta').$ 由$P$具闭值域,
$Pu\in \operatorname{Im}(P,\delta').$
因此$\,\exists\,w\in W_{\delta'}^{2,p}\subset W_\delta^{2,p},$
使得$Pu=Pw.$
即$w-u\in\ker(P,\delta)=\ker (P,\delta')\subset W_{\delta'}^{2,p}.$
进而$u\in W_{\delta'}^{2,p}.$
当$P=\Delta_g$时, $P^\ast =P.$
注意伴随由配对$(u,v)=\int uvd\mathcal{H}^n=\int uv\sqrt{g}dx$确定.
命题 18. 设$g_{ij}(x)$是$\mathbb{R}^n$上一致椭圆的, $(g_{ij}-\delta_{ij})\in W_0^{1,q},$ $n<q<\infty.$ 若$\delta$非例外, $1<p\le q,$ 则$\Delta_g:W_{\delta}^{2,p}\rightarrow W_{\delta-2}^{0,p}$是Fredholm的, $N(\Delta_g,\delta)=N_0(\delta).$
证: 由正则性条件, $\Delta_g$渐进于标准Laplace算子,
$$
\Delta_g=g^{ij}\partial^2_{ij}+\frac{\partial {}g^{ij} }{\partial {}x^j}\partial_i+\frac{g^{ik}g^{lj} }{2}\frac{\partial {}g_{jl} }{\partial {}x^l}\partial_i.
$$
又由$\Delta_g$的自伴随性, 可知$\Delta_g$可继承前述讨论, 是Fredholm算子.
由指标不变性, 只需证明当$\delta<0$时, $N(\Delta_g,\delta)=0.$
而$\Delta_gu=0,$ $\delta<0$时, 由$u\in W_{\delta}^{2,p}$出发,
反复利用前面的命题, 结合Sobolev不等式, 得到$u$在无穷远处趋于零.
这样利用极值原理即可得出结论.
由先前的命题, 自动有:
推论 19. 令$\delta$非例外, $n<q<\infty,$ $1<p\le q.$ 那么存在正常数$C,\varepsilon,$ 使得对任意度量$g_{ij},$ 满足$\Vert g_{ij}-\delta_{ij}\Vert_{1,q,0}\le \varepsilon,$ 有
$$
\Vert u-\ker \Delta_g\Vert_{2,p,\delta}\le C\Vert\Delta_g u\Vert_{0,p,\delta-2},\quad \,\forall\,u\in W_{\delta}^{2,p}.
$$
最后我们来看本节得到的最重要的一个结论:
定理 20. 设$P$渐进于$\Delta,$ 速率为$\tau>0,$ $u\in W_{\delta}^{2,q},$ $\delta$非例外, 在$E_R$上$Pu=0.$ 那么存在例外值$k\le k^-(\delta),$ $k$次齐次调和函数$h_k\in C^\infty(\mathbb{R}^n),$ 使得
$$
u-h_k=o(r^{k-\tau}),\quad r\rightarrow \infty.
$$
证: 记在$\mathbb{R}^n$中$Pu=f,$
那么$\Delta u=F\in W^{0,q}_{\delta-2-\tau}.$ 在$E_R$上,
$$
F=(\delta_{ij}-a^{ij})\partial^2_{ij}u-b^i\partial_iu-cu.
$$
由于$\Delta:{W'}_{\delta-\tau}^{2,q}\rightarrow {W'}_{\delta-\tau-2}^{0,q}$是同构,
$\,\exists\,v\in W_{\delta-\tau}^{2,q},$ 使得
$$
\Delta(u-v)=0,\quad |x|>R.
$$
由经典的调和函数展开式, 我们有:
$$
u-v=h_k+O(r^{k-1}),
$$
$k\le k^-(\delta),$ $h_k$如上所述.
现$u-h_k=v+O(r^{k-1})\in W_{\delta-\tau}^{2,q},$
对应着$F\in W_{\delta-2-2\tau}^{0,q},$
继续说明$v\in W_{\delta-2\tau}^{2,q}.$
重复直到我们得到$u-h_k\in W_{k-\tau}^{2,q}$即可.
文章最后更新于 2022-04-14 16:45:33
The Mass of an Asymptotically Flat Manifold - Robert Bartnik
Weighted Sobolev Spaces
We first discuss in $\mathbb{R}^n,$ $n\ge 3$. Note $r=|x|,$
$\sigma=\sqrt{1+r^2},$ respectively in $L_{\mathrm{loc} }^p(\mathbb{R}^n),$
The weighted Lebesgue space is defined in $L_{\mathrm{loc} }^p(\mathbb{R}^n\setminus\{0\})$:
$$
L_\delta^p:\quad \Vert u\Vert_{p,\delta}=
\left\{
\begin{aligned}
&\left(\int_{\mathbb{R}^n}|u|^p\sigma^{-\delta p-n}dx\right)^{\frac{1}{p} },&p<\infty,\\
&\operatorname{ess}\operatorname{sup}_{\mathbb{R}^n}(\sigma^{-\delta}|u|), &p=\infty,
\end{aligned}
\right.
$$
$$
{L'}_\delta^p:\quad \Vert u\Vert'_{p,\delta}=
\left\{
\begin{aligned}
&\left(\int_{\mathbb{R}^n\setminus\{0\} }|u|^pr^{-\delta p-n}dx\right)^{\frac{1}{p} },&p<\infty,\\
&\operatorname{ess}\operatorname{sup}_{\mathbb{R}^n\setminus\{0\} }(r^{-\delta}|u|), &p=\infty,
\end{aligned}
\right.
$$
Next, define the weighted Sobolev space:
$$
W_\delta^{k,p}:\quad \Vert u\Vert_{k,p,\delta}=\sum_{j=0}^k\Vert D^ju\Vert_{p,\delta-j},
$$
$$
{W'}_\delta^{k,p}:\quad \Vert u\Vert'_{k,p,\delta}=\sum_{j=0}^k\Vert D^ju\Vert'_{p,\delta-j},
$$
Note that the difference in weight here makes the dimensions of the addition terms basically the same. When $p\in [1,\infty)$,
$C_c^\infty(\mathbb{R}^n),$
$C_c^\infty(\mathbb{R}^n\setminus\{0\})$ is dense among them respectively.
The weights are selected such that for function $u_R(x)=u(Rx),$
$$
\Vert u\Vert'_{k,p,\delta}=R^{-\delta}\Vert u_R\Vert'_{k,p,\delta},
$$
$$
\Vert u\Vert_{k,p,\delta;A_R}\approx R^{-\delta}\Vert u_R\Vert_{k,p,\delta;A_1}, \quad R\ge 1.
$$
Here $A_R:=B_{2R}\setminus B_R,$ $B_R$ is a closed sphere with a radius $R$ at the origin.
"$\approx$" means that the two can be controlled by a constant multiple that is independent of $R$. Just consider controlling $\sigma$.
Next we enumerate the following properties without proof:
Theorem 1. For $1\le p\le q\le \infty,$ $\delta_2<\delta_1,$ $u\in L_{\delta_2}^q,$ there is $\Vert u\Vert_{p,\delta_1}\le c\Vert u\Vert_{q,\delta_2},$. Therefore, the higher $L_{\delta_2}^q\subset L_{\delta_1}^p.$, that is, $p$, the smaller $\delta$, the better the properties of $L_\delta^p$, and the higher the function requirements.
Theorem 2 (Hölder's inequality). $u\in L_{\delta_1}^q,$ $v\in L_{\delta_2}^r,$ $\delta=\delta_1+\delta_2,$ $1\le p,q,r\le \infty,$ $\frac{1}{p}=\frac{1}{q}+\frac{1}{r},$ then
$$
\Vert uv\Vert_{p,\delta}\le \Vert u\Vert_{q,\delta_1}\Vert v\Vert_{r,\delta_2}.
$$
Theorem 3 (Interpolation inequality). $\,\forall\,\varepsilon>0,$ $\,\exists\,C(\varepsilon)>0,$ $\,\forall\,u\in W_{\delta}^{2,p},$ $p\in [1,\infty],$
$$
\Vert u\Vert_{1,p,\delta}\le \varepsilon\Vert u\Vert_{2,p,\delta}+C(\varepsilon)\Vert u\Vert_{0,p,\delta}.
$$
Theorem 4 (Sobolev's inequality). If $u\in W_{\delta}^{k,p},$ then
$$
\Vert u\Vert_{q,\delta}\le C\Vert u\Vert_{k,p,\delta},\quad 0<\frac{n}{p}-k\le \frac{n}{q}\le \frac{n}{p};
$$
$$
\Vert u\Vert_{\infty,\delta}\le C\Vert u\Vert_{k,p,\delta},\quad \frac{n}{p}-k<0,\quad \text{且} |u(x)|=o(r^{\delta}).
$$
Define the weighted Hölder module:
$$
\Vert u\Vert_{C_\delta^{0,\alpha} }:=\sup_{x\in \mathbb{R}^n}\left(\sigma^{-\delta+\alpha}(x)\sup_{4|x-y|\le \sigma(x)}\frac{|u(x)-u(y)|}{|x-y|^{\alpha} }\right)+\sup_{x\in\mathbb{R}^n}\sigma^{-\delta}(x)|u(x)|.
$$
Theorem 5 (Negative exponential Sobolev inequality). If $u\in W_{\delta}^{k,p},$ $0<\alpha\le k-\frac{n}{p}\le 1,$ then
$$
\Vert u\Vert_{C_{\delta}^{0,\alpha} }\le C\Vert u\Vert_{k,p,\delta},\quad \Vert u\Vert_{C_{\delta}^{0,\alpha}(A_R)}=o(1).
$$
Theorem 6 (Poincaré inequality). $\,\forall\,u\in {W'}_{\delta}^{1,p},$ $1\le p<\infty,$ $\delta\neq 0,$
$$
\Vert u\Vert_{p,\delta}'\le |\delta|^{-1}\Vert u_r\Vert'_{p,\delta-1}\le |\delta|^{-1}\Vert Du\Vert'_{p,\delta-1},\quad u_r=\partial_ru=r^{-1}x\cdot Du.
$$
$$
\Vert u\Vert_{p,\delta}\le C\Vert u_r\Vert_{p,\delta-1},\quad \delta<0,\quad u\in W_{\delta}^{1,p}.
$$
Lemma 7. $k>j,$ $\delta<\varepsilon,$ $1\le p<\infty,$ $W_\delta^{k,p}\Subset W_{\varepsilon}^{j,p}$ are tight embeddings.
Asymptotic elliptic operator
Inequality estimation
Definition 8. Operator $Pu:=a^{ij}(x)\partial_{ij}^2u+b^i(x)\partial_iu+c(x)u$ is called (at rate $\tau$) asymptotic to $\Delta$, if $\,\exists\,n<q<\infty,$ $\tau\ge 0,$ $C_1,\lambda,$ makes
$$
\lambda |\xi|^2\le a^{ij}(x)\xi_i\xi_j\le \lambda^{-1}|\xi|^2,\quad \,\forall\,x,\xi\in \mathbb{R}^n;
$$
$$
\Vert a^{ij}-\delta_{ij}\Vert_{1,q,-\tau}+\Vert b^i\Vert_{0,q-1,-1-\tau}+\Vert c\Vert_{0,\frac{q}{2},-2-\tau}\le C_1.
$$
According to the previous theorem, $a^{ij}$ is Hölder continuous,
And $|a^{ij}-\delta_{ij}|=o(r^{-\tau}).$ for $1\le p\le q,$
$\delta\in \mathbb{R},$
$P:W_{\delta}^{2,p}\rightarrow W_{\delta-2}^{0,p}$ is a bounded operator.
In fact, the following estimates are true:
Proposition 9. Suppose $P$ asymptotically advances to $\Delta,$ $1<p\le q,$ $\delta\in \mathbb{R},$, then there is a constant $C=C(n,p,q,\delta,C_1,\lambda),$ such that if $u\in L_\delta^{0,p},$ $Pu\in L_{\delta-2}^{0,p},$ then $u\in W_{\delta}^{2,p},$
$$
\Vert u\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta}).
$$
To prove this proposition, just do the usual $L^p$ estimation in second-order elliptic equations.
Note that this estimate also holds for the ${W'}_{\delta}^{k,p}$ module.
Next, let’s discuss the Fredholm property of operator $P$, that is, consider whether its image is closed,
And the finiteness of the kernel and co-kernel dimensions. The above proposition is not enough to explain that it has this property.
We need to strengthen the estimate so that the right-hand error term is compact with respect to $W_\delta^{2,p}$.
The weight parameter $\delta$ is called nonexceptional,
If $\delta\in \mathbb{R}\setminus\{k\in \mathbb{Z},k\neq -1,-2,\cdots,3-n\}.$
Exceptional
value) $\{k\in \mathbb{Z}, k\neq -1,-2,\cdots,3-n\}$ corresponds to the growth order of the $\mathbb{R}^n\setminus B_1$ upharmonic function.
definition
$$
k^-(\delta)=\max\{k \text{ exceptional},k<\delta\}.
$$
Theorem 10. Assume $\delta$ is not an exception, $p\in (1,\infty),$ $s$ is a non-negative integer. Then there is an isomorphism
$$
\Delta:{W'}_\delta^{s+2,p}\rightarrow {W'}_{\delta-2}^{s,p},
$$
and there is a constant $C=C(n,p,\delta,s),$
$$
\Vert u\Vert'_{s+2,p,\delta}\le C\Vert\Delta u\Vert'_{s,p,\delta-2}.
$$
Certificate: It only needs to be proved for the case of $s=0$.
First, we explain that the distribution inverse of $\Delta$ has a convolution kernel $K(x,y):$
$$
c_nK(x,y)=
\left\{\begin{aligned}
&|x-y|^{2-n} &2-n<\delta<0,\\
&|x-y|^{2-n}-|y|^{2-n}\sum_{j=0}^k P_j^\lambda(\mu)\left(\frac{|x|}{|y|}\right)^j &k\ge 0,\\
&|x-y|^{2-n}-|x|^{2-n}\sum_{j=0}^k P_j^\lambda(\mu)\left(\frac{|y|}{|x|}\right)^j &k<2-n.
\end{aligned}\right.
$$
Here $k=k^-(\delta),$ $\mu=\frac{\left<{}x,y\right>}{|x||y|},$
$P_j^\lambda$ is an ultraspherical function, which appears in the Taylor expansion:
$$
|x-y|^{2-n}=|x|^{2-n}\sum_0^\infty P_j^\lambda(\mu)\left(\frac{|y|}{|x|}\right)^j,\quad |y|<|x|.
$$
First, let’s verify that the convolution kernel $K$ indeed defines the bounded operator of ${W'}_{\delta-2}^{0,p}\rightarrow {W'}_{\delta}^{0,p}$.
Taking the case of $k\ge 0$ as an example, we have an estimate
$$
|K(x,y)|\le c(n,k)|x-y|^{2-n}
\left\{\begin{aligned}
&\left(\frac{|x|}{|y|}\right)^{k+1} &\frac{|x|}{|y|}<\frac{1}{2},\\
&\left(\frac{|x|}{|y|}\right)^{n+k-2} & \frac{|x|}{|y|}\ge \frac{1}{2}.
\end{aligned}\right.
$$
We have the following lemma:
Lemma 11. *Fixed $p\in (1,\infty),$ $\frac{1}{p}+\frac{1}{p'}=1,$ $a,b\in \mathbb{R}$ and $a+b>0.$ have cores
$$
K'(x,y)=|x|^{-a}|x-y|^{2-n}|y|^{-b},\quad x\neq y.
$$
- Define $K'u(x)=\int K'(x,y)u(y)dy,$ for $u\in L^p(\mathbb{R}^n),$, then $K':L^p\rightarrow L^p$ is a bounded operator if and only if $a<\frac{n}{p},$ $b<\frac{n}{p'}.$
Let $K_1,K_2$ be the accounting operator $|x-y|^{2-n}\left(\frac{|x|}{|y|}\right)^\alpha,$
Corresponds to $\alpha=k+1,$ $\alpha=n+k-2$ respectively.
The above lemma shows that when $3-n+k<\delta<k+1$, the kernel
$$
K_1'(x,y)=|x|^{-\delta-\frac{n}{p} }K_1(x,y)|y|^{\delta-2+\frac{n}{p} }
$$
Defines the bounded operator of $L^p\rightarrow L^p$. Then
$$
\Vert K_1u\Vert_{p,\delta}'\le c\Vert r^{2-\delta-\frac{n}{p} }u\Vert_{L^p}\le c\Vert u\Vert'_{p,\delta-2},
$$
That is, $K_1:{W'}_{\delta-2}^{0,p}\rightarrow {W'}_{\delta}^{0,p}$ is a bounded operator.
Similarly, when $k<\delta<n+k-2$, $K_2'$ is a bounded operator on $L^p\rightarrow L^p$,
Furthermore, $K_2$ is also a bounded operator. These two estimates illustrate that when $k<\delta<k+1$,
$K$ is also a bounded operator, and the condition of $k=k^-(\delta),$ is automatically satisfied.
The other two cases in the $K$ definition are handled similarly.
We have the distribution identity:
$$
\Delta_x K(x,y)=\Delta_y K(x,y)=\delta(x-y),
$$
Therefore $K(\Delta u)=u,$
$\,\forall\,u\in C_c^\infty(\mathbb{R}^n\setminus \{0\}).$
So it is given by the boundedness of $K$ and the denseness of $C_c^\infty$
$$
\Vert u\Vert_{0,p,\delta}'=\Vert K(\Delta u)\Vert_{0,p,\delta}'\le C\Vert\Delta u\Vert'_{0,p,\delta-2},\quad \,\forall\,u\in {W'}_\delta^{k,p}.
$$
Combined with the previous $L^p$ estimation, the proof of the inequality stated in the theorem is given ($s=0$).
Since $\Delta:{W'}_{\delta}^{2,p}\rightarrow {W'}_{\delta-2}^{0,p}$ is a bounded operator,
This inequality immediately shows that it seems to be closed.
From the distribution identity, we see that,
$$
\Delta(Kf)(x)=f(x),\quad f\in C_c^\infty(\mathbb{R}^n\setminus \{0\}),\quad Kf\in {W'}_{\delta}^{0,p}.
$$
Therefore $\Delta$ is isomorphism, and the theorem is proved.
The theorem gives the following well-known Liouville theorem:
Corollary 12. Let $\Delta u=0,$ $u\in L_\delta^p,$ $1<p<\infty.$ let $k=k^-(\delta).$ if $k<0,$ then $u\equiv 0;$ if $k\ge 0,$ then $u$ be a harmonic polynomial of order $k$ and below.
Certificate: Due to $0\in {W'}_\delta^{s,p},$ $\,\forall\,s\ge 0,$
Derivation of $u\in {W'}_\delta^{s+2,p}.$ from Sobolev’s inequality,
$u\in C^\infty(\mathbb{R}^n).$ Let $h_k(x)$ be the $k$ degree Taylor expansion of $u(x)$,
That is
$$
u(x)=h_k(x)+O(|x|^{k+1}).
$$
Since $u$ is harmonic, we know that $h_k$ is a harmonic polynomial.
At this time $u-h_k\in {L'}_\delta^p,$ from the previous propositions and theorem,
$u-h_k\in {W'}_\delta^{2,p},$
And there is an estimate$\Vert u-h_k\Vert'_{2,p,\delta}\le C\Vert\Delta(u-h_k)\Vert_{0,p,\delta-2}'=0,$
Thus $u=h_k.$
Next, we give an estimate similar to the previous proposition. At first glance, the dimensions of the inequality are a bit confusing,
Hence the name scale-broken estimation (I guess).
Theorem 13. Assuming that $P$ is asymptotic to $\Delta,$ $\delta\in \mathbb{R}$ without exception, $q\ge p>1.$ in the asymptotic condition, then the kernel of the operator $P:W_\delta^{2,p}\rightarrow W_{\delta-2}^{0,p}$ is finite-dimensional and has a closed range, and for any $u\in W_{\delta}^{2,p},$, there is a constant $C,R$ controlled by $P,\delta,n,p,q$, such that
$$
\Vert u\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{L^p(B_R)}).
$$
Certificate: Take operator module
$$
\Vert P-\Delta\Vert_{op}:=\sup\{\Vert(P-\Delta)u\Vert_{0,p,\delta-2}:u\in W_{\delta}^{2,p},\Vert u\Vert_{2,p,\delta}=1\}.
$$
Define $\Vert\cdot\Vert_{op,R},$ similarly
It is required that the selected function $u$ has a support set in $E_R=\mathbb{R}^n\setminus B_R$,
That is, zero is taken in $B_R$. Then if $\operatorname{supp}(u)\subset E_R,$ is determined by $q>n,$
$$
\begin{aligned}
\Vert(P-\Delta)u\Vert_{0,p,\delta-2}\le& \sup_{|x|>R}\{|a^{ij}(x)-\delta_{ij}|\}\Vert D^2u\Vert_{0,p,\delta-2}\\
&+C\Vert b\Vert_{0,q,-1;E_R}\Vert Du\Vert_{1,p,\delta-1}+C\Vert c\Vert_{0,\frac{q}{2},-2;E_R}\Vert u\Vert_{2,p,\delta}
\end{aligned}
$$
Since $P$ is asymptotic to $\Delta,$
$$
\Vert P-\Delta\Vert_{op,R}=o(1),\quad R\rightarrow \infty.
$$
Take $\chi\in C_c^\infty(B_2),$ $\chi\in [0,1],$ $\chi|_{B_1}=1.$
Take the undetermined constant $R,$ and let $\chi_R(x)=\chi(\frac{x}{R}),$ $u_0=\chi_Ru,$
$u_\infty=(1-\chi_R)u,$ $u=u_0+u_\infty.$ From the previous theorem, we have:
$$
\begin{aligned}
\Vert u_\infty\Vert_{2,p,\delta}&\le C\Vert\Delta u_{\infty}\Vert_{0,p,\delta-2}=C\Vert(\Delta-P+P) u_{\infty}\Vert_{0,p,\delta-2}\\
&\le C(\Vert Pu_{\infty}\Vert_{0,p,\delta-2}+\Vert P-\Delta\Vert_{op,R}\Vert u_{\infty}\Vert_{2,p,\delta})
\end{aligned}
$$
Since $\Vert P-\Delta\Vert_{op,R}=o(1),$ has for sufficiently large $R,$
$$
\Vert u_{\infty}\Vert_{2,p,\delta}\le C\Vert Pu_\infty\Vert_{0,p,\delta-2}.
$$
To further estimate, there are
$$
\begin{aligned}
\Vert Pu_\infty\Vert_{0,p,\delta-2}&\le \Vert Pu\Vert_{0,p,\delta-2}+\Vert 2a^{ij}u_i\partial_j\chi_R+(a^{ij}\partial^2_{ij}\chi_R+b^i\partial_i \chi_R)u\Vert_{0,p,\delta-2;A_R}\\
&\le \Vert Pu\Vert_{0,p,\delta-2}+C\Vert u\Vert_{1,p,\delta;A_R}
\end{aligned}
$$
Then use the interpolation inequality to further transform the $\Vert\cdot\Vert_{1,p,\delta}$ term,
And $L^p$ estimate, that is, we have
$$
\Vert u_\infty\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta;A_R}).
$$
At the same time, it is estimated by $L^p$, and for $u_0$
$$
\Vert u_0\Vert_{2,p,\delta}\le C(\Vert Pu_0\Vert_{0,p,\delta-2}+\Vert u_0\Vert_{0,p,\delta})\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert Pu_{\infty}\Vert_{0,p,\delta-2}+\Vert u\Vert_{0,p,\delta;{B_{2R} } }).
$$
Using the previous estimates about $Pu_\infty$ again, we get the inequalities required by the theorem:
$$
\Vert u\Vert_{2,p,\delta}\le \Vert u_0\Vert_{2,p,\delta}+\Vert u_\infty\Vert_{2,p,\delta}\le C(\Vert Pu\Vert_{0,p,\delta-2}+\Vert u\Vert_{L^p(B_R)}).
$$
Next, we discuss the finiteness of the kernel dimension. Let $\{u_k\}$ be the function sequence in $\ker P$,
Satisfies $\Vert u_k\Vert_{2,p,\delta}=1.$ By the compact embedding theorem,
It may be assumed that $\{u_k\}$ itself converges strongly in $L^p(B_R)$. At this time, due to $Pu_k\equiv 0,$
The inequality just proved shows that $\{u_k\}$ is the Cauchy column in $W_\delta^{2,p}$.
Therefore, the unit sphere in $\ker P$ is compact and thus has finite dimensions.
Since $\ker P$ has limited dimensions, $W_\delta^{2,p}=\ker P\oplus Z,$ and $Z$ are decomposed into closed sets.
Assert that there is a constant $C$ such that
$$
\Vert u\Vert_{2,p,\delta}\le C\Vert Pu\Vert_{0,p,\delta-2},\quad \,\forall\,u\in Z.
$$
Otherwise, there exists $\{u_k\},$ $\Vert u_k\Vert_{2,p,\delta}\equiv 1,$
And $\Vert Pu_k\Vert_{0,p,\delta-2}\rightarrow 0.$ at this time, according to the compact embedding theorem,
Using the inequality in the theorem, we once again obtain that $\{u_k\}$ is the Cauchy column in $W_{\delta}^{2,p}$.
Due to the closure of $Z$, it converges to $u_0\in Z.$. However, due to the continuity of $P$, $Pu_0=0,$
$u_0\in \ker P.$ thus $u_0=0,$
But this is contradictory to $\Vert u_k\Vert_{2,p,\delta}\equiv 1$,
Because $u_k\rightarrow u_0.$, the assertion is established, immediately indicating that $P$ has a closed range.
Half Fredholm Nature
Next, we further consider the dimension of the kernel of $P$. Note
$$
N(P,\delta)=\dim\ker (P:W_\delta^{2,p}\rightarrow W_{\delta-2}^{0,p}), \quad 1<p\le q.
$$
From the $L^p$ estimation and Sobolev's inequality, it can be shown that the selection of $N(P,\delta)$ and $p$ has nothing to do.
In order to simplify the notation below, we will write $X=W_\delta^{2,p},$ $Y=W_{\delta-2}^{0,p},$
$\delta$ is not an exception, $1<p\le q.$ then according to the theorem just now,
We know that $P:X\rightarrow Y$ is Half Fredholm , that is, the kernel is finite-dimensional,
And has a closed range.
Proposition 14. Assume $P:X\rightarrow Y$ is the semi-Fredholm operator between Banach spaces, then there is a constant $C>0,$ with an estimate
$$
\Vert u-\ker P\Vert_{X}=\Vert u\Vert_{X/\ker P}\le C\Vert Pu\Vert_Y.
$$
At the same time, there are $\varepsilon>0,$
such that for any satisfaction $\Vert P-P'\Vert_{op}<\varepsilon$ The other half Fredholm operator of $P',$
$$
\dim \ker P'\le \dim \ker P.
$$
Certificate: It is known that $\operatorname{Im}P$ is closed, then it is a Banach space,
Therefore the first estimate is obtained immediately by the inverse operator theorem. Now let $\varepsilon=\frac{1}{2C},$
Assuming that the kernel dimension inequality does not hold, then there is $u\in \ker P',$ $\Vert u\Vert_X=1,$
$\Vert u-\ker P\Vert_X>\frac{1}{2}.$ At this time
$$
\frac{1}{2}<\Vert u-\ker P\Vert_{X}\le C\Vert Pu\Vert_{Y}\le C\Vert P-P'\Vert_{op}\Vert u\Vert_X<C\varepsilon,
$$
produce conflicts.
It can be seen that $\frac{1}{C}$ measures the distance between $P$ and other semi-Fredholm operators with larger kernels.
The following proposition in the opposite direction makes this point even more profound:
Proposition 15. $X,Y$ Same as above, let $\mathcal{F}$ be the set of operators that are asymptotic to $\Delta$ from $X$ to $Y$. Let $\mathcal{U}\subset \mathcal{F}$ be the set of operators that satisfy $\dim\ker P\equiv N$, then given $P\in \mathcal{U},$ $\,\exists\,\varepsilon,C>0,$ such that
$$
\Vert u-\ker P'\Vert_{X}\le C\Vert P'u\Vert_Y,\quad \,\forall\,u\in X,P'\in \mathcal{U}, \Vert P-P'\Vert_{op}\le \varepsilon.
$$
Certificate: Otherwise, there is a sequence $P_k\rightarrow P\in \mathcal{U},$
$\{u_k\}\subset X,$ Satisfy$\Vert u_k\Vert_X=1,$
$\Vert u_k-\ker P_k\Vert_{X}>\frac{1}{2},$ And
$$
\Vert u_k-\ker P_k\Vert_{X}\ge k\Vert P_ku_k\Vert_Y.
$$
This shows that $P_ku_k\rightarrow 0.$
At this time, we can continue to use the tight embedding method in the previous theorem to obtain the subsequence $u_k\rightarrow u,$
Meet $Pu=0,$ $\Vert u\Vert_X=1,$
And for sufficiently large $k$ there is $\Vert u-\ker P_k\Vert\ge \frac{1}{4}.$
Similarly, take any $v_k\in \ker P_k,$ $\Vert v_k\Vert_X=1,$
In the same way, it also has sub-sequences that converge to $v\in \ker P.$
This shows that the $N$-dimensional $\ker P_k$ converges to the $N$-dimensional subspace in $\ker P$.
However, according to the estimation in the previous paragraph, $u$ is not in such a subspace, but $u\in \ker P.$
This illustrates the $\dim \ker P\ge N+1,$ contradiction.
Finally, we give the following theorem, which shows that the set of semi-Fredholm operators with larger kernels is smaller.
Theorem 16. Assume $P:X\rightarrow Y$ is semi-Fredholm, $A:X\rightarrow Y$ is $P$-bounded, that is, $\Vert Au\Vert_Y\le C(\Vert Pu\Vert_Y+\Vert u\Vert_X),$ $\,\forall\,u\in X.$, then there is $\lambda_0>0,$ such that $|\lambda|<\lambda_0,$ $P+\lambda A$ is always semi-Fredholm, and $\dim \ker (P+\lambda A)$ is constant at $0<\lambda<\lambda_0$ (note that $\lambda=0$ is not among them).
Fredholm indicator
We have the adjoint operator $P^\ast :W^{0,p'}_{2-n-\delta}\rightarrow W^{-2,p'}_{-n-\delta},$
$\frac{1}{p'}+\frac{1}{p}=1.$ If it is also gradual,
Then we can pass Fredholm indicator Get more information. From the previous proposition,
When $1<p'\le \widetilde{q}$, $\ker P^\ast \subset W^{2,p'}_{2-n-\delta}.$
When $\delta$ is not exceptional, the adjoint is also a semi-Fredholm operator, thus:
$$
\dim \operatorname{coker}P=\dim \ker P^\ast =N(P^\ast ,2-n-\delta).
$$
Therefore $P$ becomes Fredholm operator, tools Fredholm indicator:
$$
\iota(P,\delta)=N(P,\delta)-N(P^\ast ,2-n-\delta).
$$
Since the index is locally constant on the Fredholm operator space, we have
$$
\iota(P,\delta)=\iota(\Delta,\delta)=:\iota_0(\delta).
$$
And $\iota_0(\delta)$ can be calculated.
Let $H_k$ be the set of homogeneous harmonic polynomials of degree $k$ in $\mathbb{R}^n$,
We have $n_k=\dim H_k=\frac{(n-2+2k)(n-3+k)!}{k!(n-2)!}.$
Next defining $N_0(\delta):=N(\Delta,\delta),$ we have:
$$
N_0(\delta)=n_0+\cdots+n_k=\frac{(n-1+2k)(n-2+k)!}{k!(n-1)!},\quad k=k^-(\delta)>0.
$$
Therefore, $\iota_0(\delta)=\begin{cases}
N_0(\delta),&\delta>0;\\
-N_0(2-n-\delta),&\delta<0.
\end{cases}$ the above derivation gives the following proposition:
Proposition 17. Assume $P,P^\ast $ is asymptotic to $\Delta$, $\delta$ is no exception, when $1<p\le q$, $P:W^{2,p}_\delta\rightarrow W^{0,p}_{\delta-2}$ is the Fredholm operator, $N(P,\delta)=\dim\ker P$ has nothing to do with the selection of $p$. If $k^-(\delta)<\delta'\le \delta,$ then $N(P,\delta)=N(P,\delta'),$ and if $u\in W_\delta^{2,p},$ $Pu\in W_{\delta'}^{0,p},$ then $u\in W_{\delta'}^{2,p}.$
Certificate: Due to $k^-(\delta)=k^-(\delta'),$
$\iota(P,\delta)-\iota(P,\delta')=\iota_0(\delta)-\iota_0(\delta')=0.$
Expand it
$$
N(P,\delta)-N(P,\delta')=N(P^\ast ,2-n-\delta)-N(P^\ast ,2-n-\delta').
$$
Since the left side of $L^p_{\delta'}\subset L^p_{\delta},$ is non-negative and the right side is non-positive,
So both sides are zero, $N(P,\delta)=N(P,\delta').$
This shows that $\ker(P,\delta)=\ker(P,\delta').$ is now $Pu\in W_{\delta'}^{0,p},$
Then $\left<{}Pu,v\right>=\left<{}u,P^\ast v\right>=0,$
$\,\forall\,v\in \ker(P^\ast ,2-n-\delta').$ has a closed range from $P$,
$Pu\in \operatorname{Im}(P,\delta').$
Therefore$\,\exists\,w\in W_{\delta'}^{2,p}\subset W_\delta^{2,p},$
Make$Pu=Pw.$
That is $w-u\in\ker(P,\delta)=\ker (P,\delta')\subset W_{\delta'}^{2,p}.$
Then $u\in W_{\delta'}^{2,p}.$
When $P=\Delta_g$, $P^\ast =P.$
Note that companionship is determined by pairing $(u,v)=\int uvd\mathcal{H}^n=\int uv\sqrt{g}dx$.
Proposition 18. Suppose $g_{ij}(x)$ is a uniform ellipse on $\mathbb{R}^n$, $(g_{ij}-\delta_{ij})\in W_0^{1,q},$ $n<q<\infty.$ If $\delta$ is not an exception, $1<p\le q,$ then $\Delta_g:W_{\delta}^{2,p}\rightarrow W_{\delta-2}^{0,p}$ is Fredholm's, $N(\Delta_g,\delta)=N_0(\delta).$
Certificate: Due to the regularity condition, $\Delta_g$ is asymptotic to the standard Laplace operator,
$$
\Delta_g=g^{ij}\partial^2_{ij}+\frac{\partial {}g^{ij} }{\partial {}x^j}\partial_i+\frac{g^{ik}g^{lj} }{2}\frac{\partial {}g_{jl} }{\partial {}x^l}\partial_i.
$$
From the self-adjoint property of $\Delta_g$, it can be seen that $\Delta_g$ can inherit the above discussion and is the Fredholm operator.
By index invariance, we only need to prove that when $\delta<0$, $N(\Delta_g,\delta)=0.$
And when $\Delta_gu=0,$ $\delta<0$, starting from $u\in W_{\delta}^{2,p}$,
By repeatedly using the previous proposition and combining it with Sobolev's inequality, we get that $u$ tends to zero at infinity.
In this way, conclusions can be drawn using the extreme value principle.
From the previous proposition, we automatically have:
Corollary 19. Let $\delta$ be non-exceptional, $n<q<\infty,$ $1<p\le q.$ then there exists a positive constant $C,\varepsilon,$ such that for any metric $g_{ij},$ satisfying $\Vert g_{ij}-\delta_{ij}\Vert_{1,q,0}\le \varepsilon,$ we have
$$
\Vert u-\ker \Delta_g\Vert_{2,p,\delta}\le C\Vert\Delta_g u\Vert_{0,p,\delta-2},\quad \,\forall\,u\in W_{\delta}^{2,p}.
$$
Finally, let’s look at the most important conclusion drawn in this section:
Theorem 20. Assume $P$ is asymptotic to $\Delta,$ at a rate $\tau>0,$ $u\in W_{\delta}^{2,q},$ $\delta$ is not exceptional, and on $E_R$ $Pu=0.$ then there is an exceptional value $k\le k^-(\delta),$ $k$ homogeneous harmonic function $h_k\in C^\infty(\mathbb{R}^n),$ such that
$$
u-h_k=o(r^{k-\tau}),\quad r\rightarrow \infty.
$$
Certificate: Recorded in $\mathbb{R}^n$ in $Pu=f,$
Then $\Delta u=F\in W^{0,q}_{\delta-2-\tau}.$ is on $E_R$,
$$
F=(\delta_{ij}-a^{ij})\partial^2_{ij}u-b^i\partial_iu-cu.
$$
Since $\Delta:{W'}_{\delta-\tau}^{2,q}\rightarrow {W'}_{\delta-\tau-2}^{0,q}$ is isomorphic,
$\,\exists\,v\in W_{\delta-\tau}^{2,q},$ makes
$$
\Delta(u-v)=0,\quad |x|>R.
$$
From the classic harmonic function expansion, we have:
$$
u-v=h_k+O(r^{k-1}),
$$
$k\le k^-(\delta),$ $h_k$ as above.
Now $u-h_k=v+O(r^{k-1})\in W_{\delta-\tau}^{2,q},$
Corresponds to $F\in W_{\delta-2-2\tau}^{0,q},$
Continue explaining$v\in W_{\delta-2\tau}^{2,q}.$
Repeat until we get $u-h_k\in W_{k-\tau}^{2,q}$.
The article was last updated on 2022-04-14 16:45:33
