Machine-translated from Chinese.
实指数Sobolev空间
下面希望对$H^{m,p}(\mathbb{R}^n)$将$m$为实数的情形给出定义.
考虑$p=2$的情形, 记$H^{m,2}$为$H^m.$
由于$H^m(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)\subset \mathscr{S}'(\mathbb{R}^n),$
可以对其做Fourier变换. 由变换性质, 我们有:
$$\begin{aligned}
\parallel f\parallel_{H^m}&=(2\pi)^{-n}\left(\sum_{|\alpha|\le m}\parallel\xi^\alpha \widehat{f}(\xi)\parallel_{L^2}^2\right)^\frac{1}{2}\\
&\sim \parallel(1+|\xi|^2)^{\frac{m}{2} }\widehat{f}(\xi)\parallel_{L^2}.
\end{aligned}$$
于是对$s\in \mathbb{R}^n,$
可以定义实指数Sobolev空间$H^s(\mathbb{R}^n)\subset \mathscr{S}'(\mathbb{R}^n),$
配备模
$$\parallel f\parallel_{H^s}=\parallel(1+|\xi|^2)^{\frac{s}{2} }\widehat{f}(\xi)\parallel_{L^2}.$$
在$s$为非负整数时由上述推导过程, 与原始定义的$H^{s,2}$相容.
可以证明它也是$C_c^\infty(\mathbb{R}^n)$的完备化空间,
配备与范数相容的内积. 类似地, 可以定义齐次模
$$|f|_{\dot{H}^s}=||\xi|^s\widehat{f}(\xi)|_{L^2}.$$
对实指数Sobolev空间, 我们有如下Sobolev不等式:
定理 1.1. $\,\forall\,s>\frac{n}{2},$ 我们有如下不等式:
$$\parallel f\parallel_{L^\infty}\le C\parallel f\parallel_{H^s}, \quad\,\forall\,f\in C_c^\infty(\mathbb{R}^n).$$
证: 由Fourier逆变换,
$f(x)=(2\pi)^{-n}\int_{\mathbb{R}^n} e^{ix\cdot \xi}\widehat{f}(\xi)d\xi,$
于是
$$\parallel f\parallel_{L^\infty}\le (2\pi)^{-n}\parallel\widehat{f}\parallel_{L^1}\le \parallel(1+|\xi|^2)^{\frac{s}{2} }\widehat{f}(\xi)\parallel_{L^2}\parallel(1+|\xi|^2)^{-\frac{s}{2} }\parallel_{L^2}=C\parallel f\parallel_{H^s}.$$
容易看出我们也有如下的事实:
定理 1.2. $(H^s(\mathbb{R}^n))^\ast =H^{-s}(\mathbb{R}^n).$
证: $\,\forall\,f\in H^s(\mathbb{R}^n),$
$g\in H^{-s}(\mathbb{R}^n),$ 我们有
$$\int_{\mathbb{R}^n} f\overline{g}dx=(2\pi)^{-n}\int_{\mathbb{R}^n} (1+|\xi|^2)^s\widehat{f}(\xi)(1+|\xi|^2)^{-s}\overline{\widehat{g} }(\xi)dx\le (2\pi)^{-n}\parallel f\parallel_{H^s}\parallel g\parallel_{H^{-s} }.$$
于是$H^{-s}(\mathbb{R}^n)\subset (H^s(\mathbb{R}^n))^\ast .$
反过来若$g\in (H^s)^\ast ,$ $g\in \mathscr{S}'(\mathbb{R}^n).$ 由同样的式子,
$$\left<{}f,g\right>=(2\pi)^{-n}\left<{}(1+|\xi|^2)^s \widehat{f},(1+|\xi|^2)^{-s}\widehat{g}\right>\le C\parallel f\parallel_{H^s}=C\parallel(1+|\xi|^2)^s\widehat{f}(\xi)\parallel_{L^2}.$$
由于$(1+|\xi|^2)^s \widehat{f}(\xi)$在$L^2$中稠密(包含$C_c^\infty$),
$(1+|\xi|^2)^{-s}\widehat{g}\in (L^2)^\ast =L^2,$
从而$\parallel g\parallel_{H^{-s} }=\parallel(1+|\xi|^2)^{-s}\widehat{g}\parallel_{L^2}<\infty,$
$g\in H^{-s}(\mathbb{R}^n).$
从而$(H^s(\mathbb{R}^n))^\ast =H^{-s}(\mathbb{R}^n).$
负整指数Sobolev空间
对一般的$H^{m,p},$ 定义$H^{-m,p'}:=(H_0^{m,p})^\ast \subset \mathscr{D}',$
$\frac{1}{p}+\frac{1}{p'}=1,$ 配备算子范数.
此时$C_c^\infty$也在其中稠密. 由前面的定理,
可以知道$H^s$与$H^{s,2}$的相容性在$s$为负整指数时也对.
迹定理
定理 1.3 (迹定理). $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ 记$x'=(x_2,\cdots,x_n),$ $\pi f(x')=f(0,x').$ 我们有如下不等式:
$$\parallel\pi f\parallel_{H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})}\le C\parallel f\parallel_{H^s(\mathbb{R}^n)},\quad \,\forall\,s>\frac{1}{2}.$$
证: 记$F_{n-1}f(x_1,\xi')=g(x_1,\xi')$为$f$做$n-1$次Fourier变换.
那么
$$\widehat{f}(\xi)=\int_{\mathbb{R} }e^{-ix_1\xi_1}g(x_1,\xi')dx_1,\quad g(x_1,\xi')=\frac{1}{2\pi}\int_{\mathbb{R} }e^{ix_1\xi_1}\widehat{f}(\xi)d\xi_1.$$
注意到$\widehat{\pi f}(\xi')=g(0,\xi')=\frac{1}{2\pi}\int_{\mathbb{R} }\widehat{f}(\xi)d\xi_1,$
而
$$\begin{aligned}
\int_{\mathbb{R}^n}|\widehat{f}|d\xi_1&\le \left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{-\frac{s}{2} }\right|^2d\xi_1\right)^\frac{1}{2}\\
&\le C\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\left(\int_{\mathbb{R} }\left(1+|\xi_1|+|\xi'|\right)^{-2s}d\xi_1\right)^\frac{1}{2}\\
&\le C\left(1+|\xi'|^2\right)^{\frac{-s+\frac{1}{2} }{2} }\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }
\end{aligned}$$
于是,
$$\begin{aligned}
\parallel\pi f\parallel_{H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})}&\le\frac{1}{2\pi} \parallel(1+|\xi'|^2)^{\frac{s-\frac{1}{2} }{2} }\int_{\mathbb{R} }|\widehat{f}(\xi)|d\xi_1\parallel_{L^2(\mathbb{R}^{n-1})}\\
&\le C\parallel\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\parallel_{L^2(\mathbb{R}^{n-1})}=C\parallel f\parallel_{H^s(\mathbb{R}^n)}
\end{aligned}$$
命题得证.
$\,\forall\,f\in H^s(\mathbb{R}^n),$ 改变$f$在零测集上的取值并没有意义,
因此只是取$f(0,x')$并没有实际意义, 无法反映$f$的性质. 不过由迹定理,
我们知道当$\{f_\nu\}\in C_c^\infty(\mathbb{R}^n),$
$f_\nu\xrightarrow{H^s}f$时,
$\{\pi f_{\nu}\}$构成$H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})$Cauchy列.
记其收敛到$\pi f,$ 称为$f$的迹. 迹定义了函数在边界处的取值,
反映其边界性质.
由展平技巧与单位分解, 对$f\in H^s(\Omega),$
我们也可以定义其在边界上的迹$\pi f\in H^{s-\frac{1}{2} }(\partial \Omega).$
常用$s=1$的情形,
特别地有$\parallel\pi f\parallel_{L^2(\partial\Omega)}\le C\parallel f\parallel_{H^1(\Omega)}.$
回忆空间$H_0^1(\Omega)$为$C_c^\infty(\Omega)$在$H^1$模下的完备化空间.
在给出迹的定义后, 事实上我们可以证明
$$H_0^1(\Omega)=\{f\in H^1(\Omega),\pi f=0\}.$$
这给出了$H_0^1(\Omega)$的另一种等价刻画.
Poincaré不等式
定理 1.4 (Poincaré不等式). 对有界区域$\Omega,$ $f\in H_{0}^{1,p}(\Omega),$ $p\in[1,\infty),$ 我们有如下不等式:
$$\parallel f\parallel_{L^p}\le C(\Omega)\parallel\nabla f\parallel_{L^p}$$
证:
只需对$f\in C_c^\infty(\Omega)$给出证明.
$\,\exists\,L>0$使得$\Omega\subset [-L,L]^n.$
那么$f(x)=\int_{-L}^{x_1}\partial_1f(y_1,x')dy_1.$ 于是
$$\begin{aligned}
|f(x)|&\le \int_{-L}^L |f_1(y_1,x')|dy_1\\
&\le (\int_{-L}^L|f_1|^pdy_1)^{\frac{1}{p} }(2L)^{\frac{1}{q} }\\
\parallel f\parallel_{L^p}&\le 2L\parallel f_1\parallel_{L^p}\le 2L\parallel\nabla f\parallel_{L^p}
\end{aligned}$$
紧嵌入定理
对有界区域$\Omega,$ 由Sobolev不等式与基本的分析,
$$\parallel f\parallel_{L^p(\Omega)}\le C\parallel f\parallel_{H^{m,q}(\Omega)},\quad \,\forall\,\frac{n}{p}\ge \frac{n}{q}-m.$$
因此有嵌入$H^{m,q}(\Omega)\hookrightarrow L^p(\Omega).$ 当不等式严格时,
即$\frac{n}{p}>\frac{n}{q}-m,$ 我们可以证明嵌入是紧的,
即将有界集映到列紧集.
取到严格不等号就可以说明嵌入是紧的, 大致是因为此时可以使用导数,
通过Ascoli-Arzela引理证明. 即设$\frac{n}{p}=\frac{n}{q}-s,$ 那么大致有
$$H^{m,q}\stackrel{c}\hookrightarrow H^{s,q}\hookrightarrow L^p$$
由于嵌入是连续的, 将收敛列映到收敛列, 只需说明第一个嵌入是紧的即可.
但是对一般的$q$尚未定义$H^{s,q},$ 因此上述说明只是形式上的.
具体证明时将不借助$H^{s,q},$ 直接给出证明.
我们可以对$p=2$时已定义的情形给出证明. 我们说明对$s>t>0,$ 有紧嵌入
$$H^s(\mathbb{R}^n)\stackrel{c}\hookrightarrow H^t_{loc}(\mathbb{R}^n).$$
这里$f\in H^t_{loc}(\mathbb{R}^n)$即为满足$\,\forall\,\varphi\in C_c^\infty(\mathbb{R}^n),$
$\varphi f\in H^t(\mathbb{R}^n)$的函数.
引理 1.5. $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ $s>t>0,$ 我们有如下不等式:
$$|f(x+h)-f(x)|_{\dot{H}^t(\mathbb{R}^n)}\le C|h|^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}.$$
证: 不妨设$s-t\in (0,1).$ 对一般情况在$(s,t)$间加分点即可.
$F[f(x+h)-f(x)]=(e^{-ih\cdot \xi}-1)\widehat{f}(\xi).$
注意到$|e^{-ih\cdot\xi}-1|\le 2$或$C|h\cdot\xi|,$
因此$|e^{-ih\cdot \xi}-1|\le C|h|^{s-t}|\xi|^{s-t}.$ 从而
$$|f(x+h)-f(x)|_{\dot{H}^t(\mathbb{R}^n)}\le C|h|^{s-t}\parallel|\xi|^{s}\widehat{f}(\xi)\parallel_{L^2(\mathbb{R}^n)}=C|h|^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}$$
推论 1.6. $|\alpha_\varepsilon\ast f-f|_{\dot{H}^t(\mathbb{R}^n)}\le C\varepsilon^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}$
因此设$\{f_{\nu}\}$为$\dot{H}^s(\mathbb{R}^n)$有界函数列,
那么$\,\forall\,\varepsilon>0,$
$\{\alpha_\varepsilon\ast f_\nu\}$为$C^m$有界函数列, $\,\forall\,m>t+1.$
从而由Arzela-Ascoli引理, 其有在任意紧集上$C^{m-1}$收敛的子列,
即$C_{loc}^{m-1}$紧. 自然地, 收敛列在$H^{m-1}$模意义下也收敛,
因此是$H^{m-1}_{loc}$紧的. 而$H^{m-1}\hookrightarrow H^t$为连续嵌入,
故$\{\alpha_\varepsilon\ast f_\nu\}$是$H^t_{loc}$紧的.
由推论,
对$\nu$一致地有$\alpha_\varepsilon\ast f_{\nu} \xrightarrow{\dot{H}^t} f_{\nu},$
由对角线法可证$\{f_{\nu}\}$有Cauchy子列, 从而是$H^t_{loc}$紧的.
因此$H^s(\mathbb{R}^n)$到$H^t_{loc}(\mathbb{R}^n)$的嵌入是紧的.
文章最后更新于 2021-11-13 18:16:30
Sobolev Spaces of Real Order
Below I hope to give a definition for the situation where $H^{m,p}(\mathbb{R}^n)$ and $m$ are real numbers.
Consider the situation of $p=2$, write $H^{m,2}$ as $H^m.$
Due to $H^m(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)\subset \mathscr{S}'(\mathbb{R}^n),$
Fourier transformation can be performed on it. From the transformation properties, we have:
$$\begin{aligned}
\parallel f\parallel_{H^m}&=(2\pi)^{-n}\left(\sum_{|\alpha|\le m}\parallel\xi^\alpha \widehat{f}(\xi)\parallel_{L^2}^2\right)^\frac{1}{2}\\
&\sim \parallel(1+|\xi|^2)^{\frac{m}{2} }\widehat{f}(\xi)\parallel_{L^2}.
\end{aligned}$$
So for $s\in \mathbb{R}^n,$
It is possible to define a real exponential Sobolev space $H^s(\mathbb{R}^n)\subset \mathscr{S}'(\mathbb{R}^n),$
Equipped with model
$$\parallel f\parallel_{H^s}=\parallel(1+|\xi|^2)^{\frac{s}{2} }\widehat{f}(\xi)\parallel_{L^2}.$$
When $s$ is a non-negative integer, the above derivation process is consistent with the original definition of $H^{s,2}$.
It can be proved that it is also the complete space of $C_c^\infty(\mathbb{R}^n)$,
Equipped with an inner product compatible with the norm. Similarly, one can define a homogeneous module
$$|f|_{\dot{H}^s}=||\xi|^s\widehat{f}(\xi)|_{L^2}.$$
For real exponential Sobolev spaces, we have the following Sobolev inequality:
Theorem 1.1. $\,\forall\,s>\frac{n}{2},$ We have the following inequality:
$$\parallel f\parallel_{L^\infty}\le C\parallel f\parallel_{H^s}, \quad\,\forall\,f\in C_c^\infty(\mathbb{R}^n).$$
Certificate: By Fourier inverse transformation,
$f(x)=(2\pi)^{-n}\int_{\mathbb{R}^n} e^{ix\cdot \xi}\widehat{f}(\xi)d\xi,$
So
$$\parallel f\parallel_{L^\infty}\le (2\pi)^{-n}\parallel\widehat{f}\parallel_{L^1}\le \parallel(1+|\xi|^2)^{\frac{s}{2} }\widehat{f}(\xi)\parallel_{L^2}\parallel(1+|\xi|^2)^{-\frac{s}{2} }\parallel_{L^2}=C\parallel f\parallel_{H^s}.$$
It is easy to see that we also have the following facts:
Theorem 1.2. $(H^s(\mathbb{R}^n))^\ast =H^{-s}(\mathbb{R}^n).$
Certificate: $\,\forall\,f\in H^s(\mathbb{R}^n),$
$g\in H^{-s}(\mathbb{R}^n),$ We have
$$\int_{\mathbb{R}^n} f\overline{g}dx=(2\pi)^{-n}\int_{\mathbb{R}^n} (1+|\xi|^2)^s\widehat{f}(\xi)(1+|\xi|^2)^{-s}\overline{\widehat{g} }(\xi)dx\le (2\pi)^{-n}\parallel f\parallel_{H^s}\parallel g\parallel_{H^{-s} }.$$
So$H^{-s}(\mathbb{R}^n)\subset (H^s(\mathbb{R}^n))^\ast .$
On the other hand, if $g\in (H^s)^\ast ,$ $g\in \mathscr{S}'(\mathbb{R}^n).$ has the same formula,
$$\left<{}f,g\right>=(2\pi)^{-n}\left<{}(1+|\xi|^2)^s \widehat{f},(1+|\xi|^2)^{-s}\widehat{g}\right>\le C\parallel f\parallel_{H^s}=C\parallel(1+|\xi|^2)^s\widehat{f}(\xi)\parallel_{L^2}.$$
Since $(1+|\xi|^2)^s \widehat{f}(\xi)$ is dense in $L^2$ (including $C_c^\infty$),
$(1+|\xi|^2)^{-s}\widehat{g}\in (L^2)^\ast =L^2,$
Thus $\parallel g\parallel_{H^{-s} }=\parallel(1+|\xi|^2)^{-s}\widehat{g}\parallel_{L^2}<\infty,$
$g\in H^{-s}(\mathbb{R}^n).$
Thus $(H^s(\mathbb{R}^n))^\ast =H^{-s}(\mathbb{R}^n).$
Negative Integral Exponential Sobolev Space
Define $H^{-m,p'}:=(H_0^{m,p})^\ast \subset \mathscr{D}',$ for general $H^{m,p},$
$\frac{1}{p}+\frac{1}{p'}=1,$ Equipped with operator norm.
At this time $C_c^\infty$ is also dense in it. According to the previous theorem,
It can be known that the compatibility of $H^s$ and $H^{s,2}$ is also correct when $s$ is a negative integer exponent.
trace theorem
Theorem 1.3 (Trace theorem). $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ Note $x'=(x_2,\cdots,x_n),$ $\pi f(x')=f(0,x').$ We have the following inequality:
$$\parallel\pi f\parallel_{H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})}\le C\parallel f\parallel_{H^s(\mathbb{R}^n)},\quad \,\forall\,s>\frac{1}{2}.$$
Certificate: Denote $F_{n-1}f(x_1,\xi')=g(x_1,\xi')$ as $f$ and perform $n-1$ times of Fourier transformation.
Then
$$\widehat{f}(\xi)=\int_{\mathbb{R} }e^{-ix_1\xi_1}g(x_1,\xi')dx_1,\quad g(x_1,\xi')=\frac{1}{2\pi}\int_{\mathbb{R} }e^{ix_1\xi_1}\widehat{f}(\xi)d\xi_1.$$
Notice$\widehat{\pi f}(\xi')=g(0,\xi')=\frac{1}{2\pi}\int_{\mathbb{R} }\widehat{f}(\xi)d\xi_1,$
And
$$\begin{aligned}
\int_{\mathbb{R}^n}|\widehat{f}|d\xi_1&\le \left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{-\frac{s}{2} }\right|^2d\xi_1\right)^\frac{1}{2}\\
&\le C\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\left(\int_{\mathbb{R} }\left(1+|\xi_1|+|\xi'|\right)^{-2s}d\xi_1\right)^\frac{1}{2}\\
&\le C\left(1+|\xi'|^2\right)^{\frac{-s+\frac{1}{2} }{2} }\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }
\end{aligned}$$
So,
$$\begin{aligned}
\parallel\pi f\parallel_{H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})}&\le\frac{1}{2\pi} \parallel(1+|\xi'|^2)^{\frac{s-\frac{1}{2} }{2} }\int_{\mathbb{R} }|\widehat{f}(\xi)|d\xi_1\parallel_{L^2(\mathbb{R}^{n-1})}\\
&\le C\parallel\left(\int_{\mathbb{R} }\left|\left(1+|\xi|^2\right)^{\frac{s}{2} }\widehat{f}\right|^2d\xi_1\right)^{\frac{1}{2} }\parallel_{L^2(\mathbb{R}^{n-1})}=C\parallel f\parallel_{H^s(\mathbb{R}^n)}
\end{aligned}$$
The proposition is proved.
$\,\forall\,f\in H^s(\mathbb{R}^n),$ Changing the value of $f$ on the zero test set does not make sense.
Therefore, just taking $f(0,x')$ has no practical significance and cannot reflect the properties of $f$. However, according to the trace theorem,
We know when $\{f_\nu\}\in C_c^\infty(\mathbb{R}^n),$
When $f_\nu\xrightarrow{H^s}f$,
$\{\pi f_{\nu}\}$ constitutes the $H^{s-\frac{1}{2} }(\mathbb{R}^{n-1})$ Cauchy column.
Note that it converges to $\pi f,$ and is called $f$ trace. The trace defines the value of the function at the boundary,
Reflect its boundary properties.
By flattening technique and partition of unity, for $f\in H^s(\Omega),$
We can also define its trace on the boundary $\pi f\in H^{s-\frac{1}{2} }(\partial \Omega).$
Commonly used $s=1$ situations,
Especially $\parallel\pi f\parallel_{L^2(\partial\Omega)}\le C\parallel f\parallel_{H^1(\Omega)}.$
The memory space $H_0^1(\Omega)$ is the complete space of $C_c^\infty(\Omega)$ under the module $H^1$.
After giving the definition of trace, in fact we can prove
$$H_0^1(\Omega)=\{f\in H^1(\Omega),\pi f=0\}.$$
This gives another equivalent characterization of $H_0^1(\Omega)$.
Poincaré inequality
Theorem 1.4 (Poincaré inequality). For the bounded region $\Omega,$ $f\in H_{0}^{1,p}(\Omega),$ $p\in[1,\infty),$ we have the following inequality:
$$\parallel f\parallel_{L^p}\le C(\Omega)\parallel\nabla f\parallel_{L^p}$$
Certificate:
Just give the proof for $f\in C_c^\infty(\Omega)$.
$\,\exists\,L>0$ makes $\Omega\subset [-L,L]^n.$
Then $f(x)=\int_{-L}^{x_1}\partial_1f(y_1,x')dy_1.$ So
$$\begin{aligned}
|f(x)|&\le \int_{-L}^L |f_1(y_1,x')|dy_1\\
&\le (\int_{-L}^L|f_1|^pdy_1)^{\frac{1}{p} }(2L)^{\frac{1}{q} }\\
\parallel f\parallel_{L^p}&\le 2L\parallel f_1\parallel_{L^p}\le 2L\parallel\nabla f\parallel_{L^p}
\end{aligned}$$
compact embedding theorem
For the bounded region $\Omega,$ by Sobolev's inequality and basic analysis,
$$\parallel f\parallel_{L^p(\Omega)}\le C\parallel f\parallel_{H^{m,q}(\Omega)},\quad \,\forall\,\frac{n}{p}\ge \frac{n}{q}-m.$$
Therefore there is embedding $H^{m,q}(\Omega)\hookrightarrow L^p(\Omega).$ when the inequality is strict,
That is, $\frac{n}{p}>\frac{n}{q}-m,$ we can prove that the embedding is compact,
That is to map bounded sets to compact sets.
Obtaining the strict inequality sign can indicate that the embedding is compact, roughly because derivatives can be used at this time,
Proved through the Ascoli-Arzela lemma. Assuming $\frac{n}{p}=\frac{n}{q}-s,$, then roughly
$$H^{m,q}\stackrel{c}\hookrightarrow H^{s,q}\hookrightarrow L^p$$
Since embeddings are continuous, to map convergence columns to convergence columns, it is only necessary to indicate that the first embedding is compact.
However, the general $q$ has not yet been defined $H^{s,q},$, so the above explanation is only formal.
The specific proof will be given directly without the help of $H^{s,q},$.
We can give a proof for the case defined for $p=2$. We show that for $s>t>0,$ there is a compact embedding
$$H^s(\mathbb{R}^n)\stackrel{c}\hookrightarrow H^t_{loc}(\mathbb{R}^n).$$
Here $f\in H^t_{loc}(\mathbb{R}^n)$ satisfies $\,\forall\,\varphi\in C_c^\infty(\mathbb{R}^n),$
Function of $\varphi f\in H^t(\mathbb{R}^n)$.
Lemma 1.5. $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ $s>t>0,$ We have the following inequality:
$$|f(x+h)-f(x)|_{\dot{H}^t(\mathbb{R}^n)}\le C|h|^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}.$$
Certificate: You might as well set $s-t\in (0,1).$ and add points between $(s,t)$ for general situations.
$F[f(x+h)-f(x)]=(e^{-ih\cdot \xi}-1)\widehat{f}(\xi).$
Note $|e^{-ih\cdot\xi}-1|\le 2$ or $C|h\cdot\xi|,$
Therefore $|e^{-ih\cdot \xi}-1|\le C|h|^{s-t}|\xi|^{s-t}.$ thus
$$|f(x+h)-f(x)|_{\dot{H}^t(\mathbb{R}^n)}\le C|h|^{s-t}\parallel|\xi|^{s}\widehat{f}(\xi)\parallel_{L^2(\mathbb{R}^n)}=C|h|^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}$$
Corollary 1.6. $|\alpha_\varepsilon\ast f-f|_{\dot{H}^t(\mathbb{R}^n)}\le C\varepsilon^{s-t}|f|_{\dot{H}^s(\mathbb{R}^n)}$
Therefore, let $\{f_{\nu}\}$ be $\dot{H}^s(\mathbb{R}^n)$ a bounded function sequence,
Then $\,\forall\,\varepsilon>0,$
$\{\alpha_\varepsilon\ast f_\nu\}$ is $C^m$ bounded function sequence, $\,\forall\,m>t+1.$
Therefore, according to the Arzela-Ascoli lemma, which has a subsequence that converges on any compact set $C^{m-1}$,
That is, $C_{loc}^{m-1}$ is compact. Naturally, the convergent sequence also converges in the $H^{m-1}$ modular sense,
Therefore, $H^{m-1}_{loc}$ is compact. And $H^{m-1}\hookrightarrow H^t$ is continuous embedding,
Therefore $\{\alpha_\varepsilon\ast f_\nu\}$ is tight to $H^t_{loc}$.
By inference,
Consistently have $\alpha_\varepsilon\ast f_{\nu} \xrightarrow{\dot{H}^t} f_{\nu},$ for $\nu$
It can be proved by the diagonal method that $\{f_{\nu}\}$ has a Cauchy subsequence, so $H^t_{loc}$ is compact.
Therefore the embedding from $H^s(\mathbb{R}^n)$ to $H^t_{loc}(\mathbb{R}^n)$ is tight.
The article was last updated on 2021-11-13 18:16:30
