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负指数Sobolev不等式
前面证明了如下不等式:
定理 1.1. 当$\frac{n}{p}=\frac{n}{q}-m$时, 有
$$\parallel f\parallel_{L^p(\Omega)}\le C|f|_{\dot{H}^{m,q}(\Omega)},\quad \,\forall\,f\in C_c^\infty(\Omega).$$
注意不等式仅在$p\in(1,+\infty),$ $q\in [1,+\infty)$时成立.
当$p=+\infty$时, 仅在$q=n=m=1$时成立.
也就是前面的不等式只在$\frac{n}{q}-m>0$时成立.
在$\frac{n}{q}-m<0$时, 也有相应的Sobolev不等式.
首先需要引入Hölder模的概念.
若连续函数$f$满足$|f(x)-f(y)|\le H|x-y|^{\alpha},$ $\alpha\in(0,1),$
则称$f$是Hölder连续的. 当$\alpha$取$1$时即为Lipschitz连续.
记$\Omega$上全体Hölder连续函数为$C^\alpha(\Omega),$ 定义Hölder模:
$$\parallel u\parallel_{C^\alpha}=\parallel u\parallel_{L^{\infty} }+|u|_{\dot{C}^\alpha},$$
其中$|\cdot|_{\dot{C}^\alpha}$称为齐次Hölder模, 定义为:
$$|u|_{\dot{C}^\alpha(\Omega)}:=\sup_{x\neq y\in \Omega}\frac{|u(x)-u(y)|}{|x-y|^{\alpha} }.$$
齐次Hölder模有限时$u$即为Hölder连续的, 量纲关于$u$为$1,$ $x$为$-\alpha.$
定理 1.2. 当$-k-1<-\alpha-k=\frac{n}{q}-m<-k,$ $k\in \mathbb{N}$时, $\,\forall\,\beta\in \mathbb{N}^n,$ $|\beta|=k,$ 有
$$|\partial^\beta u|_{\dot{C}^\alpha(\mathbb{R}^n)}\le C|u|_{\dot{H}^{m,q}(\mathbb{R}^n)},\quad \,\forall\,f\in C_c^\infty(\mathbb{R}^n).$$
证: 只需讨论$k=0$的情形. 对一般的情形,
对$\partial^\beta u$应用$-\alpha=\frac{n}{q}-(m-k)$的不等式即可, 即
$$|\partial^\beta u|_{\dot{C}^\alpha}
\le C|\partial^\beta u|_{\dot{H}^{m-k,q} }.
\le C|u|_{\dot{H}^{m,q} }.$$
同前面的讨论, 不妨设$m=1,$ 那么只需证$\,\forall\,x\neq y\in \Omega,$ 有
$$|u(x)-u(y)|\le C|u|_{\dot{H}^{1,q} }|x-y|^{\alpha}.$$
记$|x-y|=\rho,$ 取$B(\rho)$为含$x,y,$ 边长为$\rho$的立方体, 那么
$$|u(x)-u(y)|\le \left|\frac{1}{\rho^n}\int_{B(\rho)}u(x)-u(z)dz\right|+\left|\frac{1}{\rho^n}\int_{B(\rho)}u(y)-u(z)dz\right|$$
由于
$$u(x)-u(z)=-\int_0^1\frac{d}{ds}u(x+s(z-x))ds=-\int_0^1\nabla u(x+s(z-x))\cdot (z-x)ds,$$
$$\begin{aligned}
\left|\rho^{-n}\int_{B(\rho)}u(x)-u(z)dz\right|&\le\left|\rho^{1-n}\int_0^1\int_{B(\rho)}|\nabla u(x+s(z-x))|dzds\right|\\
&\le \left|\rho^{1-\frac{n}{q} }\int_0^1\left(\int_{B(\rho)}|\nabla u(x+s(z-x))|^qdz\right)^{\frac{1}{q} }ds\right|\\
&\le \rho^\alpha \parallel\nabla u\parallel_{L^q}\int_0^1 s^{-\frac{n}{q} }ds\le \frac{C}{2}|u|_{\dot{H}^{1,q} }|x-y|^{\alpha}
\end{aligned}$$
因此$|u(x)-u(y)|\le C|u|_{\dot{H}^{1,q} }|x-y|^\alpha$成立, 定理得证.
其它情形的Sobolev不等式
若$\frac{n}{p}=\frac{n}{q}-m,$ $p=\infty$时一般没有相应的不等式,
但是将条件放宽, 我们有:
命题 1.3. $\parallel u\parallel_{L^p(\Omega)}\le C_p\parallel u\parallel_{H^{1,n}(\Omega)},$ $\,\forall\,p\in(1,+\infty),$ $u\in C_c^\infty(\Omega)$
证: 当$\Omega$为有界区域时这是简单的, 因为取相应于$p$的$q,$ 有$q<n,$
从而$\parallel\nabla u\parallel_{L^q(\Omega)}\le C\parallel\nabla u\parallel_{L^n(\Omega)},$
$\parallel u\parallel_{L^p(\Omega)}\le C|u|_{\dot{H}^{1,q}(\Omega)}\le C|u|_{\dot{H}^{1,n}(\Omega)}.$
由此知将不等式右侧换为齐次模也对.
当$\Omega=\mathbb{R}^n$时, 由量纲分析知只能取Sobolev模.
这里我们只对$p>n+1$的情形给出证明. 取待定常数$(p_1,q_1),$
使得有Sobolev不等式$\parallel f\parallel_{L^{p_1} }\le C\parallel\nabla f\parallel_{L^{q_1} }$成立.
令$f=|u|^{\frac{n}{q_1} },$ 记$p=\frac{np_1}{q_1},$ 那么有:
$$\parallel u\parallel^{\frac{n}{q_1} }_{L^{p} }=\parallel u^{\frac{n}{q_1} }\parallel_{L^{p_1} }\le C\parallel|u|^{n-q_1}|\nabla u|^{q_1}\parallel_{L^1}^{\frac{1}{q_1} }\le C\parallel\nabla u\parallel_{L^n}\parallel u\parallel_{L^n}^{\frac{n}{q_1}-1}\le C\parallel u\parallel_{H^{1,n} }^{\frac{n}{q_1} }.$$
结合$\frac{n}{p_1}=\frac{n}{q_1}-1,$
我们即得到$\parallel u\parallel_{L^p(\mathbb{R}^n)}\le C_p\parallel u\parallel_{H^{1,n}(\mathbb{R}^n)},$
$\,\forall\,p=\frac{np_1}{q_1}=n+p_1>n+1.$
从另一个角度放宽条件, 令$p=+\infty,$ 我们有如下不等式:
命题 1.4. $\parallel u\parallel_{L^\infty}\le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)},$ $\,\forall\,q>n,$ $u\in C_c^\infty(\mathbb{R}^n).$
证: 即证$\,\forall\,x\in \mathbb{R}^n,$
$|u(x)| \le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)}.$
在前面的定理证明过程中, 令$\rho=1,$
有$|u(x)-\int_{B(1)}u(z)dz|\le C|u|_{\dot{H}^{1,q} }.$ 因此:
$$|u(x)|\le C|u|_{\dot{H}^{1,q}(\mathbb{R}^n)}+\parallel u\parallel_{L^q(B(1))}\le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)}.$$
由这一结果, 结合先前的不等式, 可以看出对$0<\alpha=1-\frac{n}{q}<1$,
有$\parallel u\parallel_{C^\alpha}\le C\parallel u\parallel_{H^{1,q} },$
将齐次模的结果换为一般模.
对$\Omega$为有界区域的情形, 若只关心其上一般模,
前面的不等式可以仅在$\mathbb{R}^n$上讨论,
而后用等价模定理将结果搬到$\Omega$上即可. 如:
$$\parallel u\parallel_{L^p(\Omega)}\le \parallel\widetilde{u}\parallel_{L^p(\mathbb{R}^n)}\le C\parallel\widetilde{u}\parallel_{H^{m,q}(\mathbb{R}^n)}, \quad \,\forall\,u\in C_c^\infty(\Omega).$$
其中$\widetilde{u}$为$u$的延拓. 因此
$$\parallel u\parallel_{L^p(\Omega)}\le C\inf_{\widetilde u|_{\Omega}=u}\parallel\widetilde{u}\parallel_{H^{m,q}(\mathbb{R}^n)}\,\approx\, \parallel u\parallel_{H^{m,q}(\Omega)}.$$
类似地对Hölder模也有同样的结果. 注意并没有齐次模形式的等价性定理,
因此不能用上述方法搬运齐次模的结果.
文章最后更新于 2021-10-31 20:22:25
Sobolev Inequalities for Negative Exponents
The following inequality was proved previously:
Theorem 1.1. When $\frac{n}{p}=\frac{n}{q}-m$, there is
$$\parallel f\parallel_{L^p(\Omega)}\le C|f|_{\dot{H}^{m,q}(\Omega)},\quad \,\forall\,f\in C_c^\infty(\Omega).$$
Note that the inequality only holds when $p\in(1,+\infty),$ $q\in [1,+\infty)$.
When $p=+\infty$, it is only true when $q=n=m=1$.
That is to say, the previous inequality only holds when $\frac{n}{q}-m>0$.
At $\frac{n}{q}-m<0$, there is also a corresponding Sobolev inequality.
First, we need to introduce the concept of Hölder module.
If the continuous function $f$ satisfies $|f(x)-f(y)|\le H|x-y|^{\alpha},$ $\alpha\in(0,1),$
Then $f$ is said to be Höldercontinuous . When $\alpha$ takes $1$, it is Lipschitzcontinuous.
Let the global Hölder continuous function on $\Omega$ be defined as $C^\alpha(\Omega),$ Hölder mold:
$$\parallel u\parallel_{C^\alpha}=\parallel u\parallel_{L^{\infty} }+|u|_{\dot{C}^\alpha},$$
where $|\cdot|_{\dot{C}^\alpha}$ is called Homogeneous Hölder model, defined as:
$$|u|_{\dot{C}^\alpha(\Omega)}:=\sup_{x\neq y\in \Omega}\frac{|u(x)-u(y)|}{|x-y|^{\alpha} }.$$
When the homogeneous Hölder module is finite, $u$ is Hölder continuous, and the dimension with respect to $u$ is $1,$ $x$ is $-\alpha.$
Theorem 1.2. When $-k-1<-\alpha-k=\frac{n}{q}-m<-k,$ $k\in \mathbb{N}$, $\,\forall\,\beta\in \mathbb{N}^n,$ $|\beta|=k,$ has
$$|\partial^\beta u|_{\dot{C}^\alpha(\mathbb{R}^n)}\le C|u|_{\dot{H}^{m,q}(\mathbb{R}^n)},\quad \,\forall\,f\in C_c^\infty(\mathbb{R}^n).$$
Certificate: We only need to discuss the case of $k=0$. For the general case,
Just apply the inequality of $-\alpha=\frac{n}{q}-(m-k)$ to $\partial^\beta u$, that is
$$|\partial^\beta u|_{\dot{C}^\alpha}
\le C|\partial^\beta u|_{\dot{H}^{m-k,q} }.
\le C|u|_{\dot{H}^{m,q} }.$$
Same as the previous discussion, suppose $m=1,$ then just prove that $\,\forall\,x\neq y\in \Omega,$ has
$$|u(x)-u(y)|\le C|u|_{\dot{H}^{1,q} }|x-y|^{\alpha}.$$
Denote $|x-y|=\rho,$ and take $B(\rho)$ as a cube containing $x,y,$ with side length $\rho$, then
$$|u(x)-u(y)|\le \left|\frac{1}{\rho^n}\int_{B(\rho)}u(x)-u(z)dz\right|+\left|\frac{1}{\rho^n}\int_{B(\rho)}u(y)-u(z)dz\right|$$
due to
$$u(x)-u(z)=-\int_0^1\frac{d}{ds}u(x+s(z-x))ds=-\int_0^1\nabla u(x+s(z-x))\cdot (z-x)ds,$$
$$\begin{aligned}
\left|\rho^{-n}\int_{B(\rho)}u(x)-u(z)dz\right|&\le\left|\rho^{1-n}\int_0^1\int_{B(\rho)}|\nabla u(x+s(z-x))|dzds\right|\\
&\le \left|\rho^{1-\frac{n}{q} }\int_0^1\left(\int_{B(\rho)}|\nabla u(x+s(z-x))|^qdz\right)^{\frac{1}{q} }ds\right|\\
&\le \rho^\alpha \parallel\nabla u\parallel_{L^q}\int_0^1 s^{-\frac{n}{q} }ds\le \frac{C}{2}|u|_{\dot{H}^{1,q} }|x-y|^{\alpha}
\end{aligned}$$
Therefore $|u(x)-u(y)|\le C|u|_{\dot{H}^{1,q} }|x-y|^\alpha$ is established and the theorem is proved.
Sobolev's Inequality in Other Cases
If $\frac{n}{p}=\frac{n}{q}-m,$ $p=\infty$, there is generally no corresponding inequality,
But relaxing the conditions, we have:
Proposition 1.3. $\parallel u\parallel_{L^p(\Omega)}\le C_p\parallel u\parallel_{H^{1,n}(\Omega)},$ $\,\forall\,p\in(1,+\infty),$ $u\in C_c^\infty(\Omega)$
Certificate: This is simple when $\Omega$ is a bounded region, because taking $q,$ corresponding to $p$ has $q<n,$
Thus $\parallel\nabla u\parallel_{L^q(\Omega)}\le C\parallel\nabla u\parallel_{L^n(\Omega)},$
$\parallel u\parallel_{L^p(\Omega)}\le C|u|_{\dot{H}^{1,q}(\Omega)}\le C|u|_{\dot{H}^{1,n}(\Omega)}.$
From this we know that it is also correct to replace the right side of the inequality with a homogeneous module.
When $\Omega=\mathbb{R}^n$, it is known from dimensional analysis that only the Sobolev module can be taken.
Here we only give the proof for the case of $p>n+1$. Take the undetermined constant $(p_1,q_1),$
So that Sobolev's inequality $\parallel f\parallel_{L^{p_1} }\le C\parallel\nabla f\parallel_{L^{q_1} }$ holds.
Let $f=|u|^{\frac{n}{q_1} },$ be $p=\frac{np_1}{q_1},$, then we have:
$$\parallel u\parallel^{\frac{n}{q_1} }_{L^{p} }=\parallel u^{\frac{n}{q_1} }\parallel_{L^{p_1} }\le C\parallel|u|^{n-q_1}|\nabla u|^{q_1}\parallel_{L^1}^{\frac{1}{q_1} }\le C\parallel\nabla u\parallel_{L^n}\parallel u\parallel_{L^n}^{\frac{n}{q_1}-1}\le C\parallel u\parallel_{H^{1,n} }^{\frac{n}{q_1} }.$$
Combine$\frac{n}{p_1}=\frac{n}{q_1}-1,$
We get $\parallel u\parallel_{L^p(\mathbb{R}^n)}\le C_p\parallel u\parallel_{H^{1,n}(\mathbb{R}^n)},$
$\,\forall\,p=\frac{np_1}{q_1}=n+p_1>n+1.$
Relaxing the conditions from another perspective, let $p=+\infty,$ we have the following inequality:
Proposition 1.4. $\parallel u\parallel_{L^\infty}\le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)},$ $\,\forall\,q>n,$ $u\in C_c^\infty(\mathbb{R}^n).$
Certificate: That’s proof$\,\forall\,x\in \mathbb{R}^n,$
$|u(x)| \le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)}.$
In the process of proving the previous theorem, let $\rho=1,$
There is $|u(x)-\int_{B(1)}u(z)dz|\le C|u|_{\dot{H}^{1,q} }.$ therefore:
$$|u(x)|\le C|u|_{\dot{H}^{1,q}(\mathbb{R}^n)}+\parallel u\parallel_{L^q(B(1))}\le C\parallel u\parallel_{H^{1,q}(\mathbb{R}^n)}.$$
From this result, combined with the previous inequality, we can see that for $0<\alpha=1-\frac{n}{q}<1$,
Yes $\parallel u\parallel_{C^\alpha}\le C\parallel u\parallel_{H^{1,q} },$
Convert the results of homogeneous modules to general modules.
For the case where $\Omega$ is a bounded area, if we only care about the general module above it,
The previous inequality can be discussed only on $\mathbb{R}^n$,
Then use the equivalent modular theorem to move the result to $\Omega$. For example:
$$\parallel u\parallel_{L^p(\Omega)}\le \parallel\widetilde{u}\parallel_{L^p(\mathbb{R}^n)}\le C\parallel\widetilde{u}\parallel_{H^{m,q}(\mathbb{R}^n)}, \quad \,\forall\,u\in C_c^\infty(\Omega).$$
Among them $\widetilde{u}$ is the continuation of $u$. Therefore
$$\parallel u\parallel_{L^p(\Omega)}\le C\inf_{\widetilde u|_{\Omega}=u}\parallel\widetilde{u}\parallel_{H^{m,q}(\mathbb{R}^n)}\,\approx\, \parallel u\parallel_{H^{m,q}(\Omega)}.$$
The same result is obtained similarly for Hölder modules. Note that there is no equivalence theorem for homogeneous modules,
Therefore, the results of homogeneous modules cannot be transported using the above method.
The article was last updated on 2021-10-31 20:22:25
