弱导数与连续性

严格证明

若一个广义函数有弱导数, 那么它是不是本身几乎处处等于一个连续函数呢? 注意这里不是指几乎处处连续, 比如$1_{[0,+\infty)}$是几乎处处连续的, 但它不几乎处处等于一个连续函数; 反过来$1_{\mathbb{Q} }$几乎处处等于连续函数$0,$ 但是几乎处处不连续.

设$u\in H^{1,p}(\Omega),$ 那么由(负指数)Sobolev不等式, 当$p>n$时, $\frac{n}{p}-1<0,$ 因此$H^{1,p}(\Omega)$可嵌入到$C^{\alpha}(\Omega)$中, 函数是Hölder连续的.

那么$p\le n$时是否还可以呢? 这时就需要加额外的条件了. 一个经典的条件是假设弱导数都是连续的. 那么任取$O\Subset \Omega,$ $\overline{O}$为紧集, 故弱导数都在$L^\infty(O)$中, 当然也在所有的$L^p(O)$中, $\,\forall\,p\ge 1.$

由Sobolev不等式, $H^{1,p}(O)$可嵌入到$L^{q}(O)$中, $0\le \frac{n}{p}-1\le \frac{n}{q}.$ 由于弱导数也在$L^{q}(O)$中, 实际上$u$也在$H^{1,q}(O)$中. 归纳地, 我们可得到$u\in H^{1,q}(O),$ $q$满足$0\le \frac{n}{p}-k\le \frac{n}{q},$ $k\in \mathbb{Z}_+.$ 令$k$充分大, 使得$\frac{n}{p}-k<1,$ 即有$q>n,$ 将问题化为上一段的情形. 由$O$的任意性即得到结论.

称前面的条件经典, 是因为这样又能接着得到$u\in C^1(\Omega)$的结论. 取磨光子$\phi_\varepsilon,$ 那么对充分小的$\varepsilon,$ $u_\varepsilon:=u\ast \phi_\varepsilon$定义在$\Omega'\Subset \Omega$上, 满足$u_\varepsilon\rightarrow u,$ $Du_\varepsilon=Du \ast \phi_\varepsilon\rightarrow Du$一致收敛. 因此$u_{\frac{1}{n} }\in C^\infty(\Omega')$是$C^1(\Omega')$模下的Cauchy序列, 从而收敛到某个函数$v\in C^1(\Omega').$ 而$u_\varepsilon\rightarrow u$即说明$v=u\in C^1(\Omega').$ 由$\Omega'$选取任意性知$u\in C^1(\Omega).$

直观说明

直观来看, 也可以尝试直接给出$u$的表达式来证明. 如对于$1$维情况, 定义

$$ v(x)=\int_{0}^x\partial u(y)dy, $$

那么$v$是连续可导的, 他的弱导数当然和它经典意义上的导数一样. 接下来我们来证$u,v$相差常数. 只需证弱导数为零能推出函数为常数即可.

设$u$的弱导数为零. 由于我们总是在分布意义下考察函数, 只需证$\int u\varphi=c\int\varphi,$ $\,\forall\,\varphi\in C_c^\infty.$ 特别地只需对所有$\int \varphi=1$的函数证明. 当$\int \phi=0$时, $\,\exists\,\psi\in C_c^\infty,$ $\phi=\partial\psi.$ 因此对这样的$\phi,$ $\int u\phi=-\int \partial u\psi=0.$ 现任取$\psi$满足$\int \psi=1,$ 那么对所有$\int \varphi=1$的函数, $\int (\varphi-\psi)=0.$ 因此

$$ \int u(\varphi-\psi)=0, \int u\varphi=\int u\psi,\quad \,\forall\,\int \varphi=1. $$

这就说明了结论, 此时$\int u\varphi$即是常数$c.$

对于高维情况, 直观说明就变得复杂了. 我们期望通过路径积分来给出$u$的表达式, 但无法先验地判断积分与路径选取无关. 我们有偏导连续的结论, 倘若能再求一阶偏导, 就能够通过格林公式(分部积分)给出证明. 但可惜的是连续函数不见得有弱导数, 甚至有界变差函数也不行, 如devil’s staircase, 尽管其几乎处处可微. 不过我们已经有了$u$的确连续可导的结论, 因此后验地我们知道的确积分与路径选取无关.

高维情况对(每个)弱导数为零的论证是类似的, 说明原函数沿所有直线为常数即可.

参考资料

https://math.stackexchange.com/questions/3550108/the-recovery-of-w1-p-omega-to-c1-omega?noredirect=1

https://math.stackexchange.com/questions/2780348/do-continuous-weak-derivatives-imply-continuity

https://math.stackexchange.com/questions/497708/is-a-continuous-function-with-continuous-weak-derivatives-of-class-c1

文章最后更新于 2022-03-30 13:22:20

Strictly prove

If a generalized function has weak derivatives, is it itself equal to a continuous function almost everywhere? Note that this does not mean continuous almost everywhere. For example, $1_{[0,+\infty)}$ is continuous almost everywhere. But it is not almost everywhere equal to a continuous function; In turn $1_{\mathbb{Q} }$ is almost everywhere equal to the continuous function $0,$ but is almost everywhere discontinuous.

Assume $u\in H^{1,p}(\Omega),$, then according to the (negative exponential) Sobolev inequality, when $p>n$, $\frac{n}{p}-1<0,$ Therefore $H^{1,p}(\Omega)$ can be embedded in $C^{\alpha}(\Omega)$, The function is Hölder continuous.

So is it still OK when $p\le n$? At this time, additional conditions need to be added. A classic condition is to assume that weak derivatives are continuous. Then take $O\Subset \Omega,$ arbitrarily $\overline{O}$ is a compact set, so the weak derivatives are all in $L^\infty(O)$, Of course, among all $L^p(O)$, $\,\forall\,p\ge 1.$

According to Sobolev's inequality, $H^{1,p}(O)$ can be embedded in $L^{q}(O)$, $0\le \frac{n}{p}-1\le \frac{n}{q}.$ Since the weak derivative is also in $L^{q}(O)$, In fact $u$ is also in $H^{1,q}(O)$. Inductively, we can get $u\in H^{1,q}(O),$ $q$ satisfies $0\le \frac{n}{p}-k\le \frac{n}{q},$ $k\in \mathbb{Z}_+.$ Let $k$ be large enough so that $\frac{n}{p}-k<1,$ has $q>n,$, which turns the problem into the situation in the previous paragraph. The conclusion is drawn from the arbitrariness of $O$.

The previous condition is called classical because it can then lead to the conclusion $u\in C^1(\Omega)$. Taking the polished photon $\phi_\varepsilon,$, then for a sufficiently small $\varepsilon,$ $u_\varepsilon:=u\ast \phi_\varepsilon$ is defined on $\Omega'\Subset \Omega$, Satisfied$u_\varepsilon\rightarrow u,$ $Du_\varepsilon=Du \ast \phi_\varepsilon\rightarrow Du$ Uniform convergence. Therefore $u_{\frac{1}{n} }\in C^\infty(\Omega')$ is the Cauchy sequence in the mode $C^1(\Omega')$, Thus converging to a certain function $v\in C^1(\Omega').$ And $u_\varepsilon\rightarrow u$ means $v=u\in C^1(\Omega').$ Select arbitrarily from $\Omega'$ to know $u\in C^1(\Omega).$

Visual explanation

Intuitively, you can also try to directly give the expression of $u$ to prove. For example, for the $1$ dimension case, definition

$$ v(x)=\int_{0}^x\partial u(y)dy, $$

Then $v$ is continuously differentiable, Its weak derivative is of course the same as its derivative in the classical sense. Next we will prove $u,v$ difference constant. It only needs to be proved that the weak derivative is zero to deduce that the function is a constant.

Let the weak derivative of $u$ be zero. Since we always examine functions in the distribution sense, Only need to prove $\int u\varphi=c\int\varphi,$ $\,\forall\,\varphi\in C_c^\infty.$ In particular, it only needs to be proved for all functions of $\int \varphi=1$. When $\int \phi=0$, $\,\exists\,\psi\in C_c^\infty,$ $\phi=\partial\psi.$ Therefore, for such $\phi,$ $\int u\phi=-\int \partial u\psi=0.$ If $\psi$ currently satisfies $\int \psi=1,$, then for all functions of $\int \varphi=1$, $\int (\varphi-\psi)=0.$ Therefore

$$ \int u(\varphi-\psi)=0, \int u\varphi=\int u\psi,\quad \,\forall\,\int \varphi=1. $$

This illustrates the conclusion. At this time $\int u\varphi$ is the constant $c.$

For high-dimensional cases, the intuitive explanation becomes complicated. We expect to give the expression of $u$ through path integral, But it cannot be judged a priori that the integral has nothing to do with path selection. We have the conclusion that the partial derivative is continuous, If we can find the first-order partial derivative, we can give the proof through Green's formula (integration by parts). But unfortunately, continuous functions do not necessarily have weak derivatives, not even bounded variograms, such as devil’s staircase, although it is differentiable almost everywhere. But we have already reached the conclusion that $u$ is indeed continuously differentiable, So a posteriori we know that indeed integration has nothing to do with path selection.

The argument for zero weak derivatives is similar for the high-dimensional case, Just show that the original function is constant along all straight lines.