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Provide details for the following alternative proof that if $u \in H^1(U)$, then $$Du = 0 \text{ a.e. } \, \text{ on the set } \{u=0\}. $$ Let $\phi$ be a smooth, bounded and nondecreasing function, such that $\phi'$ is bounded and $\phi(z)=z$ if $|z| \le 1$. Set $$u^\epsilon(x) := \epsilon \phi(u/\epsilon).$$ Show that $u^\epsilon \rightharpoonup 0$ weakly in $H^1(U)$ and therefore $$\int_U Du^\epsilon \cdot Du \, dx = \int_U \phi'(u/\epsilon) |Du|^2 \, dx \to 0.$$ Employ this observation to finish the proof.

PDE by Evans, 2nd edition: Chapter 5, Exercise 19

Here is my work so far. I tried what I can; are there any errors in this?

As $\epsilon \to 0$, \begin{align} \|u^\epsilon\|_{H^1(U)} &:= \|u^\epsilon \|_{L^2(U)} + \|D u^\epsilon \|_{L^2(U)} \\ &= \left\|\epsilon \phi\left(\frac u{\epsilon} \right) \right\|_{L^2(U)}+\|\phi'\left(\frac u{\epsilon} \right) Du \|_{L^2(U)} \\ &\le \left\|C \epsilon \frac u{\epsilon} \right\|_{L^2(U)} + \left\|C \cdot Du \right\|_{L^2(U)} \\ &\le C(\left\| u \right\|_{L^2(U)} + \left\| Du \right\|_{L^2(U)}). \end{align} Hence, $u^\epsilon \rightharpoonup 0$ weakly in $H^1(U)$. Therefore, at $u=0$, $$\require{cancel} \int_U \left. \cancelto{1}{\phi'(0/\epsilon)} |Du|^2 \right\vert_{u=0}\, dx \to 0$$ gives $\int_U \left. |Du|^2 \right\vert_{u=0} \, dx \to 0$. Hence, $\left. Du \right\vert_{u=0} =0$ a.e., or $Du=0$ a.e. on the set $\{u=0\}$.

Cookie
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    Why are you assuming $|u|\leq 1$? If you change your equality in the third line for an inequality that calculation holds for general $u$ (use the Lipschitz condition). If $\int_U |Du| \to 0$ then $u$ is constant throughout $U$, which is clearly not true in general, but you do have that $\phi'(0/\varepsilon)=1$, which says... – Jose27 Jan 18 '15 at 04:12
  • We can use the Lipschitz continutiy condition ... it was given from the fact that $\phi$ is "smooth, bounded"? – Cookie Jan 18 '15 at 04:15
  • From the fact that $\phi'$ is bounded. – Jose27 Jan 18 '15 at 04:15
  • That means $$|\phi'(u/\epsilon) Du |{L^2(U)} \le C |Du|{L^2(U)}$$ for some appropriate constant $C$? What about then for using $$\phi(u/\epsilon)=u/\epsilon$$ for $|u/\epsilon|\le 1$? I think we are also using that $\phi$ is bounded too. – Cookie Jan 18 '15 at 04:16
  • Yes but also, since $\phi(0)=0$, that $$\varepsilon |\phi(u/\varepsilon)| \leq C|u|.$$ – Jose27 Jan 18 '15 at 04:19
  • Does it have to be specifically $\phi(0)=0$? I was tempted to just have an arbitrary value of $u$ that is close to $0$ (i.e. $\phi(u/\epsilon)=u/\epsilon$). (Hence, my initial assumption of $|u| \le 1$, albeit without any good reason.) – Cookie Jan 18 '15 at 04:23
  • Well $\phi(x)=x$ for $|x|\leq 1$. In particular $\phi(0)=0$. – Jose27 Jan 18 '15 at 04:24
  • @Jose27 I had incorporated your suggestion of $\phi'(0)=1$ into the last lines of my work, so that $\int_U \left.|Du|^2 \right\vert_{u=0}, dx = 0$ (only when $u=0$), instead of having $\int_U |Du|^2 , dx = 0$ for the entire region $U$. – Cookie Jan 18 '15 at 23:22
  • @dragon When proving $u^{\epsilon} \to 0$ weakly in $H^1(U)$, why assume $u \to 0$? I think it is supposed to assume $\epsilon \to 0$. – Sherry Feb 08 '15 at 20:45
  • Can the downvote be explained please? – Cookie Feb 10 '15 at 07:48
  • @Sherry Which line in my work assumes $u \to 0$? You're probably right, but I don't see it; I only see that I did $u^\epsilon \rightharpoonup 0$. – Cookie Feb 10 '15 at 07:49
  • @dragon Hi. I saw it in the first line, after "Here is my work so far. I tried what I can; are there any errors in this?", it is "as $u \to 0$" Maybe it is a typo? – Sherry Feb 10 '15 at 08:13
  • Ohh ... yes, it was a typo! It is indeed "as $\epsilon \to 0$". Thanks, @Sherry. – Cookie Feb 10 '15 at 22:04
  • @Jose27 Excuse me if I reply on an old question, but I don't understand a couple of things. First, since by definition $u^\varepsilon(x) = \varepsilon\phi(u/\varepsilon)$ and $\phi$ is bounded, couldn't we simply bound $||u^\varepsilon||{L^2(U)}$ with a constant $C$? i.e. write $||u^\varepsilon||{L^2(U)} \le C$? – sound wave Jun 19 '21 at 15:21

1 Answers1

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Shown $u^\epsilon\to 0$ weakly in $L^2$ is the easy part, I think you already proved it. (I looked at comments, it has a good hint)

The trick part is to show $\nabla u^\epsilon\to 0$ weakly in $L^2$ as well. Applying chain rule will lead you nowhere. You need following result in Functional analysis

Let $X$ be a Banach space, $S$ be a total subset of $X^*$, i.e., the span of $S$ is dense in $X^*$, $(x_n)$ be a sequence in $X$ and $x\in X$ is given. Then $x_n\to x$ weakly as $n\to\infty$ if and only if $({x_n})$ is bounded and $$ \left<{y^*,x_n}\right>\to\left<{y^*,x}\right>\text{ for all }y^*\in S. $$

Hence, by definition we need to prove that $$ \int_U \partial_i u^\epsilon\,v\,dx\to 0\tag w$$ for all $v\in L^2(U)$, but now we only need to prove that $$ \int_U \partial_i u^\epsilon\,v\,dx\to 0\tag 1$$ for all $v\in C_c^\infty(U)$ and hence you can use IBP and have $(1)$ is equivalent to $$ \int_U u^\epsilon\,\partial_i v\,dx\to 0$$ which is true since $u^\epsilon\to 0$ weakly in $L^2$.

spatially
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  • The definition of weak $L^2$ convergence is what $(w)$ says. If you don't know what is weak convergence means, I suggest you to read some functional analysis books. – spatially Jan 18 '15 at 05:18
  • I guess I will need to go read those books. I'm sorry if I just asked you dumb questions, I just always forget that PDE theory requires a background of functional analysis, in addition to backgrounds in measure theory, multivariable calculus, linear algebra ... (the list could go on). :) (For the record, I have taken requisite classes in everything else besides functional analysis.) – Cookie Jan 18 '15 at 05:23
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    Well, the functional analysis is kind of the most important one. It looks to me we only need LDCT from measure theory. :) – spatially Jan 18 '15 at 05:33
  • Sorry, I misread the argument... – Jose27 Jan 18 '15 at 07:28
  • @wisher Could you explain why instead of all $v \in L^2(U)$, we only need to prove for all $v \in C_c^{\infty}(U)$ in $\int_U \partial_i u^{\epsilon}v dx \to 0$? Thanks! – Sherry Feb 08 '15 at 21:08
  • @Sherry It is due to the result from Functional Analysis which I mentioned in my post. – spatially Feb 09 '15 at 01:13
  • @wisher Thanks! Can I ask you a question? dragon assumes $u \to 0$ when probing the weakly convergence of $u^{\epsilon} \to 0$. But I think it is supposed to be $\epsilon \to 0$? Is that right? – Sherry Feb 10 '15 at 02:24
  • @Sherry. I think so. – spatially Feb 10 '15 at 02:26