Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove that $|\nabla u|=|\nabla|u||$ when $u$ is regular enough for example Lipschitz or $W^{1,1}_{loc}$.

Other question is about the pointwise derivative when $f:[0,1]\to R$ is BV is that possible to prove that $|f'|=||f|'|$?

share|improve this question
    
You can try starting by writing $f=f^+-f^-$ –  matgaio May 14 '12 at 22:28
    
I did that but I did not see the right combination or it is not that easy. –  checkmath May 14 '12 at 22:37
    
I'm consulting here Evans's "Partial Differential Equations". There is (p. 292 - ex 17) a hint wich helps solving this. It is not just write $f$ properly. Are you reading this book? –  matgaio May 14 '12 at 23:19
2  
The first question is basically the same as here –  Jose27 May 14 '12 at 23:45
    
@matgaio Not now but I have book and I can check it. –  checkmath May 15 '12 at 3:59

1 Answer 1

up vote 1 down vote accepted

Here's how the proof goes for $W^{1,1}(\mathbb{R}^n)$. It should be adaptable to the Sobolev space of your choice.

First we establish the chain rule:

Proposition. Suppose $\psi \in C^1(\mathbb{R})$ satisfies $\psi(0) = 0$ and $|\psi'| \le C$. If $f \in W^{1,1}$, then $\psi \circ f \in W^{1,1}$, and $\nabla (\psi \circ f) = (\psi' \circ f) \nabla f$.

Proof. Take $f_n \in C^\infty_c(\mathbb{R}^n)$ with $f_n \to f$ in $L^1$ and $\nabla f_n \to \nabla f$ in $L^1$. Passing to a subsequence, we can also assume both convergences hold almost everywhere.

Since $\psi$ is continuous, we have $\psi \circ f_n \to \psi \circ f$ almost everywhere. Additionally, since $\psi$ is Lipschitz, we have $|\psi \circ f_n| \le C |f_n|$; since the $f_n$ are converging in $L^1$, a version of the dominated convergence theorem gives $\psi \circ f_n \to \psi \circ f$ in $L^1$.

Now by calculus we have $\nabla (\psi \circ f_n) = (\psi' \circ f_n) \nabla f_n$. Since $\psi'$ is continuous, $\nabla (\psi \circ f_n) \to (\psi' \circ f) \nabla f$ pointwise, and since $|\nabla(\psi \circ f_n)| \le C |\nabla f_n|$, where $\nabla f_n$ converges in $L^1$, as before we have $\nabla(\psi \circ f_n) \to (\psi' \circ f) \nabla f$ in $L^1$.

In particular, $\{\psi \circ f_n\}$ is Cauchy in $W^{1,1}$, and it converges to $\psi \circ f$ in $L^1$, so by the completeness of $W^{1,1}$ we must have $\psi \circ f_n \to \psi \circ f$ in $W^{1,1}$ (in particular $\psi \circ f \in W^{1,1}$). The gradient operator is continuous from $W^{1,1}(\mathbb{R}^d)$ to $L^1(\mathbb{R}^d, \mathbb{R}^d)$, so we must have $\nabla (\psi \circ f_n) \to \nabla (\psi \circ f)$ in $L^1$. But we have just shown $\nabla(\psi \circ f_n) \to (\psi' \circ f) \nabla f$ so it must be that $\nabla (\psi \circ f) = (\psi' \circ f)\nabla f$ almost everywhere. QED chain rule.

Notation. Let's let $s(t)$ denote the sign of $t$, except that we take $s(0) = 1$.

Let $f \in W^{1,1}$, and take a sequence $\psi_n$ of $C^1(\mathbb{R})$ functions such that:

  1. $\psi_n(t) \to |t|$ pointwise;
  2. $\psi_n'(t) \to s(t)$ pointwise;
  3. $\psi_n(0) = 0$;
  4. $|\psi_n'| \le 2$.

(Constructing such a sequence is left as an exercise. Hint: construct $\psi_n'$ first and then integrate.) We have $\psi_n \circ f \to |f|$ pointwise, and $|\psi_n \circ f| \le 2|f|$ so by dominated convergence the convergence is also in $L^1$. We also have $\nabla (\psi_n \circ f) = (\psi_n' \circ f) \nabla f \to (s \circ f) \nabla f$ pointwise, and $|\nabla (\psi_n \circ f)| \le 2 |\nabla f|$ so this convergence is in $L^1$ as well. By the same argument as above, we find that $|f| \in W^{1,1}$ and $\nabla |f| = (s \circ f)\nabla f$ almost everywhere. In particular, since $|s(t)| = 1$, we have $|\nabla |f| | = |\nabla f|$ almost everywhere.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.