# Weak derivatives

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 $|f'|=||f|'|$?

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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
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

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.

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