Positive part of Cauchy sequence of Sobolev functions is again Cauchy Let $p\geq 1$ and consider the space $W^{1,p}(B)$ where $B\subset \mathbb{R}^{n}$ is the standard unit ball. Moreover, let $f_{k} \in C^{\infty}(B)$ be a Cauchy sequence in $W^{1,p}(B)$ of smooth function. How can one deduce that also $f_{k}^{+}$ is a Cauchy sequence in $W^{1,p}(B)$, where $f_{k}^{+}$ are defined by
$$
f_{k}^{+}(x) = \text{max}\{f_{k}(x),0\}?
$$
Greetings.
 A: You have
$
\bigl\vert\vert f_{k+l}(x)\vert-\vert f_{k}(x)\vert\bigr\vert\le 
\vert f_{k+l}(x)-f_{k}(x)\vert
$
and thus
$$
\Vert\vert f_{k+l}\vert-\vert f_{k}\vert\Vert_{W^{1,1}}
\le 
\Vert f_{k+l}- f_{k}\Vert_{L^{1}}+\Vert \nabla{\vert f_{k+l}\vert}
-\nabla{\vert f_{k}\vert}\Vert_{L^{1}}.
$$
Now, let us calculate $\nabla \vert f\vert$, say for $f\in C^1$. We claim that we have
$$
\nabla \vert f\vert=S(f) \nabla f, \quad S(f)=\mathbf 1(f>0)-\mathbf 1(f<0).
\label{1}\tag{1}$$
Note that the function $S(f)$ is bounded measurable and that the product makes sense, even if you have only $\nabla f\in L^1$. Let us prove our claim: with brackets of duality and $\phi\in C^\infty_c$, we have
\begin{multline}
\langle \nabla \vert f\vert, \phi\rangle=-\int\vert f\vert \nabla \phi dx=-\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{1/2} \nabla \phi dx=
\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f)  \phi dx
\\
=\lim_{\epsilon \rightarrow 0_+}\int(f^2+\epsilon^2)^{-1/2} f(\nabla f)  \phi dx
=\int S(f) (\nabla f) \phi dx,
\end{multline}
proving \eqref{1}, using Lebesgue's Dominated Convergence Theorem. I feel a bit uncomfortable to extend this formula to $f\in W^{1,1}$, but it might be a first step.
