# Does strong convergence in $W_p^1$ imply strong convergence of derivatives of absolute values in $L_p$?

Let $G$ be a Lipshitz domain and $u_n \to u$ in $W_p^1(G)$. Is it correct, that $\frac{\partial |u_n|}{\partial x_m} \to \frac{\partial |u|}{\partial x_m}$ in $L_p(G)$?

I know, that $\frac{\partial |u|}{\partial x_m} = \operatorname{sgn} u\frac{\partial u}{\partial x_m}$, so I have

$$\int\limits_G \left|\frac{\partial |u_n|}{\partial x_m} - \frac{\partial |u|}{\partial x_m}\right|^p\,dx = \int\limits_G \left|\operatorname{sgn} u_n\frac{\partial u_n}{\partial x_m}-\operatorname{sgn} u\frac{\partial u}{\partial x_m}\right|^p\,dx=\\=\int\limits_G \left|\operatorname{sgn} u_n\left(\frac{\partial u_n}{\partial x_m}-\frac{\partial u}{\partial x_m}\right)+\frac{\partial u}{\partial x_m}\left(\operatorname{sgn} u_n -\operatorname{sgn} u\right)\right|^p\,dx \leq \int\limits_G \left|\frac{\partial u_n}{\partial x_m}-\frac{\partial u}{\partial x_m}\right|^p\,dx+\\+\int\limits_G \left|\frac{\partial u}{\partial x_m}\left(\operatorname{sgn} u_n -\operatorname{sgn} u\right)\right|^p\,dx$$

So, the last inequality is not easy to check, whether it converges to 0.

• I think it is sufficient to use that $u_{x_m}$ is 0 almost everywhere on $\{ u=0\}$ and that $\text{sgn}(u_n)$ converges almost everywhere to $\text{sgn}(u)$ on $\{ u\not= 0\}$.
– O.G.
Jun 30, 2015 at 19:27
• For the last term, you use Lebesgue's dominated convergence theorem. Jul 3, 2015 at 9:46

Yes, this is correct; in fact, you could replace the map $|\cdot|$ with an arbitrary (uniformly) Lipschitz map $f$ (with $f(0) = 0$ if $G$ is unbounded) [1].