> [!theorem] > > Let $U \subset \real^d$ be a bounded [[Open Set|open]] set and $f \in UC^1(U)$ be a [[Space of Uniformly Continuously Differentiable Functions|UC]] function. If $p \in [1, \infty)$, then > $ > \abs{f}^p \in \begin{cases} > UC^1(U) &p > 1 \\ > W^{1, p}(U) &p = 1 > \end{cases} > $ > where $\partial^{e_j}\abs{u}^p = p\sgn{(u)}\abs{u}^{p - 1} \cdot \partial^{e_j}u$. > > *Proof*. If $p > 1$, then the mapping $P: \real \to \real$ with $x \mapsto \abs{x}^p$ is differentiable with $DP(x) = p\sgn(x)\abs{x}^{p - 1}$. From here, applying the chain rule yields that $\partial^{e_j}\abs{f}^p = \sgn(f)\abs{f}^{p - 1} \cdot p\partial^{e_j}f$. Since $P$ is uniformly continuous on bounded sets, $\partial^{e_j}\abs{f}^p \in UC(U)$. > > If $p = 1$, then by breaking off the derivative $\abs{f} \in UC(U \cap \bracs{f > 0})$ and $\abs{f} \in UC(U \cap \bracs{f < 0})$, for any $\phi \in \cd(U)$, > $ > \begin{align*} > \angles{\abs{f}, \partial^{e_j}\phi} &= \angles{\one_{\bracs{f > 0}}\abs{f}, \partial^{e_j}\phi} - \angles{\one_{\bracs{f < 0}}\abs{f}, \partial^{e_j}\phi} \\ > &= -\angles{\one_{\bracs{ f> 0}}\partial^{e_j}f, \phi} + \angles{\one_{\bracs{f < 0}}\partial^{e_j}f, \phi} \\ > &= -\angles{\sgn(f) \cdot \partial^{e_j}f, \phi} > \end{align*} > $