I trying to show that if $u\in W^{1,p}_{loc}(\Omega)~~u>0$ then $\log u\in W^{1,p}_{loc}(\Omega)$.
My attempt
Let $D\subset\subset\Omega$ and $\varphi\in C^{\infty}_0(D)$ $$\int_{D}\frac{\partial \log u}{\partial x_i}\varphi dx=\int_{D}\frac{\partial u}{\partial x_i}\frac{1}{u}\varphi dx=\int_{D}\frac{\partial u}{\partial x_i}\frac{\varphi}{u} dx~~(1)$$
Since $u\in W^{1,p}_{loc}(\Omega)$ then $\exists~~g_1\ldots g_n\in L^{p}(D)$ such that $\forall \varphi\in C^\infty_0(D)$ $$\int_{D}\frac{\partial u}{\partial x_i}\varphi dx=-\int_{D}g_i\frac{\partial \varphi}{\partial x_i}dx$$
In (1) this yields
$$\int_{D}\frac{\partial \log u}{\partial x_i}\varphi dx=-\int_{D}g_i\frac{\partial \frac{\varphi}{u}}{\partial x_i}dx$$
i got stuck here, any remark or hint would be helpful. Thank you
The assertion is untrue. Consider $\Omega = (-1, 1)$ and $u(x) := \lvert x \rvert$. Clearly $u>0$ (in an a.e. sense) and $u \in L^1(\Omega)$. Since $u'(x) = \frac{x}{\lvert x \rvert}$ a.e. in a weak sense (this is can easily be computed using integration by parts), we have, because of $\log\circ u$ being $C^1$ in $\Omega \setminus \lbrace 0 \rbrace$, that $\log(u(x))' = \frac{x}{\lvert x \rvert^2}$ is the only possible candidate for the weak derivative. But $\log(u(x))'$ does not live in $L^1_{loc}(\Omega)$: $$ \int_{-\frac{1}{2}}^\frac{1}{2} \frac{\lvert x \rvert}{\lvert x \rvert^2}~\mathrm{d}x = 2 \int^{\frac{1}{2}}_0 \frac{1}{x}~\mathrm{d}x = 4 - 2\lim_{x \downarrow 0} \log(x) = \infty $$
Quick sidenote: If $u>0$ and $u \in W^{1, p}(\Omega)$ and $f \in C^1([0, \infty))$ with $f'$ bounded, then $f(u) \in W^{1, p}(\Omega)$. About the $loc$ case I am not quite sure.