If we have $$(f-g)\in W^{1,p}_0(E)$$
where $f,g$ are both positive, $p\geq 1$ and $E\subset\mathbb{R}^n$ is a bounded domain, how do we then proof that also
$$\left(f^\alpha - g^\alpha\right) \in W^{1,\frac{p}{\alpha}}_0(E)$$
for arbitrary $p\geq \alpha \geq 1$? Intuitively this should be true because of "$f=g$" on the boundary so also "$f^\alpha = g^\alpha$" but of course this is not yet a proof.
By definition of $W^{1,p}_0(\Omega)$ there is a sequence of $\varphi_n \in C^\infty_c(\Omega)$ so that $ \varphi_n \to f-g$ in $W^{1,p}(\Omega)$. Hence
$$ \varphi_n + g \to f$$ in $W^{1,p}(\Omega)$ and
$$ (\varphi_n + g)^\alpha - g^\alpha \to f^\alpha - g^\alpha$$
in $W^{1,p/\alpha}(\Omega)$. Since $(\varphi_n+g)^\alpha - g^\alpha$ has compact support (indeed, the support is inside that of $\varphi_n$), we have $(\varphi_n+g)^\alpha - g^\alpha \in W^{1,p/\alpha}_0(\Omega)$ (see here). Since $W^{1,p/\alpha}_0(\Omega)$ is by definition a closed subspace of $W^{1,p/\alpha}(\Omega)$ we conclude that $f^\alpha - g^\alpha \in W^{1,p/\alpha}_0(\Omega)$.