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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.

HelloEveryone
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1 Answers1

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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 $f^\alpha - g^\alpha \in W^{1,p/\alpha}_0(\Omega)$.

Arctic Char
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    I'd add that the convergence in $W^{1,p/\alpha}(\Omega)$ is not immediate, and is a good exercise for the OP to verify. In particular it's crucial that we assume $\alpha \geq 1$ in this step, as the result can fail when $\alpha < 1$. – ktoi Mar 07 '23 at 10:04