I am currently trying to establish the local Lipschitz continuity of the mapping $$ F : H_0^1 (\Omega,\mathbb{R}) \rightarrow L^2 (\Omega, \mathbb{R}), \ u \mapsto u^\gamma, $$ where $\Omega \subset \mathbb{R}^d$ is assumed to be a bounded domain with nice boundary and $\gamma > 1$. In the crucial case $d >2$ welldefinedness is given for $ \gamma \leq \frac{d}{d-2}$.
I was hoping for an estimate of the type "$ \Vert w \Vert_{L^q} \Vert u-v \Vert_{L^p}$" with $w \in H_0^1 (\Omega, \mathbb{R})$ and $q$, $p$ suitable for the application of the Sobolev embeddings. I tried to play with the Mean Value Theorem applied to $F$ pointwise, but in this case I end up with a function $w(\theta, x)$ on $[0,1] \times \Omega$, where $\theta$ depends on the spatial variable $x$. I failed in proving that this mapping is in $H_0^1 (\Omega, \mathbb{R})$.
Do you have easier alternatives how to get started with the estimate? Any help would be appreciated!
Best regards :)
Following the Hint that daw gave you, by the first order expansion of $u\mapsto u^\gamma$ you get $$ u^\gamma-v^\gamma = (u-v) \,\gamma\int_0^1 \left((1-t)\,u+t\,v\right)^{\gamma-1}\,\mathrm d t $$ therefore by Hölder's and Minkowski's inequalities, for $(p,q)\in [2,\infty]^2$ such that $1/p + 1/q = 1/2$, $$ \|u^\gamma-v^\gamma\|_{L^2} \leq \gamma\,\|u-v\|_{L^p} \int_0^1 \|(1-t)\,u+t\,v\|_{L^{q(\gamma-1)}}^{\gamma-1}\,\mathrm d t \\ \leq \gamma\,\|u-v\|_{L^p} \int_0^1 \left((1-t)\,\|u\|_{L^{q(\gamma-1)}} + t\,\|v\|_{L^{q(\gamma-1)}}\right)^{\gamma-1}\,\mathrm d t $$ and so since $\gamma-1 > 0$, using the fact that $(a+b)^{\gamma-1} \leq 2^{(\gamma-2)_+} (a^{\gamma-1}+b^{\gamma-1})$, you get $$ \|u^\gamma-v^\gamma\|_{L^2} \leq \gamma\,\|u-v\|_{L^p} \int_0^1 \|(1-t)\,u+t\,v\|_{L^{q(\gamma-1)}}^{\gamma-1}\,\mathrm d t \\ \leq \gamma\,2^{(\gamma-2)_+-1}\,\|u-v\|_{L^p} \left(\|u\|_{L^{q(\gamma-1)}}^{\gamma-1} + \|v\|_{L^{q(\gamma-1)}}^{\gamma-1}\right) $$ Now you just want to use Sobolev's inequalities to find $p$ and $q$ so that both these norms are controlled by the $H^1$ norm. Since the domain is bounded, if If $d\leq 2$, then the $H^1$ norm controls all $L^p$ norms with $p<\infty$, so it is always true in this case. If $d>2$, then it is sufficient to find them such that $p\leq \frac{2d}{d-2}$ and $q\,(\gamma-1)\leq \frac{2d}{d-2}$. Taking for instance $p=\frac{2d}{d-2}$, then this works as soon as $$ \gamma -1 \leq \frac{2d}{d-2}\, \frac{1}{q} = \frac{2d}{d-2}\left(\frac{1}{2}-\frac{1}{p}\right) = \frac{d}{d-2}-1. $$
To conclude: As soon as $d\leq 2$ or $\gamma \leq \frac{d}{d-2}$, then $$ \|u^\gamma-v^\gamma\|_{L^2} \leq \frac{\gamma\,2^{(\gamma-2)_+}}{2}\,\|u-v\|_{H^1} \left(\|u\|_{H^1}^{\gamma-1} + \|v\|_{H^1}^{\gamma-1}\right) $$