# Composition with Lipschitz map is Lipschitz on Sobolev spaces

Suppose that $$F: \mathbb{R}^d \rightarrow \mathbb{R}^d$$ is Lipschitz with some constant $$L$$ and that $$F(0)=0$$. Then it is clear that $$F$$ defines a Lipschitz continuous map $$L^2(\mathbb{R}^d) \rightarrow L^2(\mathbb{R}^d)$$ by $$u \mapsto F(u)$$ with the same Lipschitz constant $$L$$.

We can extend this to Sobolev spaces $$H^k(\mathbb{R}^d)$$ when $$k \in \mathbb{N}$$, by requiring in addition that $$F \in C^k$$ with the first $$k$$ derivatives of $$F$$ Lipschitz and that the first $$k$$ derivatives vanish at $$0$$. For instance, if $$k=1$$, we can compute \begin{align} \| \partial_i(F(u)-F(\tilde{u})) \|_{L^2(\mathbb{R}^d)} &= \| \nabla F(u) \partial_i u - \nabla F(\tilde{u}) \partial_i \tilde{u} \|_{L^2} \\ &\leq \| \nabla F(u) \|_{L^2} \|\partial_i u - \partial_i \tilde{u} \|_{L^2} + \|\partial_i \tilde{u} \|_{L^2} \|\nabla F(u) - \nabla F(\tilde{u}) \|_{L^2} \\ &\leq C \left( \|u\|_{L^2} \|u-\tilde{u} \|_{H^1}+\| \tilde{u} \|_{H^1} \| u- \tilde{u}\|_{L^2} \right) \\ &\leq C (\|u\|_{H^1} + \| \tilde{u} \|_{H^1}) \| u - \tilde{u} \|_{H^1} \end{align} where we have used that $$\nabla F$$ is again a Lipschitz map (and I denoted all constants by $$C$$). Thus we obtain that the map $$u \mapsto F(u)$$ is (at least) locally Lipschitz in $$H^1$$. Similarly of course for $$H^k$$, when $$k \in \mathbb{N}$$.

Now here is my question: (how) is it possible to extend this result to Sobolev spaces $$H^s$$ with real index $$s \in \mathbb{R}$$? Is the map $$u \mapsto F(u)$$ well-defined and locally Lipschitz continuous on the space $$H^s(\mathbb{R}^d)$$? Is there some interpolation argument that easily accomplishes this? Finally, did I miss something, and is the map perhaps even globally Lipschitz?