Need a source for the following result: From $v \in H^1$ and $f$ Lipschitz it follows that $f(v) \in H^1$.

Question

I am looking for a source of the following (or a similar) result:

If $v \in H^1(\Omega,\mathbb{C})$ on a bounded domain $\Omega$ and $f: R(v) \to \mathbb{C}$ is Lipschitz continuous, then $f(v) \in H^1(\Omega, \mathbb{C})$

Here $H^1$ denotes the Sobolev space $W^{1,2}$ and $R(v)$ denotes the range of $v$.

In various answers on StackExchange this is used (e.g. here or here) and is referred to as a "standard fact".

I can't seem to produce a proof of it and I don't find the result in my books about Sobolev Spaces. Does anybody have a source for it?

EDIT: In my case I can't suppose that $f(0)=0$!

Answer 1

It is easiest to approach this using difference quotients. Let $\tau_h^i u(x) = u(x + h e_i)$ where $e_i$ is one of the standard unit vectors. Define the difference quotient $\Delta_h^iu = (\tau_h^i u - u)/h$.

You need the two propositions:

Proposition 1 If $u\in W^{1,p}(\Omega)$, then for every $\Omega' \Subset \Omega$ with $\mathrm{dist}(\Omega',\Omega^c) > h$, the difference quotient $\Delta_h^iu \in L^p(\Omega')$ and $$\|\Delta_h^i u\|_{L^p(\Omega')} \leq \|u\|_{W^{1,p}(\Omega)}$$ is true.

Proposition 2 If $u\in L^p(\Omega)$ for $p \in (1,\infty)$. If there exists a constant $K$ so that $\|\Delta_h^iu\|_{L^p(\Omega')} \leq K$ for every index $i$ and for every $h > 0$, and every $\Omega'\Subset \Omega$ satisfying $\mathrm{dist}(\Omega',\Omega^c) > h$; then $u$ is weakly differentiable with its weak derivative in $L^p(\Omega)$.

(Those are propositions 16 and 17 in my lecture notes on Sobolev spaces. Alternatively you can also find them in Section 7.11 of Gilbarg and Trudinger.)

To conclude the proof it suffices to observe that if $f$ is Lipschitz, then

$$ |\Delta_h^i (f(v))| \leq \mu |\Delta_h^i v| $$

where $\mu$ is the Lipschitz constant of $f$.