Consider the Anisotropic Sobolev Space defined by: $$W^{1,\overrightarrow{p},\epsilon}(\Omega) := \{ u \in L^{1+\frac{1}{\epsilon}}(\Omega), \frac{\partial u}{\partial x_{i}} \in L^{p_{i}}(\Omega), i=1,...,n \}$$ where $p_{i} > 1,~~ \forall i \in \{1,...,n\}$. Let $W^{1,\overrightarrow{p},\epsilon}_{0}(\Omega)$ be defined as the closure of $C^{\infty}_{c}(\Omega)$ in $W^{1,\overrightarrow{p},\epsilon}(\Omega)$ and let it be endowed with norm $\Vert u \Vert := \sum_{i=1}^{n}\Vert \frac{\partial u}{\partial x_{i}} \Vert$.
If you have a truncation $T_{k}u$ defined as:
$$ T_{k}u := \begin{cases} u(x),& \text{ if }~ |u(x)| \leq 1\\ k\frac{u(x)}{|u(x)|}, & \text{ if }~|u(x)| > k \end{cases} $$
Prove that $T_{k}(u) \rightarrow u$ in $W^{1,\overrightarrow{p},\epsilon}_{0}(\Omega)$
Proposed proof: Observe that $\frac{\partial T_{k}(u(x))}{\partial x_{i}} \rightarrow \frac{\partial u(x)}{\partial x_{i}}$ pointwise a.e. and also $|\frac{\partial T_{k}(u(x))}{\partial x_{i}}| \leq |\frac{\partial u(x)}{\partial x_{i}}|$ for all $k \in \mathbb{N}$. Therefore by the Dominated Convergence Theorem it follows that for each $i \in \{1,...,n\}$ we have $\Vert \frac{\partial T_{k}u}{\partial x_{i}} - \frac{\partial u }{\partial x_{i}} \Vert_{L^{p_{i}}} \rightarrow 0$. Therefore $$T_{k}(u) \rightarrow u ~~~\text{in } W^{1,\overrightarrow{p},\epsilon}_{0}(\Omega)$$
$\square$
Is this a good proof? Are there any recommended changes?
Thanks.
It may be helpful to notice that if function $f$ is lipschitz and $f(0)=0$ then $f\circ u$ is sobolev with $\partial_i (f\circ u)(x)=f'(u)\partial_i u$.
Hence, you could use $$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases} $$ as your truncation function. Of course $f$ satisfies all conditions I wrote above and of course the weak derivative is well defined, as you wrote in your post.
For your questions.
Q1: Yes.
Q2: This is the result of general chain rule. For prove details please refer to Leoni's book, exercise 10.37, for the idea of the proof, please refer to E&G, Theorem 4.2.2
Q3: I think this question will be clear once you understand the proof of Theorem 4.2.2 above