# Proposed proof for convergence in Sobolev space

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.

• You should give a better justification for the inequality $$\left\lvert \frac{\partial T_k u}{\partial x_i}\right\rvert\le \left\lvert\frac{\partial u}{\partial x_i}\right\rvert.$$This is true (up to a set of measure zero), but I am not so sure it is obvious. – Giuseppe Negro Dec 17 '14 at 22:43
• @GiuseppeNegro Does this follow since $|\frac{\partial T_{k}u}{\partial x_{i}}| = |\frac{\partial u(x)}{\partial x_{i}}|$ or $|\frac{\partial T_{k}u}{\partial x_{i}}| = 0$ (depending on $u(x)$)? – Lucio D Dec 17 '14 at 22:53
• @LucioD: Of course, but it is not clear that $T_ku$ is differentiable at almost all points. – Giuseppe Negro Dec 17 '14 at 23:02
• @GiuseppeNegro I meant "Are you saying this is because it is not clear that $u$ is differentiable at almost all points" in the above comment. Since if $u$ is differentiable I think differentiability of $T_{k}u$ follows easily, what do you think? – Lucio D Dec 17 '14 at 23:35
• @JohnDoe: I know that the result is true, what I mean is that it was not properly justified. You need the Stampacchia's version of the chain rule (i.e. the one wisher mentions) to do that. – Giuseppe Negro Dec 18 '14 at 22:47

## 1 Answer

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

• I don't understand what exactly the problem is with the way I proved it? If we assume that $u \in W^{1,\overrightarrow{p},\epsilon}_{0}(\Omega)$ does it not follow that weak derivative is defined almost everywhere. – user100431 Dec 18 '14 at 7:50
• Well, the problem is, how can you make sure a combined function is weak differentiable? – spatially Dec 18 '14 at 15:20
• Yeah I see that now, thought it followed easily, intuitively it does. – user100431 Dec 18 '14 at 15:22
• I updated my answer. Please check it out. – spatially Dec 20 '14 at 0:55
• It is page 129 - 130 – spatially Feb 7 '15 at 19:44