1
$\begingroup$

Fix $r>0$. For $h>0$ let $u_h\in W^{1,2}(B_h,B_r)$, where $B_h$ and $B_r$ are the ball of radius $h$ and $r$ centered at the origin in $\mathbb R^n$, respectively.
I want to approximate the $u_h$ with Lipschitz function $f_h^\lambda\in C^{0,1}(B_h,B_r)\cap W^{1,2}(B_h,B_r)$.
There is a theorem which says (https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.10048, Appendix A.1)

Theorem: Let $U\subset\mathbb R^n$ be a bounded Lipschitz domain.
Then there exists a constant $C(U)$ with the following property:
For each $u\in W^{1,2}(U,\mathbb R^n)$ and each $\lambda>0$ there exists $v:U\rightarrow\mathbb R^n$ such that
1. $\|dv\|_{L^\infty(U)}\leq C\lambda$,
2. $|\{x\in U: u(x)\neq v(x)\}|\leq\frac{C}{\lambda^2}\int_{\{x\in U:|du(x)|>\lambda\}}|du|^2 dx$,
3. $\|du-dv\|_{L^2(U)}^2\leq C\int_{\{x\in U:|du(x)|>\lambda\}}|du|^2 dx$.

If I fix $h>0$ and take the $f_h^\lambda$ as in the theorem, then as $\lambda$ grows I get the problem that $\|df_h^\lambda\|_{L^\infty}$ could blow up, and I could no longer guarantee that the image of $f_h^\lambda$ is contained in $B_r$.
For me it would be enough that as $h\rightarrow 0$ I get that $$\frac{1}{\mathrm{Vol}(B_h)}\int_{B_h}\mathrm{dist}(du(x),SO(n))-\mathrm{dist}(df(x),SO(n))dx$$ tends to zero, where the distance is taken w.r.t. the Frobenius norm (almost everywhere).
Although I have 2. in the theorem, it does not solve my problem, or does it?
I mean even if the set where $u_h$ and $f_h^\lambda$ differ shrinks, I cannot say that the image of $f_h^\lambda$ is contained in $B_r$, since $|df_h^\lambda|$ potentially becomes very large.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.