# approximating functions in $W^{1,2}$ with Lipschitz functions with image contained in fixed ball

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.