# Is a locally invertible weak limit of injective maps injective almost everywhere?

This is a cross-post.

Let $$\Omega_1,\Omega_2 \subseteq \mathbb R^2$$ be open, connected, bounded, with non-empty $$C^1$$ boundaries.

Let $$f_n:\bar\Omega_1 \to \bar\Omega_2$$ be Lipschitz injective maps with $$\det(df_n)>0$$, and suppose that $$f_n$$ converges to a $$C^1$$ function $$f: \bar\Omega_1 \to \bar\Omega_2$$ weakly in $$W^{1,2}$$, and that $$\det(df)>0$$ everywhere on $$\bar\Omega_1$$.

Is it true that $$|f^{-1}(y)| \le 1$$ a.e. on $$\Omega_2$$?

Does the answer change if we assume in addition that $$f_n|_{K} \to f|_{K}$$ strongly in $$W^{1,2}$$ for every $$K \subset \subset \Omega_1$$?

The condition $$\det(df)>0$$ rules out degenerate counterexamples such as $$f_n(x)=\frac{x}{n}$$ which converge to a constant.

• What does it mean "Lipschitz injective"? Oct 24, 2020 at 13:27
• I just mean that each $f_n$ is a Lipschitz map (with a Lipschitz constant which might depend on $n$) and that it's injective on $\Omega_1$. I am fine with replacing the injectivity assumption with the requirement $|f_n^{-1}(y)| \le 1$ for almost every $y \in \Omega_2$. And of course by $\det(df_n)>0$ I mean "almost everywhere", since $df$ might not be defined on all $\Omega_1$. Oct 24, 2020 at 16:45
• I am pretty sure that what you ask is true as it is essentially just degree theory for Sobolev spaces in conjunction with the fact that non-negative determinants converge slightly better than one would expect. You don't even need to exclude the degenerate case of constants, as for those $|f^{-1}(y)|=0$ a.e. anyway. I'll try to give a full answer once I find the time.
– mlk
Oct 26, 2020 at 11:24
• Thanks, this sounds interesting. I know that the Jacobians $Jf_n$ converge weakly in $L^1(K)$ for $K \subset \subset \Omega_1$; Unfortunately, I am not sufficiently familiar with degree theory in the Sobolev context; it sounds like the right tool tough. I would be happy to see the details. Oct 26, 2020 at 11:55
• @LeoMoos Thanks, but I am not sure how do you continue from here. Also is it clear that $\int Jf$ is an integer multiple of the image's volume? (We have manifolds with boundary here; does that covering argument requires assuming that $f(\partial \Omega_1) \subseteq \partial \Omega_2$?) Oct 26, 2020 at 15:43

Okay, let me try a writeup of the comment chain. For any reasonable subset $$A\subset \Omega_2$$ and $$B := f^{-1}(A)$$ you get $$\int_A |f^{-1}(y)| dy = \int_B \det df dx \leq \liminf_{n\to\infty} \int_B \det df_n dx = \liminf_{n\to\infty} \mathcal{H}^2(f_n(B)).$$

Then if we know that $$\mathcal{H}^2(f_n(B)) \to \mathcal{H}^2(f(B)) \leq \mathcal{H}^2(A)$$ ($$A$$ can have points with no preimage), we get that $$|f^{-1}(y)| \leq 1$$ a.e. as $$A$$ was arbitrary.

Now using the existence of a pointwise a.e. converging subsequence (never relabeled) and Egorov's theorem, for any $$\epsilon > 0$$ there is $$B_\epsilon$$ such that $$\mathcal{H}^2(B \setminus B_\epsilon) < \epsilon$$ and $$f_n$$ converges uniformly on $$B_\epsilon$$. But then a quick argument shows that $$f_n(B_\epsilon)$$ converges in the Hausdorff-sense and thus $$\mathcal{H}^2(f_n(B_\epsilon)) \to \mathcal{H}^2(f(B_\epsilon))$$. Now the leftover set is small and Müller's famous result gives us that $$\det df_n$$ converges weakly in $$L^1$$ (see ¹). So in particular as $$\chi_{B\setminus B_\epsilon} \in (L^1)^*$$ $$\mathcal{H}^2(f_n(B\setminus B_\epsilon)) = \int_{B\setminus B_\epsilon} \det d f_n dy \to \int_{B\setminus B_\epsilon} \det df dy$$ which is small for small enough $$\epsilon$$ as $$f \in C^1$$. Similarly $$\mathcal{H}^2(f(B\setminus B_\epsilon)) \leq \int_{B\setminus B_\epsilon} \det df dy.$$

¹As remarked by Asaf in the comments, the result gives convergence on compact subsets. However as $$\Omega_1$$ is $$C^1$$, there exists an extension operator to a larger domain $$\Omega \supset \Omega_1$$ and thus $$\tilde{f}_n,\tilde{f} \in W^{1,2}(\Omega)$$ such that $$\tilde{f}_n \to \tilde{f}$$ in the same sense and $$\tilde{f}_n|_{\Omega_1} = f_n, \tilde{f}|_{\Omega_1} = f$$. Now $$\overline{\Omega_1} \subset \Omega$$ is the required compact set.

• Thanks, that looks like a nice solution! I have one question: Müller's result gives weak convergene of the Jacobians on $L^1(K)$ for compactly contained subsets $K \subset \subset \Omega_1$. So, for your argument to work you need to make sure that $B=f^{-1}(A)$ is compactly contained in the interior $\Omega_1$. (I actually think that you need it also for the first $\liminf$ inequality $\int_B \det df dx \leq \liminf_{n\to\infty} \int_B \det df_n dx$.) Oct 26, 2020 at 17:40
• I don't think that it's trivial to guarantee this, since even though $f \in C^1$, we don't know that it's injective (yet), so $A$ may well have some weird pre-images which are close to the boundary of $\Omega_1$ Oct 26, 2020 at 17:42
• @AsafShachar Ah, yeah I forgot that detail. However the fix is easy. Since $\Omega_1$ has $C^1$ boundary there is an extension operator for $W^{1,2}$. So we can extend $f_n$ to $\Omega \supset \Omega_1$ with the same convergence and then $\overline{\Omega_1}$ is compact within $\Omega$.
– mlk
Oct 26, 2020 at 18:51