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.

  • 1
    $\begingroup$ What does it mean "Lipschitz injective"? $\endgroup$ Oct 24, 2020 at 13:27
  • $\begingroup$ 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 $\Omega_1$. $\endgroup$ Oct 24, 2020 at 16:45
  • $\begingroup$ 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. $\endgroup$
    – mlk
    Oct 26, 2020 at 11:24
  • $\begingroup$ 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. $\endgroup$ Oct 26, 2020 at 11:55
  • $\begingroup$ @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 $f(\partial \Omega_1) \subseteq \partial \Omega_2$?) $\endgroup$ Oct 26, 2020 at 15:43

1 Answer 1


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.

  • $\begingroup$ 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 $.) $\endgroup$ Oct 26, 2020 at 17:40
  • $\begingroup$ 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 $\Omega_1$ $\endgroup$ Oct 26, 2020 at 17:42
  • $\begingroup$ @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 $\Omega$. $\endgroup$
    – mlk
    Oct 26, 2020 at 18:51

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