# Jacobian, inverse function theorem and continuously differentiable functions

Question: Let $$f \colon \Omega \to \mathbb{R}^n$$ be such that $$f$$ is continuously differentiable where $$\Omega$$ is a bounded connected set in $$\mathbb{R}^n$$. For each $$t \in \mathbb{R}$$ define $$f_t(x) := f(x) + t x$$. Let $$D$$ be any open connected set in $$\Omega$$ such that $$\overline{D} \subset \Omega$$. Show that $$f_t$$ is injective on $$D$$ for sufficiently large $$t$$ and $$f_t(\partial D) = \partial(f_t(D))$$. Also, suppose $$g\colon \Omega \to \mathbb{R}^n$$ that is continuously differentiable with $$f(x) = g(x)$$ for $$x \in \partial D$$. Show that $$\int_D |J_g|\, dx = \int_D |J_f|\, dx$$ where $$J_g, J_f$$ are Jacobians of $$f$$ and $$g$$ respectively.

This questions seems to require the use of inverse function theorem due to the injectivity involved, however I not too sure how to tackle $$f_t(\partial D) = \partial(f_t(D))$$ and well as the second part of the question involving Jacobians. Also, is it necessary for $$\Omega$$ to be connected?

## 2 Answers

I will just give you some hints and leave the details to you.

The inverse function theorem will only give you local invertibility and not global invertibility. I assume that $$\Omega$$ is open. Since the closure of $$D$$ is compact, it has positive distance from the boundary of $$\Omega$$. If $$2d$$ is the distance, we can enlarge $$D$$, and the open set $$A=\{x\in \Omega: dist(x,\partial D) has closure contained in $$\Omega$$. Since $$f$$ is continuously differentiable, it’s gradient is bounded in the closure of $$A$$. Use this to prove that $$f$$ is Lipschitz continuous in the smaller set $$D$$ with Lipschitz constant $$L$$.

Now assume by contradiction that $$f_t$$ is not injective. Then there exist $$x,y$$ in $$D$$ such that $$f_t(x)=f_t(y)$$. It follows that $$t(x-y)=-f(x)+f(y).$$ Taking the norm on both sides, and using the fact that $$f$$ is Lipschitz, you get a contradiction provided $$t>L$$.

The fact that $$\partial f_t(D)=f_t(\partial D)$$ comes from the fact that, by the inverse function theorem, the function $$f_t$$ is open, so it maps open sets into open sets.

The last part should follow by using $$f_t$$ as a change of variables.

• Thank you so much for the help. What is the significance of the set $A$ in this solution and how do you show that $\overline{A}$ is contained in $\Omega$? Also could you elaborate more on how to solve the integrals involving the integrals and is the connectedness of $\Omega$ necessary for this question? I have added a sketch of my solution following your hints, hope it's useful for my clarification.
– L-JS
Jun 29 at 9:32
• The distance between the boundary of $D$ and the boundary of $\Omega$ is $2d$, so if you are at distance $d$ from the boundary of $D$, you are at distance $d$ from the boundary of $\Omega$, which means that you are still well inside $\Omega$ Jun 29 at 19:33
• If you try to prove carefully that $f$ is Lipschitz continuous in $D$, you will see why you need the st $A$. Jun 29 at 19:34

Partial solution following the hint provided by @Gio67:

Assuming that $$\Omega$$ is a open bounded set in $$\mathbb{R}^n$$ then the distance between $$\partial \Omega$$ and $$\overline{D}$$ is positive. Suppose $$r>0$$ is the radius of $$D$$ then there exists $$\epsilon >0$$ so that $$\overline{D} \subsetneq D_{\epsilon + R} \subsetneq \Omega$$ where $$D_{\epsilon + r}$$ is the disk with same center as $$D$$ but having $$\epsilon + r$$ as its radius. Given that $$\overline{D}$$ is a closed and bounded (hence compact) set in $$\mathbb{R}^n$$ and $$f$$ being a continuously differentiable function on $$\overline{D}$$ then $$f$$ is Lipschitz continuous on $$\overline{D}$$ for all $$t$$, that is there exists $$L > 0$$ so that $$x,y \in \overline{D} \implies |f(x) - f(y)| \leq L |x-y|$$ If for a contradiction that $$f_t$$ is not injective for $$t$$ sufficiently large then there exists $$x\neq y \in D$$ so that \begin{align} |f(x) - f(y)| = |t| |y-x| \end{align} which is a contradiction whenver $$|t|> L$$. This concludes that $$f_t$$ is injective on $$D$$ for sufficently large $$t$$. By a similar argument, we have $$f_t$$ to be injective and continuously differentiable on $$D_{\epsilon + r}$$ for sufficiently large $$t$$. Also, for sufficiently large $$t$$, $$f_t'(x)$$ can be made invertible for all $$x \in D_{\epsilon + r}$$ so by inverse function theorem, there exists open sets $$V, W\subset \mathbb{R}^n$$ so that $$V \supset D_{\epsilon + r} \supset \overline{D}$$ and $$f_t \colon V \to W$$ is invertible continuously differentiable for sufficiently large $$t$$. Also, $$f_t^{-1} \colon W \to V$$ is also invertible continuously differentiable for sufficiently large $$t$$. Thus, $$f_t$$ is both open and closed map where both $$f_t(\partial D)$$ and $$\partial f_t(D)$$ are closed sets in $$\mathbb{R}^n$$.

Since $$\partial f_t(D) = \overline{f_t(D)} - Int(f_t(D)) = \overline{f_t(D)} - f_t(D)$$ and $$f_t(\partial D) = f_t(\overline{D} - Int(D)) = f_t(\overline{D}) - f_t(D)$$, it suffices to show that $$f_t(\overline{D })=\overline{f_t(D)}$$. For the forward inclusion, let $$y \in f_t(\overline{D})$$, if $$y = f_t(\partial D)$$ then $$y = f(x)$$ where $$x \in \partial D$$ and so there exists a sequence $$(x_k)_k \subset D$$ so that $$x_k \to x$$ as $$k \to \infty$$. By continuity of $$f_t$$ we have $$y = f_t(y) = \lim_{k\to \infty} f_t(x_k)\implies y \in \overline{f_t(D)}.$$ So, $$f_t(\overline{D })\subset\overline{f_t(D)}$$. For the reverse inclusion, it follows a similar argument as above, this concludes that $$f_t(\overline{D })=\overline{f_t(D)}$$ and so $$\partial f_t(D) = f_t(\partial D)$$.