Jacobian, inverse function theorem and continuously differentiable functions

Question

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?

Answer 1

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 $$A=\{x\in \Omega: dist(x,\partial D)<d\}$$ 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.

Answer 2

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 $\partial f_t(D) = f_t(\partial D)$.