Strongly differentiable function has inverse satisfying Lipschitz condition

Question

Problem Statement: Let $U\subset \mathbb{R}^{m}$ be open, and $K\subset U$, compact. If $f\in C^{1}(U;\mathbb{R}^{n})$ and $df\in \mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$ is injective for each $\mathbf{x}\in K$, then $\exists c,r>0$ so that $\lVert f(\mathbf{x})-f(\mathbf{y})\rVert \geq c\lVert \mathbf{x}-\mathbf{y}\rVert$ whenever $\mathbf{x}\in K$, $\mathbf{y}\in U$, and $\lVert \mathbf{x}-\mathbf{y}\rVert <r$. ($\mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$ is the set of bounded linear maps $T:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$)

I am working on this problem, and I have a theorem that I think will be helpful, but I am unsure how to apply it.

First, the $f$ given in the problem statement is strongly differentiable.

Also, the theorem states that if $f:U\rightarrow \mathbb{R}^{n}$ is strongly differentiable at $\mathbf{a}\in U\subset \mathbb{R}^{m}$, $U$ open, and $df(\mathbf{a})\in \mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$ is injective, then $\exists c,r>0$, such that $\mathbf{x},\mathbf{y}\in B_{r}(\mathbf{a})$ implies $\lVert f(\mathbf{x})-f(\mathbf{y})\rVert \geq c\lVert \mathbf{x}-\mathbf{y}\rVert$.

The difference between this case and the problem statement is that we must show that for any $\mathbf{x}\in K\subset U$, $\mathbf{y}\in U$, the same inequality holds, while in the theorem, these two points $\mathbf{x}$ and $\mathbf{y}$ must be inside a ball of finite radius centered at $\mathbf{a}$.

We know that $df(\mathbf{x})$ is injective for $\mathbf{x}\in K$, but it is not true for any $\mathbf{y}\in U$. I was thinking to consider $B_{r}(\mathbf{x})$, but we must show that the distance between any point in $U$ and another in $K$ is finite, which I am not sure how to do, since $U$ is not necessarily bounded.

I am also confused what it means for $df(\mathbf{a})$ to be injective. Would that not be an explicit point in $\mathbb{R}^{n}$, since it is the differential of $f$ evaluated at $\mathbf{a}$? Or is the theorem essentially saying the conclusion holds if it is injective at any point in which the function is strongly differentiable?

Any suggestions on how to approach this problem are appreciated!

Answer 1

If the theorem fails, then for each choice of $c,r>0,$ there are $x\in K,y\in U$ with $|y-x|<r$ and $|f(y)-f(x)|<c|y-x|.$ In particular, for $c=r=1/k,k=1,2,\dots,$ there exist $x_k\in K,y_k\in U$ with $|y_k-x_k|<1/k$ and $|f(y_k)-f(x_k)|<(1/k)|y_k-x_k|.$

Now $K$ is compact, so there exists a subsequence $x_{k_j}$ converging to some $x_0\in K.$ Because $|y_{k_j}-x_{k_j}|<1/k_j,$ we also have $y_{k_j}\to x_0.$ By the theorem we are given, there exist $c,r>0$ that "work" in $B(x_0,r).$ But for large $j$ we will have $1/k_j$ smaller than both $c$ and $r/2.$ For such $j$ we then have $x_{k_j},y_{k_j}\in B(x_0,r),$ while $|f(y_{k_j})-f(x_{k_j})|<(1/k_j)|y_{k_j}-x_{k_j}|<c|y_{k_j}-x_{k_j}|.$ That's a contradiction, proving the desired result.

Note: The above replaces an earlier answer to an earlier question.

Answer 2

To apply the known theorem, you can use a general trick for compact sets: if you have a result that holds at every point in this set, and if one of the things that this result gives you at each point is a neighbourhood of that point, then you've got an open cover of the compact set, so pass to a finite subcover so that you can focus attention on finitely many points; that might be enough.

So in this case, take the collection of all triples $(\mathbf{a},c,r) \in K \times (0,\infty) \times (0,\infty)$ such that for $\mathbf{x}, \mathbf{y} \in U$, $\lVert{f(\mathbf{x})-f(\mathbf{y})}\rVert \geq c\,\lVert{\mathbf{x}-\mathbf{y}}\rVert$ whenever $\lVert{\mathbf{a}-\mathbf{x}}\rVert < r$ and $\lVert{\mathbf{a}-\mathbf{y}}\rVert < r$; each such triple defines an open set $B_{r/2}(\mathbf{a})$. Your theorem guarantees that each $\mathbf{a} \in K$ (indeed each $\mathbf{a} \in U$) shows up in at least one such triple; since $\mathbf{a} \in B_{r/2}(\mathbf{a})$, this means that these open sets form an open cover of $K$. So there is a finite subcover, indexed by a finite collection of triples $(\mathbf{a}_i,c_i,r_i)$ for $i = 1, \ldots, n$. Let $c$ be $\min_i c_i$, and let $r$ be $\min_i (r_i/2)$ (which is why we need a finite subcover). Then if $\lVert{\mathbf{x}-\mathbf{y}}\rVert < r$ for some $\mathbf{x} \in K$ and $\mathbf{y} \in U$, we have $\mathbf{x} \in B_{r_i/2}(\mathbf{a}_i)$ for some $\mathbf{a}_i$ (which is why we need a finite subcover), so $\lVert{\mathbf{a}_i-\mathbf{x}}\rVert < r_i/2 < r_i$ and $$\lVert{\mathbf{a}_i-\mathbf{y}}\rVert \leq \lVert{\mathbf{a}_i-\mathbf{x}}\rVert + \lVert{\mathbf{x}-\mathbf{y}}\rVert < r_i/2 + r \leq r_i/2 + r_i/2 = r_i$$ (which is why I had divide $r$ by $2$ all the time), so $\lVert{f(\mathbf{x})-f(\mathbf{y})}\rVert \geq c_i\,\lVert{\mathbf{x}-\mathbf{y}}\rVert \geq c\,\lVert{\mathbf{x}-\mathbf{y}}\rVert$, QED.