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While arguing about (essentially a prelude) of the fact that embeddings are open in $C^r$, Hirsch makes the following claim about $C_1$ functions $g_n$ such that $|g_n(x)-f(x)| \to 0 $ and $\Vert Dg_n(x) - Df(x) \Vert\to 0$, where $f$ is also $C^1$:

By uniformity of Taylor expansion (with remainder in integral form), $$\frac{|g_n(a_n)-g_n(b_n)-Dg_n(b_n)(a_n-b_n)|}{|a_n-b_n|} \to 0.$$

(here, $a_n,b_n \to p \in K$, a compact set on which each $g_n$ is defined). However, it is not clear to me what he means by "integral form", because the integral form I know would need the second derivative of $g_n$. And even if we suppose $g_n$ is $C^2$, we could have the second derivative of $g_n$ blowing up, and the estimate would not hold.

My question is: What exactly is happening, and how to make this work?

OBS: For the sake of completeness, I put the entire lemma and hypotheses:

Let $U \subset \mathbb{R}^n$ be an open set and $W \subset U$ an open set with compact closure $\overline{W} \subset U$. Let $f:U \to \mathbb{R}^n$ be a $C^1$ embedding. There exists $\epsilon>0$ such that if $g:U \to \mathbb{R}^n$ is $C^1$ and $$\Vert Dg(x)-Df(x) \Vert < \epsilon ~\text{ and }~|g(x)-f(x)| < \epsilon$$ for all $x \in W$ then $g|_W$ is an embedding.

The $g_n$ arise by supposing the result isn't true and using the fact that immersions are open.

Aloizio Macedo
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2 Answers2

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What about using that $$ g_n(a_n) - g_n(b_n) = \int_0^1Dg_n(ta_n + (1 - t)b_n)(b_n - a_n)dt? $$ Then \begin{eqnarray*} \frac{|g_n(a_n) - g_n(b_n) - Dg_n(b)(a_n - b_n)|}{|a_n - b_n|} &\leq& \bigg(\int_0^1 ||Dg_n(ta_n + (1 - t)b_n) - Dg_n(b_n)||dt\bigg) \bigg|\frac{a_n - b_n}{|a_n - b_n|}\bigg| \to 0. \end{eqnarray*} Recall that on Hirsch's argument $(a_n - b_n)/|a_n - b_n|$ converges.

João Caminada
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  • Hi, I'm reading through the same lemma. I know this is a silly question, but I guess your integral is valid because $g$ is a map from $U$ open in $\mathbb{R}^n$ to $\mathbb{R}^m$. At the beginning I missunderstood the lemma cause I was thinking it was about general manifolds, instead it is specifically about euclidean spaces. Am I right? (i.e. your integral is a standard integral of vector function of vector variables). – user8469759 Sep 14 '20 at 11:52
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    @user8469759 you're right. – João Caminada Sep 14 '20 at 11:58
  • A remark: the function $g_n$ is defined on the $U$ and $a_n,b_n \in W$. But $W$ or $U$ may not be convex, so we cannot do integration along the segment $[a_n,b_n]$ directly. One way to fix it is to replace $g_n$ by $h_n = g_n \chi$ where $\chi \in C^{\infty}_{c}(\mathbb{R}^m, [0,1])$ such that $\chi =1$ on a nbhd of $\bar{W}$ and $\chi =0$ on a nbhd of $\mathbb{R}^m-U$. – Skyskie Sep 22 '23 at 15:22
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Those results, lemma 1.3 and theorem 1.4, have little problems for me. In the lemma, Hirsch show that $g_n$ is injective immersion for large $n$. But it's not clear that $g_n$ is embedding for larger $n$.

In the theorem, you can consider that $g_{\vert U_i}$ is embedding for the argumentation.

Armando j18eos
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Elizeu França
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