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I'm trying to understand the proof of Lemma 5 from Terence Tao's blog, i.e.,

Let $X$ be an open subset of $\mathbb R^n$ and $f:X \to \mathbb R^n$ be differentiable such that $\partial f (x)$ is invertible for all $x\in X$. Fix $x_0 \in X$ and $y_0 := f(x_0)$ and let $$ K := \{x \in X: f(x)=y_0\}. $$

Lemma 5: Let $H$ be the connected component of $K$ that contains $x_0$. Then $H = \{x_0\}$.

Could you have a check on my attempt?

Proof: Assume the contrary that $H \neq \{x_0\}$. Then there is a path $\gamma:[0,1] \to H$ such that $\gamma(0) = x_0$ and $\gamma(1) \in H \setminus \{x_0\}$. We have $f \circ \gamma = y_0$. Then $$ \begin{align} \lim_{t \to 0^+} \frac{f \circ \gamma (t) - f(x_0) - \partial f (x_0)(\gamma(t)-x_0)}{|\gamma(t)-x_0|} &= -\lim_{t \to 0^+} \frac{ \partial f (x_0)(\gamma(t)-x_0)}{|\gamma(t)-x_0|} \\ &= - \partial f (x_0) \left (\lim_{t \to 0^+}\frac{\gamma(t)-x_0}{|\gamma(t)-x_0|} \right ). \end{align} $$

Because $\partial f (x_0)$ is invertible, it is bijective. It follows that $$ \lim_{t \to 0^+}\frac{\gamma(t)-x_0}{|\gamma(t)-x_0|} = 0, $$ which is a contradiction.

Akira
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  • Are you deducing the existence of the path $\gamma$ from the connectedness of $H$ alone? Note that $H$ being connected doesn't imply it is path-connected. – plop Oct 01 '22 at 20:05
  • @user85667 You're right! Do you have any idea for a fix? – Akira Oct 01 '22 at 20:15
  • Well, I would be lazy and apply the inverse function theorem. There are neighborhoods of $x_0$ and $y_0$ on which $f$ has an inverse. In particular, the restriction of $f$ to the neighborhood of $x_0$ is injective and therefore there are no other points of $H$ in it. – plop Oct 01 '22 at 20:20
  • @user85667 but $f$ is not $C^1$... – Akira Oct 01 '22 at 20:21
  • @user85667 I may be wrong, but Rolle's theorem applies to only real-valued functions. – Akira Oct 01 '22 at 21:23
  • Let $A$ be the differential of $f$ at $x_0$. Since it is non-singular then there is some positive constant $c>0$ (the norm of its inverse) such that $|Ax|\geq c|x|$. Let $x_n\to x_0$ be points where $f(x_n)=f(x_0)$. We have that $0=f(x_n)-f(x_0)=A(x_0-x_n)+R(x_0-x_n)$, where $R$ satisfies $|R(x)|/|x|\to0$ as $x\to0$. Then $|R(x_0-x_n)|=|A(x_0-x_n)|\geq c|x_0-x_n|$. If we divide by $|x_0-x_n|$ and let $n\to\infty$, we would get $0\geq c$, which is a contradiction with $c>0$. Therefore $x_0$ is an isolated point of $K$. – plop Oct 02 '22 at 01:03
  • @user85667 Thank you so much for your help! Could you check my answer below to see if I understand your idea correctly? – Akira Oct 02 '22 at 09:12

1 Answers1

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I formulate @user85667's idea below to better understand it.

Assume the contrary that $H$ has an element other than $x_0$.

Lemma 1: Let $E, F$ be Banach spaces and $f:E \to F$ linear continuous. Then $f$ is bounded from below, i.e., $$ \exists c>0, \forall x \in E: |f(x)| \ge c |x|, $$ if and only if $f$ is injective and has closed range.

Let $A := \partial f (x_0)$. Then $A$ is invertible. By Lemma 1, there is $c>0$ such that $|A (x)| \ge c|x|$ for all $x \in \mathbb R^n$.

Theorem: A connected metric space $(E, d)$ with at least $2$ points has no isolated points.

By Lemma 2, $H$ has no isolated points. So there is $(x_n)_{n\ge 1} \subset H \setminus \{x_0\}$ such that $x_n \to x_0$. Because $f$ is differentiable at $x_0$, $$ f(x)-f(x_0) = A(x-x_0) + o (|x-x_0|) \quad \text{as} \quad x \to x_0. $$

Notice that $x_n \in H$ and thus $f(x_n)=y_0$ for all $n \in \mathbb N$, so $$ A(x_n-x_0) =- o (|x-x_0|) \quad \text{as} \quad n \to \infty. $$

On the other hand, $$ \lim_{n \to \infty} \frac{|A(x_n-x_0)|}{|x_n -x_0|} \ge c>0, $$ which is a contradiction.

Akira
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    Yes, that's fine, although Lemma 1 is more general than needed. We only need the bound in finite dimension, which we can just get from something as simple as the triangle inequality $|A^{-1}x|^2\leq n^2\max_{i,j}(|a_{i,j}|^2)|x|^2$. – plop Oct 02 '22 at 12:46
  • @user85667 Thank you so much for your dedicated help! – Akira Oct 02 '22 at 13:01