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In the proof of Inverse Function Theorem, Spivak states enter image description here

Lemma 2-10 is given below. Can anyone explain why the statement above is correct? Thanks!

Lemma 2-10 : Let $A \subset \mathbb{R}^n$ be a rectangle and let $f : A \to \mathbb{R}^n$ be continuously differentiable. If there is a number $M$ such that $| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$, then $$ |f(x)-f(y)| \leq n^2 M |x-y| $$ for all $x,y \in A$.

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2 Answers 2

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Recall right at the beginning of the proof Spivak shows that $Df(a)$ is the identity $\pi$, so that $$|D_jg^i(x)|=|D_jf^i(x)-D_j\pi^i(x)|=|D_jf^i(x)-D_jf^i(a)|\lt\frac{1}{2n^2}.$$ So, letting $M=1/2n^2$, we can apply Lemma 2-10 to conclude $$\big|f(x_1)-x_1-\big(f(x_2)-x_2\big)\big|=|g(x_1)-g(x_2)|\leq\frac{1}{2}|x_1-x_2|.$$

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I finally understand it now. The explanation by @Moed Pol Bollo is not accurate. Inequality (3) is correct since $f$ is $C^1$. Because $\lambda=Df(a)=I$, we have $D_jg^i(x)=D_jf^i(x)-D_jf^i(a)$. Hence, $M$ is actually $\frac{1}{2n^2}$ here. Applying Lemma 2-10 yields, $$|g(x_1)-g(x_2)|\le n^2 M|x_1-x_2|=\frac{1}{2}|x_1-x_2|.$$

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  • $\begingroup$ Yeah sorry, was typing on mobile and omitted the 2’s. $\endgroup$
    – Mo Pol Bol
    Commented Oct 21, 2021 at 6:58
  • $\begingroup$ @MoedPolBollo Thank you. I accepted your answer. $\endgroup$ Commented Oct 21, 2021 at 13:34

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