Spivak's proof of Inverse Function Theorem.

Question

This question concerns the Inverse Function Theorem done in the book.

Spivak states that

$$|D_jf^i(x) - D_jf^i(a)| < 1/2n^2$$ for all $i,j$ and $x \in U$

Here $U$ is a closed rectangle set, $n$ is an integer.

The lemma 2.10 (on the previous page) states that

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 $|D_jf^i(x)| \leq M$ for all $x$ in the interior of $A$, then $$|f(x) - f(y)| \leq n^2 M|x - y|$$

He define $g(x) := f(x) - x$ and states that, for $x_1, x_2 \in U$, we have

$|f(x_1) - x _1 -(f(x_2) - x_2)| \leq \frac{1}{2} |x_1 - x_2|$

and it's ok for me, because

$|D_jg^i(x) - D_jg^i(a)| \leq \frac{1}{2n^2} n^2 |x_1 - x_2| = \frac{1}{2} |x_1 - x_2|$

by the lemma, but we can do this only if $|D_jg^i(x)| \leq M$, my doubt is why is this valid?

Answer 1

Correct me if I am wrong - the broader context is we have the following results already:

(a) $Df(a)= I$, the identity matrix, by assumption.

(b) $|D_jf^i(x) - D_jf^i(a)| < \frac{1}{2n^2} $ for all $i,j$ and $x \in U$.

So $g=f-I$. From (b) we have almost what you wrote, $$ |D_jg^i(x)| = |D_jf^i(x) - D_jI^i(x)| = |D_jf^i(x) - D_jf^i(a)| <\frac{1}{2n^2} =M $$ since $D_jI^i(x) = D_jI^i(a) = D_jf^i(a)$ ($I$ is a linear map). We can then apply lemma to deduce, $$ |g(x_1 ) - g(x_2) | \le \frac{1}{2n^2} n^2 |x_1 - x_2| = \frac{1}{2} |x_1 - x_2 |.$$