Spivak’s proof on the Inverse Function Theorem

multivariable-calculus

In Spivak's proof of the Inverse Function Theorem (in his calculus on manifolds book), As one of the steps in proving that
$$(f^{-1})'(y)=[f'(f^{-1}(y))]^{-1}$$
He writes that "Since $f$ is continuously differentiable in an open set containing $a$, we can assume that"

  1. $\det f'(x)\ne0$ for $x\in U$
  2. $|D_jf^i(x)-D_jf^i(a)|<\frac{1}{2n^2}$ for all $i,j$, and $x\in U$

I'm trying to figure out why Spivak chooses the factor $\frac{1}{2n^2}$ at point 3 (point 2 here). As far as I understand, this is just a restatement of the continuity of the first derivative (taking the rectangle $U$ to be the "size" $\frac{1}{2n^2}$). Initially I thought this was an arbitrary choice, but I realised the factor in front of $1/n^2$ would effect the factor in point 4 ($|x_1-x_2|\le2|f(x_1)-f(x_2)|$). I have two questions

  1. Is my reasoning for point 3 correct? Is it simply some arbitrary constant used to restate the continuity of the first derivative (and also cancels into Lemma 2 nicely e.g. 1/3 would give a factor of $3/2$ in point 4)?
  2. If the factor of $1/2$ is not arbitrary, why does it specifically have to be $1/2$?

Best Answer

Your reasoning for point 3 correct. The precise values of the factors are not really important, but the choice is very convenient. Given $\epsilon > 0$, we can choose $U$ such that

  1. $\det f'(x)\ne0$ for $x\in U$
  2. $|D_jf^i(x)-D_jf^i(a)|<\epsilon$ for all $i,j$, and $x\in U$

Then 2-10 applies to show that $\lvert (f(x_1) - x_1) - (f(x_2) - x_2) \rvert \le n^2\epsilon\lvert x_1- x_2 \rvert$ so that $$\lvert x_ 1 - x_2 \rvert - \lvert f(x_1) - f(x_2) \rvert \le \lvert (f(x_1) - x_1) - (f(x_2) - x_2) \rvert \le n^2\epsilon \lvert x_1- x_2 \rvert .$$ Hence $$(1- n^2 \epsilon)\lvert x_ 1 - x_2 \rvert \le \lvert f(x_1) - f(x_2) \rvert .$$ This is a non-trivial inequality only when $\epsilon < \frac 1 {n^2}$. If this is satisfied we get $$\lvert x_ 1 - x_2 \rvert \le M \lvert f(x_1) - f(x_2) \rvert $$ with $M = \frac{1}{1- n^2 \epsilon} $. Note that $M> 1$ because $1- n^2 \epsilon < 1$.

Spivak's choice gives $M = 2$, but everything in Spivak's proof also works with any other $M$. Especially 6. reads as $$\lvert f^{-1}(y_1) - f^{-1}(y_2) \lvert \le M \lvert y_1 - y_2 \rvert \text{ for } y_1,y_2 \in W .$$

Related Question