On inverse function theorem in Spivak's CoM

[psie]

TL;DR

I have two questions on Spivak's proof of the inverse function theorem in Calculus on Manifolds.

I know of a thread on this site with a similar question, but no definite answer. I will not state the whole proof, as it is quite long.

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and $\det f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$ Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since $\det f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$. Therefor we may assume at the outset that $\lambda$ is the identity.

... ... ...

Note that (3) and Lemma 2-10 applied to $g(x)=f(x)-x$ imply for $x_1,x_2\in U$ that $$|f(x_1)-x_1-(f(x_2)-x_2)|\leq\frac12|x_1-x_2|.$$ Since \begin{align*}
|x_1-x_2|-|f(x_1)-f(x_2)|&\leq |f(x_1)-x_1-(f(x_2)-x_2)| \\ &\leq\frac12 |x_1-x_2|,
\end{align*}
... ... ...

1. Why can we assume $f$ to have the identity map as its derivative? I understand how if the theorem is true for $g = \lambda^{-1} \circ f$, then its true for $f$ and I also understand that $Dg(a)=\mathrm{id}$, where $\mathrm{id}$ is the identity map. If $\lambda=\mathrm{id}$, then simply $g=f$ and this is how the proof proceeds. But what when $\lambda\neq \mathrm{id}$?

2. Why is $$|x_1-x_2|-|f(x_1)-f(x_2)|\leq |f(x_1)-x_1-(f(x_2)-x_2)|,$$ true? I'm tempted to say it is the reverse triangle inequality, but shouldn't there be an extra pair of absolute value signs around the LHS of this inequality?

 

[Euge]

Hi @psie,

1. Suppose the assertion holds true for $g$. Then there are open neighborhoods $V \ni a$ and $U\ni g(a)$ such that $g : V \to W$ is a $C^1$-diffeomorphism. Since $\lambda : \mathbb{R}^n \to \mathbb{R}^n$ is a $C^1$-diffeomorphism, the image $W := \lambda(U)$ is an open neighborhood of $f(a)$ and the composition $V \xrightarrow[\approx]{g} U \xrightarrow[\approx]{\lambda} W$ is a $C^1$-diffeomorphism. But this composition is $f : V \to W$. The inverse $f^{-1} : W \to V$ is $g^{-1} \circ \lambda^{-1}$, so the formula for $(f^{-1})'(y)$ is deduced from the formula for $(g^{-1})'(z)$ and the chain rule.

2. You are right that this is due to the reverse triangle inequality. Absolute value signs around the LHS are not needed since $|x| \ge x$ for all real numbers $x$.

 

[psie]

Thank you, @Euge. Regarding 1, I understand what you’re writing, but shouldn’t we then prove the theorem for $g$? Spivak only seems to prove a very special case, namely when $g=f$, i.e. when $\lambda=\mathrm{id}$. How do we modify the proof when $\lambda\neq\mathrm{id}$?

 

[Office_Shredder]

If you multiply your function by a linear transformation the derivate is also multiplied by that linear transformation. So if you multiply it by the inverse of its derivative, you get a new function g whose derivative is the identity map.

 

[Euge]

Thank you, @Euge. Regarding 1, I understand what you’re writing, but shouldn’t we then prove the theorem for $g$?

That is what he did. Specifically, he proves the result when $\lambda = \text{id}$. The general result is then obtained by considering $\lambda^{-1} \circ f$. Since $D(\lambda^{-1}\circ f)(a)$ is the identity, by what he has shown, $\lambda^{-1}\circ f$ is $C^1$ locally invertible. Thus $f = \lambda \circ (\lambda^{-1}\circ f)$ is also $C^1$ locally invertible.

 

[stan_s]

Hi,
I don't want to create a new thread because my question is related to this one. One of the steps in the proof of this theorem is the following relation:
2\cdot|f(x_{1})-f(x_{2})|\geq|x_{1}-x_{2}|
Let's assume the simplest case n=1, and the function:
f(x)=\frac{1}{3}x
It is 𝐶^{1} and invertible: f^{-1}(y)=3y
Using this relation, we get, for example when dx = 1 that \frac{2}{3} >= 1. How to understand it?

 

[Euge]

Hi,
I don't want to create a new thread because my question is related to this one. One of the steps in the proof of this theorem is the following relation:
2\cdot|f(x_{1})-f(x_{2})|\geq|x_{1}-x_{2}|
Let's assume the simplest case n=1, and the function:
f(x)=\frac{1}{3}x
It is 𝐶^{1} and invertible: f^{-1}(y)=3y
Using this relation, we get, for example when dx = 1 that \frac{2}{3} >= 1. How to understand it?

It was assumed in the proof that $Df(a)$ is the identity, from which the inequality $2|f(x_1) - f(x_2)| \ge |x_1 - x_2|$ was obtained. Your function has derivative $1/3$, not $1$.