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This is a question about a step in the proof of the inverse function theorem.

Say we have a function $f\colon V\to W$, that is $C^1$ and where $V,W\subset R^n$. The derivative of $f$ is assumed to be invertible.

My book says the following: let $y,b\in W$, with $y=f(x)$ and $b=f(a)$. Then $$ f^{-1}(y)-f^{-1}(b)=x-a=f’(a)^{-1}(y-b)+f’(a)^{-1}o(\Vert x-a\Vert). $$ I don't understand how they got the second equality. I know that $$ f(x)-f(a)=f’(a)(x-a)+o(\Vert x-a\Vert), $$ and it seems that they kind of applied this, but it doesn’t seem exactly the same to me. First of all, why do they multiply $o(\Vert x-a\Vert)$ by $f’(a)^{-1}$? And why don't they just write something like $$ f^{-1}(y)-f^{-1}(b)=(f(b)^{-1})'(y-b)+o(\Vert y-b\Vert). $$ Maybe it's equivalent. It seems we need to show that $f'(a)^{-1}=(f(b)^{-1})'$ and $f'(a)^{-1}o(\Vert x-a\Vert)=o(\Vert y-b\Vert)$. I know the second is true. So I'm only having problems with $$ f'(a)^{-1}=(f(b)^{-1})'. $$ The inverse of the derivative is the derivative of the inverse?

EDIT

The last question can be answered by noting that $f^{-1}f(x)=x$, so $f^{-1}(f(x))'f'(x)=I$, which implies that $(f'(x))^{-1}=f^{-1}(y)'$. So that basically answers everything.

Sha Vuklia
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1 Answers1

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$f'(a)^{-1}$ means the inverse matrix of $f'(a)$.

I think they have just rearranged $$f(x)-f(a)=f’(a)(x-a)+o(\Vert x-a\Vert)$$ to get $$\begin{aligned}x-a&=(f'(a))^{-1}(f(x)-f(a))-(f'(a))^{-1}o(\Vert x-a\Vert)\\ &=(f'(a))^{-1}(y-b)-(f'(a))^{-1}o(\Vert x-a\Vert)\\ &=(f'(a))^{-1}(y-b)+(f'(a))^{-1}o(\Vert x-a\Vert) \end{aligned}$$ The last step follows from the fact that $\phi\in o(\Vert x-a\Vert)$ if and only if $-\phi\in o(\Vert x-a\Vert)$. This is true because $\frac{\phi}{||x-a||}\to 0$ if and only if $-\frac{\phi}{||x-a||}\to 0$.

A. Goodier
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