1
$\begingroup$

I was studying the Inverse Function Theorem, and I found this proof on the internet:

http://virtualmath1.stanford.edu/~andras/174A-2.pdf

In the proof, there is this line about $C^k$ functions:

If $F$ is $C^k$, $k > 1$, then $DF$ is $C^{k−1}$, hence $(DF)^{−1}$ is $C^{k−1}$, hence $F^{-1}$ is $C^k$.

Now what I don't get is the last "hence" part, since $(DF)^{-1}\neq D(F^{-1})$. Is there any reasoning in why this is true?

$\endgroup$
0
$\begingroup$

Yes, the Inverse Function Theorem, which tells you precisely that $$ D(F^{-1})=(DF\circ F^{-1})^{-1} $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You can express the derivative $D (F^{-1})$ of $F^{-1}$ in terms of $F$ and the derivative of $F$. If these are both differentiable then so is $D (F^{-1})$, just because it can be expressed through the composition of differentiable maps. The general case is just by induction.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.