The theorem is given below:

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And here is the question:

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Could anyone give me a hint on how to prove the required in the question please?

  • $\begingroup$ In order to show that the inverse is also a $ \mathcal{C}^R $ mapping you can use Cramer's rule to explicitly calculate the partial derivatives of the inverse function. This gives you a relation between the partial derivatives of h and the partial derivatives of $h^{-1}$. $\endgroup$ – C. Grant Apr 2 at 2:21
  • $\begingroup$ Could you provide more details please?@C.Grant $\endgroup$ – hopefully Apr 2 at 3:09
  • $\begingroup$ My impression is that, following C. Grant's advice, you should ignore everything other than part (e) of the theorem. $\endgroup$ – CJD Apr 2 at 3:25
  • $\begingroup$ @CJD could you provide more details please :) $\endgroup$ – hopefully Apr 2 at 3:32
  • $\begingroup$ @hopefully Say the Jacobian of $f$ is $\left( \begin{array}{cc} a(x) & b(x) \\ c(x) & d(x) \end{array} \right)$, and we know that its determinant is non-zero in $B$ (because an inverse exists) and we know the component functions have continuous partial derivatives to I guess $k-1$ order. For example, the top left entry of the inverse matrix would be $d(x)/(a(x)d(x) - b(x)c(x))$. $\endgroup$ – CJD Apr 2 at 3:41

There is a more general form of Cramer's rule than the one used for solving system's of linear equations. See the section on inverting matrices here on Wikipedia for example: https://en.wikipedia.org/wiki/Cramer%27s_rule#Finding_inverse_matrix.

In this question in particular we have that $Dh(v)= (Df(h(v)))^{-1}$ by part (e) of the inverse function theorem. To show that the inverse function is also a $\mathcal{C}^{r}$ function we must show that the partial derivatives $\frac{\partial h_i}{\partial x_j}$ of the function are $\mathcal{C}^{r-1}$ (as then the original function will be $\mathcal{C}^r$.)

Using the version of Cramer's rule above we have that the ij-th entry of $(Df(h(v)))^{-1}$ (and hence the partial derivative $\frac{\partial h_i}{\partial x_j}$) will be given by:

$$ ((Df(h(v))^{-1})_{ij} = \frac{\det(A)}{\det(Df(v(h)))} $$ Where $A$ is the matrix obtained from $Df(v(h))$ by co-factor expanding on the ij-th entry of the matrix.

This expression gives $\frac{\partial h_i}{\partial x_j}$ in terms of the partial derivatives of $f$. In particular we see that the partial derivative of $h_i$ with respect to $x_j$ is a quotient of a sum and product of $\mathcal{C}^{r-1}$ functions and a no-where zero $\mathcal{C}^{r-1}$ function on the ball.

We conclude from this that the partial derivates of the inverse function are $\mathcal{C}^{r-1}$ and hence the inverse function is $\mathcal{C}^r$.

  • $\begingroup$ Does not the question asks for the proof of the existence only ...... is not their a proof of the existence without calculation? $\endgroup$ – Smart Apr 2 at 18:52
  • $\begingroup$ How is the original function is $C^k$ while the partial derivatives are $C^{k-1}$ could please give me an example? $\endgroup$ – Intuition Apr 2 at 19:31
  • $\begingroup$ How are you sure that there exist an open n-ball such that the inverse is $C^k$ $\endgroup$ – Intuition Apr 2 at 20:02
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    $\begingroup$ Taking the transpose of A is equivalent to taking A as determinants are invariant under taking the transpose. $\endgroup$ – C. Grant Apr 3 at 0:18
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    $\begingroup$ The set f(B) is open and thus there is an open ball contained in f(B) around any point f(b) for $ b \in B$. The argument given above shows that in any neighbourhood where the determinant of $Df(h(v))$ is non zero that the inverse function has partial derivatives that are $\mathcal{C}^{r-1}$. $\endgroup$ – C. Grant Apr 3 at 0:22

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