# An additional assumption to the inverse function theorem.

The theorem is given below:

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

• 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 $h^{-1}$. – C. Grant Apr 2 at 2:21
• Could you provide more details please?@C.Grant – hopefully Apr 2 at 3:09
• My impression is that, following C. Grant's advice, you should ignore everything other than part (e) of the theorem. – CJD Apr 2 at 3:25
• @CJD could you provide more details please :) – hopefully Apr 2 at 3:32
• @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 $d(x)/(a(x)d(x) - b(x)c(x))$. – CJD Apr 2 at 3:41

## 1 Answer

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 $$\mathcal{C}^r$$.

• Does not the question asks for the proof of the existence only ...... is not their a proof of the existence without calculation? – Smart Apr 2 at 18:52
• How is the original function is $C^k$ while the partial derivatives are $C^{k-1}$ could please give me an example? – Intuition Apr 2 at 19:31
• How are you sure that there exist an open n-ball such that the inverse is $C^k$ – Intuition Apr 2 at 20:02
• Taking the transpose of A is equivalent to taking A as determinants are invariant under taking the transpose. – C. Grant Apr 3 at 0:18
• 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 $\mathcal{C}^{r-1}$. – C. Grant Apr 3 at 0:22