Clarification in a step for a proof of the Inverse Function Theorem.

Question

This is from Tao's Analysis II pp.153:

Theorem 6.7.2 (Inverse function theorem).
Let $E$ be an open subset of $\mathbb{R}^n$, and let $f:E\to\mathbb{R}^n$ be a function which is continuously differentiable on E. Suppose $x_0\in E$ is such that the linear transformation $f'(x_0):\mathbb{R}^n\to \mathbb{R}^n$ is invertible. Then there exists an open set $U$ in $E$ containing $x_0$, and an open set $V$ in $\mathbb{R}^n$ containing $f(x_0)$, such that $f$ is a bijection from $U$ to $V$. In particular, there is an inverse map $f^{-1}:V\to U$. Furthermore, this inverse map is differentiable at $f(x_0)$ and, \begin{equation} (f^{-1})'(f(x_0))=(f'(x_0))^{-1}. \end{equation}

In the proof, he then goes on to say:

It suffices to prove the theorem under the additional assumption $f(x_0)=0$. The general case then follows from the special case by replacing $f$ by a new function $\tilde{f}(x):=f(x)-f(x_0)$ and then applying the special case to $\tilde{f}$.

I'm having quite a bit of trouble seeing how to go from $(\tilde{f}^{-1})'(\tilde{f}(x_0))=(\tilde{f}'(x_0))^{-1}$ to $(f^{-1})'(f(x_0))=(f'(x_0))^{-1}$.

Any help would be greatly appreciated!

Answer 1

• By definition, $\tilde{f}(x) := f(x) - f(x_0)$.
• Show that $\tilde{f}'(x) = f'(x)$.
• Show that $\tilde{f}^{-1}(y) = f^{-1}(y + f(x_0))$ and thus $(\tilde{f}^{-1})'(y) = (f^{-1})'(y + f(x_0))$.
• Plug the above three results into $(\tilde{f}^{-1})'(\tilde{f}(x_0)) = (\tilde{f}'(x_0))^{-1}$.

Answer 2

It follows from $f^{-1}(y)=\tilde f^{-1}(y-f(x_0)),$ by differentiating at $y=f(x_0):$ applying the chain rule, you prove that this differential exists and find its value: $$(f^{-1})'(f(x_0))=(\tilde f^{-1})'(0)\circ{\rm id}=(\tilde f^{-1})'(\tilde f(x_0))=(\tilde f'(x_0))^{-1}.$$