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In proving the inverse function theorem using the approximation characterization of the derivative, we are given $F:\mathbb{R}^n \to \mathbb{R}^n$ such that 

$$F(p_0 + h) - F(p_0) = DF_{p_0}(h) + o(\|h\|).$$

Applying $(DF_{p_0})^{-1}$ to both sides, we get

$$(DF_{p_0})^{-1}(\eta) = h + (DF_{p_0})^{-1}(o(\|h\|))$$ $$\Rightarrow F^{-1}(q_0 + \eta)-F^{-1}(q_0) = (DF_{p_0})^{-1}(\eta) - (DF_{p_0})^{-1}(o(\|h\|)), \tag{1}$$

where $F(p_0) = q_0$ and $F(p_0 + h) - F(p_0) = \eta$. The proof is over if we can show that $(DF_{p_0})^{-1}(o(\|h\|)) = o(\|\eta\|)$.

My Question: Is the following proof that $(DF_{p_0})^{-1}(o(\|h\|)) = o(\|\eta\|)$ correct? 

Because $(DF_{p_0})^{-1}$ is invertible, we have for all $x\in \mathbb{R}^n$ $\exists m, M>0$ s.t.

$$m \|x\| \leq \| (DF_{p_0})^{-1} (x) \| \leq M \|x\|.$$

Now $(1)$ implies that 

$$(DF_{p_0})^{-1}(\eta) = h + (DF_{p_0})^{-1}(o(\|h\|)) \\ \Rightarrow M\|\eta\| \geq \|h\| - M \epsilon \|h\| \\ \Rightarrow \frac {M \|\eta\|} {1 - M\epsilon} \geq \|h\|$$

So $\|h\| = O(\|\eta\|)$. (Here we've chosen $h$ small enough so that $|o(h)| \leq \epsilon \|h\|.$) Now

$$(DF_{p_0})^{-1}(o(\|h\|)) \leq M(o(\|h\|))= M(o(O(\|\eta\|)))=o(\|\eta\|).$$

Does that work?

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    $\begingroup$ This works perfectly well, provided that you also start with $\varepsilon<1/M$. But why put a bounty on this question? $\endgroup$ – Etienne Sep 2 '13 at 19:27

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