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I am stuck on the following problem, and I need any kind of help that leads to solve it:

Let $L:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be an isomorphism and let: $f(x)=L(x)+g(x)$, where: $\left \| g(x) \right \|\leq M\left \| x \right \|^{2}$ and $f\in C^{1}$. Show that $f$ is locally invertible near $0$

What I was trying to do is to show that $Jf(0)\neq 0$. Obviously: $f(0)=L(0)+g(0)=g(0)$ because $L(0)=0$. That's all what I could deduce. Any help?

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up vote 1 down vote accepted

$\bf Hint$: First show that $g(0)=0$ then prove that $Df(0)=L$.

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Hint: Show that $Dg(0)=0$.



Note that for assumption we have $||g(0)||\leq||0||=0$ then $g(0)=0$ if we choose $T=0$ let us prove that

$$\frac{||r(h)||}{||h||}\to 0\ as\ h\to 0.$$

But substitute $\tag{1}$ in and you will have $g(h)=r(h)$ then for the property of $g$ we get

$$\frac{||r(h)||}{||h||}=\frac{||g(h)||}{||h||}\leq \frac{M||h||^2}{||h||}\leq M||h|| \to 0\ as\ h\to 0.$$

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That is what the "Application of Inverse function theorem" gives you, exactly the "invertibility" of the $f$. Take a deep look at the inverse Application Theorem. – checkmath Mar 9 '12 at 3:39
This is what I understood: $f$ is invertible at $0$ if $Jf(0)=det(Df(0))\neq 0$. Since $Df(0)=DL(0)$, then $detDf(0)=detDL(0)$. And since $L$ is an isomorphism, it is invertible any point and specifically at $0$, hence $det(DL(0))$ is not zero. Is that correct? – M.Krov Mar 9 '12 at 3:50

We have $g(0)=0$ by sandwich rule. Then $Dg(0)=0$ by definition of derivative. It's a famous exercise in Spivak which has terrorized students on calculus exams for ages; and it is very instructive.

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