Real Analysis II: Application of Inverse Function Theorem

Question

I missed two weeks' worth of classes in my Real Analysis II course die to personal issues, and while going over past exam questions for midterm revision, I came across some problems that I had trouble even attempting to try, due to lack of background knowledge.

Let $L$ be a vector field from $\mathbb{R}^n$ to $\mathbb{R}^n$. Let $f(x)=L(x)+g(x)$ with $L$ being a linear isomorphism and $g$ of class $C^1$ satisfying $||g(x)|| \leq M||x||^2$ for some fixed positive $M$. Is $f$ locally invertible near $0$? (that is, does some open neighborhood $U$ of $0$ exist, with $f$ restricted to $U$ being invertible from $f(U)$)

Thanks in advance for any help given, may it be hints, guidance on which material to look at, a rough sketch of a solution, et cetera.

Answer 1

Note that the Inverse Function Theorem states:

Theorem. Let $U \subseteq \def\R{\mathbf R}\R^n$ be open and $f \in C^1(U, \R^n)$, $x \in U$. If $Df(x)$ is invertible, then $f$ is locally invertible near $x$.

That is, we have to check that $Df(0)$ is invertible. We have $$ Df(0) = DL(0) + Dg(0) $$ as $f = L + g$. Since $L$ is linear, $DL(x) = L$ for all $x \in \R^n$, hence $DL(0) = L$. For $g$, we will show that $Dg(0) = 0$, note that \begin{align*} \def\norm#1{\left\|#1\right\|}\frac{\norm{g(x) - 0x}}{\norm x} &= \frac{\norm{g(x)}}{\norm x}\\ &\le \frac{M \norm x^2}{\norm x}\\ &= M \norm x \to 0, \qquad x \to 0 \end{align*} Hence, $Dg(0) = 0$ and therefore $$ Df(0) = DL(0) + Dg(0) = L. $$ As $L$ is invertible by assumption, $f$ is locally invertible near $0$.