Perturbation of identity still surjective?

Question

Let $f : \mathbb{R^n}\mapsto \mathbb{R^n}$ a smooth function. Can we give a condition for the application $F=I+f$, where $I$ is the identity, to be surjective ? I know that if the operator norm of the differential of $f$ is small, $\vert Df\vert <= 1/2$, say, then $F$ is a submersion, but it does not tell me that it is surjective.

On the other hand, if we look at the problem for $n=1$, then we have $F'>1/2$, which of course implies that $F$ is surjective. In other words we avoid counter-examples like $arctan$, which are not surjective submersions. Can such a result be extended to higher dimensions ?

Edit : Another way to formulate this, is that $f$ is $1/2$-Lipschitz. Any additionnal hypothesis on $f$ is welcomed

Edit 2 : More precisely, wlog suppose that $f(0)=0$, then the function $F$ is a quasi isometry in the sens that $\frac{1}{2}\vert\vert F(x) - F(y)\vert\vert \leq \vert\vert x - y\vert\vert \leq 2\vert\vert F(x) - F(y)\vert\vert$. It is known that an isometry of $\mathbb{R^n}$ must be surjective, see for instance Isometries of $\mathbb{R}^n$

Answer 1

$\renewcommand{\Im}{\operatorname{Im}}$ From the inverse function Theorem, $F$ is a diffeomorphism onto its open image. To show that $F$ is surjective, one only has to show that the image of $F$ is closed. Indeed, the image would be clopen, and by connectedness of $\Bbb R^n$, the result would follow.

Let us consider $(y_n)_{n\in \Bbb N}$ a sequence in $\Im (F)$ that converges in $\Bbb R^n$, to say $\bar{y}$. Let $(x_n)_{n\in \Bbb N}$ be defined such that $F(x_n) = y_n$. For all $n$, we have $$ y_n - x_n = f(x_n), $$ and from the fact that $f(0)=0$ and $|df|\leqslant \frac{1}{2}$, it follows that $$ \|y_n-x_n\| \leqslant \frac{1}{2}\|x_n\|. $$ Let $\varepsilon>0$ be small. From the triangle inequality, and from the fact that $y_n \to \bar{y}$, there exists some $N$ such that if $n\geqslant N$, then $$ \left|\|\bar{y}\| - \|x_n\|\right|\leqslant\|\bar{y}-x_n\|\leqslant \frac{1}{2}\|x_n\| + \varepsilon $$ From there, I claim that there exists a bounded subsequence in $(x_n)$. By contradiction, suppose that $\|x_n\|\to \infty$. Then for $n$ large enough, we have $$ \left| \frac{\|\bar{y}\|}{\|x_n\|} -1\right| \leqslant \frac{1}{2} + \frac{\varepsilon}{\|x_n\|}, $$ and taking the limit as $n\to \infty$ gives $1 \leqslant \frac{1}{2}$, a contradiction.

Therefore, $(x_n)$ has a bounded subsequence, and hence, a converging subsequence in $\Bbb R^n$, say to $\bar{x}$. By a continuity argument, $F(\bar{x}) = \bar{y}$, and $\bar{y}\in \Im(f)$.

Answer 2

To complete, there is actually a general result by Hadamard which states that if $DF$ is everywhere invertible and $F$ is proprer (the pre-image of each compact is compact) then $F$ is indeed a bijection.