I encountered the following exercise in calculus course:
Let $U$ be a convex open subset of $\mathbb{R}^n$ and $f \colon U \to \mathbb{R}^n$ differentiable. Suppose that $\lVert Df(\mathbf{x}) - I_n \rVert < 1$ for all $\mathbf{x} \in U$ where $I_n \colon \mathbb{R}^n \to \mathbb{R}^n$ is an identity map. Show that $f$ is injective. Furthermore, show that $f(U)$ is open.
$\lVert \cdot\rVert$ is an operator norm, i.e. $\lVert A \rVert = \max_{x\in S}\lVert Ax\rVert$ where $S$ is an unit sphere.
I think there is a convexity condition to show that $f$ is injective. If $f(a) = f(b)$ for some $a,b \in \mathbb{R}^n$, then consider a path $c(t) = ta + (1-t)b$ and do something with this. Since only condition given is that $\lVert Df(\mathbf{x}) - I_n \rVert < 1$, I tried to find some point such that $Df(\mathbf{x})$ is "far away" from $I_n$. I have a vague idea to use MVT, or Rolle's theorem, but I am seriously stuck. How can I use those conditions to prove injectivity of $f$ and that $f(U)$ is open?
Thanks in advance!
The convex condition allows us to turn this question into a single parameter calculus. Assume $g(x_1)=g(x_2)$ and $x_1\neq x_2$,
Let $g(t)=f(x_1+t(x_2-x_1))$.
then $g$ is a function from $[0,1]$ to $\mathbb{R}^n$, satisfying $g(0)=g(1)$, and the operator norm condition works when we consider
$g'(t)=Df(x_2+t(x_2-x_1))(x_2-x_1)=Df(whatever)v$, where $v=x_2-x_1$
so that we have
$\parallel g'(t)-v\parallel <\parallel v\parallel $
which means $g'(t)$ have positive component on the direction of $v$ for every $t$. But on the other hand $g(0)=g(1)$, so that is a contradiction.