If $\langle f'(x) \cdot v , v \rangle > 0$ then $f$ is injective Question: 

Let $f: U \to \mathbb R^m$ differentiable at the convex set $U \subseteq \mathbb R^m$. If $$\langle f'(x) \cdot v , v \rangle > 0 , \,\,\, \forall\,\, x \in U, v \neq 0 \in \mathbb R^m $$
  then $f$ is injective. If $f \in C^1$ then $f$ is a diffeomorphism of $U$ over a subset of $\mathbb R^m$. Give an example such that $U = \mathbb R^m$, but $f$ is not surjective. 

Attempt: The idea is to show $$|f(x+v) - f(x)| > 0$$
As $U$ is convex and $f$ id differentiable in $U$ then $[x,x+v] \subseteq U$ for any $x, x + v \in U$, by the Mean Value Theorem there exists $\theta \in (0,1)$ such that $$f(x + v) - f(x) = \frac{\partial f}{\partial v}(x + \theta v) = f'(x + \theta v) \cdot v$$
Then $$\langle f(x + v) - f(x) , v\rangle  = \langle f'(x + \theta v) \cdot v , v\rangle > 0 $$
and I couldn't conclude anything. 
The second part is o.k.
Any thoughts?
Edit: I  can't use the Mean Value Theorem here.
 A: After a while I figured it out.
Consider the function $$\begin{align}\psi: [0,1] &\to \mathbb R\\ t &\mapsto \langle f(a + tv), v \rangle\end{align}$$
defined at $[0,1]$, differentiable with
$$\begin{align}\psi' (t)  &= \langle f'(a + tv)\cdot v ,v\rangle + \langle f(a+tv),0 \rangle \\ &= \langle f'(a + tv)\cdot v,v \rangle > 0\end{align} \tag {*}$$
for $v = b -a$ and for all $t \in [0,1]$, then $\psi$ is strictly increasing at this closed interval. Now suppose that for $a \neq b$ we have $f(a) = f(b)$.
$$\begin{align}\psi (1) - \psi (0) &= \langle f(b), v \rangle - \langle f(a),v \rangle\\&= \langle f(b) - f(a), v \rangle\\ &= 0\end{align}$$
As $\psi$ is continuous and $\psi (1) = \psi (0)$ by Rolle's Theorem there exists $c \in (0,1)$ such that $\psi ' (c) = 0$, which contradicts $(*)$ .Thus $f(a) \neq f(b)$ and it follows that $f$ is injective.
A: I think that you have already solved the problem. Pick $x, y\in U$, define your $v=y-x$, and proceed. Note that $||f(y)-f(x)||\cdot||y-x||>0$. As you picked $x\neq y$, then $f$ must be injective. Sorry for my english.
A: What about the second part? How did you prove that $f$ is a homeomorphism over $f(U)$? 
