Positive-definite derivative implies injective function

Question

Suppose $f:\mathbb R^n\to \mathbb R^n$ is differentiable and that $Df$ is positive-definite at every point. As homework, I need to prove $f$ is injective.

I thought about proving by contradiction. If $f(a) = f(b) ,a\neq b$, consider the straight line $\gamma:a\to b$ in $\mathbb R^n$, the composition $f\circ \gamma:\mathbb R\to \mathbb R$ (we take the codomain as $\mathbb R$ only) allows using the mean value theorem to find some $c\in (a,b)$ for which $(f\circ \gamma)^\prime (c)=0$. Then, by the chain rule $(f\circ \gamma)^\prime(c)=Df|_{\gamma c} \gamma^\prime(c)=0$. At this point I'm stuck. I don't see how to get to an inner product to contradict positive definiteness.

Positive-definiteness implies invertibility whence $\gamma^\prime(c)=0$, which can't happen for a straight line, so it looks like the weaker hypothesis that $Df$ is everywhere invertible implies $f$ is injective. But that doesn't sound right...

Answer 1

Hint: You are almost there. You just have to take the scalar product with $(b-a)$, i.e. look at $\langle f(b)-f(a),b-a \rangle$, apply your argument and show that this quantity is strictly positive whenever $b\neq a$.

More details: Let $\gamma(t)=a+t(b-a)$, $t\in [0,1]$ be our path from $a$ to $b$ and consider the real-valued function: $$F(t)=\langle f(\gamma(t))-f(a), b-a \rangle$$$\langle f(b)-f(a),b-a \rangle$ Suppose for a contradiction that $f(b)=f(a)$, whence implying that $F(1)=F(0)=0$. Then by the MVT there is $00$, i.e. that $$\langle f(b)-f(a),b-a \rangle > 0$$