Simple to understand calculus question that involved change of variable theorem in integration.
Suppose $B$ is some open ball in $\mathbb R^n$, and $f: \mathbb R^n \to \mathbb R^n$ is a diffeomorphism. Meaning $f$ is bijective and continuous and the differential of $f$ is also continuous.
Let $V(B)$ be the volume of the ball $B$.
Show there is no such diffeomorphism $f$ such that $V(B) < V(f(B)) <V(B)+V(B)^2$ for any open ball $B$.
I know that the idea of the proof is to consider a ball with a very very very small radius $\epsilon$. If we can show that for a ball with a very small radius there is no such diffeomorphism, then we found a ball where $V(B) < V(f(B)) <V(B)+V(B)^2$ and we're done.
But how can i show theres no such $f$ for a small ball?