Let $U\subset \mathbb{R}^2$ and $f:U\rightarrow \mathbb{R}^2$ be an everywhere differentiable short map (i.e. a Lipschitz function with upper Lipschitz constant =1, the upper Lipschitz constant is defined as $\sup\limits_{x\in U}\sup\limits_{v\in\mathbb{R}^2,||v||=1}||df_xv||$). Suppose that $f$ is also bi-Lipschitz and that its lower Lipschitz constant is equal to $c$, $0<c<1$ (the lower Lipschitz constant is defined as $\inf\limits_{x\in U}\inf\limits_{v\in\mathbb{R}^2,||v||=1}||df_xv||$).
Let $|J_f(p)|$ be the determinant of the Jacobian of $f$ in $p\in U$.
In this case I've read that it should be $|J_f(p)|\ge c$ for every $p\in U$, but I can't understand why. Can you explain it to me?
UPDATE: The argument below shows that pointwise the Jacobian determinant is the product of the upper and lower local Lipschitz constants, but that actually does not quite adress the question.
With the definitions and assumptions given, the conclusion is wrong, the best lower bound for the Jacobian determinant is $c^2$. One simple counterexample to the stated conclusion is given by the conformal map $f(z) = e^z$ where $U = [-1,0] \times [0,\pi]$ (and where $\mathbb{R}^2$ is identified with the complex plane $\mathbb{C}$ in the usual way.) Pointwise upper and lower Lipschitz constants are $|f'(z)| = |e^z| = e^x$, which has maximum $1$ and minimum $1/e$ in $U$. The Jacobian determinant is $|f'(z)|^2 = e^{2x}$ which has minimum $1/e^2$ in $U$.
The way you defined the upper and lower Lipschitz constants is not standard, typically they are the optimal positive constants $c$ and $C$ such that $c|x-y| \le |f(x)-f(y)| \le C|x-y|$. However, going with your definition this is basically just a local result which comes down to linear algebra. For fixed $p$ with $A = df_p$, the directional derivative in the direction $v$ with $\| v \|=1$ is just $Av$, and the maximal and minimal norms of these are the maximal and minimal singular values of $A$, i.e., the square roots of the maximal and minimal eigenvalues of $A^t A$. In your case where the dimension is $2$ there are just two eigenvalues, and the determinant is the product of those, so under your assumptions $\det A^t A = c^2$, so the Jacobian determinant is $\det A = \pm c$.