Bounding the Jacobian determinant for "sublipschitz" function

Question

I'm practicing for my upcoming exam in calculus 3. I came across the following question in a practice paper:

$ f:\mathbb R^3 \rightarrow \mathbb R^3 $, f is $C^1$. Also, $K|p-p'| \le |f(p)-f(p')|$ for all $p, p' \in \mathbb R^3$ (where $|\cdot|$ is the Euclidean norm and $K>0$ is a real constant). Prove that $K^3 v(\Omega) \le v(f(\Omega))$ for every Jordan measurable $\Omega\subset\mathbb R^3$ ($v$ is the Jordan measure on $\mathbb R^3$)

I approached the question as follows. First, I noticed I can prove $f$ is a diffeomorphism $\mathbb R^3 \rightarrow f(\mathbb R^3)$ such that $D_f (x)$ is an invertible matrix for every $x\in\mathbb R^3$. Provided this is true, using variable substitution theorem, I can obtain the equality $v(f(\Omega))=\int_{f(\Omega))} 1 = \int_{\Omega} |J_f(x)|dx$, where $J_f (x) = det(D_f(x))$ is the Jacobian determinant.

Now, I wanted to prove $(*)$ $|J_f (x)| \ge K^3$. From this I'll get $\int_{\Omega} |J_f(x)|dx \ge K^3 \int_{\Omega} 1 = K^3 v(\Omega)$, thus completing the proof.

However, though $(*)$ seems intuitively true to me, I couldn't manage to prove it. In particular, I tried primarily to rely on the equality $det(D_f(x)) = lim_{Q\downarrow x} \frac{v(f(Q))}{v(Q)}$, where $Q$ is a cube centered around $x$ and $Q\downarrow x$ means $Q$'s area approaches zero. Regardless, nothing I tried worked.

Any help in proving $(*)$ will be much appreciated, particularly if it was done with the equality I mentioned above. Thanks!

Answer 1

Still interested? Here's a working approach (modulo errors of mine):

The assumptions give that $$ \epsilon K\lvert \vec h\rvert \le \lvert f(p+\epsilon\vec h)-f(p)\rvert,$$ so letting $\epsilon\to 0$ one gets $$ \lvert Df(p)h\rvert\ge K\lvert h\rvert,\qquad \forall h\in \mathbb{R}^n.$$ From now on it is a matter of linear algebra:

Linear Algebra Problem. Suppose that the matrix $A\in\mathbb{R}^{n\times n}$ satisfies the relation $$\tag{1} \lvert Ah\rvert^2\ge K^2 \lvert h\rvert^2,\qquad \forall h\in\mathbb{R}^n.$$ Then $\lvert \det A\rvert \ge K^n$.

Sketch of proof. As you rightly point out the determinant is the product of all (complex) eigenvalues. So one needs to prove:

1. Relation (1) holds for $h\in\mathbb{C}^n$ as well;
2. Relation (1) implies that $\lvert \lambda\rvert\ge K$ for any eigenvalue $\lambda\in\mathbb{C}$ of $A$.

$\square$

Hope this helps