I am studying Calculus 3 and I am trying to understand the relation between volume of a cube and its image under a Lipschitz function.

Assume a Lipschitz function $f:Q→\mathbb{R}^n$ has Lipschitz constant $K$, where $Q$ is a cube in $\mathbb{R}^n$. I believe there is some upper bound to $\operatorname{Vol}(f(Q))$ that can expressed with $K$ and $\operatorname{Vol}(Q)$, but I am struggling getting it. Any help would be appreciated!


1 Answer 1


Vol$(f(Q))=\displaystyle\int_{f(Q)}1 dx^1dx^2\cdots dx^n=\int_Q \det \Big(\frac{\partial{f}}{\partial{x}}\Big) dx^1dx^2\cdots dx^n$

$\displaystyle\Big| \det\Big(\frac{\partial{f}}{\partial{x}}\Big) \Big|\leq K^n$ implies Vol$(f(Q))\leq K^n \cdot$Vol$(Q)$

Remark) The differentiability of $f$ is guaranteed by Rademacher's theorem.

For the inequality $\displaystyle\Big| \det\Big(\frac{\partial{f}}{\partial{x}}\Big) \Big|\leq K^n$, Hadamard's inequality is applied.

For the answer to the comment, I finally conclude that the opposite equality condition implies the similar result as above.

First, assume $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ satisfies $|f(x)-f(y)|\geq K|x-y|$ for some constant $K>0$.

Then, it is easily known that $f$ is injective, so we consider $f^{-1}:f(Q)\rightarrow Q$, and we get $|f^{-1}(x)-f^{-1}(y)|\leq \frac{1}{K}|x-y|$.

Now, we apply the above result, thereby Vol$\displaystyle (Q)\leq \Big(\frac{1}{K}\Big)^n \cdot$Vol$(f(Q))$.

Thus, we get Vol$\displaystyle (f(Q))\geq K^n \cdot$Vol$(Q)$.

  • $\begingroup$ Thank you! Can I say the same thing if the inequality of the lipschitz definition was opposite (i.e |f(x)-f(y)|>=K|x-y|), and I just change the inequality at the end? $\endgroup$ Jun 24, 2018 at 12:32
  • $\begingroup$ @TamirShalev I added the answer to your comment, I hope it may help! $\endgroup$ Jun 24, 2018 at 15:02
  • $\begingroup$ Thank you very much! I truely appreciate the time and effort. $\endgroup$ Jun 24, 2018 at 16:58

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