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I have a question about an estimate of the surface area of a set.

Let $B(r)$ denotes the open ball of $\mathbb{R}^{d}$ centered at origin with radius $r>0$. Let $F:\mathbb{R}^{d} \to \mathbb{R}^{d}$ be a Lipschitz continuous function and define \begin{equation} \text{Lip}(F)=\inf\{L>0:|F(x)-F(y)| \le L|x-y|,x,y \in \mathbb{R}^{d} \}. \end{equation}

Question

Define $B^{\ast}(r)=F(B(r))$. I want to obtain an upper estimate of $\sigma(B^{\ast}(r))$, where $\sigma(B^{\ast}(r))$ is the surface area of $B^{\ast}(r)$. Strictly speaking, $\sigma(B^{\ast}(r))$ is the $d-1$ dimensional Hausdorff measure of $\partial B^{\ast}(r)$. That is, $\sigma(B^{\ast}(r))=\mathcal{H}^{d-1}(\partial B^{\ast}(r))$.

My attempt

It is known that the following inequality: \begin{equation*} \mathcal{H}^{d-1}(F(E)) \le \text{Lip}(F)^{d-1}\mathcal{H}^{d-1}(E), \end{equation*} for all $E \subset \mathbb{R}^{d}$.

But can we show $\partial B^{\ast}(r) \subset F \bigl( \partial B(r) \bigr)$ (if necessary, we may assume $F$ is smooth)? If this is true, we have \begin{align*} \sigma(B^{\ast}(r))&=\mathcal{H}^{d-1}(\partial B^{\ast}(r)) \\ & \le \text{Lip}(F)^{d-1}\mathcal{H}^{d-1}(\partial B(r))\\ &= \text{Lip}(F)^{d-1}\sigma( B(r)). \end{align*}

Can we have an upper estimate like as $ \sigma(B^{\ast}(r)) \le Cr^{d-1} ? $

Thank you in advance.

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    $\begingroup$ In addition to what Mikhail Katz said, see for example this duplicate : math.stackexchange.com/questions/917494/…. Furthermore, while $\partial B^\ast(r)$ is $(d-1)$-rectifiable and has finite $\mathcal{H}^{d-1}$ measure, I don't know why you should have the same decay with $r^{d-1}$. $\endgroup$ – Paul-Benjamin Oct 26 '16 at 16:04
  • $\begingroup$ Could you please show me a reference for the inequality \begin{equation*} \mathcal{H}^{d-1}(F(E)) \le \text{Lip}(F)^{d-1}\mathcal{H}^{d-1}(E), \end{equation*}? $\endgroup$ – Humed Feb 1 at 20:59
  • $\begingroup$ Please see Section 2.4.1 in this book: crcpress.com/Measure-Theory-and-Fine-Properties-of-Functions/…. $\endgroup$ – sharpe Feb 3 at 10:50
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There is no reason to assume that the boundary of the image is contained in the image of the boundary, even in the case $n=1$; consider for example the function $y=x^2$ on $[-1,1]$.

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