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.



On the surface area of a set
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 $[-1,1]$.


