Bi-Lipschitz application

239 Views Asked by At

A function $f: U_1, \left\| \cdot \right\|_1 \rightarrow U_2, \left\| \cdot \right\|_2 $is defined to be Lipschitz if there is a constant $K>0$ such that for all $a,b\in U_1$. $$ \left\| f(a) - f(b) \right\|_2 \le K \left\| a - b \right\|_1 $$

I want to find a function $f$ between the cube and the unit ball of $\mathbb R^d$: $$f : Q=B(0,1),\left \| \cdot \right \|_\infty\rightarrow B=B(0,1),\left \| \cdot \right \|_2, .$$ such that : $f$ is Lipschitz, bijective, and $f^{-1}$ Lipschitz.

Do you have any idea, or a reference about this problem?

I will appreciate all your help ( it's URGENT!).

1

There are 1 best solutions below

1
On

The following definition of $f$ is a simple bijection from the unit cube to the unit ball. It might also be Bi-Lipschitz as required (although a proof alludes me at this moment).

Define $$ \begin{equation} \begin{split} \varphi(\textbf{x})&:=\,\sup\{\,t\in\mathbb{R}\,:\,\left\|\,t\,\frac{\textbf{x}}{\|\textbf{x}\|_2}\right\|_\infty\le1\,\}\;\;\;\left(\textbf{x}\in\mathbb{R}^d,\textbf{x}\neq0\right) \end{split} \end{equation} $$ Essentially, $\varphi(\textbf{x})$ represents a ratio $\frac{\|\textbf{x}_1\|_2}{\|\textbf{x}_2\|_2}$, where $\textbf{x}_1$ is some point on the unit cube, and $\textbf{x}_2$ is some point on the unit sphere, and the points $0, \textbf{x}, \textbf{x}_1, \textbf{x}_2$ all lay on the same line. With that, we define: $$ \begin{equation} \begin{split} f(\textbf{x})& :=\varphi\left(\textbf{x}\right)^{-1}\textbf{x} & \;\;\;\;\;\;\left(\textbf{x}\in\mathbb{R}^d,\textbf{x}\neq0\right) \\ f(\textbf{0}) &:= 0 \end{split} \end{equation} $$ First, we note that $f$ is a bijection, since $f^{-1}(\textbf{x})=\varphi\left(\textbf{x}\right)\textbf{x}$. Moreover, $\|\textbf{x}\|_\infty=1$ iff $\|f(\textbf{x})\|_2=1$, and $\|\textbf{x}\|_\infty<1$ iff $\|f(\textbf{x})\|_2<1$.