Observe $\gamma : [-1,1]\times [-1,1]\to D^2$ given by$$\gamma : \begin{cases}0, \quad \quad \quad \, \,\,(x,y)= (0,0) \\ \frac{||(x,y)||_{2}}{||(x,y)||_\infty}\cdot (x,y), \quad (x,y)\neq(0,0)\end{cases}$$
This function maps for each $0 < d \leq 1$, the border of each square $\{ (x_1,x_2) \in \mathbb{R}^2 : \text{max}( \lvert x_1 \rvert, \lvert x_2 \rvert) \leq d \}$ into the circle of radius d centered at the origin.
Since we can write down the inverse explictly:$$\gamma^{-1} : \begin{cases}0, \quad \quad \quad \, \,\,(x,y)= (0,0) \\ \frac{||(x,y)||_\infty}{||(x,y)||_{2}}\cdot (x,y), \quad (x,y)\neq (0,0)\end{cases}$$ we can conclude that $\gamma$ is bijective.
I am having trouble showing that $\gamma$ and $\gamma^{-1}$ are continuous however. Can someone please help me?
Best Answer
You have the two norms $\lVert - \rVert_2$ and $\lVert - \rVert_\infty$. The "unit disk" with respect to the first norm is $D^2$ and with respect to the second norm is $[-1,1]^2$. All norms on $\mathbb R^2$ (or more generally $\mathbb R^n$) are equivalent and are thus continuous real-valued functions with respect to the standard Euclidean topology on $\mathbb R^n$. For the above two norms you can prove this also completely elementary without invoking the general norm-equivalence theorem.
Therefore $\gamma$ and $\gamma^{-1}$ are continuous in all $\xi \ne 0$. But they are also continuous in $0$ since $$\lVert \gamma(\xi) - \gamma(0) \rVert_\infty = \lVert \gamma(\xi) \rVert_\infty = \lVert x \rVert_2 = \lVert x - 0 \rVert_2 ,$$ $$\lVert \gamma^{-1}(\xi) - \gamma^{-1}(0) \rVert_2 = \lVert x - 0 \rVert_\infty .$$