Computing natural pseudo distance.

Consider the functions $\varphi , \ \psi: \, D^2= \{ (x,y) \in \mathbb R^2 : x^2+y^2 \leq 1 \} \rightarrow \mathbb R$ defined by $$\varphi (x,y):= \frac{5(1-x^2-y^2)}{2}$$ and $$\psi (x,y):= 1 + \cos(3 \pi(x^2+y^2))$$ for all $(x,y) \in D^2$. Compute the natural pseudodistance $d_G(\varphi, \psi)$ for $G=\text{Homeo}(D^2)$.

I've studied the problem in one dimension.

I found that $d_G(\varphi,\psi)\geq \frac{1}{2}$ studying persistence diagrams $D^1_\varphi$ and $D^1_\psi$ and computing the $d_{match}$. For this reason, I'd like to find out that $\frac{1}{2}$ is also an upper bound for $d_G$ in order to conclude that it is exactly $\frac{1}{2}$.

I tried to find a sequence $(g_n):[-1,1] \rightarrow [-1,1]$ of homeomorfisms such that $$\lim_n \ \lvert\lvert \varphi - \psi \circ g_n \lvert\lvert_\infty = \frac{1}{2} \ \ .$$

In doing so I consider $g_n(x)=x^n$, and now I have $$\lim\limits_n \ \lvert\lvert \varphi - \psi \circ g_n \lvert\lvert_\infty \ \leq \ \limsup\limits_n \ \lvert\lvert \varphi - \psi \circ g_n \lvert\lvert_\infty = \ \lvert\lvert \varphi - \limsup\limits_n \ \psi \circ g_n \lvert\lvert_\infty \ \ .$$ And $$\lvert\lvert \varphi - \limsup\limits_n \ \psi \circ g_n \lvert\lvert_\infty \ = \lvert\lvert \frac{5}{2} (1-x^2) - \limsup\limits_n \ ( 1 + \cos(3\pi x^{2n})) \lvert\lvert_\infty = \frac{1}{2} \ \ .$$

Is it correct?

• Did you attempt to start solving the problem? – Umberto P. Feb 11 '18 at 15:53
• I found that $d_G(\varphi,\psi) \geq \frac{1}{2}$ studying persistence diagrams $D_\varphi^1$ and $D_\psi^1$ and computing the $d_{match}$. For this reason, I'd like to find out that $\frac{1}{2}$ is also an upper bound for $d_G$ in order to conclude that it is exactly $\frac{1}{2}$. – Badigallo Feb 23 '18 at 21:52