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Let $E$ be a completely metrizable separable topological space and $\mathscr E$ be its Borel $\sigma$-algebra. Consider a measurable map $F:E\to E$ such that

  • if $f:E\to \mathbb R$ is continuous and bounded, then $f\circ F:E\to\mathbb R$ is continuous and bounded.

Let us say that the function $g$ is nice if:

  1. $g$ is continuous and $\inf\limits_E g = 0$;

  2. $\{g = 0\}:=\{x\in E: g(x) = 0\}$ is not empty;

  3. there is $\delta>0$ such that for all $g(F(x))<g(x)$ whenever $g(x)<\delta$.

Claim 1: if $g$ is a nice function and $\{g = 0\}$ is compact, then $\rho(\cdot,\{g = 0\})$ is also a nice function for any metric $\rho$ on $E$ which agree with the given topology.

Unfortunately, I couldn't neither prove nor disprove this claim. I had an idea that if this claim can be disproved, it is sufficient to disprove a conequence of that:

Claim 2: if $A$ is a compact set s.t. $\rho(\cdot,A)$ is a nice function (i.e. it verifies the property 3.) then $\rho'(\cdot,A)$ is also nice whenever $\rho'\sim \rho$.

but for that one I also didn't manage to construct a counterexample. Any help is appreciated.

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I'm happy also to hear any suggestion about how to tag this question appropriately. – Ilya May 11 '12 at 11:35
up vote 1 down vote accepted

I think Claim 1 is false in general. This is the counterexample I have in mind. Let $E = \mathbb{C}$, and let $F\colon E\to E$ be the map $F(z) = .99e^{i\pi/4}z$. Define $g\colon E\to \mathbb{R}$ by $g(z) = |z|$. This $g$ is a "nice" function (in fact any $\delta>0$ will work for $g$).

Now, let $\rho$ be the metric on $E$ which is induced by the $1$-norm $\|z\|_1 = |\mathrm{Re}\,{z}| + |\mathrm{Im}\,{z}|$. Suppose for contradiction that the function $\rho(\cdot, \{g = 0\})$ is "nice" with respect to $\delta>0$. Since $$\rho(\delta/2, \{g = 0\}) = \|\delta/2\|_1 = \delta/2,$$ it should follow that $\rho(F(\delta/2), \{g = 0\})< \delta/2.$ However, upon actually computing, we see that $$\rho(F(\delta/2),\{g = 0\}) = \|F(\delta/2)\|_1 = \|.99e^{i\pi/4}\delta/2\|_1 = .99\frac{\delta}{\sqrt{2}}>\frac{\delta}{2}.$$

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