I have a concrete question about continuous functions on $X = [0,1]^\omega$ (with the product topology).


The map $f:X\to [0, 1]$ given by $(x\_i)\mapsto \prod x\_i$ is well-defined and Borel but not continuous.


Suppose instead that a *continuous* function $g: X\to [0, 1]$ satsifies $g \ge f$ on $X$. What can be said about $g$?


In particular, I would like to know if the following statements are true (it is clear that 1 implies 2):


1. For $S$ a subset of $\omega$, let $f\_S(x) = \prod\_{s\in S} x\_s$. Then $g\ge c\cdot f\_S$ everywhere for some finite set $S$ (EDIT: and some $c > 0$).
2. If $x$ satisfies $x\_i > 0$ for all $x$, then $g(x) > 0$.



Continuous functions on $[0,1]^\omega$ and a product lower bound
The second question also has the negative answer:


Take any sequence $z=(z\_n)\_{n\in\omega}\in[0,1]^\omega$ with $f(z)=0$. On the Hilbert cube $[0,1]^\omega$, consider the metric $d(x,y)=\max\_{n\in\omega}\frac{|x\_n-y\_n|}{2^n}$.


Let $c=(c\_n)\_{n\in\omega}$ be the constant sequence with $c\_n=\frac12$ for all $n$. It is clear that $f(c)=0$. For every $n\in\mathbb Z$, let $$B\_n=\{x\in [0,1]^\omega:d(x,c)< 2^{-n}\}=\{x\in[0,1]:\forall k\le n;(|x\_k-\tfrac12|<2^{k-n}\}$$ be the open ball of radius $2^{-n}$ centered at the point $c$. Observe that $B\_n=[0,1]^\omega$ for $n\le 0$.


For every $n\in\omega$ consider the continuous function $$\lambda\_n:[0,1]^\omega\to[0,1],\quad \lambda\_n:x\mapsto \inf\{d(x,y):y\in B\_{n+1}\cup([0,1]^\omega\setminus B\_{n-1})\},$$ and observe that for every $x\in [0,1]^\omega\setminus\{c\}$ the set $\{n\in\omega:\lambda\_n(x)>0\}$ is not empty and contains at most three numbers. This fact can be used to show that the function $\lambda=\sum\_{n\in\omega}\lambda\_n$ is continuous and $\lambda(x)>0$ for all $x\in [0,1]^\omega\setminus\{c\}$.


For every $n\in\omega$ consider the continuous function $f\_n:[0,1]^\omega\to[0,1]$, $f\_n:(x\_k)\_{k\in\omega}\mapsto\prod\_{k\in n}x\_k$. It is clear that $f(x)=\lim\_{n\to\infty}f\_n(x)=\inf\_{n\in\omega}f\_n(x)$ for every $x\in[0,1]^\omega$.


It is easy to see that $\sup\_{x\in B\_n}f\_n(x)\le \prod\_{k\le n}(\frac12+2^{k-n})\to 0$. Then the function $g:[0,1]^\omega\to[0,1]$ defined by $$g(x)=\begin{cases}
\sum\_{n\in\omega}\frac{\lambda\_n(x)f\_n(x)}{\lambda(x)}&\mbox{if $x\ne c$};\\
0&\mbox{if $x=c$};
\end{cases}
$$
is continuous, and $g\ge f$ because for every $x\in[0,1]^\omega\setminus\{c\}$ the value $g(x)$ belongs to the convex hull of the set $\{f\_n(x):n\in\omega\}\subseteq [f(x),1]$.


