Continuous functions on $[0,1]^\omega$ and a product lower bound

Question

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 $g(x) > 0$.

Answer 1

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 $\{f_n(x):n\in\omega\}\subseteq [f(x),1]$.

Answer 2

Let $$ p_n(t) = \begin{cases} 1-2^{-n} & 0\le t \le 1-2^{-n}\\ t & 1-2^{-n}<t\le 1 \end{cases} . $$ Then $g(x)=\prod p_n(x_n)$ is continuous since $1\ge p_n\ge 1-2^{-n}$, so $x_n$ for large $n$ doesn't matter much when computing $g(x)$. Also, $g\ge f$ since $p_n(t)\ge t$.

If $x_n=1-2^{-(N+1)}$ ($n\le N+1$), $x_n=1$ ($n>N+1$), then $$ g(x)=\prod_{n=1}^{N+1} x_n<\prod_{n=1}^N x_n , $$ so this function does not satisfy (1). (It does satisfy (2), though, so that part of the question remains open.)