When is the Infinite product of continuous functions a continuous function? Assume that the product is convergent.

Question

Is there any theorem about the continuity of an infinite product of continuous real valued functions on compact Housdorff spaces, if the product is convergent? I mean, for each natural number $n$, let $X_n$ be a compact Housdorff space and $f_n: X_n\to R$ be a continuous function ($R$ denotes the set of real numbers). Assume that the product $∏_{n=1}^∞ f_n$ converges to a function $f$. Then what can we say about the continuity of $f$? By the convergence of the product $∏_{n=1}^∞ f_n$ I mean the following:

Answer 1

If all the $f_n$ are continuous and the convergence is uniform then the limit $f(x)$ is continuous. If that's not the case nothing much because you can easily set a counter example by looking at the set $ C$, where {$C \in R : f(x) \neq 0$}and taking $\log|f|$.

Proof : $\epsilon ,\delta \ge 0$

$g_n∏_{m=1}^n f_n$, clearly each $g_n$ is continuous

$\lim_{n \to \infty} g_n = f(x)$

Uniform convergence implies that for some large $N$, such that for all $n \ge N$, we have $|f(x)-g_n(x)| \le \epsilon$ for all $x$

By continuity of $g_n$, we have $|g_n(x+dx)-g_n(x)| \le \delta$

$|f(x+dx)-f(x)| \le |f(x+dx)-g_n(x)| +|g_n(x)-f(x)| \le |f(x+dx)-g_n(x+dx)|+|g_n(x+dx)-g_n(x)| +|g_n(x)-f(x)| \le 2\epsilon+\delta $

Since $\epsilon , \delta$ are arbitrary it implies $f(x)$ is continuous.