Uniform convergence of the product without uniform-absolute convergence of the series.

Question

Consider a sequence of functions $a_n:A \subset \mathbb{R} \to [0,\infty)$ (note nonnegativity) such that $\sum_{n=1}^\infty (-1)^na_n(x)$ converges uniformly on $A$ but $\sum_{n=1}^\infty a_n(x)$ converges pointwise but not uniformly on $A$. An example is $a_n(x) = \frac{x^n}{n}$ with $A = [0,1)$.

What can we say about $\sum_{n=1}^\infty \log\left(\,1+(-1)^na_n(x)\,\right)$ and equivalently $\prod_{n=1}^\infty (\,1 + (-1)^na_n(x)\,)$? Must the convergence be uniform or if not what is a counterexample -- $a_n(x) = \frac{x^n}{n}$ perhaps?

Some context: In every presentation on uniform convergence of infinite products I have seen there is a theorem:

If the series $\sum |a_n(x)|$ is uniformly convergent, then so too is $\prod (1 + a_n(x)).$

The proof is straightforward, noting that $0 < |a_n(x)| < 1/2$ for sufficiently large $n$ and

$$|\log(1+a_n(x)| \leqslant |a_n(x)| + \frac{1}{2}|a_n(x)|^2 + \frac{1}{3} |a_n(x)|^3 + \ldots \leqslant \frac{|a_n(x)|}{1- |a_n(x)|} \leqslant 2 |a_n(x)|$$

However, nothing is said about the case where $\sum a_n(x)$ is uniformly and absolutely convergent but not uniformly-absolutely convergent -- that is, $\sum a_n(x)$ is uniformly convergent, $\sum |a_n(x)|$ is convergent, but $\sum|a_n(x)|$ is not uniformly convergent.

Answer 1

Consider your example $a_k(x)=\frac {x^{k}} k$ on $[0,\infty)$ instead of $[0,1]$.

$t-\log (1+t)\geq \frac 1 3t^{2}$ for $|t|$ sufficently small. Note that $a_n(x) \to 0$ uniformly. So we can put $t=(-1)^{n}a_n(x)$ in above inequality if $n$ is sufficiently large. Suppose $\sum \log (1+(-1)^{n})a_n(x)$ converges uniformly. Then our inequality implies that $\sum a_n (x)^{2}$ converges uniformly but this is false. Hence $\sum \log (1+(-1)^{n})a_n(x)$ does not converge uniformly.