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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.

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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.

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  • $\begingroup$ Thanks, but I think $t - \log(1+t) < \frac{1}{2}t^2$ for small $t > 0$. There is an inflection point in $t - \log(1+t) - \frac{1}{2}t^2$ at $t=0$. SInce $(-1)^na_n(x)$ changes sign this brings me back to problems I had on my own. $\endgroup$
    – RRL
    Commented Aug 18, 2021 at 0:54
  • $\begingroup$ @RRL $\frac {t-\log (1+t)} {t^{2}} \to \frac 1 2$ as $ t \to 0$. So $t-\log (1+t) \geq \frac 1 3 t^{2}$ for $|t|$ sufficiently small. I have changed $\frac 1 2$ to $\frac 1 3$ and I think my proof works fine now. $\endgroup$ Commented Aug 18, 2021 at 5:00
  • $\begingroup$ That fixes the inequality and it would seem the argument leads to $\sum_{k=n+1}^m |a_k(x)|^2 \leqslant 3\left| \sum_{k=n+1}^m (-1)^ka_k(x)\right| + 3 \left| \sum_{k=n+1}^m \log(1 + (-1)^ka_k(x))\right| $. That leads to a contradiction of the Cauchy criterion if the series on the LHS is not uniformly convergent. It does not help with the specific example $a_k(x) = x^k/k$ on $[0,1)$ because $|a_k(x)|^2 \leqslant 1/k^2$ and $\sum |a_k(x)|^2$ converges uniformly by the M test. $\endgroup$
    – RRL
    Commented Aug 18, 2021 at 16:19
  • $\begingroup$ However, using this idea I can find $a_k(x) = x^k/\sqrt{k}$ as a counterexample. Thanks. $\endgroup$
    – RRL
    Commented Aug 18, 2021 at 16:20

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