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

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.

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.

• 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.
– RRL
Commented Aug 18, 2021 at 0:54
• @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. Commented Aug 18, 2021 at 5:00
• 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.
– RRL
Commented Aug 18, 2021 at 16:19
• However, using this idea I can find $a_k(x) = x^k/\sqrt{k}$ as a counterexample. Thanks.
– RRL
Commented Aug 18, 2021 at 16:20