Rearrangement of uniformly convergent power series is uniformly convergent?

Question

Is rearrangement of uniformly convergent power series again uniformly convergent?

Suppose we have $\sum a_nx^n$ converging unifomly on $[0,1]$. Is its rearrangement $\sum a_{\sigma(n)}x^{\sigma(n)}$ again uniformly convergent?

Answer 1

Not necessarily. Suppose that the series $\sum_{n=0}^\infty a_n$ doesn't converge absolutely. By Riemann's rearrangement theorem, there is a bijection $\sigma\colon\mathbb{Z}^+\longrightarrow\mathbb{Z}^+$ such that the series $\sum_{n=0}^\infty a_{\sigma(n)}$ converges to a number which is different from $\sum_{n=0}^\infty a_n$. But $\sum_{n=0}^\infty a_{\sigma(n)}x^{\sigma(n)}=\sum_{n=0}^\infty a_nx^n$ when $x\in[0,1)$, because then the convergence is absolute. But then the functions$$x\mapsto\sum_{n=0}^\infty a_nx^n\text{ and }x\mapsto\sum_{n=0}^\infty a_{\sigma(n)}x^{\sigma(n)}$$take the same values in $[0,1)$ and distinct values in $1$. Therefore they cannot be both continuous. So, the second series cannot converge uniformly: if a series of continuous functions converges uniformly, then its sum is also continuous.

Answer 2

I don't think so. Consider $$\log(1+x)=\sum_{n=1}^\infty (-1)^{n+1}{x^n\over n}$$ which converges uniformly on $(0,1].$ If we rearrange the sum as $$1+{x^3\over3}-{x^2\over 2} + {x^5\over 5} +{x^7\over 7} -{x^4\over 4}...$$ where each two positive terms are followed by a negative term, the series still converges to $\log(1+x)$ on $(0,1)$ because the series is absolutely convergent there, and so can be rearranged without altering the sum. But at $x=1$ the series converges to something other than $\log{2}.$ (If I recall correctly, it's $3\log{2}$.) Anyway, the convergence can't be uniform, because the limit function isn't continuous.