Extending uniform convergence

Question

Let $\sum_{n=0}^{\infty} f_n(x)$ converge to $f(x)$ on $[0, a)$ and converge uniformly on evey closed sub-interval of $[0, a)$. If $\lim_{x \to a^-} f(x) = L < \infty$ does it follow that the series converges uniformly on $[0, a)$? I'm looking for either a proof or a counterexample as I'm unable to come up with either.

Answer 1

It's well-known that the functions $F_{n} : [0,1] \to \mathbb{R}$ given by $F_{n}(x) = x^{n}$ converge to zero uniformly in $[0,\zeta]$ for each $\zeta \in (0,1)$, but not in $[0,1]$.

Define a sequence of functions $\{f_{n}\}_{n \in \mathbb{N}}$ in $[0,1]$ by $f_{n}(x) = F_{1}(x)$ if $n = 1$ and $f_{n}(x) = F_{n}(x) - F_{n - 1}(x)$, otherwise. It follows that $\sum_{j = 1}^{N} f_{j}(x) = F_{N}(x)$ for all $N \in \mathbb{N}$. Hence $\sum_{j = 1}^{\infty} f_{j} = 0$ uniformly in $[0,\zeta]$ for each $\zeta \in (0,1)$, but not uniformly in $[0,1]$.

Answer 2

Consider the example $f(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$ for $-1<x<1$. In fact, the series converges even for $x=\pm1$ in the sense of the question, that is, $x\to1^-$, $x\to(-1)^+$, however the convergence is not uniform (see picture below). The reason is that there is a singularity at the imaginary number $i$.

In general, for any power series, uniform convergence at a boundary point of a ball $B_r(a)\subset\mathbb{C}$ implies uniform convergence on smaller closed balls $\overline{B_s(a)}$, $s<r$. If there is a singularity at the boundary, then there cannot be uniform convergence on the ball. Nevertheless, there can still be convergence at some other points.