A question about series convergence at a given point if functional series uniform convergence on interval

Question

Could you give me some hint how to prove this statement: If $f_n(x)$ is continuous sequence on [0,1] and the series $\sum_{n\ge1}f_n(x)$ converges uniformly on [0,1) then the series $\sum_{n\ge1}f_n(1)$ converges.

I have no idea how to start.

Thanks.

Answer 1

(Hint) Use in some way the inequality$$|s_m(1)-s_n(1)|\le$$$$|s_m(1)-s_m(x)|+|s_m(x)-s_n(x)|+|s_n(x)-s_n(1)|$$ where $s_n$ is a partial sum.

Answer 2

Fix $\epsilon > 0$. For each $n$, $s_n = \sum_{i=1}^n f_i$ is continuous at $1$. Therefore, for each $n$ we can pick $\delta_n > 0$ such that for any $|y - 1| \le \delta_n$, $|s_n(y) - s_n(1)| < \frac{\epsilon}{3}$. We can further guarentee that $\delta_n$ is decreasing.

Now, choose $N$ such that for $m, n > N$, $|s_m(x) - s_n(x)| < \frac{\epsilon}{3}$ for any $x \in [0,1)$. Then for $m > n > N$, \begin{align*} \quad &\left| s_m(1) - s_n(1) \right| \\ &\le \left| s_m(1) - s_m(1 - \delta_m) \right| + \left| s_m(1 - \delta_m) - s_n(1 - \delta_m) \right| + \left| s_n(1 - \delta_m) - s_n(1) \right| \\ \end{align*} All terms are $< \frac{\epsilon}{3}$, the first two terms directly and the third since $\delta_m < \delta_n$.