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I'd like your help with proving or refuting the following claim:

If for every $n \in \mathbb{N}$, $f_n:[a,b] \to \mathbb{R} $ is an increasing function, and if $f_n \to f$ pointwise in $[a,b]$ and $\sum \limits_{n=1}^{\infty}f_n(a)$ and $\sum \limits_{n=1}^{\infty}f_n(b)$ converge, then $f_n\to f$ uniformly in $[a,b]$.

I didn't come to any smart conclusion.

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    $\begingroup$ This is kind of a strange question, but if it's written correctly, then here's a hint: if $\sum_{n=1}^\infty f_n(a)$ converges, what can you say about $\lim_{n \to \infty} f_n(a)$? $\endgroup$ – Nate Eldredge Dec 13 '11 at 22:34
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    $\begingroup$ Here's two more hints: what can you say about $\lim_{n\rightarrow\infty }f_n(b)$? Then, what would $f_n$ increasing tell you? $\endgroup$ – David Mitra Dec 13 '11 at 22:56
  • $\begingroup$ It says that $\lim_{n \to infty}f_n(a)=0$, $\lim_{n \to infty}f_n(b)=0$, Does it tell me that $f_n=0$? $\endgroup$ – Jozef Dec 14 '11 at 7:40
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The statement is true. In fact, $\{f_n\}$ converges uniformly to the 0 function on $[a,b]$:

Let $\epsilon>0$. Since $\sum\limits_{n=1}^\infty f_n(a)$ converges, $f_n(a)\rightarrow 0$. Thus, there is an $N_1$ such that $|f_n(a)|<\epsilon$ for all $n\ge N_1$. A similar argument gives us an $N_2$ such that $|f_n(b)|<\epsilon$ for all $n\ge N_2$.

Let $N=\max\{N_1,N_2\}$. Let $n\ge N$ and $x\in[a,b]$. Then, since $f_n$ is increasing: $$ -\epsilon< f_n(a) \le f_n(x) \le f_n(b)<\epsilon; $$ and so, $|f_n(x)|<\epsilon$ for all $n\ge N$ and $x\in[a,b]$.

Thus, $\{f_n\}$ converges uniformly to 0 on $[a,b]$.

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