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Theorem: Let $f_n$ continuous functions defined on a set $A\subseteq\mathbb{R}$, and assume $\sum_{n=1}^{\infty}f_n$ converges uniformly on $A$ to a function $f$. Then $f$ is continuous on $A$.

My Approach: Let $\epsilon>0$ and fix $c\in A$. Choose $N$ so that $\forall x\in A$ $$\underbrace{\left|f(x)-\sum_{n=1}^{\infty}f_N(c)\right|}_{\text{using uniform convergence}}<\frac{\epsilon}{3}$$ Because $f_N$ is continuous, there exists a $\delta>0$ for which $$|f_N(x)-f_n(c)|<\frac{\epsilon}{3n}$$ is true whenever $|x-c|<\delta$. But this implies,

$$\begin{aligned} |f(x)-f(c)| &= \left|f(x)-\sum_{n=1}^{\infty}f_N(x)+\sum_{n=1}^{\infty}f_N(x)-\sum_{n=1}^{\infty}f_N(c)+\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\ &\leq \left|f(x)-\sum_{n=1}^{\infty}f_N(x)\right|+\left|\sum_{n=1}^{\infty}f_N(x)-\sum_{n=1}^{\infty}f_N(c)\right|+\left|\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\ &= \left|f(x)-\sum_{n=1}^{\infty}f_N(x)\right|+\sum_{n=1}^{\infty}\left|f_N(x)-f_N(c)\right|+\left|\sum_{n=1}^{\infty}f_N(c)-f(c) \right| \\&\leq \frac{\epsilon}{3}+n.\frac{\epsilon}{3n}+\frac{\epsilon}{3} \\&=\epsilon \end{aligned}$$

Thus, $f$ is continuous on $A$.

falamiw
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  • Do you know that the same is true (and easier to prove) for sequences? In fact, one usually proves that the uniform limit of a sequcnes of continuous functions is continuous and then, extend (using partial sums) to series. – Tito Eliatron Jan 30 '21 at 19:12
  • No, Can you provide the statement? I want to see the extension version of that proof. Can you give me hints? @TitoEliatron. And is my proof OK? – falamiw Jan 30 '21 at 19:13
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    See this question. – José Carlos Santos Jan 30 '21 at 19:17
  • Thanks @JoséCarlosSantos But I was curious about the fact, one usually proves that the uniform limit of a sequcnes of continuous functions is continuous and then, extend (using partial sums) to series. the extension part what was suggested by TitoEliatron – falamiw Jan 30 '21 at 19:20
  • Indeed, that's the usual (and natural) way of dealing with this. – José Carlos Santos Jan 30 '21 at 19:21
  • The extension is obvious since $f(x)=\sum_{n=1}^\infty f_n(x)\iff f(x)=\lim_{N\to\infty} \sum_{n=1}^N f_n(x)$ and any FINITE sum of continuous functios is continuous – Tito Eliatron Jan 30 '21 at 19:23
  • OK I got intuition how to prove the uniform limit of a sequence of continuous functions is continuous which seems similar work of mine (Just without the sum). But how to use that in order to say $f(x)$ is continuous? @TitoEliatron. Sorry but my head didn't hit any clue. – falamiw Jan 30 '21 at 19:29
  • $f(x)=\sum f_n(x)\iff f(x)=\lim S_n(x)$ where $S_n(x)$ denotes the SEQUENCE of partial sums – Tito Eliatron Jan 30 '21 at 19:30

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