Prove that a sum of continuous functions is continuous

$\forall n\in N$, let $f_n(x): [0,1]\rightarrow \mathbb{R}$ be continuous functions and $M$ a positive integer.

If $\forall x \in [0,1]$, $\lvert\sum_{n=1}^{\infty} f_n(x)\rvert \lt M$, then $\sum_{n=1}^{\infty} f_n(x)$ is continuous on $[0,1]$.

In my opinion this function $\sum_{n=1}^{\infty} f_n(x)$ is continuous on [0,1] but it is just a conjecture.

How to prove this problem? Or find a counterexample?

• I assume you mean that $|\sum_{n=1}^{\infty} f_n(x)| \lt M$? – Zestylemonzi Jul 14 '16 at 10:04
• Maybe you want to add that $f_n(x)$ is always $\ge 0$, or that $|\sum f_n(x) | < M$ – user258700 Jul 14 '16 at 10:05
• I suppose the sum must be continuous in the event that it converges, isn't it? – cronos2 Jul 14 '16 at 10:15
• if all the $f_n$ are non-negative, or if $\sum_n |f_n(x)| < M$, then yes. As I said before this is the en.wikipedia.org/wiki/Uniform_limit_theorem – reuns Jul 14 '16 at 13:04

I assume $|\sum_{n=1}^{\infty}f_n(x)|<M$ in your question. This should be bounded upward and downward or the following argument is useless.
Define continuous functions $P_n(x)$ on $[0,1]$ for every $n\in \mathbb{N}$ to be $P_n(x)=\sum_{i=1}^{n}f_n(x)$. These $P_n(x)$'s are definitely continuous since they are linear combinations of finite continuous functions. Then the goal inclined to prove that $P(x):=\lim_{n\rightarrow \infty}P_n(x)$ is continuous.
First of all, you defined this $P(x)$ pointwisely, and you know it converges since $P(x)$ is bounded by $M$. Therefore, we have Cauchy sequences defined on evert point in $[0,1]$. Now all you need to do is the "tri-$\varepsilon$" argument on the limit.
• No. Let $f_n(x)=x^{n+1}-x^n$ and $g(x)=\sum_1^{\infty}f_n(x).$ Then $g(x)=-x$ for $x\in [0,1)$ and $g(1)=0.$ The Q is equivalent to asking whether a bounded function that is a point-wise limit of continuous functions is necessarily continuous. But that is not so. – DanielWainfleet Jul 15 '16 at 2:26
• Thank you. You are right. It will be true if the $P_n(x)$'s, in my argument, should converges uniformly. – Hilaire Jul 16 '16 at 15:26
• Isn't that the convergence is considered in a different way? For the Fourier series, one usual consider $L^2$-convergence, which do not care about discontinuity for points of measure 0. – Hilaire Jul 16 '16 at 15:51