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Let $(X,d)$ be a metric space. Let $f_n : X \rightarrow \mathbb{R} $ be continuous for each $n \geq 1$. Assume that $|f_n(x)|\leq a_n$ and assume that the series $\sum_n a_n$ converges. Show that $F(x) = \sum_{n=1}^\infty f_n(x) $ defines a continuous function. My attempt: Since $|f_n(x)|\leq a_n$ for all $n$ and since $\sum_n a_n$ converges, we know that $\sum f_n(x)$ converges. What do I do from here? |
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$\left(\sum_{k=1}^n f_k \right)_n$ converges uniformly to $F$ since $$\left| F(x)-\sum_{k=1}^n f_k(x) \right| = \left| \sum_{n+1}^\infty f_k(x) \right| \leq \sum_{k=n+1}^\infty |f_k(x)| \leq \sum_{k=n+1}^\infty \underbrace{\|f_k\|_\infty}_{\leq a_k} \qquad (x \in X)\\ \Rightarrow \left\|F-\sum_{k=1}^n f_k\right\|_{\infty} \leq \sum_{k=n+1}^\infty a_k \to 0 \quad (n \to \infty)$$ Since uniform limits of continuous functions are continuous, we conclude that $F$ is continuous. |
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We know more than just $\sum f_n(x)$ converges, since $|f_n(x)| \leq |a_n| \forall x \in X$ we have that $\sum f_n(x)$ converges uniformly to a function $f$. It is a well known result that the uniform limit of continuous functions is also continuous, and so we are done. |
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It suffices to show that $\sum_1^N f_n(x)$ converges to $\sum f_n(x)$ uniformly. (which I leave to you.) It suffices to show this because if a sequence of continuous functions converges uniformly then the limiting function is also continuous. This can be seen via the triangle inequality. Break up $\lvert f(x) - f(y) \rvert$ into $$\lvert f(x) - f(y) \rvert \leq \lvert f(x) - f_n(x) \rvert + \lvert f_n(x) - f_n(y) \rvert + \lvert f_n(y) - f(y) \rvert$$ for some sufficiently large $n$ using the uniform convergence and then conclude using continuity of $f_n$. |
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