# Understanding when convergence implies uniform convergence for sequences of non-continuous functions

I am working on the following problem:

Let $(f_n)$ be a sequence of functions $[a,b] \rightarrow \mathbb{R}$ such that: (i) $f_n(x)≤0$ if $n$ is even, $f_n(x)≥0$ if $n$ is odd; (ii) $|f_n(x)|≥|f_{n+1}(x)|$ for all $x$; (iii) $f_n$ converges to $0$ uniformly. Prove that then $\sum_n f_n$ is uniformly convergent.

I take as given the following theorem (3.43 in Rudin's Principles of Mathematical Analysis):

Suppose (a) $|c_1|≥|c_2|≥|c_3|≥...$; (b) $c_{2m-1}≥0$, $c_{2m}≤0$ ($m=1,2,3,...$); (c) $lim_{n\rightarrow\infty} c_n=0$ Then $\sum_n c_n$ converges.

Now, it follows immediately that for given $x$, $\sum_n f_n(x)$ converges, say, to $F_n(x)$, a constant function.

I am wondering how to show rigorously that this convergence is in fact uniform. We are dealing with a compact set, but the $\sum_n f_n(x)$ are not continuous (are they?), so we can't use Dini's theorem.

I have tried saying something along these lines:

• Choose $\epsilon>0$. Then $\forall x \in [a,b]$, $\exists N_x$ s.t. $n≥N_x \implies |\sum_nf_n(x)-F_n(x)|<\epsilon.$
• Put $M=sup_{x\in [a,b]}N_x$. Then $n≥M \implies|\sum_nf_n(x)-F_n(x)|<\epsilon$ $\forall x \in [a,b]$.

But I worry that the reasoning is faulty. ($\forall x \exists N$ does not usually imply $\exists N \forall x$ - but I wonder if it does in this case because we're working on a compact set?) If anyone has any thoughts I'd be very grateful. Thanks!

• Show that the sum is uniformly Cauchy. Note $|\sum_{k=n}^m f_n(x)|\le |f_n(x)|$ for all $x$ and all $n,m$ with $m\ge n$. – David Mitra May 19 '13 at 18:09
• Thanks! I've posted a full solution using your hint. – mww May 19 '13 at 19:14

• We are given that $f_n \rightarrow 0$ uniformly, so for any $\epsilon > 0$ $\exists N$ s.t. $n≥N \implies |f_n(x)|<\epsilon$ $\forall x$.
• We have, then, for $m>n>N$, $|\sum_1^n f_n(x) - \sum_1^m f_n(x)|=|\sum_{n+1}^m f_n(x)|≤|f_n(x)|≤\epsilon$ $\forall x$, where the penultimate inequality follows immediately from conditions (i) and (ii).
• Thus, $\sum_n f_n$ is uniformly convergent.