Uniform continuity of series

Question

Let $F= \sum_{j=1}^{\infty} f_j$ where $f_j : [a,b] \to \mathbb R$ are continuous functions and the series is uniformly convergent. Show that $F$ is uniformly continuous.

I'm unsure how to prove this, I first showed that $f_j$ is uniformly continuous since it is continuous on a closed, bounded interval, but then can we deduce that $F$ is also uniformly continuous from this?

Do you know the theorem "if a sequence of continuous functions converges uniformly, its limit is continuous"? (Yes, this is basically the same thing, if you squint; have you studied it already and can you use it?) — Clement C., May 24 '18 at 16:56
In fact, I think you don't need to actually use the compactness of the domain. If you know the proof that a sequence of continuous functions which converges uniformly has a continuous limit, then a similar proof should show the same thing about uniformly continuous functions - and it should work with any metric space as a domain. — Daniel Schepler, May 24 '18 at 17:30
@DanielSchepler $f_n(x)=\frac{1}{x}+\frac{1}{n}$ on $(0,1)$. — The Phenotype, May 24 '18 at 17:35
@ThePhenotype I don't see how any $f_n$ in that sequence is uniformly continuous. Ohh... but now I see the original problem statement only assumes $f_j$ continuous. So, yes, I guess you have the choice of either using each $f_j$ being uniformly continuous along with the "uniform limit of uniformly continuous functions" result, or else showing that $f$ is continuous and thus uniformly continuous. — Daniel Schepler, May 24 '18 at 17:39

score 1 · Answer 1 · edited May 24 '18 at 17:46

I would prove first that $F$ is continuous, then only easily deduce it's uniformly continuous on $[a,b]$. For that first part, it's the very important uniform limit theorem, which you can find in any good analysis book. If you just want a hint on how to prove this theorem, consider the Cauchy criterion, and the inequality $$ |f(x) - f(y)| \leq |f(x) - f_{N_\varepsilon}(x)| + |f_{N_\varepsilon}(x) - f_{N_\varepsilon}(y)| + |f(y) - f_{N_\varepsilon}(y)| $$ for some $N_\varepsilon$ large enough, which does not depend on $x$ or $y$ since the convergence is uniform.

Note: You're using series, I'm using sequences. This doesn't change much by using $F_n = \sum_{j=1}^{n}f_j$, noting that $F$ is the uniform limit of that sequence and taking the argument from there.

Uniform continuity of series

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