Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like your help proving that

If $\sum\limits_{n=1}^{\infty}|f_n|$ converges uniformly, so does $ \sum\limits_{n=1}^{\infty}f_n$.

There a Weierstrass theorem saying that if there's a positive sum $ \sum\limits_{k=1}^{\infty}b_k$ which converges and $|u_k(x)|\leq b_k$ for every $x$ so $ \sum\limits_{k=1}^{\infty}u_k$ converges uniformly. Can it goes the opposite way too? And then we can claim that there $ \sum\limits_{k=1}^{\infty}b_k$ such that $\|f_n\|\leq b_k$ and so does $|f_n|\leq b_k$ and we are done?

If not, we know that in general if $ \sum\limits_{k=1}^{\infty}|b_k|$ converges, so does $ \sum\limits_{k=1}^{\infty}b_k$, how can apply the uniform convergence here?

Thanks a lot!

share|cite|improve this question
up vote 10 down vote accepted

I don't think the converse of the Weierstrass test holds. However, you can use the following

Fact: A series $\sum\limits_{i=1}^\infty f_i(x)$ of real-valued functions converges uniformly on a set $E$ if and only if it is uniformly Cauchy. That is, if and only if, given $\epsilon>0$, there is an $N$ such that $$\Bigl|\sum_{i=n}^m f_i(x)\,\Bigr |<\epsilon$$ for all $m\ge n\ge N$ and for all $x\in E$.

Now think about this and the triangle inequality.

$\color{maroon}{\text {Warning! Solution follows:}}$

We now show that your series $\sum\limits_{i=n}^m f_i(x)$ converges uniformly:

Let $\epsilon>0$. Since, $\sum\limits_{i=1}^\infty |f_i(x)|$ converges uniformly, there is an $N$ such that for all $m\ge n\ge N$ and for all $x$: $$ \sum_{i=n}^m |f_i(x)| <\epsilon.$$

Using this and the triangle inequality, it follows that for all $m\ge n\ge N$ and for all $x$: $$ \Bigl|\sum_{i=n}^m f_i(x)\,\Bigr| \le \sum_{i=n}^m |f_i(x)| <\epsilon. $$

So $\sum\limits_{i=1}^\infty f_i(x)$ is uniformly Cauchy, and thus uniformly convergent.

For completeness:

The Fact above follows (by looking at the sequence of partial sums of the series) from the following standard theorem:

Theorem: A sequence of real-valued functions $\{f_n\}$ is uniformly convergent on a set $E$ if and only if it is uniformly Cauchy on $E$; that is, given $\epsilon>0$, there is an $N$ so that $$ \tag{1}|f_n(x)-f_m(x)|<\epsilon,\quad \text{ for all }n,m\ge N\text{ and for all }x\in E $$


To prove the forward implication, suppose $\{f_n\}$ converges uniformly to $f$ on the set $E$. Then, given $\epsilon>0$, there is an $N$ so that $$ |f_n(x)-f(x)|<\epsilon/2 $$ for all $n\ge N$ and for all $x\in E$. Thus, if $m,n\ge N$ and $x\in E$ $$ |f_n(x)-f_m(x)|<|f_n(x)-f(x)|+|f_m(x)-f(x)|<{\epsilon\over2}+{\epsilon\over2}=\epsilon. $$ From this, it follows that $\{f_n\} $ is uniformly Cauchy on $E$.

To prove the reverse implication, suppose $\{f_n\}$ is uniformly Cauchy on $E$. Then for each $x\in E$, the sequence $\{f_n(x)\}$ is Cauchy and thus converges to some number $f(x)$.

We claim that $\{f_n\}$ converges uniformly to $f$, as defined above, on $E$.

Towards proving the claim, let $\epsilon>0$ and choose $N$ a positive integer that verifies equation (1).

Then if $m\ge N$ is fixed, $n\ge N$, and $x\in E$: $$ \tag{2}|f_m(x)-f_n(x)|<\epsilon $$ Taking the limit as $n\rightarrow\infty$ in (2) gives $$ \tag{3}|f_m(x)-f (x)|\le\epsilon . $$ Since $\epsilon$ was an arbitrary positive number, and since (3) holds for all $m\ge N$ and all $x\in E$, it follows that $\{f_n\}$ converges uniformly to $f$ on $E$.

share|cite|improve this answer
You didn't have to show me the triangle inequality, I managed with the first version of the answer.Thanks a lot! – Jozef Dec 15 '11 at 9:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.