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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!

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1 Answer 1

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 $$ \tag{1}|f_n(x)-f_m(x)|<\epsilon,\quad \text{ for all }n,m\ge N\text{ and for all }x\in E $$

Proof:

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 $E$.

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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

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