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The Weierstrass M-test tells that given a function sequence $(u_{n}(x))$ where $x \in I$, if there exists a convergent series $\sum a_{n}$ such that $|u_{n}(x)|\leq a_{n}$ for all $n$ and $x\in I$, then $\sum u_{n}(x)$ converges uniformly in $I$.

What about the opposite of it?

If $\sum u_{n}(x)$ converges uniformly in $I$, then there exists a convergent series $\sum g_{n}$ such that $|u_{n}(x)|\leq g_{n}$ for all $n$ and $x \in I$.

Since the theorem isn't in the form of 'if and only if', I'm trying to think of an example to counter the above:

My attempt was to define a function sequence such as this:


Then taking $u_{1}(x)=f_{1}(x),\ u_{n}(x)=f_{n}(x)-f_{n-1}(x)$. It can be shown that $f_{n}(x)$ converges uniformly in $[1, \infty)$, therefore $u_{n}(x)$ does as well.

But $|u_{n}(1)|=|f_{n}(1)|$ which is the constant sequence: $1, 1, 1, ...$. If we assume by contradiction that there exists such a sequence $1\leq g_{n}$ that $\sum g_{n}$ converges then by the comparison test $\sum 1$ converges which is obviously not true.

First, I'd like to know if the above is true. It took me quite a while to come up with something and I'm not even sure it's true.

Also, It's very difficult for me to visualize these complicated functions (like the one above) where both $x$ and $n$ play a role. Is there an easier way to deal with these questions?

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Can't you pick one that converges uniformly but not absolutely? – Jonas Teuwen Feb 5 '11 at 12:37
@daniel.jackson: Why not just a constant one? Then it converges certainly uniformly. – Jonas Teuwen Feb 5 '11 at 13:07
@daniel.jackson: Independent on $x$. Maybe I'm missing something? – Jonas Teuwen Feb 5 '11 at 13:53
@Jonas: I see, something like $\sum \frac{(-1)^n}{n}$? – daniel.jackson Feb 5 '11 at 14:23
@daniel.jackson: that is exactly the example I thought of upon reading your question, yes. Jonas T's answer is dead on: you need to remember the difference between uniform convergence and absolute convergence. – Pete L. Clark Feb 5 '11 at 14:41
up vote 8 down vote accepted

The first thing I would think of is to consider a constant series, that is one that does not depend on any variable. This will certainly converge uniformly.

So the next step is to think about your second condition, for this to fail it would work if the series does not converge absolutely.

So, this leads to the most simple example I can think of:

$$\sum_{n \geq 1} \frac{(-1)^n}{n}$$

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$u_n = f_n - f_{n-1}$ and $u_n(1) = 0$ for n>1. However, if $a_n = \sup_{x\in I} |u_n(x)|$, you have $a_n \geq u_n(n) = 1/n$ is still a divergent sequence.

Your idea to "spread" where the functions $u_n$ are adding things along the x axis is good : You never add too much at a specific x so that the sum converges, but you are still able to add enough "big" things at each n so that $\Sigma a_n$ doesn't converge.

You can use this idea to make a more clear counter example :

Pick $B(x)$ a bump function on $\mathbb{R}$ with support in [0,1], for example a piecewise affine function with $B(0)=B(1) = 0$ and $B(1/2) = 1$.

Then define $u_n(x) = B(x-n)/n$.

$\Sigma_{n\geq 1} u_n$ converges uniformly because at every x, $\Sigma_{k\geq n} u_k(x) \leq 1/n$. However, $\Sigma_{n\geq 1} \sup_x |u_n(x)| = \Sigma_{n\geq 1} 1/n$ will diverge

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Thanks. One question though: why is $\sup_{x\in I} |u_n(x)| \geq u_n(n) = 1/n$? – daniel.jackson Feb 5 '11 at 13:03

Almost a hundred years ago G.H. Hardy showed that there are power series over $\mathbb{C}$ that converge uniformly on the closed unit disk but not absolutely on the boundary, which is the same as saying that the hypothesis of the Weierstrass M-test doesn't hold (A theorem concerning Taylor's series, Quart. J. Pure Appl. Math. 44 (1913), 147-160). I posted some references including links to examples here.

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+1, interesting information! – Jonas Teuwen Feb 5 '11 at 20:30

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