# Does the Weierstrass M-Test work for all spaces?

1. Oct 1, 2011

### lugita15

The http://en.wikipedia.org/wiki/Weierstrass_M-test" [Broken] states roughly that if you have an infinite series of bounded functions, and the sum of the bounds of all those functions is finite, then your infinite series is a uniformly convergent series.

My question is, can those functions, which are of course real- or complex- valued, be defined on any kind of space? Specifically, instead of the functions being defined on R or R^n, can they be defined on the natural numbers, so that instead of functions you have sequences? Would the theorem still hold in this case? My guess is that it would work, because I don't see anywhere in the proof where assumptions are made about the domain of the functions.

Any help would be greatly appreciated.

Last edited by a moderator: May 5, 2017
2. Oct 3, 2011

### Eynstone

I think the codomain must be a complete metric space for the series to converge pointwise (otherwise the convergence wouldn't make sense).

3. Oct 3, 2011

### lugita15

But I don't want to change the domain of the functions, not the codomain. They would still be real-valued, I just want them defined on the natural numbers rather than the real numbers. In other words, I want to talk about sequences, which are of course functions of natural numbers, instead of ordinary functions.

4. Oct 4, 2011

### lugita15

I should add that I'm not interested in the Weierstrauss M-test as such, I just want to use it in order to interchange a limit and an infinite sum.

5. Oct 9, 2011

### micromass

Staff Emeritus
The Weierstrass M-test works for any domain. It also works for any codomain, provided that the codomain is a Banach space.

6. Oct 9, 2011

### lugita15

Thank you, micromass. And does the theorem still hold that if you have a uniformly convergent sum, you can interchange the sum with a pointwise limit of functions, even if the domain of the functions are the natural numbers rather than the real numbers?

7. Oct 9, 2011

### micromass

Staff Emeritus
Yes, uniform convergence always allows you to switch limits. You might still want to prove that though, it's not that difficult.

8. Oct 9, 2011

### lugita15

OK, that's what I wanted. And yes, perhaps I will try proving that uniform convergence implies interchange.

I should also probably give the context where I'm dealing with this: proving the bounded convergence theorem for the Lebesgue integral. I am dealing with the expression $lim_{k→\infty}\sum_{i}\int_{F_{i}} f_{k}$ (where the $F_{i}$ are disjoint measurable sets and the $f_{k}$ are real-valued measurable functions defined on $ℝ^{n}$). I have shown that $\left|\int_{F_{i}} f_{k}\right| < M_{i}$ and that $\sum_{i}M_{i} < \infty$. So what I want to do is apply the Weierstrauss M-test and say that the infinite sum is uniformly convergent, and after that I want to interchange limits and get $\sum_{i}lim_{k→\infty}\int_{F_{i}} f_{k}$. Is that mathematically valid?

(You can see why I was asking about the domain being the natural numbers, because my functions are $\int_{F_{i}} f_{k}$ and my variable is $i$.)