# Problem with function of sequence simple convergence

1. Mar 29, 2013

### trenekas

Hello. I have a problem. I dont know how to finish my proof. Maybe someone of you will be able to help me.

Suppose that fn and gn are function of the sequence. fn simply convergence(? dont know if english term is same) to f, and gn simply convergence to g. Need to prove that gn*fn convergence to g*f.

my attempt:

for every ε1 there is N1 that |gn-g|<ε1
for every ε2 there is N2|that fn-f|<ε2
so i need to prove that |fn*gn-f*g|<ε.
So I make a few alterations |fn*gn-f*g|=|(gn-g)*fn+g(fn-f)|< |(gn-g)fn|+|g(fn-f)|<|gn-g|*|fn|+|g|*|fn-f|...
and dont know how to finish that... Maybe someone have a tip for me? Thanks.

2. Mar 29, 2013

### Fredrik

Staff Emeritus
It's just a matter of making clever choices of N1 and N2. It may also help to do rewrite $f_n$ as $(f_n-f)+f$. If you don't know the term for the type of convergence you have in mind, you should post the definition. Then maybe we can tell you what it's called.

3. Mar 29, 2013

### trenekas

As you can see something like that i tried to do. |fn*gn-f*g|=|(gn-g)*fn+g(fn-f)| plus and minus fn*g.

definition of simply convergence is that N(ε) dependending not only from n, but also from x. For example in uniform convergence(? also dont know if its correct in english) N(ε) depend only from n, and for every x.

I found english term: and it is pointwise convergence, not simply :D

Last edited: Mar 29, 2013
4. Mar 29, 2013

### Fredrik

Staff Emeritus
I meant that you may need to do what I suggested to deal with the factor of $f_n$ that appears in the result you got, without an $f$ next to it.

$f_n\to f$ pointwise if and only if for all x, $f_n(x)\to f(x)$. Is this what you had in mind?

5. Mar 29, 2013

### trenekas

ok understand. but nothing goes on for me :(
I got that
$|g_n*f_n|*|f_n| + |g|*|f_n-f|=|g_n-g|*|f_n-f+f|+|g|*|f_n-f|< |g_n-g|*(|f_n-f|+|f|)+|g|*|f_n-f|$
$= |f_n-f|*|g_n-g|+|f|*|g_n-g|+|g|*|f_n-f|$
I do not think that progressed... or maybe i misunderstand you.

6. Mar 29, 2013

### trenekas

Here is definition:
Let D be a subset of R and let {fn} be a sequence of real valued functions
deﬁned on D. Then {fn} converges pointwise to f if given any x in D and
given any ε > 0, there exists a natural number N = N(x, ε) such that
|fn(x) − f(x)| < ε for every n > N.
Note: The notation N = N(x, ε) means that the natural number N depends
on the choice of x and ε.

7. Mar 29, 2013

### Fredrik

Staff Emeritus
That is what I meant that you should do. It's an improvement because the assumption that $f_n\to f$ pointwise makes it easy to deal with $|f_n(x)-f(x)|$, while it's not so easy to deal with $|f_n(x)|$.

As in almost all convergence proofs, you will need to choose the N's of the sequences that are known to be convergent, and use them to define the N of the sequence that we want to prove is convergent. In other words, make a choice of $N_1$ and $N_2$. Define $N=\operatorname{max}\{N_1,N_2\}$. Prove that for all x and all positive integers n such that $n\geq N$, we have $|f_n(x)g_n(x)-f(x)g(x)|<\varepsilon$.

OK. That definition is equivalent to saying that $f_n\to f$ pointwise on D if $f_n(x)\to f(x)$ for all x in D.

8. Mar 29, 2013

### trenekas

or maybe: $|f_n-f|*|g_n-g| < \epsilon_1*\epsilon_2= \epsilon_5$
$|f|*|g_n-g|< \epsilon_2*|f|= \epsilon_3$
$|g|*|f_n-f|< \epsilon_1*|g| = \epsilon_4$
So let $\epsilon:= \epsilon_3+ \epsilon_4+ \epsilon_5$
Maybe correct? :D

9. Mar 29, 2013

### Fredrik

Staff Emeritus

Let $x\in D$ be arbitrary. Let $\varepsilon>0$ be arbitrary.

Since $\varepsilon$ is arbitrary, you can't define it in terms of the other "epsilons". You have to choose the other epsilons so that they add up to something no greater than this arbitrary one.

10. Mar 29, 2013

### trenekas

ok understand. thanks for help

11. Mar 29, 2013

### trenekas

can you help me to choose x, y, z that sum of them would be ε? or its possible of this =>? i think should be something with root of epsilon... but cant concoct
$|f_n-f|*|g_n-g| < z$
$|f|*|g_n-g|< y$
$|g|*|f_n-f|< x$

12. Mar 29, 2013

### Fredrik

Staff Emeritus
We have three terms and we want them to add up to something less than $\varepsilon$. So we can try to make each of them less than $\varepsilon/3$. This is very easy with the two terms that only have one n each. Unfortunately the obvious way to do this doesn't automatically make the term with two n's small enough. Similarly, there's an obvious way to make the term with two n's small enough, but that doesn't automatically ensure that the other two terms are small enough.

It's actually not hard to combine these two ideas. You just need to realize that if you have one choice of $N_1,N_2$ that takes care of two terms, and another choice that takes care of the third term, then the larger of the two choices takes care of all three.

It's also perfectly fine to e.g. choose $N_1$ such that $|f_n(x)-f(x)|<\operatorname{max}\{\varepsilon,\varepsilon^2\}$ when $n\geq N_1$. I'm not saying that this is a good choice here. The point is just that the right-hand side of the inequality doesn't have to be a simple expression like $\varepsilon$ or $\varepsilon^2$ but can be anything that gets the job done.

Last edited: Mar 29, 2013
13. Mar 29, 2013

### trenekas

Im found easier way i think. Found the fact that convergence sequence is smaller than some number. let |fn|<M
so |gn-g|<ε/2M and |gn-g|*|fn|<ε/2M*M
|fn-f|*|g|<ε/2|g|*g
so ε/2+ε/2=ε. i think that is enough?

14. Mar 29, 2013

### Fredrik

Staff Emeritus
That's a good strategy. But you will of course have to specify an N such that this holds for all integers n such that $n\geq N$.

15. Mar 29, 2013

### micromass

Staff Emeritus
Do you know that if $x_n\rightarrow x$ and $y_n\rightarrow y$ as sequence of real numbers, that then $x_ny_n\rightarrow xy$?