Problem with function of sequence simple convergence

[trenekas]

Hello. I have a problem. I don't know how to finish my proof. Maybe someone of you will be able to help me.

So task is:
Suppose that fn and gn are function of the sequence. fn simply convergence(? don't 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 don't know how to finish that... Maybe someone have a tip for me? Thanks.

 

[Fredrik]

Hello. I have a problem. I don't know how to finish my proof. Maybe someone of you will be able to help me.

So task is:
Suppose that fn and gn are function of the sequence. fn simply convergence(? don't 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 don't know how to finish that... Maybe someone have a tip for me? Thanks.

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.

 

[trenekas]

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.

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 don't 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

 

[Fredrik]

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.

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.

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

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

 

[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.

 

[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.

 

[Fredrik]

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.

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 $|f_n(x)g_n(x)-f(x)g(x)|<\varepsilon$.

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.

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.

 

[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

 

[Fredrik]

The proof should start with these statements:

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.

 

[trenekas]

The proof should start with these statements:

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.

ok understand. thanks for help

 

[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 can't concoct
$|f_n-f|*|g_n-g| < z$
$|f|*|g_n-g|< y$
$|g|*|f_n-f|< x$

 

[Fredrik]

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.

 

[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?

 

[Fredrik]

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 $n\geq N$.

 

[micromass]

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