# Math Help - Continuous sum function?

1. ## Continuous sum function?

is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it

2. Originally Posted by cp05
is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it
What about $f_n(x)=\sum_{m=1}^{n}\frac{f(x)}{2^m}$ the sum is uniformly convergent since $\|f\|_{\infty}$ exists, etc.

3. How do you know it's uniformly convergent, because you don't know what f is?

Also, how does that show every continuous function on the interval [0,1] is a limit of a sequence of continuous functions? Or is it showing the statement is false?
I don't understand.
Don't I need to start with a continuous function on [0,1] and show it's not the limit of a sequence of continuous functions to prove it false?

4. Originally Posted by cp05
is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it
Actually you can say even more :Stone?Weierstrass theorem - Wikipedia, the free encyclopedia (look for the original Wierstrass approximation theorem).

That said, Drexel's argument works fine since $\sum_{k=1}^{\infty } \frac{1}{2^k} =1$ and you know $f$ is continous.

5. Read these posts again having in mind that you were wrong, the statement is true.

The proof should come out easily, take f continuous on [0,1],

use what drexel and jose said and prove f is the uniform limit of f_n.

Dont forget that since f's domain is compact then f is bounded there, it will come in handy somewhere.

Good luck!

6. Originally Posted by cp05
is every continuous function on [0,1] the uniform limit of a sequence of continuous functions??

I feel like that's not true...but I can't find a counterexample or figure out how to prove it
I might be missing something here, but what about $f_n = f$?

7. I can't prove something saying fn=f can I? Because that's being specific.

And why am I picking ? Isn't that being specific also?

8. Oh wait i think I might understand it.

So I'm taking an f continuous on [0,1]
And I have to show that there is A (like...ONE at least) sequence of continuous functions that converges to that uniformly

So taking , sup|f/2^m| = 1/2^m (I'm a little confused about this...because how do I know what f is? Would the supremum be f/2^m instead? )
Either way
then SUM(1/2^m) or SUM(f/2^m)...whichever one it is, will be geometric with r = 1/2, so it converges
So by Wierstrass M-test, fn is uniformly convergent.

So how do I show it converges to f(x)?? because as m -> infinity, the sum goes to f/[1-(1/2)], not f right?

But for continuity...we know f is continuous and 2^m is continuous, so division of two continuous functions is continuous and then the sum of continuous functions is continuous so fn(x) is continuous.

Am I anywhere on the right track? I'm just not sure about the supremum part of the M-test and how to show fn(x) converges to f.

9. I think I get the fn=f better now

f is continuous on [0,1]
so fn is continuous on [0,1] also since fn=f

and the sup|fn-f| = 0 which ->0 so fn converges to f uniformly on M

Is that right?

10. first,

what bruno suggests is "trivially" true (because he proposes a constant sequence) so i think what cp05 needs to prove is that there exists a "non constant" sequence of continuous f_n... bla bla bla

second,

good try cp05, still you need to be sure exactly what and why you are doing the things you are doing :

take f continuous on [0,1].

Define a sequence by $f_n(x)=\sum_{k=1}^n \frac{f(x)}{2^k}$

Since f is continuous and f_n is a finite sum of continuous functions then each f_n is continuous on [0,1].

By what i said earlier f is bounded on [0,1] by, lets say, M. So $|f(x)| \leq M$ $\forall x\in [0,1]$

Now lets see the uniform convergence

Given $\epsilon$ we need to find an adequate N which does not depend on x, only on $\epsilon$.

Take ANY x\in [0,1]

$|f(x)-f_n(x)|=|f(x)||1-\sum_{k=1}^n \frac{1}{2^k}|$

We know that $\sum_{k=1}^n \frac{1}{2^k}\to 1$ when $n\to \infty$
so for $\frac{\epsilon}{M}$ there is an N for which

$|1-\sum_{k=1}^n \frac{1}{2^k}|<\frac{\epsilon}{M}$ if $n\geq N$

Using the first equalities we have that for $n\geq N$

$|f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$

where we used that f is bounded by M. Take a good look, N does not depend on x, only on M (and on the bound but that is fixed anyway).

If you want to get the supremum way of seing it then think like this:

We just showed that for any $\epsilon$ there is an $N(\epsilon)$ such that
$|f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$ if $n\geq N$ for any $x\in [0,1]
$

Taking supremum over [0,1] grants you that

$||f-f_n||_{sup}<\epsilon$ if $n\geq N$

i hope i didnt make any mistake

11. Originally Posted by mabruka
first,

what bruno suggests is "trivially" true (because he proposes a constant sequence) so i think what cp05 needs to prove is that there exists a "non constant" sequence of continuous f_n... bla bla bla

second,

good try cp05, still you need to be sure exactly what and why you are doing the things you are doing :

take f continuous on [0,1].

Define a sequence by $f_n(x)=\sum_{k=1}^n \frac{f(x)}{2^k}$

Since f is continuous and f_n is a finite sum of continuous functions then each f_n is continuous on [0,1].

By what i said earlier f is bounded on [0,1] by, lets say, M. So $|f(x)| \leq M$ $\forall x\in [0,1]$

Now lets see the uniform convergence

Given $\epsilon$ we need to find an adequate N which does not depend on x, only on $\epsilon$.

Take ANY x\in [0,1]

$|f(x)-f_n(x)|=|f(x)||1-\sum_{k=1}^n \frac{1}{2^k}|$

We know that $\sum_{k=1}^n \frac{1}{2^k}\to 1$ when $n\to \infty$
so for $\frac{\epsilon}{M}$ there is an N for which

$|1-\sum_{k=1}^n \frac{1}{2^k}|<\frac{\epsilon}{M}$ if $n\geq N$

Using the first equalities we have that for $n\geq N$

$|f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$

where we used that f is bounded by M. Take a good look, N does not depend on x, only on M (and on the bound but that is fixed anyway).

If you want to get the supremum way of seing it then think like this:

We just showed that for any $\epsilon$ there is an $N(\epsilon)$ such that
$|f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon$ if $n\geq N$ for any $x\in [0,1]
$

Taking supremum over [0,1] grants you that

$||f-f_n||_{sup}<\epsilon$ if $n\geq N$

i hope i didnt make any mistake
Or...just invoke the Weierstrass M-test.

12. Originally Posted by Jose27
Actually you can say even more :Stone?Weierstrass theorem - Wikipedia, the free encyclopedia (look for the original Wierstrass approximation theorem).

That said, Drexel's argument works fine since $\sum_{k=1}^{\infty } \frac{1}{2^k} =1$ and you know $f$ is continous.
Even more generally Muntz's theorem.

13. Originally Posted by Drexel28
Or...just invoke the Weierstrass M-test.

yeah, well i found out that when i was learning analysis, getting "full" proofs "while you can" its a better way to learn.

"while you can" meaning that later you must rely entirely on more complex theorems

14. Wow thanks so much guys for all the help! I'm terrible at this stuff. Still a little confused but I think I'm getting there.

This might be a stupid question...but why does . It's geometric right? So it should go to 1/1-r = 1/1-(1/2) = 2?

And also, I know I'm supposed to prove this using the Weierstrass M-test, where I have to show SUMsup|fn| converges => fn converges...just not sure how to show it specifically converges to f. If were true, that would make sense because then as n->infinity, the limit would just be f(x)*1....but I don't understand why it's 1

15. Originally Posted by cp05
Wow thanks so much guys for all the help! I'm terrible at this stuff. Still a little confused but I think I'm getting there.

This might be a stupid question...but why does . It's geometric right? So it should go to 1/1-r = 1/1-(1/2) = 2?

And also, I know I'm supposed to prove this using the Weierstrass M-test, where I have to show SUMsup|fn| converges => fn converges...just not sure how to show it specifically converges to f. If were true, that would make sense because then as n->infinity, the limit would just be f(x)*1....but I don't understand why it's 1
$\sum_{n=\color{red}{0}}^{\infty}x^n=\frac{1}{1-x}\cdots$

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