Continuous sum function?

Apr 2010
43
1
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 :(
 

Drexel28

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Nov 2009
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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 \(\displaystyle f_n(x)=\sum_{m=1}^{n}\frac{f(x)}{2^m}\) the sum is uniformly convergent since \(\displaystyle \|f\|_{\infty}\) exists, etc.
 
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Apr 2010
43
1
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?
 

Jose27

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Apr 2009
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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 \(\displaystyle \sum_{k=1}^{\infty } \frac{1}{2^k} =1\) and you know \(\displaystyle f\) is continous.
 
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Jan 2010
150
29
Mexico City
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!
 
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Bruno J.

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Jun 2009
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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 \(\displaystyle f_n = f\)?
 
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Apr 2010
43
1
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?
 
Apr 2010
43
1
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.
 
Apr 2010
43
1
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?
 
Jan 2010
150
29
Mexico City
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 \(\displaystyle f_n(x)=\sum_{k=1}^n \frac{f(x)}{2^k}\)

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 \(\displaystyle |f(x)| \leq M\) \(\displaystyle \forall x\in [0,1]\)

Now lets see the uniform convergence

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

Take ANY x\in [0,1]

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

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

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

Using the first equalities we have that for \(\displaystyle n\geq N\)


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

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

Taking supremum over [0,1] grants you that


\(\displaystyle ||f-f_n||_{sup}<\epsilon \) if \(\displaystyle n\geq N\)

i hope i didnt make any mistake
 
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