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
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?
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.
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!
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.
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
Even more generally Muntz's theorem.
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