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Math Help - Continuous sum function?

  1. #1
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    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
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    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.
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    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?
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  4. #4
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    Quote Originally Posted by cp05 View Post
    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.
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  5. #5
    Member mabruka's Avatar
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    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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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by cp05 View Post
    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?
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    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?
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    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.
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  9. #9
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    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?
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  10. #10
    Member mabruka's Avatar
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    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]<br />

    Taking supremum over [0,1] grants you that


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

    i hope i didnt make any mistake
    Last edited by mabruka; May 11th 2010 at 03:56 PM.
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    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]<br />

    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.
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    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.
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  13. #13
    Member mabruka's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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
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  14. #14
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    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
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    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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