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Math Help - Series of continuous functions

  1. #1
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    Series of continuous functions

    "Prove that if a series of continuous functions converges uniformly, then the sum function is also continuous."

    I'm trying to interpret what this is saying so that I can know what to prove. Does the given portion mean that the individual f_n's uniformly converge to f, or that \sum_{n=1}^{\infty} f_n \to \sum_{n=1}^{\infty} f?
    And what does "the sum function is continuous" mean? Simply that \sum_{n=1}^{\infty} f is continuous?
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    Re: Series of continuous functions

    Quote Originally Posted by phys251 View Post
    "Prove that if a series of continuous functions converges uniformly, then the sum function is also continuous."

    I'm trying to interpret what this is saying so that I can know what to prove. Does the given portion mean that the individual f_n's uniformly converge to f, or that \sum_{n=1}^{\infty} f_n \to \sum_{n=1}^{\infty} f?
    And what does "the sum function is continuous" mean? Simply that \sum_{n=1}^{\infty} f is continuous?

    This has the definition you need. Right at the top.
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    Re: Series of continuous functions

    Quote Originally Posted by romsek View Post
    The part about "A series...converges uniformly"? OK, thanks.

    Again, though, what does "the sum function is continuous" mean? S_n(x) --> ?
    Last edited by phys251; March 16th 2014 at 10:09 PM.
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    Re: Series of continuous functions

    Quote Originally Posted by phys251 View Post
    The part about "A series...converges uniformly"? OK, thanks.

    Again, though, what does "the sum function is continuous" mean? S_n(x) --> ?
    if the $f_k(x)$ are continuous then

    let $f(x)=\displaystyle{\lim_{n \to \infty}}S_n(x)\text{ where }S_n=\displaystyle{\sum_{k=0}^n} f_k(x)$

    if $S_n(x)$ converges uniformly then $f(x)$ is continuous.
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    Re: Series of continuous functions

    Quote Originally Posted by romsek View Post
    if the $f_k(x)$ are continuous then

    let $f(x)=\displaystyle{\lim_{n \to \infty}}S_n(x)\text{ where }S_n=\displaystyle{\sum_{k=0}^n} f_k(x)$

    if $S_n(x)$ converges uniformly then $f(x)$ is continuous.
    OK. Can we just use the fact that if (1) f_n is continuous for all n's, and (2) f_n -> f uniformly, then (3) f is continuous? I.e., does the result I'm trying to prove immediately follow, since the sum of continuous functions on an interval is also continuous on that interval?
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    Re: Series of continuous functions

    Quote Originally Posted by phys251 View Post
    OK. Can we just use the fact that if (1) f_n is continuous for all n's, and (2) f_n -> f uniformly, then (3) f is continuous? I.e., does the result I'm trying to prove immediately follow, since the sum of continuous functions on an interval is also continuous on that interval?
    Yes you can.
    Use what Ken Ross calls the famous $\dfrac{\varepsilon }{3}>0$ argument.

    $|f(x)-f(x_0)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x)| $

    You select an $n$ large enough to insure that on RHS the first and third term is less that the magic number from the continuity of $f_n$.

    The middle comes from the uniform convergence of the sequence.
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    Re: Series of continuous functions

    Thanks, Plato, but I think you have a few errors in there. Isn't that last term supposed to be f_n(x_0) - f(x_0)? And uniform convergence produces the first and third terms, not the second one, right?
    Last edited by phys251; March 17th 2014 at 01:44 PM.
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    Re: Series of continuous functions

    Quote Originally Posted by phys251 View Post
    Thanks, Plato, but I think you have a few errors in there. Isn't that last term supposed to be f_n(x_0) - f(x_0)? And uniform convergence produces the first and third terms, not the second one, right?
    You are correct. But the same idea holds.
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