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Thread: uniform convergence series

  1. #1
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    uniform convergence series

    Suppose $\displaystyle \sum f_n(x)$ converges uniformly on $\displaystyle A $and $\displaystyle g :A \rightarrow B$ is a continuous function.
    Do we have any conclusion about the series $\displaystyle \sum g(f_n(x)) $?
    ($\displaystyle A$ and $\displaystyle B$ are subsets of real number)
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  2. #2
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    Quote Originally Posted by problem View Post
    Suppose $\displaystyle \sum f_n(x)$ converges uniformly on $\displaystyle A $and $\displaystyle g :A \rightarrow B$ is a continuous function.
    Do we have any conclusion about the series $\displaystyle \sum g(f_n(x)) $?
    ($\displaystyle A$ and $\displaystyle B$ are subsets of real number)
    No.

    $\displaystyle \sum g(f_n(x))$ might diverge or converge. Just take $\displaystyle f_n(x)=2^{-n}$ and $\displaystyle g(x)=1$, in this case we have divergence. However, if $\displaystyle g(x)=x$ we have convergence. So we need to know more about $\displaystyle g$
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    putnam120,if I change the g to be a uniformly continuous function,can I claim that \sum g(f_n) converges uniformly?

    Or it will be the case that convergence/uniform convergence of a series is not invariant under any continuous/uniform continuous function?
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    Senior Member Shanks's Avatar
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    Notice that g(x)=1 and g(x)=x are both uniformly continious function, the claim still can't be true.
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    So is there any properties that a function should have so that the uniform convergence of a series is invariant under the particular function?
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  6. #6
    Senior Member Shanks's Avatar
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    I think, the answer is negative.
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  7. #7
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    Quote Originally Posted by problem View Post
    putnam120,if I change the g to be a uniformly continuous function,can I claim that \sum g(f_n) converges uniformly?

    Or it will be the case that convergence/uniform convergence of a series is not invariant under any continuous/uniform continuous function?
    Going by what I think you mean you're misunderstanding your question:

    If $\displaystyle f_n : A\rightarrow B$ and $\displaystyle g: B \rightarrow \mathbb{R}$ are such that $\displaystyle f_n \rightarrow f$ unif. on $\displaystyle A$ and $\displaystyle g$ is unif. cont. on $\displaystyle B$ then $\displaystyle g(f_n)\rightarrow g(f)$ unif. on $\displaystyle A$. Applying this result to the partial sums of a series, say $\displaystyle \sum_{k=1}^{\infty } h_k = h$ (which converges unif.) we get that $\displaystyle g(\sum_{k=1}^{n} h_k) \rightarrow g(h)$ uniformly. Thus unif. continuity does preserve unif. convergence. What you're asking however is convergence of the series given by $\displaystyle \sum_{k=1}^{\infty } g(h_k)$. Do you see the difference?

    PS. I think this works for your original question:

    Let $\displaystyle h_k,h : A \rightarrow B$ such that $\displaystyle \sum_{k=1}^{\infty } h_k(x) = h(x)$ be a unif. conv. series of functions such that it converges absolutely for all $\displaystyle x \in A$, and $\displaystyle g: B\rightarrow \mathbb{R}$ be unif. continous and $\displaystyle \vert g(x) \vert \leq M \vert x \vert$ for all $\displaystyle x\in B$ then $\displaystyle \sum_{k=1}^{\infty } g(h_k)$ converges unif. on $\displaystyle A$ and absolutely for all $\displaystyle x\in A$
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