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Math Help - Analysis Help - 1

  1. #1
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    Analysis Help - 1

    IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)


    IV. 2. Give an example of a function series that converges (pointwise) at every point of (a, b), each summand is continuous at every point in [a, b], but the series does not converge at every point of [a, b]. Formally prove that your function series satisfies all these conditions.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pila0688 View Post
    IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)
    For the series \{s_n(x)\} to converge uniformly to s(x) on (a,b) for all \varepsilon>0 there exists an N_{\varepsilon} such that for all n>N_{\varepsilon}:

    |s_n(x)-s(x)|<\varepsilon

    and N_{\varepsilon} is independednt of x.

    As the series converges at x=s and x=b, for all \varepsilon>0 there exist N_{1,\varepsilon} and N_{2,\varepsilon} such that for all n>N_{1,\varepsilon}:

    |s_n(a)-s(a)|<\varepsilon

    and for all n>N_{2,\varepsilon}:

    |s_n(b)-s(b)|<\varepsilon.

    Now for \varepsilon>0 let M_{\varepsilon}=\max(N_{\varepsilon}, N_{1,\varepsilon},N_{2,\varepsilon}).

    Now for all x in [a,b] we have for all n>M_{\varepsilon}:

    |s_n(x)-s(x)|<\varepsilon

    which proves that \{ s_n(x) \} is uniformly convergent on [a,b]

    CB
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