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Math Help - Pointwise convergence.

  1. #1
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    Pointwise convergence.

    I'm ask to prove whether or not the sequence of functions {f_n} defined on [-1,1] is pointwise convergent, where f_n = (1-|x|)^n for each natural number n. I was hoping someone could look over my proof and see if I made any mistakes. The cases where x=0 and |x|=1 are obvious, so consider those as already proven.

    Proof: Let \epsilon >0 be given and let x \in (-1, 1) where x \ne 0. Then, 0<1-|x|<1. Let N be a positive integer such that N> \frac{|ln(\epsilon)|}{|ln(1-|x|)|}. Then, \epsilon >(1-|x|)^N (I'm not 100% sure this is right, so please correct me if I'm wrong). So, for any positive integer n>N , we have |(1-|x|)^n-0|=(1-|x|)^n<(1-|x|)^N<\epsilon. Therefore, {f_n} converges pointwise to the zero function on (-1,1)-{0}.
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    Quote Originally Posted by spoon737 View Post
    I'm ask to prove whether or not the sequence of functions {f_n} defined on [-1,1] is pointwise convergent, where f_n = (1-|x|)^n for each natural number n. I was hoping someone could look over my proof and see if I made any mistakes. The cases where x=0 and |x|=1 are obvious, so consider those as already proven.
    If x\not = 0 then 0<1-|x| < 1 and so (1-|x|)^n \to 0. Otherwise, if x=0 then (1-|x|)^n = 1^n \to 1. Thus, f_n\to f where f(x) = \left\{ \begin{array}{c} 1\mbox{ if }x=0\\0\mbox{ if }-1\leq x\leq 1 \mbox{ and }x\not = 0 \end{array} \right..
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  3. #3
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    It was easy enough for me to see that (1-|x|)^n converges to zero where 0<|x|\le 1, but my professor wants us to provide more details. I guess my main concern is if I made the right choice for N. Will that work, or did I make a mistake?
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