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Math Help - Completeness of Vector Space

  1. #1
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    Completeness of Vector Space

    Let H be the space of sequences (x(1),x(2),x(3),...) such that \sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty, with inner product <x,y>=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)}  }{j^2} and norm \|x\|=<x,x>^{1/2}.

    I need to show that H is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that \mathbb{C} is complete.

    I know that I need to show that every Cauchy sequence in H is convergent. I can see that this follows relatively easily from the definition of H.

    So I have taken a sequence x_{n}(j) but for this to be Cauchy |x_{n}(j)-x_{m}(j)|<\epsilon. But I can't see how to show that, given that the inner product on H has a j^2 below it. Any help would be great. Thanks
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    Let H be the space of sequences (x(1),x(2),x(3),...) such that \sum^{\infty}_{j=1}\frac{|x(j)|^2}{j^2}<\infty, with inner product <x,y>=\sum^{\infty}_{j=1}\frac{x(j)\overline{y(j)}  }{j^2} and norm \|x\|=<x,x>^{1/2}.

    I need to show that H is complete with respect to the inner product i.e. a Hilbert Space. I am assuming that \mathbb{C} is complete.

    I know that I need to show that every Cauchy sequence in H is convergent. I can see that this follows relatively easily from the definition of H.

    So I have taken a sequence x_{n}(j) but for this to be Cauchy |x_{n}(j)-x_{m}(j)|<\epsilon. But I can't see how to show that, given that the inner product on H has a j^2 below it. Any help would be great. Thanks
    You can just get rid of the j. Suppose that x_n is Cauchy, then
    <br />
\sum\frac{|x_n(j)-x_m(j)|^2}{j^2} \rightarrow 0

    In particular for each j
    <br />
\frac{|x_n(j)-x_m(j)|^2}{j^2} \leq ||x_n-x_m|| \rightarrow 0

    which can only happen if the top part of the LHS goes to zero (I am taking m,n to infinity).
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  3. #3
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    Thanks for the reply, okay so I was already assuming that x_n was Cauchy, the bit I was confused about was the inequality

    \frac{|x_n(j)-x_m(j)|}{j}\leq\|x_n-x_m\|<\epsilon

    Which is need to show that x_n(j) is also Cauchy, for each j.

    My question is: If you have the \frac{1}{j} in the inequality, does that still satisfy the definition of Cauchy?
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  4. #4
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    Quote Originally Posted by ejgmath View Post
    Thanks for the reply, okay so I was already assuming that x_n was Cauchy, the bit I was confused about was the inequality

    \frac{|x_n(j)-x_m(j)|}{j}\leq\|x_n-x_m\|<\epsilon

    Which is need to show that x_n(j) is also Cauchy, for each j.

    My question is: If you have the \frac{1}{j} in the inequality, does that still satisfy the definition of Cauchy?
    Why would it? You are considering j fixed, so if you really are worried about it just pick  \frac{\epsilon}{j}.

    A formal way would be to say, fix j, and let epsilon be greater than zero, then there exists an N such that for all n,m>N;
    <br />
\frac{|x_n(j)-x_m(j)|^2}{j^2} \leq ||x_n-x_m||< \frac{\epsilon^2}{j^2}

    Thus for n,m>N
    <br />
|x_n(j)-x_m(j)|< \epsilon

    i.e. x_n(j) is Cauchy.
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