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Math Help - scwharz inequality

  1. #1
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    scwharz inequality

    when does equality hold in the scwharz inequality. When ai, bi are real numbers, then there is equality when ai=kbi, for i =1,...,n. But when they are complex sequences I tried setting ai=kb*i, (where b* is b conjugate) for i=1,...,n and it didn't work.

    Thanks. I am not showing my work because I have not got the time to do latex atm.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: scwharz inequality

    In general, for an inner product space V over \mathbb{K} ( \mathbb{K}=\mathbb{R} or \mathbb{K}=\mathbb{C} ) the equality |<x,y>|=||x||\;||y|| holds iff x and y are linearly dependent.
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    Re: scwharz inequality

    Yes I know for vectors but if ai and bi are complex sequences, then is the condition ai=kb*i (b* is b conjugate) or is it ai=kbi
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    MHF Contributor FernandoRevilla's Avatar
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    Re: scwharz inequality

    Quote Originally Posted by Duke View Post
    Yes I know for vectors but if ai and bi are complex sequences, then is the condition ai=kb*i (b* is b conjugate) or is it ai=kbi
    The complex sequences are also vectors so, for any inner vector space whose elements are complex sequences, you only have to translate the condition of linear dependence. For example, translate it to the vector space V=\{(a_n)\} of all finite non null sequences with <(a_n),(b_n)>=\sum_{n=1}^{+\infty}a_nb_n^* . What do you obtain?
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    Re: scwharz inequality

    so it should be ai.bi not ai.bi*in the sum?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: scwharz inequality

    Quote Originally Posted by Duke View Post
    so it should be ai.bi not ai.bi*in the sum?
    No, it is a_ib_i^* .
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    Re: scwharz inequality

    Right in that case, a few questions.

    1) How can a vector space be a set of sequences (I thought it was a set of vectors)?
    2) In that sum, should it be k instead of infinity since we are dealing with finite sequences?
    3) So you are saying this is how we define the dot product for vectors with complex components. So in that case I can set ai=kb*i and get equality?

    Thanks, you have been very helpful.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: scwharz inequality

    Quote Originally Posted by Duke View Post
    1) How can a vector space be a set of sequences (I thought it was a set of vectors)?
    Define (a_n)+(b_n)=(a_n+b_n) and \lambda (a_n)=(\lambda a_n) . With these operations, V is a vector space over \mathbb{C} . For that reason we call vector to any sequence (a_n) .

    2) In that sum, should it be k instead of infinity since we are dealing with finite sequences?
    It is irrelevant, the sum is finite: a_1b_1^*+\ldots+a_kb_k^*+0\cdot 0+0\cdot 0+\ldots

    3) So you are saying this is how we define the dot product for vectors with complex components. So in that case I can set ai=kb*i and get equality?
    We get the equality iff (a_n) and (b_n) are linearly dependent, i.e. iff there exists \alpha \in\mathbb{C} such that a_n=\alpha b_n for all n or b_n=\alpha a_n for all n.
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  9. #9
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    Re: scwharz inequality

    Thanks very much. Yes I worked out number 3 last night. So a vector doesn't have to be a column of numbers?
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: scwharz inequality

    Quote Originally Posted by Duke View Post
    So a vector doesn't have to be a column of numbers?
    Right, any element of a vector space is called vector.
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