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Math Help - Bessel's inequality proof

  1. #1
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    Bessel's inequality proof

    Hi guys.
    I need to prove Bessel's inequality for a finite set of orthonormal vectors \left \{ e_1, e_2,...e_n \right \}
    I looked around the internet and found this on Wikipedia:

    "Bessel's inequality follows from the identity:
    0\leq \left \| x-\sum_{i=1}^{n}<x,e_i>e_i \right \|^2=\left \| x \right \|^2-2\sum_{i=1}^{n}|<x,e_i>|^2+\sum_{i=1}^{n}|<x,e_i>|  ^2=\left \| x \right \|^2-\sum_{i=1}^{n}|<x,e_i>|^2
    "

    but something just does not make sense to me.
    how did they make the first transition from \left \| x-\sum_{i=1}^{n}<x,e_i>e_i \right \|^2 to \left \| x \right \|^2-2\sum_{i=1}^{n}|<x,e_i>|^2+\sum_{i=1}^{n}|<x,e_i>|  ^2?
    you can't just use the binomial expanding formula here, can you?

    Is there another, more simple proof for this inequality?

    Thanks in advanced!
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  2. #2
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    Re: Bessel's inequality proof

    Quote Originally Posted by Stormey View Post
    Hi guys.
    I need to prove Bessel's inequality for a finite set of orthonormal vectors \left \{ e_1, e_2,...e_n \right \}
    This is one I have used.

    Let \alpha_j=<x,e_j>.

     \begin{align*} 0 &\le\left\langle {x - \sum\limits_{j = 1}^n {{\alpha _j}{e_j}} ,x - \sum\limits_{i = 1}^n {{\alpha _i}{e_i}} } \right\rangle \\ &\le{\left\| x \right\|^2} + \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {{\alpha _j}\overline {{\alpha _i}} \left\langle {{e_j},{e_i}} \right\rangle } }  - \sum\limits_{i = 1}^n {\overline {{\alpha _i}} \left\langle {x,{e_i}} \right\rangle }  - \sum\limits_{j = 1}^n {{\alpha _j}\left\langle {{e_j},x} \right\rangle } \\&\le {\left\| x \right\|^2} + \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}}  - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}}  - \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}}\\&\le {\left\| x \right\|^2} - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}} \end{align*}
    Thanks from Stormey
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  3. #3
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    Re: Bessel's inequality proof

    Great. thank you.
    that's more clear.
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  4. #4
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    Re: Bessel's inequality proof

    Hi,
    The proof that you indicate is not hard to understand. It mainly uses distribution of inner product (dot product) over addition.

    Bessel's inequality proof-mhfbessellinequality.png
    Thanks from Stormey
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