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Math Help - Inner Product Space

  1. #1
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    Inner Product Space

    Hi guys, I'm trying to show that for x,y\in S where S is an inner product space that

    ||x+y||\cdot||x-y||\leq ||x||^2+||y||^2

    This looks simple but I just can't crack it at all
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  2. #2
    Senior Member roninpro's Avatar
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    I haven't worked it out, but it may be easier to manipulate the square of a norm. That is, it would be enough to show

    \|x+y\|^2\cdot \|x-y\|^2\leq (\|x\|^2+\|y\|^2)^2

    This way, you could write \|x+y\|^2=\langle x+y,x+y\rangle and \|x-y\|^2=\langle x-y,x-y\rangle. The inner products are much nicer to play with.
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  3. #3
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    Quote Originally Posted by kenndrylen View Post
    Hi guys, I'm trying to show that for x,y\in S where S is an inner product space that

    ||x+y||\cdot||x-y||\leq ||x||^2+||y||^2

    This looks simple but I just can't crack it at all
    Denote by \langle,\rangle the corresponding inner product, so:

    ||x+y||^2\cdot ||x-y||^2=\langle x+y, x+y\rangle\cdot\langle x-y,x-y\rangle=

    \left(||x||^2+||y||^2+2Re\langle x,y\rangle\right)\left(||x||^2+||y||^2-2Re\langle x,y\rangle\right)=

    ||x||^4+||y||^4-4(Re\langle x,y\rangle )^2+2||x||^2||y||^2\leq ||x||^4+||y||^4+2||x||^2||y||^2 , and we're done

    Tonio
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  4. #4
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    Thanks so much, both of you
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