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Inner Product Space Proof

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    First I'd like to state the meaning of my notations
    x = (x0,x1,x2...xn)
    y = (y0,y1,y2...yn)
    |x| = absolute value of x
    ||x|| = Normal of x
    <x,y> = Inner Product of x and y

    I have to prove the following

    |<x1,y1> - <x2,y2>| ≤ ||x1 - x2||*||y1|| + ||x2||*||y1-y2||

    2. Relevant equations

    Applicable Axioms of Normals and Inner Products
    ||x|| = √(<x,x>)
    <x + z,y> = <x,y> + <z,y>

    3. The attempt at a solution

    I tried expanding the right hand side as such:
    ||x1 - x2|| = √(<x1-x2,x1-x2>) = √(<x1,x1> + 2*<x1,-x2> + <-x2,-x2>)
    ||x2|| = √(<x2,x2>)

    I did similarly for the y values, and I'm not seeing anything that pops out to me as a solution to this proof, nothing seems to cancel, and no axioms seem to make this work in a general sense. I guess what I need...is a HINT.
     
  2. jcsd
  3. Apr 8, 2013 #2

    LCKurtz

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    Hint: Try subtracting and adding ##\langle x_2,y_1\rangle##.
     
  4. Apr 8, 2013 #3
    Thank you, but I still don't see how that's applicable, everything's inside an absolute value or a square root so I'm not sure how to get in there? Can I have a double hint?
     
  5. Apr 8, 2013 #4

    LCKurtz

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    Don't you have the Cauchy-Schwartz inequality ##|\langle x,y\rangle|\le \|x\|\|y\|## for an inner product space? That plus the first hint should do it...

    [Edit, added] And, of course, the triangle inequality
     
    Last edited: Apr 8, 2013
  6. Apr 9, 2013 #5
    LCKurtz there's no easy way to put this, but you are THE MAN. Thank you so much, here's the solution I came to thanks to your hints. I really should've came to it easier, but what can you do.

    |<x1,y1> - <x2,y2>| ≤ ||x1 - x2||*||y1|| + ||x2||*||y1-y2||

    Using the first hint so graciously given me by LCKurtz, I added and subtracted <x2,y1> on the left hand side.

    This gives:
    |<x1,y1> - <x2,y1> - <x2,y2> + <x2,y1>|
    Then by an axiom listed above, we can bring this to:
    |<x1 -x2,y1> + <y1-y2,x2>|

    Then working on the right hand side, and using the memory jump-starter so humbly given by Zeus Incarnate, Mr. LCKurtz (Talking about the Cauchy-Schwartz inequality).
    ||x1-x2||*||y1|| >= |<x1-x2,y1>|
    ||y1-y2||*||x2|| >= |<y1-y2,x2>|

    This means the right hand side as it is is greater than or equal to |<x1-x2,y1> + <y1-y2,x2>|, also known as... DUN DUN DUNNNNN the left hand side! So clearly, this is the end of the proof. A shout out to my boy LCKurtz for his help, thanks.
     
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