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Math Help - Complex Schwarz's Inequality

  1. #1
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    Complex Schwarz's Inequality

    Prove Schwarz's for complex numbers: (x,y)^2 <= |x|^2 |y|^2
    Under what conditions does equality hold?

    Hint:
    | lx + my |^2 = l^2 (x,x) + 2lm (x,y) + m^2 (y,y)

    is positive definite wrt l and m

    The hint is really confusing me...
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  2. #2
    Newbie MarceloFantini's Avatar
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    Re: Complex Schwarz's Inequality

    Let z = y - \frac{(y,x)}{|x|^2} \cdot x. Now, we have (z,x) = 0 and all you have to do is compute 0 \leq |z|^2.
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  3. #3
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    Re: Complex Schwarz's Inequality

    Can you explain how (z,x)=0?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex Schwarz's Inequality

    Quote Originally Posted by monomoco View Post
    Can you explain how (z,x)=0?

    (z,x)=\left(y-\dfrac{(y,x)}{|x|^2}x,x\right)=(y,x)-\dfrac{(y,x)}{|x|^2}(x,x)=

    (y,x)-\dfrac{(y,x)}{|x|^2}|x|^2=(y,x)-(y,x)=0\quad (\forall x\neq 0)
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  5. #5
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    Re: Complex Schwarz's Inequality

    I don't understand how you take the inner product like that. Is x a vector and its elements are also being called x? And can you explain how it distributes that way?
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  6. #6
    Newbie MarceloFantini's Avatar
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    Re: Complex Schwarz's Inequality

    I believe the problem is a notation one. Allow me to change inner products from (y,x) to \langle y \, , \,x \rangle. So, we have:

    \langle z \, , \, x \rangle = \left\langle y - \frac{\langle y \, , \, x \rangle}{|x|^2} \, , \, x \right\rangle = \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \langle x \, , \, x \rangle =

    = \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \cdot |x|^2 = \langle y \, , \, x \rangle - \langle y \, , \, x \rangle = 0

    I hope it's clearer now.
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  7. #7
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    Re: Complex Schwarz's Inequality

    I get it now. Thanks very much!
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