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How to prove Schwartz Inequality?

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove the Schwartz Inequality

    /<u, v>/ is < or = //u//*//v//

    for the Hermitian product.

    2. Relevant equations



    3. The attempt at a solution

    Assume u, v are non-zero. For any r, s exist in set of complex numbers, //rv + su// is > or = 0. Then, //rv + su//^2 = <rv+su, rv+su> = <rv, rv> + <rv, su> + <su, rv> + <su, su> = r^2<v, v> + rs <v, u> + sr <u, v> + s^2 <u, u> is > or = 0

    Not sure where to go from there...set r = <u u> and s = ....?? I don't know.

    Note: I was working off of the proof of the inequality for the dot product. I am just not sure how the Hermitian properties affect it. is <v, u> = <u, v>, etc.
     
  2. jcsd
  3. Mar 14, 2009 #2
    We say (u.v)2 ≤ |u|2 |v|2, then if u or v = 0 then both sides of both sides are zero. But if they are not zero, the vector v can be rewritten as the sum of some scalar multiple of u, ku, and a vector w that is orthogonal to u. The k is obtained by w = v-ku & the orthogonality condition u.w=0

    0 = u.w = u.(v-ku) = (u.v)-k(u.u), it follows that k = u.v/u.u, now apply the pythagoras theorem:

    |v|2 = |ku|2 + |w|2 = k|u|2 + |w|2

    |u|2|v|2 = (u.v) + |u|2|w|2

    since |u|2|w|2 ≥ 0, you have (u.v)2 ≤ |u|2 |v|2
    This establishes what we said first and hence the cauchy shwartz inequality.
     
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