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What is Schwarz inequality

  1. Jul 23, 2014 #1

    the Schwarz inequality (also called Cauchy–Schwarz inequality and Cauchy inequality) has many applications in mathematics and physics.

    For vectors [itex]a,b[/itex] in an inner product space over [itex]\mathbb C[/itex]:

    [tex] \|a\|\|b\| \geq |(a,b)|[/tex]

    For two complex numbers [itex] a,b [/itex] :
    [tex] |a|^2|b|^2 \geq |ab|^2 [/tex]

    In bra-ket notation, which is commonly used in Quantum Mechanics:
    [tex] | \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 [/tex]

    For functions [itex] a, b [/itex] which maps a real numbers into complex: [itex] x \in \mathbb{R} [/itex], [itex] a(x), \: b(x) \in \mathbb{C}[/itex]:
    [tex] \int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a^*bdx \right|^2 ,[/tex]
    if the integrals have finite values.


    The Schwarz inequality may be written in a number of equivalent ways:

    [tex] \|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

    [tex] |a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

    [tex] | \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]

    [tex] \int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)[/tex]

    Extended explanation


    If a and b are vectors in an inner product space X over [tex]\mathbb C[/itex], then

    [tex]|(a,b)| \leq \|a\|\|b\|[/tex]​

    where the norm is the standard norm on an inner product space.

    Proof #1:

    Let t be an arbitrary complex number.

    [tex]0 \leq (a+tb,a+tb)=\|a\|^2+t(a,b)+t^*(b,a)+|t|^2\|b\|^2[/tex]​

    [tex]=\|a\|^2+2\mbox{ Re}(t(a,b))+|t|^2\|b\|^2[/tex]​

    The inequality is obviously satisfied when the real part of t(a,b) is non-negative, so we can only learn something interesting when it's negative. Let's choose Arg t so that it is


    Now let's choose |t| so that it minimizes the sum of the last two terms. (This should give us the most interesting result).

    [itex]s=|t|[/itex], [itex]A=\|b\|^2[/itex], [itex]B=2|(a,b)|[/itex]​


    [tex]f'(s)=2As-B=0 \implies s=\frac{B}{2A} = \frac{|(a,b)|}{\|b\|^2}[/tex]​


    Continuing with this value of |t|...



    Thus, [itex]0 \leq \|a\|^2\|b\|^2 - |(a,b)|^2[/itex]

    Proof #2:

    A proof for Ineq(3) (the last inequality in the "equations" section) will now be given, but first some basic properties of complex numbers...

    [tex] z = \text{Re}z + i \text{Im}z [/tex]
    [tex] z^* = \text{Re}z - i \text{Im}z[/tex]
    [tex] |z|^2 = z^*z \geq 0 [/tex]
    [tex] |zy|^2 = (zy)^*zy [/tex]

    Now let [itex] c(x) = a(x) + d \, b(x) [/itex], [itex] d \in \mathbb{C} [/itex].

    [tex] \int |c(x)|^2 dx \geq 0 [/tex]

    [tex] \int |a(x)|^2dx + d^*d \int |b(x)|^2dx + d \int a(x)^*b(x) dx + d^* \int a(x)b(x)^* dx \geq 0 [/tex]

    if the integrals have finite value, this inequality must hold for all [itex] d [/itex], we can choose:

    [tex] d = - \left( \int b(x)^*a(x) dx \right) / \left( \int |b(x)|^2 dx \right) [/tex]

    [tex] \int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2 [/tex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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