Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is Schwarz inequality

  1. Jul 23, 2014 #1
    Definition/Summary

    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.

    Equations

    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

    Theorem:

    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

    [tex]=\|a\|^2-2|t||(a,b)|+|t|^2\|b\|^2[/tex]​

    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)=As^2-Bs[/tex]​

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

    [tex]f''(s)=2A>0[/tex]​

    Continuing with this value of |t|...

    [tex]=\|a\|^2-2\frac{|(a,b)|}{\|b\|^2}|(a,b)|+\frac{|(a,b)|^2}{\|b\|^4}\|b\|^2[/tex]​

    [tex]=\|a\|^2-\frac{|(a,b)|^2}{\|b\|^2}[/tex]​

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: What is Schwarz inequality
  1. An inequality (Replies: 11)

Loading...