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# Schwarz inequality

 Definition/Summary the Schwarz inequality (also called Cauchy–Schwarz inequality and Cauchy inequality) has many applications in mathematics and physics. For vectors $a,b$ in an inner product space over $\mathbb C$: $$\|a\|\|b\| \geq |(a,b)|$$ For two complex numbers $a,b$ : $$|a|^2|b|^2 \geq |ab|^2$$ In bra-ket notation, which is commonly used in Quantum Mechanics: $$| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2$$ For functions $a, b$ which maps a real numbers into complex: $x \in \mathbb{R}$, $a(x), \: b(x) \in \mathbb{C}$: $$\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a^*bdx \right|^2 ,$$ if the integrals have finite values.

 Equations The Schwarz inequality may be written in a number of equivalent ways: $$\|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$ $$|a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$ $$| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$ $$\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)$$

 Scientists Augustin Cauchy (1789-1857) Hermann Amandus Schwarz (1843-1921) Viktor Bunyakovsky (1804-1889)

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 Breakdown Mathematics > Calculus/Analysis >> Inequalities

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 Extended explanation Theorem: If a and b are vectors in an inner product space X over $$\mathbb C[/itex], then [tex]|(a,b)| \leq \|a\|\|b\|$$where the norm is the standard norm on an inner product space. Proof #1: Let t be an arbitrary complex number. $$0 \leq (a+tb,a+tb)=\|a\|^2+t(a,b)+t^*(b,a)+|t|^2\|b\|^2$$$$=\|a\|^2+2\mbox{ Re}(t(a,b))+|t|^2\|b\|^2$$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 $$=\|a\|^2-2|t||(a,b)|+|t|^2\|b\|^2$$Now let's choose |t| so that it minimizes the sum of the last two terms. (This should give us the most interesting result). $s=|t|$, $A=\|b\|^2$, $B=2|(a,b)|$$$f(s)=As^2-Bs$$$$f'(s)=2As-B=0 \implies s=\frac{B}{2A} = \frac{|(a,b)|}{\|b\|^2}$$$$f''(s)=2A>0$$Continuing with this value of |t|... $$=\|a\|^2-2\frac{|(a,b)|}{\|b\|^2}|(a,b)|+\frac{|(a,b)|^2}{\|b\|^4}\|b\|^2$$$$=\|a\|^2-\frac{|(a,b)|^2}{\|b\|^2}$$Thus, $0 \leq \|a\|^2\|b\|^2 - |(a,b)|^2$ 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... $$z = \text{Re}z + i \text{Im}z$$ $$z^* = \text{Re}z - i \text{Im}z$$ $$|z|^2 = z^*z \geq 0$$ $$|zy|^2 = (zy)^*zy$$ Now let $c(x) = a(x) + d \, b(x)$, $d \in \mathbb{C}$. $$\int |c(x)|^2 dx \geq 0$$ $$\int |a(x)|^2dx + d^*d \int |b(x)|^2dx + d \int a(x)^*b(x) dx + d^* \int a(x)b(x)^* dx \geq 0$$ if the integrals have finite value, this inequality must hold for all $d$, we can choose: $$d = - \left( \int b(x)^*a(x) dx \right) / \left( \int |b(x)|^2 dx \right)$$ $$\int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2$$

Commentary

 Gokul43201 @ 01:16 PM Feb11-09 included Fredrik's proof and made some small edits.

 malawi_glenn @ 08:55 AM Feb8-09 Fredrik, please feel free to add your derivation if you like :-)