# What is Schwarz inequality

1. Jul 23, 2014

### Greg Bernhardt

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:

$$\|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$

$$|a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$

$$| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$

$$\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)$$

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:

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

$$\int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2$$

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