What is the Schwarz Inequality and Its Applications?

  • Thread starter Greg Bernhardt
  • Start date
  • Tags
    Inequality
In summary, the Schwarz inequality, also known as the Cauchy-Schwarz inequality, has various applications in mathematics and physics. It states that for vectors a and b in an inner product space over complex numbers, the product of their norms is greater than or equal to the absolute value of their inner product. This inequality can also be written in other equivalent forms, such as for complex numbers and functions. The proof for this inequality involves choosing an optimal complex number and using basic properties of complex numbers. This concept is often used in Hilbert spaces and in distinguishing operations, operators, functionals, and representations.
  • #1
19,445
10,025
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!
 
Mathematics news on Phys.org

Related to What is the Schwarz Inequality and Its Applications?

What is Schwarz inequality?

Schwarz inequality, also known as Cauchy-Schwarz inequality, is a mathematical concept that states the relationship between the dot product of two vectors and the magnitude of those vectors. It is commonly used in linear algebra and analysis.

What is the formula for Schwarz inequality?

The formula for Schwarz inequality is |A · B| ≤ ||A|| ||B||, where A and B are vectors and ||A|| and ||B|| represent the magnitude of those vectors.

What is the significance of Schwarz inequality?

Schwarz inequality is significant because it provides a way to measure the relationship between two vectors. It also has many applications in mathematics, physics, and engineering, such as in proving the convergence of series and in optimization problems.

What is an example of Schwarz inequality?

An example of Schwarz inequality is in the dot product of two vectors, where the absolute value of the dot product is always less than or equal to the product of the magnitudes of the two vectors. For instance, if A = (1,2,3) and B = (4,5,6), then |A · B| ≤ ||A|| ||B|| becomes |1·4 + 2·5 + 3·6| ≤ √(1^2+2^2+3^2) √(4^2+5^2+6^2), which is true since 32 ≤ 91.

How is Schwarz inequality related to other mathematical concepts?

Schwarz inequality is related to other concepts such as the triangle inequality, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. It is also related to the concept of orthogonality, where two vectors are perpendicular to each other and have a dot product of zero.

Similar threads

Replies
3
Views
306
Replies
13
Views
1K
  • General Math
Replies
2
Views
2K
Replies
4
Views
1K
Replies
6
Views
2K
  • General Math
Replies
4
Views
845
Replies
1
Views
458
Replies
2
Views
630
  • Advanced Physics Homework Help
Replies
10
Views
636
  • Advanced Physics Homework Help
Replies
3
Views
913
Back
Top