Cauchy–Schwarz inequality

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In mathematics, the Cauchy–Schwarz inequality, also known as the Schwarz inequality, the Cauchy inequality, or the Cauchy–Bunyakovski–Schwarz inequality, named after Augustin Louis Cauchy, Viktor Yakovlevich Bunyakovsky and Hermann Amandus Schwarz, is a useful inequality encountered in many different settings, such as linear algebra applied to vectors, in analysis applied to infinite series and integration of products, and in probability theory, applied to variances and covariances.

The inequality is represented in a simple manner by

$(a_1 b_1 + \cdots + a_n b_n)^2 \le (a_1^2 + \cdots + a_n^2) (b_1^2 + \cdots + b_n^2).$

Equality occurs exactly when

$\frac {a_1}{b_1} = \frac {a_2}{b_2} = \cdots = \frac {a_n}{b_n}.$

The inequality states that if x and y are elements of real or complex inner product spaces then

$|\langle x,y\rangle|^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle.$

The two sides are equal if and only if x and y are linearly dependent (or in geometrical sense they are parallel). This contrasts with a property that the inner product of two vectors is zero if they are orthogonal (or perpendicular) to each other.

The inequality hence confers the notion of "the angle between the two vectors" to an inner product, where concepts of Euclidean geometry may not have meaningful sense, and justifies that the notion that inner product spaces are generalizations of Euclidean space.

An important consequence of the Cauchy–Schwarz inequality is that the inner product is a continuous function.

Another form of the Cauchy–Schwarz inequality is given using the notation of norm, as explained under norms on inner product spaces, as

$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.\,$

The finite-dimensional case of this inequality for real vectors was proved by Cauchy in 1821, and in 1859 Cauchy's student V.Ya.Bunyakovsky noted that by taking limits one can obtain an integral form of Cauchy's inequality. The general result for an inner product space was obtained by K.H.A.Schwarz in 1885.

1 Proof
- 1.1 First proof
- 1.2 Second proof
2 Notable special cases
3 Usage
4 References

[edit] Proof

[edit] First proof

As the inequality is trivially true in the case y = 0, we may assume <y, y> is nonzero. Let $λ$ be a complex number. Then,

$0 \leq \left\| x-\lambda y \right\|^2 = \langle x-\lambda y,x-\bar{\lambda} y \rangle = \langle x,x \rangle - \bar{\lambda} \langle x,y \rangle - \lambda \langle y,x \rangle + |\lambda|^2 \langle y,y\rangle.$

Choosing

$\lambda = \langle x,y \rangle \cdot \langle y,y \rangle^{-1}$

we obtain

$0 \leq \langle x,x \rangle - |\langle x,y \rangle|^2 \cdot \langle y,y \rangle^{-1}$

which is true if and only if

$|\langle x,y \rangle|^2 \leq \langle x,x \rangle \cdot \langle y,y \rangle$

or equivalently:

Q.E.D.

[edit] Second proof

Given the polynomial

$(a_1 x + b_1)^2 + \cdots + (a_n x + b_n)^2 = 0$

it is quick to note that it's a quadratic polynomial and its discriminant is not greater than zero, because it does not have any roots (unless all the ratios a_i/b_i are equal), thus

$( \sum ( a_i \cdot b_i ) )^2 - \sum {a_i^2} \cdot \sum {b_i^2} \le 0$

which yields the Cauchy-Schwarz inequality.

[edit] Notable special cases

In the case of the Euclidean space Rⁿ, we get

$\left(\sum_{i=1}^n x_i y_i\right)^2\leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right).$ In particular, in the Euclidean space of 2 or 3, if the dot product is defined in terms of the angle between two vectors, then one can immediately see the inequality: $|\mathbf{x} \cdot \mathbf{y}| = |\mathbf{x}| |\mathbf{y}| |\cos \theta| \le |\mathbf{x}| |\mathbf{y}|$ . Also, in this case the Cauchy–Schwarz inequality can be deduced from Lagrange's identity by omitting a term. In three dimensions n = 3, Lagrange's identity takes the form

$\langle x,x\rangle \cdot \langle y,y\rangle = |\langle x,y\rangle|^2 + |x \times y|^2.$

For the inner product space of square-integrable complex-valued functions, one has

A generalization of these two inequalities is the Hölder inequality.

[edit] Usage

The triangle inequality for the inner product is often shown as a consequence of the Cauchy–Schwarz inequality, as follows: given vectors x and y,

$\\|x + y\\|^2$	$= \langle x + y, x + y \rangle$
	$= \\|x\\|^2 + \langle x, y \rangle + \langle y, x \rangle + \\|y\\|^2$
	$\le \\|x\\|^2 + 2\|\langle x, y \rangle\| + \\|y\\|^2$
	$\le \\|x\\|^2 + 2\\|x\\|\\|y\\| + \\|y\\|^2$
	$\le \left(\\|x\\| + \\|y\\|\right)^2$

Taking the square roots gives the triangle inequality.

The Cauchy–Schwarz inequality is usually used to show Bessel's inequality.

The general formulation of the Heisenberg uncertainty principle is derived using the Cauchy-Schwarz inequality in the inner product space of physical wave functions.

[edit] References

J.M. Steele, The Cauchy-Schwarz Master Class, Cambridge University Press, 2004.

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Categories: Inequalities | Linear algebra

$\\|x + y\\|^2$	$= \langle x + y, x + y \rangle$
	$= \\|x\\|^2 + \langle x, y \rangle + \langle y, x \rangle + \\|y\\|^2$
	$\le \\|x\\|^2 + 2\|\langle x, y \rangle\| + \\|y\\|^2$
	$\le \\|x\\|^2 + 2\\|x\\|\\|y\\| + \\|y\\|^2$
	$\le \left(\\|x\\| + \\|y\\|\right)^2$

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Cauchy–Schwarz inequality

From Wikipedia, the free encyclopedia

Contents

[edit] Proof

[edit] First proof

[edit] Second proof

[edit] Notable special cases

[edit] Usage

[edit] References

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