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In Mathematical, the inequality of Cauchy-Schwarz , also called inequality of Schwarz , or inequality of Cauchy-Bunyakovski-Simart-Schwarz , meets in many fields such as the Linear algebra with the Vecteur S, the analyzes with the series and in Intégration with the integrals of products.

It owes its name with Hermann Amandus Schwarz and Augustin Louis Cauchy.

The inequality is stated in the following way:

For all X and there elements of a real Space préhilbertien or complex

$|\backslash \; langle\; X,\; there\; \backslash \; rangle|\backslash \; the\; \backslash |X\; \backslash |.\; \backslash |there\; \backslash |$

The two members are equal if and only if X and is there linearly dependant S.

Consequences

A consequence of the inequality of Cauchy-Schwarz is that the scalar Produit is a continuous function .

In the case of the Euclidean Space $\backslash \; quad\; \backslash \; mathbb\; R\; ^n$ provided with the canonical scalar product, the inequality of Cauchy-Schwarz is written:

$\backslash \; left|\backslash \; sum\_\; \{i=1\}\; ^n\; x\_\; \{I\}\; y\_\; \{I\}\; \backslash \; right|\backslash \; the\; \backslash \; left\; (\backslash \; sum\_\; \{i=1\}\; ^n\; x\_\; \{I\}\; ^\; \{2\}\; \backslash \; right)\; ^\; \{1/2\}.\; \backslash \; left\; (\backslash \; sum\_\; \{i=1\}\; ^n\; y\_\; \{I\}\; ^\; \{2\}\; \backslash \; right)\; ^\; \{1/2\}$

In the case of the functions with complex values of Square integrable, the inequality of Cauchy-Schwarz is written:

$\backslash \; left|\backslash \; int\; \backslash \; overline\; \{F\}.\; G\; \backslash ,\; \backslash \; textrm\; \{D\}\; X\; \backslash \; right|\; \backslash \; Leq\; \backslash \; left\; (\backslash \; int\; |F|^2\; \backslash ,\; \backslash \; textrm\; \{D\}\; X\; \backslash \; right)\; ^\; \{1/2\}.\; \backslash \; left\; (\backslash \; int\; |G|^2\; \backslash ,\; \backslash \; textrm\; \{D\}\; X\; \backslash \; right)\; ^\; \{1/2\}$

These two last formulations are generalized by the Inégalité of Hölder.

Demonstration

Let us show the result in the case of a complex préhilbertien.

Inequality

That is to say (X, there) a couple of vectors. Even if it means to multiply X by a scalar of the form $e^\; \{I\; \backslash \; theta\}$, with $\backslash \; theta$ real, one can suppose that the product $\backslash \; langle\; X,\; \backslash \; rangle$ is real there, and one obtains:

$\backslash \; forall\; T\; \backslash \; in\; \backslash \; mathbb\; \{R\}\; \backslash \; quad\; 0\; \backslash \; Leq\; \backslash |x+t.y\; \backslash |^2\; =\; \backslash |X\; \backslash |^2+\; T\; \backslash \; left\; (\backslash \; overline\; \{\backslash \; langle\; X,\; there\; \backslash \; rangle\}\; +\; \backslash \; langle\; X,\; there\; \backslash \; rangle\; \backslash \; right)\; +\; t^2\; \backslash |there\; \backslash |^2=\; \backslash |X\; \backslash |^2\; +\; 2t\; \backslash \; langle\; X,\; there\; \backslash \; rangle\; +\; t^2\; \backslash |there\; \backslash |^2$

Thus, the polynomial with real coefficients $P\; (X)\; =\; \backslash |there\; \backslash |^2\; X^2\; +\; 2\; \backslash \; langle\; X,\; there\; \backslash \; rangle\; X\; +\; \backslash |X\; \backslash |^2$ of unknown factor X, is positive on $\backslash \; mathbb\; \{R\}$ according to the preceding relation. It cannot thus have two distinct real roots. This implies that its discriminant is negative. One obtains:

$4|\backslash \; langle\; X,\; there\; \backslash \; rangle\; |^2\; -\; 4\; \backslash |X\; \backslash |^2\; \backslash |there\; \backslash |^2\; \backslash \; the\; 0$

What involves the announced inequality well.

Case of equality

If vectors X and are bound there, one can without loss general information suppose that $y=\; \backslash \; alpha\; X\; \backslash \; quad\; (\backslash \; alpha\; \backslash \; in\; \backslash \; mathbb\; \{C\})$. One from of deduced immediately: $\backslash \; quad\; |\backslash \; langle\; X,\; there\; \backslash \; rangle\; |=|\backslash \; alpha|\backslash |X\; \backslash |^2=|\backslash \; alpha|\backslash |X\; \backslash |\backslash |X\; \backslash |=\; \backslash |X\; \backslash |\backslash |there\; \backslash |$

Reciprocally, let us suppose that one has the equality $\backslash \; quad\; |\backslash \; langle\; X,\; there\; \backslash \; rangle\; |=\; \backslash |X\; \backslash |\backslash |there\; \backslash |$ If y=0, the vectors are dependant. If y$\backslash \; not=$ 0, the polynomial above is written:

$P=\backslash |there\; \backslash |^2\; X^2\; +\; 2\; \backslash |X\; \backslash |\backslash |there\; \backslash |\; X\; +\; \backslash |X\; \backslash |^2\; =\; \backslash \; left\; (\backslash |there\; \backslash |\; X\; +\; \backslash |X\; \backslash |\backslash \; right)\; ^2$

He admits for real root doubles $t\_0=-\; \backslash \; frac\; \{\backslash |X\; \backslash |\}\; \{\backslash |there\; \backslash |\}\; \backslash \; in\; \backslash \; mathbb\; \{R\}$, from where $P\; (t\_0)\; =\; 0\; \backslash \; geq\; \backslash |x+t\_0\; there\; \backslash |^2\; \backslash \; geq\; 0$, then $x+t\_0\; y=0$. From where the result.

Real case

In the case of a real space the demonstration is similar. One can also propose a slightly different proof:

The proof for $y\; =\; 0$ is commonplace, one thus considers $y\; \backslash \; neq\; 0$. For $\backslash \; lambda\; \backslash \; in\; \backslash \; mathbb\; \{R\}$ one a:

$0\; \backslash \; Leq\; \backslash \; langle\; X\; \backslash \; lambda\; there,\; X\; \backslash \; lambda\; there\; \backslash \; rangle\; =\; \backslash \; langle\; X\; \backslash \; lambda\; there,\; X\; \backslash \; rangle\; -\; \backslash \; lambda\; \backslash \; langle\; X\; \backslash \; lambda\; there,\; there\; \backslash \; rangle$

\ langle X, X \ rangle -2 \ lambda \ langle X, there \ rangle + \ lambda^2 \ langle there, there \ rangle.

Let us take $\backslash \; lambda\; =\; \backslash \; langle\; X,\; there\; \backslash \; rangle\; \backslash \; cdot\; \backslash |there\; \backslash |^\; \{-\; 2\}$ so that:

$0\; \backslash \; Leq\; \backslash |X\; \backslash |\; ^2\; -\; \backslash \; langle\; X,\; there\; \backslash \; rangle^2\; \backslash \; cdot\; \backslash |there\; \backslash |^\; \{-\; 2\},$

Thus

$\backslash \; langle\; X,\; there\; \backslash \; rangle^2\; \backslash \; Leq\; \backslash |X\; \backslash |^2\; \backslash |there\; \backslash |^2.$

Then

$\backslash \; big|\; \backslash \; langle\; X,\; there\; \backslash \; rangle\; \backslash \; big|\; \backslash \; Leq\; \backslash |X\; \backslash |\; \backslash |there\; \backslash |.$

This proof can easily be adapted to the complex case.

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