Cauchy-Schwarz inequality

The Cauchy-Schwarz inequality , also known as the Schwarz inequality or Cauchy-Bunjakowski-Schwarz inequality , is an inequality that is used in many areas of mathematics , e.g. B. in linear algebra ( vectors ), in analysis ( infinite series ), in probability theory and in the integration of products. It also plays an important role in quantum mechanics , such as in the proof of Heisenberg's uncertainty principle .

The inequality is named after the mathematicians Augustin-Louis Cauchy , Hermann Amandus Schwarz and Wiktor Jakowlewitsch Bunjakowski .

General case

The inequality says: If and are elements of a real or complex vector space with an inner product , then the relation applies to the scalar product or inner product ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle \langle x,y\rangle }$

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

Equality applies if and only if and are linearly dependent . ${\displaystyle x}$ ${\displaystyle y}$

Equivalent formulations are obtained using the norm induced by the scalar product : ${\displaystyle \|x\|:={\sqrt {\langle x,x\rangle }}}$

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

or.

${\displaystyle \left|\langle x,y\rangle \right|\leq \|x\|\cdot \|y\|.}$

In the real case, you can do without the amount bars :

${\displaystyle \langle x,y\rangle \leq \|x\|\cdot \|y\|}$

Special cases

Applied to the space with the standard scalar product, we get: ${\displaystyle \mathbb {R} ^{n}}$

${\displaystyle \left(\sum x_{i}\cdot y_{i}\right)^{2}\leq \left(\sum x_{i}^{2}\right)\cdot \left(\sum y_{i}^{2}\right)}$

In the case of quadratically integrable complex-valued functions one obtains:

${\displaystyle \left|\int f(x)\cdot {\overline {g(x)}}\,dx\right|^{2}\leq \left(\int \left|f(x)\right|^{2}\,dx\right)\cdot \left(\int \left|g(x)\right|^{2}\,dx\right)}$

For random variables that can be integrally square one obtains:

${\displaystyle \left(\operatorname {E} (XY)\right)^{2}\leq \operatorname {E} (X^{2})\cdot \operatorname {E} (Y^{2})}$

These three inequalities are generalized by the Hölder inequality .

On square matrices used are obtained for the track :

${\displaystyle \vert \mathrm {Spur} (AB^{*})\vert \leq (\mathrm {Spur} (AA^{*}))^{\frac {1}{2}}(\mathrm {Spur} (BB^{*}))^{\frac {1}{2}}}$

In , the statement of the Cauchy-Schwarz inequality can be specified in the form of an equation : ${\displaystyle \mathbb {R} ^{3}}$

${\displaystyle \langle x,x\rangle \cdot \langle y,y\rangle =|\langle x,y\rangle |^{2}+\|x\times y\|^{2}}$

The summand is always non-negative. It is zero if and only if and are linearly dependent. ${\displaystyle \|x\times y\|^{2}}$ ${\displaystyle x}$ ${\displaystyle y}$

history

The inequality is named after Augustin Louis Cauchy , Wiktor Jakowlewitsch Bunjakowski and Hermann Amandus Schwarz . Cauchy found the sum form of the inequality in his analysis algébrique (1821). The integral form of the inequality was first published historically in 1859 by Bunjakowski in a work on inequalities between integrals; Schwarz published his work only in 1884 without reference to the work of Bunjakowski. Corresponding to this development, there is sometimes only the designation as Cauchy inequality for the discrete, finite case and as Bunjakowski inequality or Schwarz's inequality in the integral case.

Applications

In a vector space with an inner product, the triangle inequality for the induced norm can be derived from the Cauchy-Schwarz inequality

${\displaystyle \|x\|={\sqrt {\langle x,x\rangle }}}$

and thus show that a norm so defined fulfills the norm axioms .

Another consequence of the Cauchy-Schwarz inequality is that the inner product is a continuous function .

The Cauchy-Schwarz inequality ensures that in the expression

${\displaystyle \cos \varphi ={\frac {\langle x,y\rangle }{\|x\|\cdot \|y\|}}}$

the amount of the fraction is always less than or equal to one, so that it is well-defined and thus the angle can be generalized to any spaces with an inner product. ${\displaystyle \varphi \;}$

In physics , the Cauchy-Schwarz inequality is used to derive the Heisenberg uncertainty principle.

Proof of the inequality

If one of the vectors is the zero vector , then the Cauchy-Schwarz inequality is trivially fulfilled. The following evidence is therefore sometimes without notice and provided. ${\displaystyle x\neq 0}$ ${\displaystyle y\neq 0}$

Special case of real standard scalar product

Proof from the inequality of the arithmetic and geometric mean

A proof of the Cauchy-Schwarz inequality can be done, for example, with the help of the inequality of the arithmetic and geometric mean :

Defined for the values ${\displaystyle i=1,\dots ,n}$

${\displaystyle \xi _{i}:={\frac {|x_{i}|}{\sqrt {\sum _{j}x_{j}^{2}}}}}$  and  ${\displaystyle \eta _{i}:={\frac {|y_{i}|}{\sqrt {\sum _{j}y_{j}^{2}}}},}$

thus the relation results from the inequality of the arithmetic and geometric mean

${\displaystyle \sum _{i}\xi _{i}\eta _{i}=\sum _{i}{\sqrt {\xi _{i}^{2}\eta _{i}^{2}}}\leq \sum _{i}\left({\frac {\xi _{i}^{2}}{2}}+{\frac {\eta _{i}^{2}}{2}}\right)=1}$

The Cauchy-Schwarz inequality follows directly from this.

Proof from the rearrangement inequality

Another proof of the Cauchy-Schwarz inequality results from the rearrangement inequality . If you set

${\displaystyle S={\sqrt {\sum _{i}x_{i}^{2}}}}$ and ${\displaystyle T={\sqrt {\sum _{i}y_{i}^{2}}}}$

as well as and so applies ${\displaystyle \xi _{i}={\tfrac {x_{i}}{S}}}$ ${\displaystyle \xi _{n+i}={\tfrac {y_{i}}{T}}}$

${\displaystyle 2=\sum _{i=1}^{n}{\frac {x_{i}^{2}}{S^{2}}}+\sum _{i=1}^{n}{\frac {y_{i}^{2}}{T^{2}}}=\sum _{i=1}^{2n}\xi _{i}^{2}.}$

Because of the rearrangement inequality, now

${\displaystyle \sum _{i=1}^{2n}\xi _{i}^{2}\geq \xi _{1}\xi _{n+1}+\xi _{2}\xi _{n+2}+\dots +\xi _{n}\xi _{2n}+\xi _{n+1}\xi _{1}+\xi _{n+2}\xi _{2}+\dots +\xi _{2n}\xi _{n}.}$

In summary, you get

${\displaystyle 2\geq {\frac {2\sum _{i=1}^{n}x_{i}y_{i}}{ST}}.}$

This results in the Cauchy-Schwarz inequality.

General scalar product

The proofs given above only prove the special case of the Cauchy-Schwarz inequality for the standard scalar product im . However, the proof for the general case of the scalar product in a vector space with inner product is simple. ${\displaystyle \mathbb {R} ^{n}}$

Real case

Under the condition applies . Applies to everyone ${\displaystyle y\neq 0}$ ${\displaystyle \langle y,y\rangle \neq 0}$ ${\displaystyle \lambda \in \mathbb {R} }$

${\displaystyle 0\leq \langle x-\lambda y,x-\lambda y\rangle =\langle x-\lambda y,x\rangle -\lambda \langle x-\lambda y,y\rangle =\langle x,x\rangle -2\lambda \langle x,y\rangle +\lambda ^{2}\langle y,y\rangle .}$

If you now choose specifically this results ${\displaystyle \lambda :={\tfrac {\langle x,y\rangle }{\langle y,y\rangle }}=\langle x,y\rangle \cdot \|y\|^{-2}}$

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

so

${\displaystyle \langle x,y\rangle ^{2}\leq \|x\|^{2}\|y\|^{2}.}$

Taking the square root now gives exactly the Cauchy-Schwarz inequality

${\displaystyle {\big |}\langle x,y\rangle {\big |}\leq \|x\|\|y\|.}$

Complex case

The proof in the complex case is similar, but it should be noted that the scalar product in this case is not a bilinear form , but a Hermitian form . The proof is given for the variant linear in the first and semilinear in the second argument; if the reverse variant is chosen, the complex conjugate is to be used at the appropriate points .

Is , the statement is clear. Be . Applies to everyone ${\displaystyle y=0}$ ${\displaystyle y\neq 0}$ ${\displaystyle \lambda \in \mathbb {C} }$

${\displaystyle 0\leq \langle x-\lambda y,x-\lambda y\rangle =\langle x-\lambda y,x\rangle -{\overline {\lambda }}\langle x-\lambda y,y\rangle =\langle x,x\rangle -\lambda \langle y,x\rangle -{\overline {\lambda }}\langle x,y\rangle +{\big |}\lambda {\big |}^{2}\langle y,y\rangle .}$

Here are the special election results on ${\displaystyle \lambda ={\tfrac {\langle x,y\rangle }{\langle y,y\rangle }}={\langle x,y\rangle }\cdot \|y\|^{-2}={\overline {\langle y,x\rangle }}\cdot \|y\|^{-2}}$

${\displaystyle 0\leq \langle x,x\rangle -\lambda \cdot {\overline {\lambda }}\|y\|^{2}-{\overline {\lambda }}\cdot \lambda \,\|y\|^{2}+{\big |}\lambda {\big |}^{2}\langle y,y\rangle =\|x\|^{2}-{\big |}\langle x,y\rangle {\big |}^{2}\cdot \|y\|^{-2},}$

so

${\displaystyle {\big |}\langle x,y\rangle {\big |}^{2}\leq \|x\|^{2}\|y\|^{2}.}$

Here, semilinearity was assumed in the second argument and linearity in the first argument. In the other case one uses ${\displaystyle \lambda ={\frac {\langle y,x\rangle }{\langle y,y\rangle }}.}$

Generalization for positive semidefinite, symmetric bilinear forms

The proof of the theorem can be reformulated in such a way that the positive definiteness of the scalar product is not used. The statement therefore also applies to every positive, semidefinite, symmetrical bilinear form (or Hermitian sesquilinear form ) . ${\displaystyle b}$

Proof for the real case

One takes the same approach as in the proof using the scalar product, but here the choice is made

${\displaystyle \lambda ={\frac {b(x,y)}{b(y,y)+\varepsilon }}.}$

This means that you no longer have to demand that it is not 0. That makes ${\displaystyle b(y,y)}$

${\displaystyle 0\leq b(x-\lambda y,x-\lambda y)=b(x,x)-2\lambda b(x,y)+\lambda ^{2}b(y,y).}$

Similar to the proof above, one concludes

${\displaystyle 2b(x,y)^{2}-b(x,y)^{2}{\frac {b(y,y)}{b(y,y)+\varepsilon }}\leq b(x,x)(b(y,y)+\varepsilon ).}$

and the assertion is shown when converging to zero. For follows . ${\displaystyle \varepsilon }$ ${\displaystyle b(y,y)=0}$ ${\displaystyle b(x,y)=0}$

Conditions for equality

Here, too, the situation is conceivable that the inequality becomes an equality, for example if (as with the scalar product) are linearly dependent. However, cases are also conceivable where equality occurs without there being a linear dependency. Consider a degenerate bilinear form . Then there is a such that is for all of the vector space . Now let us be arbitrary from the vector space. You then get ${\displaystyle x,y}$ ${\displaystyle b}$ ${\displaystyle x\neq 0}$ ${\displaystyle y}$ ${\displaystyle b(x,y)=0}$ ${\displaystyle y}$

${\displaystyle |b(x,y)|^{2}=0}$

and

${\displaystyle b(x,x)\,b(y,y)=0\cdot b(y,y)=0,}$

so

${\displaystyle |b(x,y)|^{2}=b(x,x)\,b(y,y),}$

also for the case that and are linearly independent. ${\displaystyle x}$ ${\displaystyle y}$

swell

1. ↑ Cauchy, Augustin-Louis. Analysis algébrique, page 455f
2. ↑ VI Bityutskov: Bunyakovskii inequality . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).
3. ↑ Eric W. Weisstein : Schwarz's Inequality . In: MathWorld (English).

Web links

Wikibooks: Proof of the Cauchy-Schwarz Inequality  - Learning and Teaching Materials

• Earliest Uses of Some of the Words of Mathematics: Cauchy-Schwarz inequality.

literature

• Peter Schreiber: The Cauchy-Bunyakovsky-Schwarz inequality , in: Ders .: Hermann Grassmann, work and effect. International conference on the occasion of the 150th anniversary of the first publication of the "linear expansion theory" , University of Greifswald, 1995, pp. 64–70