Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.


Then:

$\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map g t}^2 \rd t$


Proof

\(\ds \forall x \in \R: \ \ \) \(\ds 0\) \(\le\) \(\ds \paren {x \map f t + \map g t}^2\)
\(\ds 0\) \(\le\) \(\ds \int_a^b \paren {x \map f t + \map g t}^2 \rd t\) Relative Sizes of Definite Integrals
\(\ds \) \(=\) \(\ds x^2 \int_a^b \paren {\map f t}^2 \rd t + 2 x \int_a^b \map f t \, \map g t \rd t + \int_a^b \paren {\map g t}^2 \rd t\) Linear Combination of Integrals
\(\ds \) \(=\) \(\ds A x^2 + 2 B x + C\)

where:

\(\ds A\) \(=\) \(\ds \int_a^b \paren {\map f t}^2 \rd t\)
\(\ds B\) \(=\) \(\ds \int_a^b \map f t \map g t \rd t\)
\(\ds C\) \(=\) \(\ds \int_a^b \paren {\map g t}^2 \rd t\)

The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$.

It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\paren {2 B}^2 - 4 A C$ of this polynomial must be non-positive.

Thus:

$B^2 \le A C$

and hence the result.

$\blacksquare$


Also known as

This theorem is also known as the Cauchy-Schwarz inequality.

Some sources give it as the Cauchy-Schwarz-Bunyakovsky inequality.


Source of Name

This entry was named for Augustin Louis CauchyKarl Hermann Amandus Schwarz and Viktor Yakovlevich Bunyakovsky.


Historical Note

The Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals was first stated in this form by Bunyakovsky in $1859$, and later rediscovered by Schwarz in $1888$.


Sources