Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

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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 Definite 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

The Cauchy-Bunyakovsky-Schwarz Inequality in its various form is also known as:

the Cauchy-Schwarz-Bunyakovsky Inequality
the Cauchy-Schwarz Inequality
Schwarz's Inequality or the Schwarz Inequality
Bunyakovsky's Inequality or Buniakovski's Inequality.

For brevity, it is sometimes referred to by the abbreviations CS inequality or CBS inequality.


Source of Name

This entry was named for Augustin Louis Cauchy, Karl 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

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