Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$ be $\mu$-square integrable functions, that is $f, g \in \map {\LL^2} \mu$, Lebesgue $2$-space.
Then:
- $\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$
where $\norm {\, \cdot \,}_2$ is the $2$-norm.
Equality
Equality in the above, that is:
- $\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$
holds if and only if for almost all $x \in X$:
- $\dfrac {\size {\map f x}^2} {\norm f_2^2} = \dfrac {\size {\map g x}^2} {\norm g_2^2}$
Proof
Follows directly from Hölder's Inequality for Integrals with $p = q = 2$.
$\blacksquare$
Also known as
This theorem is also known as the Cauchy-Schwarz Inequality.
Source of Name
This entry was named for Augustin Louis Cauchy, Karl Hermann Amandus Schwarz and Viktor Yakovlevich Bunyakovsky.
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Cauchy-Schwarz inequality: $(1)$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $12.3$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Cauchy-Schwarz inequality: $(1)$