Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

Theorem
Let $$f$$ and $$g$$ be real functions which are continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Then:
 * $$\left({\int_a^b f \left({t}\right) g \left({t}\right) dt}\right)^2 \le \int_a^b \left({f \left({t}\right)}\right)^2 dt \int_a^b \left({g \left({t}\right)}\right)^2 dt$$.

Proof

 * $$\forall x: 0 \le \left({x f \left({t}\right) + g \left({t}\right)}\right)^2$$.

$$ $$ $$

where:
 * $$A = \int_a^b \left({f \left({t}\right)}\right)^2 dt$$;
 * $$B = \int_a^b f \left({t}\right) g \left({t}\right) dt$$;
 * $$C = \int_a^b \left({g \left({t}\right)}\right)^2 dt$$.

As the Quadratic Equation $$A x^2 + 2 B x + C$$ is positive for all $$x$$, it follows that (using the same reasoning as in Cauchy's Inequality) $$B^2 \le 4 A C$$.

Hence the result.

Alternative names
This theorem is also known as the Cauchy-Bunyakovsky-Schwarz Inequality.

It was first stated in this form by Bunyakovsky in 1859, and later rediscovered by Schwarz in 1888.