Cauchy-Bunyakovsky-Schwarz Inequality

= Inner Product Spaces =

(AKA: Schwartz inequality or Cauchy–Bunyakovsky–Schwarz inequality)

Theorem
Let $$V$$ be an inner-product space over $$\mathbb{K}$$ where $$\mathbb{K} = \R$$ or $$\C$$.

Let $$x, y$$ be vectors in $$V$$.

Then $$\left|{\left \langle {x, y} \right \rangle}\right|^2 \leq \left\|{x}\right\| \times \left\|{y}\right\|$$.

Proof
Let $$\lambda \in \mathbb{K}$$. Since an inner-product is generated by a norm on the underlying normed linear space we may expand as follows:

$$ $$ $$ $$

where $$\lambda^*$$ is the complex conjugate of $$\lambda$$.

(If $$\mathbb{K} = \R$$, then $$\lambda^* = \lambda$$.)

If we let $$\lambda = \left \langle {x, y} \right \rangle \times \left \langle {y, y} \right \rangle^{-1}$$ then we obtain:

$$0 \le \left \langle {x, x} \right \rangle - \left|{\left \langle {x, y} \right \rangle}\right|^2 \times \left \langle {y, y} \right \rangle^{-1} $$

Solving this for $$\left|{\left \langle {x, y} \right \rangle}\right|^2 $$, we see that

$$\left|{\left \langle {x, y} \right \rangle}\right|^2 \le \left \langle {x, x} \right \rangle * \left \langle {y, y} \right \rangle = \left\|{x}\right\| \times \left\|{y}\right\|$$

as desired.

= Cauchy's Inequality =

The special case of the Cauchy-Schwarz Inequality in a Euclidean space is called Cauchy's Inequality. It was Cauchy who first published this result in 1821.

It is usually stated as:


 * $$\sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$$

where all of $$r_i, s_i \in \R$$.

It is proved here.

= Definite Integrals =

The version for integrals was first stated by Bunyakovsky in 1859, and later rediscovered by Schwarz in 1888.

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.