Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 1

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Theorem

Let $\mathbb K$ be a subfield of $\C$.

Let $V$ be a semi-inner product space over $\mathbb K$.

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


Then:

$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$


Proof

Let $x, y \in V$.

Let $\lambda \in \mathbb K$.

Then:

\(\ds 0\) \(\le\) \(\ds \innerprod {x - \lambda y} {x - \lambda y}\) Semi-inner product axioms $\paren {3}$: Non-Negative Definiteness
\(\ds \) \(=\) \(\ds \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}\) Semi-inner product axioms $\paren {2}$: Sesquilinearity
\(\ds \) \(=\) \(\ds \innerprod x x - \lambda^* \innerprod x y - \lambda \innerprod x y^* + \lambda \lambda^* \innerprod y y\) Semi-inner product axioms $\paren {2}$ and $\paren {1}$

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

(If $\mathbb K$ is a subfield of $\R$ , then $\lambda^* = \lambda$.)


First, suppose $\innerprod y y \ne 0$.

Insert $\lambda = \innerprod x y \innerprod y y^{-1}$ in the inequality:

\(\ds 0\) \(\le\) \(\ds \innerprod x x - \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod x y - \innerprod x y \innerprod y y^{-1} \innerprod x y^* + \innerprod x y \innerprod y y^{-1} \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod y y\) Product of Complex Conjugates
\(\ds \) \(=\) \(\ds \innerprod x x - \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^* - \size {\innerprod x y}^2 \innerprod y y^{-1} + \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^*\) Modulus in Terms of Conjugate
\(\ds \) \(=\) \(\ds \innerprod x x - \size {\innerprod x y}^2 \innerprod y y^{-1}\)


Reorder the inequality to get:

$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$

$\Box$


Next, suppose $\innerprod y y = 0$.

By Number Field has Rational Numbers as Subfield, $\Q \subset \mathbb K$.

Let $n \in \N \subset \Q \subset \mathbb K$ be a natural number.

Insert $\lambda = n \innerprod x y$ in the inequality:

\(\ds 0\) \(\le\) \(\ds \innerprod x x - n^* \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^* + 0\) Product of Complex Conjugates
\(\ds \) \(=\) \(\ds \innerprod x x - n \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^*\) Complex Number equals Conjugate iff Wholly Real
\(\ds \) \(=\) \(\ds \innerprod x x - 2 n \size {\innerprod x y}^2\) Modulus in Terms of Conjugate


Rearrange the inequality to get:

$\innerprod x x \ge 2 n \size {\innerprod x y}^2$

If $\size {\innerprod x y}^2 \ne 0$, then:

$\dfrac {\innerprod x x} {2 \size {\innerprod x y}^2} \ge n$

for every $n \in \N$, contradicting the Axiom of Archimedes.

Therefore, by Proof by Contradiction:

$\size {\innerprod x y}^2 = 0$.

Then:

$\size {\innerprod x y}^2 \le 0 = \innerprod x x \innerprod y y$

$\blacksquare$

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