Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 2
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
This proof assumes that $V$ is a semi-inner product space over $\R$.
Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of semi-inner products.
Define $f_{x, y}: \R \to \R_{\ge 0}$ by:
- $\map {f_{x, y} } \lambda = \innerprod {x - \lambda y} {x - \lambda y}$
Then by property $(4)$ of semi-inner product:
- $\forall \lambda \in \R: \map {f_{x, y} } \lambda \ge 0$
For all $\lambda \in \R$, it follows that:
\(\ds \map {f_{x, y} } \lambda\) | \(=\) | \(\ds \innerprod {x - \lambda y} {x - \lambda y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}\) | Property $(3)$ of Semi-Inner Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x x - 2 \lambda \innerprod x y + \lambda^2 \innerprod y y\) | Properties $(1')$ and $(2)$ of Semi-Inner Product | |||||||||||
\(\ds \) | \(=\) | \(\ds a \lambda^2 + b \lambda + c\) |
where we have put $a = \innerprod y y$, $b = -2 \innerprod x y$, and $c = \innerprod x x$.
Then $f_{x, y}$ is a quadratic polynomial which satisfies $\map {f_{x, y} } \lambda \ge 0$.
Hence $f_{x, y}$ has at most one distinct real root.
From Solution to Quadratic Equation, it follows that the discriminant $\Delta$ satisfies:
- $\Delta = b^2 - 4 a c \le 0$
Therefore:
- $4 \innerprod x y^2 - 4 \innerprod x x \innerprod y y \le 0$
which we rearrange as:
- $\size {\innerprod x y}^2 = \innerprod x y^2 \le \innerprod x x \innerprod y y$
$\blacksquare$