# Properties of Semi-Inner Product

## Theorem

Let $V$ be a vector space over $\Bbb F \in \set {\R, \C}$.

Let $\innerprod \cdot \cdot$ be a semi-inner product on $V$.

Denote, for $x \in V$, $\norm x := \innerprod x x^{1 / 2}$.

Then, $\forall x, y \in V, a \in \Bbb F$:

$(1): \quad \norm {x + y} \le \norm x + \norm y$
$(2): \quad \norm {a x} = \size a \norm x$

## Proof

### Proof of $(1)$

For $x, y \in V$, compute:

 $\ds \norm {x + y}^2$ $=$ $\ds \innerprod {x + y} {x + y}$ Definition of $\norm \cdot$ $\ds$ $=$ $\ds \innerprod x x + \innerprod x y + \innerprod y x + \innerprod y y$ Linearity of $\innerprod \cdot \cdot$ $\ds$ $\le$ $\ds \innerprod x x + \sqrt {\innerprod x x \innerprod y y} + \sqrt {\innerprod y y \innerprod x x} + \innerprod y y$ Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces $\ds$ $=$ $\ds \norm x^2 + 2 \norm x \norm y + \norm y^2$ Definition of $\norm \cdot$ $\ds$ $=$ $\ds \paren {\norm x + \norm y}^2$

Taking square roots on either side gives the result.

$\Box$

### Proof of $(2)$

For $x \in V$, $a \in \Bbb F$, compute:

 $\ds \norm {a x}^2$ $=$ $\ds \innerprod {a x} {a x}$ Definition of $\norm \cdot$ $\ds$ $=$ $\ds a \innerprod x {a x}$ Linearity of $\innerprod \cdot \cdot$ $\ds$ $=$ $\ds a \overline {\innerprod {a x} x}$ Conjugate symmetry of $\innerprod \cdot \cdot$ $\ds$ $=$ $\ds a \overline a \overline {\innerprod x x}$ Linearity of $\innerprod \cdot \cdot$ $\ds$ $=$ $\ds \size a^2 \norm x^2$ Modulus in Terms of Conjugate

Taking square roots on either side gives the result.

$\blacksquare$