Properties of Semi-Inner Product

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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.

$\blacksquare$


Sources

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