Properties of Semi-Inner Product

Theorem

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

Let $\left \langle{\cdot, \cdot}\right \rangle$ be a semi-inner product on $V$.

Denote, for $x \in V$, $\left\Vert{x}\right\Vert := \left\langle{x,x}\right\rangle^{1/2}$.

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

$(1): \quad \left\Vert{x + y}\right\Vert \le \left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert$

$(2): \quad \left\Vert{a x}\right\Vert = \left|{a}\right| \left\Vert{x}\right\Vert$

Proof

Proof of $(1)$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\Vert{x + y}\right\Vert^2\)	\(=\)	\(\displaystyle \left\langle{x + y, x + y}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of $\left\Vert{\cdot}\right\Vert$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\langle{x, x}\right\rangle + \left\langle{x, y}\right\rangle + \left\langle{y, x}\right\rangle + \left\langle{y, y}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Linearity of $\left\langle{\cdot, \cdot}\right\rangle$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\Vert{x}\right\Vert^2 + \sqrt{\left\langle{x, x}\right\rangle \left\langle{y, y}\right\rangle} + \sqrt{\left\langle{y, y}\right\rangle \left\langle{x, x}\right\rangle} + \left\Vert{y}\right\Vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cauchy-Schwarz Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\Vert{x}\right\Vert^2 + 2 \left\Vert{x}\right\Vert \left\Vert{y}\right\Vert + \left\Vert{y}\right\Vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of $\left\Vert{\cdot}\right\Vert$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Taking square roots on either side gives the result.

$\Box$

Proof of $(2)$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\Vert{a x}\right\Vert^2\)	\(=\)	\(\displaystyle \left\langle{a x, a x}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of $\left\Vert{\cdot}\right\Vert$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \left\langle{x, a x}\right\rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Linearity of $\left\langle{\cdot, \cdot}\right\rangle$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \overline{\left\langle{a x, x}\right\rangle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Conjugate symmetry of $\left\langle{\cdot, \cdot}\right\rangle$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \overline a \overline{\left\langle{x, x}\right\rangle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Linearity of $\left\langle{\cdot, \cdot}\right\rangle$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{a}\right\vert^2 \left\Vert{x}\right\Vert^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Modulus in Terms of Conjugate

Taking square roots on either side gives the result.

$\blacksquare$

Sources

• John B. Conway: A Course in Functional Analysis (1990)... (previous)... (next) $I.1.5$