Properties of Semi-Inner Product

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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:

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

Taking square roots on either side gives the result.

$\Box$


Proof of $(2)$

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

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

Taking square roots on either side gives the result.

$\blacksquare$


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