Review inner product: an inner product \(\langle x, y \rangle \to \mathbb{R}\) if and only if it satisfies
\(\langle x, y \rangle = \langle y, x \rangle\)
\(\langle x, x \rangle \ge 0\)
\(\langle x, x \rangle = 0 \quad \implies \quad x = 0\). Confusion: do we really need this?
\(\langle x, \alpha y + \beta z \rangle = \alpha \langle x, y \rangle + \beta \langle x, z \rangle\)
\(\langle \alpha x + \beta y, z \rangle = \alpha \langle x, z \rangle + \beta \langle y, z \rangle\)
Cauchy Inequality
\[\vert \langle x, y \rangle \vert \le \sqrt{\langle x, x\rangle \langle y, y \rangle}\]Proof
\[g(\lambda) = \langle \lambda x + y, \lambda x + y \rangle = \lambda^2 \langle x,x \rangle + 2 \lambda \langle x, y \rangle + \langle y, y \rangle \ge 0\]the above we use proerty 1, 2, 4, 5.
if \(\langle x, x \rangle = 0\), then \(\langle x, y \rangle\) must be 0. Otherwise we can take \(\lambda = \dfrac{-1 + \langle y, y \rangle}{2 \langle x , y \rangle}\) such that \(g(\lambda) = -1\).
When \(\langle x, x \rangle = 0\), \(0 \le 0\) of course holds.
When \(\langle x, x \rangle \ne 0\), \(g(\lambda)\) is a quadratic equation with 1 or zero real root. Thus its discriminate \(b^2 - 4 ac < 0\)
\[(2\langle x, y \rangle )^2 - 4 \langle x, x \rangle \langle y, y \rangle \le 0\]it gives
\[\vert \langle x, y \rangle \vert \le \sqrt{\langle x, x\rangle \langle y, y \rangle}\]