# Shae's Ramblings

Stuff I find slightly meaningful and close to achievable

## A well-known proof of cauchy-schwartz

Suppose $\langle\cdot, \cdot\rangle$ is any real-valued inner product, be it on a space $V$ of vectors, matrices, continuous functions, or even Sobolev spaces.
We can prove using simple algebra that we always have

$$\lvert \langle\phi, \psi\rangle \vert \leq \lVert \phi \rVert\,\lVert \psi \rVert$$

for any two elements $\phi, \psi$ in $V$. The norm $\lVert \cdot \rVert$ is the usual inner-product associated norm $\lVert \phi \rVert = {\langle\phi,\phi\rangle}^{1 / 2}$

## The algebraïc lemma

It is well-known that the quadratic $q(x) \triangleq ax^2 + bx + c$ with can be factorized as

$$q(x) = a\left(x + \dfrac{b}{2a}\right)^2 – \dfrac{b^2 – 4ac}{4a}$$

therefore, the equation $q(x) = 0$ has no solution if and only if $b^2 – 4ac < 0$.

## The proof

Let us assume $\phi \neq \lambda\psi$ and compute $\langle\phi – \lambda\psi, \phi – \lambda\psi\rangle > 0$, we get the inequality

$$\lambda^2\langle\psi, \psi\rangle – 2\lambda\langle\phi, \psi\rangle + \langle\phi, \phi\rangle$$
which is a strictly positive quadratic form in the variable $\lambda$.
Using our lemma, we therefore must have
$$4\langle\phi, \psi\rangle^2 < 4\langle\psi, \psi\rangle\langle\phi, \phi\rangle$$
this yields the cauchy-schwartz inequality which turns into an equality when $\phi = \lambda\psi$.