Shae's Ramblings

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 \(\phi = \lambda\psi\).