Pythagorean theorem does require the Cauchy-Schwarz inequality?

Question

In Wikipedia I found a proof of Schwarz inequality making use of the Pythagorean Theorem. Anyway I'm wondering... isn't that a petitio principii? Isn't the Pythagorean Theorem in Hilbert spaces proved by the Cauchy-Schwarz inequality? Or effectively is the Pythagorean Theorem a stronger if not equivalent notion?

Answer 1

$\newcommand{\ip}[2]{\left\langle #1,#2\right\rangle}$ $\newcommand{\norm}[1]{\left|\left| #1\right|\right|}$

I think your confusion is due to notation used in Pythagorean theorem: we don't actually require any norm-property of $||\cdot||$ induced by the inner product in proving it. It's only after we recognise $||\cdot||$ as a norm that it has its usual realization.

Consider an inner product space $V$ over real field for simplicity.

Let's denote

$$ \norm{x}^2=\ip{x}{x} $$

The important thing to realize here is that although notation $||\cdot||$ is used, at this stage we don't recognize $||\cdot||$ as a "norm". It's just a convenient notation to denote $\langle x,x\rangle$.

Suppose now that $x,y$ are orthogonal. We prove the "Pythagorean Theorem"

Claim : If $x,y$ are orthogonal then $$ ||x+y||^2=||x||^2+||y||^2 $$ Proof : $$\ip{x+y}{x+y}= \norm{x}^2+\norm{y}^2+2\ip{x}{y}=\norm{x}^2+\norm{y}^2$$ by orthogonality, bilinearlity.

So that didn't require CS inequality.

So this implies CS inequality, and this CS inequality ensures that the function $\norm{\cdot}$ is indeed a norm. In short,

\begin{align*} IP &\implies \text{Pythagorean Theorem} \implies \text{Cauchy-Schwarz} \\ &\implies \text{"norm function" is indeed a norm} \end{align*}