Abraham Bradley

2022-05-03

Pythagorean theorem does require the Cauchy-Schwarz inequality?

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?

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?

Landyn Whitney

Beginner2022-05-04Added 19 answers

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

${\left|\left|x\right|\right|}^{2}=\u27e8x,x\u27e9$

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 $\u27e8x,x\u27e9$.

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 :

$\u27e8x+y,x+y\u27e9={\left|\left|x\right|\right|}^{2}+{\left|\left|y\right|\right|}^{2}+2\u27e8x,y\u27e9={\left|\left|x\right|\right|}^{2}+{\left|\left|y\right|\right|}^{2}$

by orthogonality, bilinearlity.

So that didn't require CS inequality.

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

$\begin{array}{rl}IP& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{Pythagorean Theorem}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{Cauchy-Schwarz}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{"norm function" is indeed a norm}\end{array}$

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

Let's denote

${\left|\left|x\right|\right|}^{2}=\u27e8x,x\u27e9$

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 $\u27e8x,x\u27e9$.

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 :

$\u27e8x+y,x+y\u27e9={\left|\left|x\right|\right|}^{2}+{\left|\left|y\right|\right|}^{2}+2\u27e8x,y\u27e9={\left|\left|x\right|\right|}^{2}+{\left|\left|y\right|\right|}^{2}$

by orthogonality, bilinearlity.

So that didn't require CS inequality.

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

$\begin{array}{rl}IP& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{Pythagorean Theorem}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{Cauchy-Schwarz}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{"norm function" is indeed a norm}\end{array}$

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