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Recently I've learn about the Cauchy-Schwarz inequality. So, as we all know this this inequality states that for an inner product vector space for all vectors $x, y$ we have that:

$|\langle x,y\rangle|\le ||x||\text{ }||y||$.

The proof of this inequality is well known but I have trouble with understanding something. I mean if this inequality works for any inner product vector space we just have to proof it once, yes? What I mean is that for different vector spaces this inequality takes different form but these are just special cases, right? If we proved that this inequality works in any inner product vector space then we, know that it works for example in $\mathbb{R}^n$ with dot product right? This is just special case of general result so we don't need a special proof, right? So all the proofs that can be found for example here Proofs of the Cauchy-Schwarz Inequality? for the lack of a better world are made for "fun"?

Do I get it right?

Alice211
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2 Answers2

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Yes you only have to prove it once for a vector space with inner product, as most proofs don't use any specific properties of such specific vector spaces (for example https://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality). Although, it can be intresting to look at the inequality in certain vector spaces with inner product, as you might be able to simplify it or rewrite it in a more usefull way (when your trying to solve some certain exercise)

algebrah
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Yes, you get it right. What we prove here is that the inequality holds in every inner product vector space. The reason why it is enough to prove it only "once" is because you're only making use of the properties of inner product vector spaces (which on the other hand is the only thing you can do, because it is the definition you are given), so every space you can think of with this property will, as you say, be a particular case of what has been proven.

This happens all the time in mathematics. You don't have to prove every single particular case you come up with of a certain statement when you find a general proof for it.

And yes, as it happens here, one can sometimes find alternative proofs for the same result.

salomi
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