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I am trying to understand that the Euclidean norm $\|x\|_2 = \left(\sum|x_i|^2\right)^{1/2}$ is in fact a norm and having trouble with the triangle inequality.

All the proofs I have referred to involve the Cauchy-Schwarz inequality. But it seems that this inequality is proved in an inner product space, which has additional properties to a normed space.

So, my question is whether starting with any (possibly infinite dimensional) vector space over $\mathbb{C}$ and taking any algebraic basis for it, can the triangle inequality be proved for the Euclidean norm without making assumptions about an inner product or an orthonormal basis ?

(I don't think that infinite dimensionality should be a problem as any two vectors have finite representations in an algebraic basis).

Addendum after 2 answers and comments.

Can one take the Cauchy-Schwarz inequality "out of context" as an algebraic statement about two finite lists $(x_i) $ and $(y_i)$ and then apply it to the complex coefficients of any algebraic basis to say that $\sum \left|x_iy_i^*\right|\leq (\sum|x_i|^2)^{1/2} (\sum|y_i|^2)^{1/2}$ and then complete the proof of the triangle inequality ?

Tom Collinge
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  • The Euclidean norm you have described is the norm associated to the inner product space where the inner product is given by $\left<x,y\right>=\sum x_iy_i$. – John Gowers Oct 14 '15 at 09:40
  • @Donkey_2009. Yes, that's the essence of my question: Can you prove that $(V, |.|_2 )$ is a normed space for any algebraic basis without having to establish an orthonormal basis for an inner product ? – Tom Collinge Oct 14 '15 at 09:48
  • As an answer to your addendum, I don't see why not. Wikipedia has a proof of this inequality. – Tunococ Oct 14 '15 at 10:39
  • @TomCollinge Well, yes, but I don't see why you'd want to. Defining the inner product in this case is easy to do, and the proof using the C-S inequality then has a natural geometric interpretation. – John Gowers Oct 14 '15 at 11:27
  • @Donkey_2009 I was taking the view that a plain vector space, a normed space and an inner product space have increasing structure, and attempting to associate various properties at the lowest necessary level . Having established the Euclidean norm as a valid norm, one can go on to show it is also a foundation for an inner product. – Tom Collinge Oct 14 '15 at 11:55
  • @Tom Collinge I don't get what you mean by 'foundation for an inner product'. Rather, the Euclidean norm is the norm associated to the standard inner product on $\mathbb R^n$. Though, 'inner product space' is a strictly stronger notion than 'normed space', it's not a simple matter of 'An inner product space is a normed space with extra structure put on it', because in this case the norm comes from the inner product. – John Gowers Oct 14 '15 at 13:43
  • @Donkey_2009. By 'foundation for an inner product' I mean that the Euclidean norm satisfies the parallelogram identity (I would have said "basis" for an inner product, but that term is far too confusing). As far as I can see, I can define this norm in the 'weaker' normed space without considering inner products, and then go on to show that there is an inner product associated with it (rather than starting from the 'stronger' inner product and going the other way). – Tom Collinge Oct 14 '15 at 13:52
  • Not on first sight, but this is relevant: https://math.stackexchange.com/q/247425/96384. Via choice of basis, any (real or complex) vector space can not only be given your "Euclidean" norm, but in fact that "standard" inner product, which induces that "Euclidean" norm. Of course can insist on proving the triangle inequality without making use of that stronger construction if you feel so inclined. – Torsten Schoeneberg Nov 30 '21 at 18:23

2 Answers2

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Hint:

Note that the Euclidean norm is a particular case of a $p$-norm and for these norms the triangle inequality can be proved using the Minkowky inequality.

Anyway, the Euclidean norm is the only $p$-norm that satisfies the parallelogram identity ( see: Determining origin of norm), so it is coming from an inner product.

About the addendum.

In an $n$ dimensional real space we can prove the C-S inequality with simply algebraic methods (see here). So, yes, in this case we can proof the triangle inequality without explicitly using an inner product space.

Community
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Emilio Novati
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  • Thanks, but don't you have to prove it's a norm before you can consider the parallelogram identity ? – Tom Collinge Oct 14 '15 at 09:44
  • Good point!! I've changed my answer. – Emilio Novati Oct 14 '15 at 09:55
  • Thanks: I changed the question (added to it actually) - any opinion on my addendum ? – Tom Collinge Oct 14 '15 at 10:29
  • I've added something to my answer. – Emilio Novati Oct 14 '15 at 18:55
  • I think the proof extends to a Euclidean norm on an algebraic basis of an infinite dimensional space (not that such basis is necessarily easy to find) since CS relates two vectors, which will have finite representations in such a basis. But does the wiki reference cover complex coefficients ? I think that since any two lists of n complex numbers could represent vectors in $C^n$ then CS should apply to them: is this correct in your opinion ? – Tom Collinge Oct 14 '15 at 19:19
  • About the first question. Since two vectors stay always in a $2$-dimensional subspace, it seems reasonable that we can find a finite basis to represent them, but i don't easely see how find this basis without having as starting point an orthonormal basis. – Emilio Novati Oct 14 '15 at 19:33
  • About the second question. Note that $\mathbb{C}^n$ and $\mathbb{R}^{2n}$ are isomorphic as real vector spaces, but not as complex vector spaces (see http://math.stackexchange.com/questions/619298/is-mathbbc2-isomorphic-to-mathbbr4). – Emilio Novati Oct 14 '15 at 19:38
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Community wiki: Just to show the C-S proof. We have $$||x+y||_2^2=\sum|x_i+y_i|^2 = \sum \left|x_i^2+2x_iy_i+y_i^2\right|.$$ Then by the triangle inequality (over $|\cdot|$) $$\sum \left|x_i^2+2x_iy_i+y_i^2\right|\leq \sum|x_i|^2+2\sum|x_i||y_i|+\sum|y_i|^2.$$ But, by the Cauchy-Schwarz inequality, $$\sum|x_i|^2+2\sum|x_i||y_i|+\sum|y_i|^2\leq \sum|x_i|^2+2\left(\sum |x_i|^2\right)^{1/2}\left(\sum|y_i|^2\right)^{1/2}+\sum|y_i|^2.$$ But this is just equal to $$\left(\left(\sum |x_i|^2\right)^{1/2}+\left(\sum |y_i|^2\right)^{1/2}\right)^2=\left(||x||_2+||y||_2\right)^2.$$ Hence, $$||x+y||_2\leq ||x||_2+||y||_2.$$ Note

Concerning the comments on the triangle inequality using $|\cdot |$. There is a trivial proof: $$|a+b|^2=(a+b)^2=a^2+b^2+2ab=|a|^2+|b|^2+2ab\leq(|a|+|b|)^2\implies|a+b|\leq |a|+|b|.$$ Hence, $$|x^2+2xy+y^2|\leq|x|^2+|2xy+y^2|\leq|x|^2+|2xy|+|y|^2=|x|^2+2|x||y|+|y|^2.$$

pshmath0
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  • But you used triangle inequality here... – luka5z Oct 14 '15 at 09:33
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    @luka5z They used the triangle inequality for the absolute value, but not the 2-norm. – Santeri Oct 14 '15 at 09:40
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    Yes, that's correct - I've updated the proof to reflect this. – pshmath0 Oct 14 '15 at 09:41
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    Thanks for the response. Actually you use the CS inequality to prove the triangle inequality where you say that $\sum \left|2x_iy\right|\leq 2\sum|x_i||y_i|$ . The CS inequality is proved in an inner product space and question is about the validity of this without assuming an inner product space in the first place. – Tom Collinge Oct 14 '15 at 09:42
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    How does $\sum \left|2x_iy\right|\leq 2\sum|x_i||y_i|$ use CS? – luka5z Oct 14 '15 at 10:00
  • @luka5z. Yes you're correct, it's in the next line, but the answer now reflects this. The question remains the same - the CS inequality is proved in an inner product space and question is about the validity of this without assuming an inner product space in the first place - but I've added a possible solution as an addendum for feedback. – Tom Collinge Oct 14 '15 at 10:16
  • That's why I made this community wiki. Just though it might give some background. And exercise my little grey cells ! – pshmath0 Oct 15 '15 at 09:25