# Vector space, norm and metric

This is a follow up question to my previous question.

It is perfectly clear to me what a metric space is. Basically, one can see that notions like convergence and continuity only depend on some sort of closeness or distance, so we can define a distance and then talk about these things.

Now it might happen that we have a vector space and we would like to talk about these concepts. Now in some cases we want the metric to respect the vector space operations because otherwise we can't really make use of them.

This means that the distance should have some additional properties: $$(i) d(w,v)=d(w+u,v+u)$$ $$(ii) d(\alpha u,\alpha v)=\lvert\alpha\rvert d(u,v)$$

In most cases we introduce a norm which in turn induces a metric with these properties. The norm itself can be seen as a generalization of length, size or magnitude, so it is also an inuitive concept.

Now I've got 2 questions:

1)I am wondering if the only reason to introduce a norm is to be able to talk about convergence and continuity or are there other reasons why one would want a generalization of length for arbitrary vector spaces. Are there other things we can do once we have a norm? If yes, please provide an example. If not why don't we just define a norm as a metric with the additional properties (i) and (ii). I am guessing it´s because the properties of the norm are sufficient and we want our definitions to be short.

2) What is the intuition for the triangle inequality for an arbitrary norm in any vector space? In $$\mathbb{R^{n}}$$ it makes sense to require that the length of a sum of two vectors is smaller than the sum of the length since the directions might be different. For example, why should this be true if I have some norm for a function space? For a distance this is intuitive since it basically says that the shortest path between 2 points is a line between the points.

Thanks very much!

Edit:

Regarding 2) it might make sense to view the norm as a generalization of the absolute value rather than the length in $$\mathbb{R^{n}}$$. Intuitively the absolute value gives the magnitude of a number. The triangle inequality can be seen as the property of subadditivity. In higher dimensions it then makes sense to say the magnitude of a vector is its length. Of course it satisfies the properties of a norm.

• I think you are using the word "generalisation" back to front – Ali Feb 12 '20 at 11:22
• I am not sure what you mean. But I've just corrected an error. I meant to say "The norm itself can be seen as a generalization of length, size or magnitude...". – DerivativesGuy Feb 12 '20 at 11:27
• I've only said that a norm induces a metric. The norm itself is not a metric and as I've pointed out it generalizes the notion of length, size or magnitude while a metric is a generalization of distance. I see what you mean, but I am interested in the norm viewed as generalization of length. – DerivativesGuy Feb 12 '20 at 11:32
• Sorry I think you had said that the norm was a generalisation of the distance – Ali Feb 12 '20 at 11:33
• I guess, condition (ii) would go like $d(\alpha u,\, \alpha v) =\alpha\cdot d(u,v)$ for $\alpha>0$, right? Is it clear that a metric on an $\Bbb R$-vector space is induced by a norm iff (i) and (ii) hold? – Berci Feb 12 '20 at 11:40

## 1 Answer

(1) Well, apart from convergence and continuity, it allows us to talk about completeness. Complete normed spaces are called Banach spaces and are vastly important. But we can talk about completeness in the general setting of a metric space, so this doesn't answer your question. My guess for defining norms as we do, instead of requiring structure-preserving properties out of a metric, is that norms themselves can be induced out of inner products in a natural way: in some way, norms are at the "structural midpoint" between a purely topological notion (a metric) and a purely linear one (an inner product).

(2) One reason for requiring that metrics (and thus norms) satisfy the triangle inequality is that, otherwise, you can't really deal with convergence in the way we are used to, and this would make the whole edifice of analysis come crumbling down. Convergent sequences wouldn't be Cauchy, so completeness would be harder to define. Convergent sequences wouldn't be bounded. Differentiable functions wouldn't be continuous. The Banach fixed point theorem would not hold, so you wouldn't be able to prove ODE solution uniqueness. From a theoretical point of view, it would be a mess.

This is ultimately because, out of all the axioms in the definition of a metric, the triangle inequality is the one encoding the notion of "nearness" required for the basic idea of convergence.

• Thanks for your answer. Actually, I am looking at inner products as well now, but I wanted to clear up my understanding about norms first. So it seems like you agree that we only define norms because it is useful and because of their relation to inner products? I haven't quiet figured out what an inner product does. Usually I try to understand it in Euclidean space first, but so far I haven't understood the intuition behind the fact that the Euclidean norm is the square root of the scalar product. – DerivativesGuy Feb 12 '20 at 12:10
• As for the triangle inequality I know that it's needed for convergence and all that and if I think in terms of distance then all of it makes sense, but once I think of magnitude or size as a generalization of the Euclidean norm then my intuition for the triangle inequality kind of breaks down. In Euclidean space length must be subadditive because two vectors can have different directions, but I can't seem to make sense of this in more general spaces since there is not really a direction. – DerivativesGuy Feb 12 '20 at 12:12