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I've already studied real analysis and I've just finished studying linear algebra (the source I've used did not cover norms, but I have some basic understanding about them).

Now I know that there are normed vector spaces and they have a lot of applications. From my understanding the reason for defining them is that it is a way to give a vector space some additional structure to be able to consider things like convergence and continuity. This is because a norm induces a metric, and therefore all the metric space theorems are applicable.

Now I've got two questions:

1) Although I can mathematically understand that a norm induces a metric and it also intuitively makes sense in euclidean spaces since the norm can be interpreted as length which makes the connection to the metric or distance obvious (We can just draw two vectors in $\mathbb{R}^{2}$ and then it is easy to see the that the relation follows by the Pythagorean Theorem.) However, I was wondering why this holds for any normed vector space. In general, the norm can be seen as magnitude or size of an object while the metric measures similarity. Can someone give me an intuition about the connection between norm and metric in a broader context?

2) As mentioned above the ultimate goal of defining the norm is to introduce a metric space structure. I've read different posts on this topic and it seems that we want "the metric space structure to play nice with the vector space structure" (Metric spaces and normed vector spaces). Can someone give me an example of an application where this goes wrong and what the consequences are? Translation invariance and homogeneity seem to be important properties for this (What's the need of defining notion of distance using norm function in a metric space?).

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However, I was wondering why this holds for any normed vector space. In general, the norm can be seen as magnitude or size of an object while the metric measures similarity. Can someone give me an intuition about the connection between norm and metric in a broader context?

If you can measure the size of an object and you can subtract objects, then you can produce a measure of similarity. More precisely, if $\|\cdot\|$ is a norm (measure of size), then your measure of similarity is the "size of the difference", i.e. $$ d(x,y) = \|x-y\|. $$

We want "the metric space structure to play nice with the vector space structure". Can someone give me an example of an application where this goes wrong and what the consequences are?

Here is an example of a metric on $\Bbb R$. We define $$ d(x,y) = \begin{cases} 0 & x=y\\ \min\{|x-y|,1\} & x=0 \text{ or } y = 0\\ 1 & \text{otherwise} \end{cases} $$ This defines a metric. The difficult thing to prove here is the triangle inequality when $x=0$ but $y,z$ are non-zero; we find $$ \min\{|z|,1\} = d(x,z) \leq d(x,y) + d(y,z) = \min\{|y|,1\} + 1. $$ Here's something that goes wrong: we would expect that for $f:(\Bbb R, d) \to (\Bbb R,|\cdot|)$ and any $c \in \Bbb R$, $f(x - c)$ is continuous if and only if $f(x)$ is continuous. However, this is not the case.

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  • $\begingroup$ Thanks for your answer! This really cleared things up regarding norms and metrics for me. It also makes sense that you say a measure of similarity (meaning that of course we can define other ways of measuring similarity if we do not care about the magnitude) since there are various ways to define a metric and some are induced by a norm while other's are not. $\endgroup$ Jan 12 '20 at 9:40
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1) Starting from a normed vector space $V$, then if $v\in V$ we write the norm as $\|v\|$ and it should be thought of as the magnitude of the vector $v$, i.e. the distance from the origin. Now the vector space has some symmetries which we want the metric to preserve. Think of translation in Euclidean space: if we shift the two objects we are comparing in the same way, their distance stays the same. In an arbitrary vector space, the condition on the metric is that $d(a+x,b+x)=d(a,b)$. If we choose $x=-b$ then we get $d(a,b)=d(a-b,0)$. But we already said that the norm $\|x\|$ is the distance from the origin, i.e. $d(x,0)$, so this means the metric must be given by $d(a,b)=d(a-b,0)=\|a-b\|$.

2) In general nothing "goes wrong", it depends on the application, although one of the lessons of physics is that losing symmetries is not something that should be done lightly.

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  • $\begingroup$ Thanks for your answer. The problem is that in general we don't really have an origin for abstract vector spaces, say some function space. So that's why I find it hard to talk about symmetries and the like. How would you understand symmetry in a function spaces for example? The point about translation invariance is that the space looks the same no matter where in the space you are. Of course, this does not always pose problems. But it would be helpful if you could give me an example where it does. $\endgroup$ Jan 12 '20 at 8:57
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    $\begingroup$ @DerivativesGuy In a function space, you do have an origin: the function $f = 0$. Likewise, the zero-vector is the origin of any vector space $\endgroup$ Jan 12 '20 at 8:58
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    $\begingroup$ @DerivativesGuy having an origin is one of the defining properties of a vector space and is fundamental to the concept of "linearity", which vector spaces formalize. I think you may want to review some of the fundamentals of vector space theory before moving on to more complicated structures like normed vector spaces, as that is (I believe) where your confusion stems from. $\endgroup$
    – pre-kidney
    Jan 12 '20 at 9:01
  • $\begingroup$ Fair enough, but I'm having a hard time thinking about symmetries in such spaces. I guess it makes sense though if you recall that in vector spaces we can define coordinate vectors with respect to the bases and then only work with the coordinates and the coordinate space looks like a euclidean space, at least for finite spaces. $\endgroup$ Jan 12 '20 at 9:01
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    $\begingroup$ Just think of addition. That is the most important symmetry in a vector space. It works the same way for function spaces. There is nothing mysterious about adding functions... $\endgroup$
    – pre-kidney
    Jan 12 '20 at 9:02

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