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Do these three kinds of vector spaces, those with an inner-product, those with a norm and those with a metric, are the same sets of vector spaces? At least for finite dimensional vector spaces all of these coincide?

It would be great to know finite dimensional counter-examples if they exist any and if anyone can link to some lecture notes explaining this point.

Like are there examples of inner product spaces which cannot have a metric or metric spaces which cannot have a norm or various other such possible conflicts between these 3 properties.

guest
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  • You have : inner-product $\implies$ norm $\implies$ metric, but the converses are wrong in general. – Watson Feb 12 '16 at 13:36
  • Yes, but with finite dimensional vector spaces will the converses also work? – guest Feb 12 '16 at 13:38
  • Not even in finite-dimensional vector spaces. – John B Feb 12 '16 at 13:42
  • in the finite dimensional case, all the topologies are equivalent, that is if a sequence $(\phi_n)_n$ is convergent $\to \phi$ for a certain norm, it will also converge $\to \phi$ for all the other possible norms on the vector space. this is because a norm asks for $||a u|| = |a| , ||u||$ for any constant $a$ and $||u+v|| \le ||u||+||v||$ – reuns Feb 12 '16 at 14:04

5 Answers5

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Even in finite dimension, the converses are not true.

For instance, the $p$-norm ($p≠2$) on $\Bbb R^n$ doesn't come from an inner product. This can be proven by contradiction, using the parallelogram law or the polarization identity. The polarization identity $$2 \|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2 \qquad \forall x,y \in V$$ precisely gives a necessary and sufficient condition for a normed vector space $V$ to have an inner product that induces that norm.

The discrete metric on $\Bbb R$ doesn't come from a norm, because it takes only $0$ and $1$ as values.

If $d$ is a metric such that there is a norm $\|\cdot\|$ on $V$ with $d(x,y)=\|x-y\|, \; \forall x,y \in V$, then two necessary conditions are that $$d(ax,0)=|a|d(x,0) \tag 1 $$ $$d(x,y)=d(x+z,y+z) \tag 2$$ for every $x,y,z \in V$ and $a \in \Bbb R$. Conversely, if the metric $d$ satisfies $(1)$ and $(2)$, then $\|x\|:=d(x,0)$ yields a norm on $V$.

You can also see this question.

Watson
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  • Thanks! Is there any reference for the proofs for the tests you said for the converses to work? Typically I would think of a metric on a vector space to be coming from a positive-semidefinite symmetric bilinear map/form (but I guess not all metrics come like this?) and these particular class of metrics will always then I guess induce a norm and such norms that they infact induce induce inner-products? – guest Feb 12 '16 at 14:17
  • For inner products/norms you can see here. For norms/metrics, you can try to prove by yourself, this is not difficult. A metric is already symmetric and "definite-positive". – Watson Feb 12 '16 at 14:22
  • @guest : Yes, if $(.,.)$ is a positive-definite symmetric bilinear form, then $d(x,y)=(x-y,x-y)$ is a metric, as I said before. All metrics don't come like this, as I said before (e.g. discrete metric). – Watson Feb 12 '16 at 14:25
  • but these psd bilinear form metrics will necessarily induce norms which will necessarily induce an inner product? – guest Feb 12 '16 at 14:28
  • Oops it was $d(x,y)=\sqrt{(x-y,x-y)}$ in my previous comment. – Watson Feb 12 '16 at 14:32
  • @guest : Yes, in that case, defining $|x|=d(x,0)=\sqrt{(x,x)}$ (as I did before) yields exactly the norm induced by the positive-definite symmetric bilinear form you started with. – Watson Feb 12 '16 at 14:35
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Take $X=\mathbb R^n$ and equip it with the topology induced by the the distance function $$ {\rm d}(x,y) = \begin{cases} 0 &\text{if } x=y\\1&\text{if }x\neq y\end{cases} $$ Then $X$ is a metric space, specifically the discrete metric space (in which all points are isolated) However, $(X,{\rm d})$ fails to be a normed space in that ${\rm d}$ does not induce any norm, as you can easily verify: $$ {\rm d}(\lambda x,\lambda y) \neq |\lambda|{\rm d}(x,y) \quad\text{whenever $x\neq y$ and $\lambda\neq\pm1$} $$

Let now $$ \|x\|_p = \left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} \quad\text{for some $p>0$} $$ Then you can easily verify that $\|{}\cdot{}\|_p$ is a norm for any $p>0$, hence $(X,\|{}\cdot{}\|_p)$ is a normed space, but for $p\neq 2$ it is not induced by a scalar product. Therefore, for $p\neq 2$ $(X,\|{}\cdot{}\|_p)$ is a normed space which is not induced by an inner product.

AndreasT
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You can always use an inner product to define a norm and you can always use a norm to define a metric. Specifically, if $<a,b>$ is the inner product of $a$ and $b$, then $\sqrt{<a,a>}$ is a norm, i.e. you can say $\Vert a\Vert = \sqrt{<a,a>}$ and it has all the properties required of a norm. If you have a norm $\Vert \cdot \Vert$, then you can define a metric by $d(a, b)=\Vert a-b\Vert$ and that metric has all the properties required.

However, as pointed out in the comments, the reverse is not true. I.e. having a metric does not imply that you have an inner product.

TravisJ
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  • But on finite dimensional vector spaces will the converses also always work? (..like trivially an "unit ball" is a metric space that is not a vecctor space and hence it makes no sense to talk of an inner-product or a norm structure on it anyway!..) And if one goes beyond vector spaces then do we know of criteria for the converses to work? – guest Feb 12 '16 at 13:43
  • @guest, What I'm saying is that any time you have an inner product you automatically have a norm and a metric. – TravisJ Feb 12 '16 at 13:44
  • I am asking about conditions when the converse would work.. – guest Feb 12 '16 at 13:45
  • @guest : the polarization identity gives a condition for the "norm $\implies$ inner product" (you can see my answer). – Watson Feb 12 '16 at 13:49
  • Just a side note: the norm should be $\sqrt{ <a,a> }$, I think. – Watson Feb 12 '16 at 14:33
  • @TravisJ : I'm not sure to understand your previous comment. I may be wrong, but $|2a|=<2a,2a>=4<a,a>=4|a|$ contradicts the homogeneity of a norm. This is why I think you should add a square root. – Watson Feb 12 '16 at 19:44
  • @Watson, Thanks you are correct. This is my mistake. I will fix it. – TravisJ Feb 12 '16 at 20:28
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I don't know if this helps, but a metric space is not necessarily a vector space.

"The set of positive real numbers, ℝ+=(0,∞), with the metric given by d(x,y):=|x−y| is a metric space, but it is not a linear space, since it contains neither an additive identity (0) nor additive inverses (−x)."

Source: https://wiki.math.ntnu.no/users/ehrnstro/teaching/linearmethods/metricspaces

alf
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In the plane the ordinary Euclidean metric comes from the norm that comes from the usual inner product.

In the taxicab metric the distance from $(a,b)$ to $(c,d)$ is $|a-c| + |b-d|$. That metric does come from a norm, but the norm doesn't come from an inner product.

The metric for which the distance between any two distinct points is $1$ doesn't come from a norm.

Ethan Bolker
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  • On vector spaces is there an "iff" test which one can apply to a metric to know if it can come from a norm? Similarly an "iff" test one can do on a norm to test if it comes from an inner product? – guest Feb 12 '16 at 13:56
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    @guest : you can see my answer : the second one is solved thanks to the polarization identity, whereas the first one is just a homogeneity condition on the metric… – Watson Feb 12 '16 at 13:58
  • @Watson I was just about to point the OP to your fine answer. – Ethan Bolker Feb 12 '16 at 13:59