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I know that when we have a normed vector space the metric is, by definition, $d(x,y) = ||x - y||$. Why is it that, when we want to induce a norm from a metric, we have to satisfy two properties, namely translation invariance and scaling property? How are those two properties deduced and why are they required for this converse induction?

Michael Munta
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1 Answers1

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$\require{cancel}$Because, by the definition of norm,$$\bigl\lVert(x+w)-(y+w)\bigr\rVert=\lVert x+\cancel w-y-\cancel w\rVert=\lVert x-y\rVert$$and$$\lVert\lambda v-\lambda w\rVert=\bigl\lVert\lambda(v-w)\big\rVert=\lvert\lambda\rvert\lVert v-w\rVert.$$

José Carlos Santos
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  • Is the first line in your answer derived from the triangle inequality property of a norm? If so, can you please show it? – Michael Munta Oct 09 '19 at 12:22
  • No, it is not. There is no inequality there. – José Carlos Santos Oct 09 '19 at 13:23
  • But you said by the definition of the norm. Such a property is not in the definition of the norm. What is this derived from then? – Michael Munta Oct 09 '19 at 14:20
  • I only needed the properties of the norm in my second line of equalities. On the other hand, given any function $f$ from a vector space into $[0,\infty)$, the map $(x,y)\mapsto f(x-y)$ is translation invariant. This is what I proved (when $f(x)=\lVert x\rVert$) in my first line of equalities. – José Carlos Santos Oct 09 '19 at 14:31
  • I understand, but I want to know why is this translation invariance important to induce a norm? – Michael Munta Oct 09 '19 at 15:11
  • Because if you have a norm $\lVert\cdot\rVert$, then the distance $(x,y)\mapsto\lVert x-y\rVert$ is translation invariant. Therefore, if you have a distance on a vector space and if want to decide whether or not it is induced by a norm then if it is not translation invariant, you can be sure that the answer is negative. – José Carlos Santos Oct 09 '19 at 15:14
  • So only by first recognizing the properties of a metric induced by norm ($d(x,y)=||x-y||$) we are able to induce a norm from a metric? What I mean is, first we need to induce a metric, then from that metric we can see that the induced norm can be $d(x,0)=||x||$ by definition as well. – Michael Munta Oct 09 '19 at 16:05
  • Yes. We first check the the distance induced by a norm is translation-invariant and has the scaling property. So, if we have a distance defined on a metric space, those two properties must hold if we want to prove that the distance is induced from a norm. And it turns out that no other condition is required. – José Carlos Santos Oct 09 '19 at 16:38
  • Ok, I know now that itranslation invariance has to hold to induce a norm. My real question is why? Of course the scaling property must hold since it is a property of norm by definition. What about translation invariance? How does that even appear in the induced metric? – Michael Munta Oct 09 '19 at 17:06
  • The answer to that question lies in the first sequence of equalities from my answer. – José Carlos Santos Oct 09 '19 at 17:13
  • Maybe I am not forming my question correctly? Let's put it like this, if the metric did not have translation invariance property, why would that invalidate the resulting norm? – Michael Munta Oct 09 '19 at 19:47
  • If the metric did not have the translation invariance property, then no norm could possibly induce it. – José Carlos Santos Oct 09 '19 at 20:09
  • Yes, but can you please give a concrete example how exactly it would fail? – Michael Munta Oct 10 '19 at 12:19
  • For instance if, in $\mathbb R$, you define $d(x,y)=\lvert x^3-y^3\rvert$, then $d$ is a distance which is not translation invariant (for instance, $d(0,1)\neq d(0+1,1+1)$). Therefore, there is no norm $\lvert\cdot\rVert$ such that$$(\forall x,y\in\mathbb R):d(x,y)=\lVert x-y\rVert.$$ – José Carlos Santos Oct 10 '19 at 12:32
  • Well, I am not sure I understand. If we define this $d(x,y)=\lvert x^3-y^3\rvert$ as you say, then of course that $d(x,y)$ won't be equal to $\lVert x-y\rVert$ because it is equal to $\lvert x^3-y^3\rvert$. We need a translation invariant metric to induce a norm $\lVert x\rVert = d(x,0)$. Why is translation invariance preventing us from simply defining such a norm? – Michael Munta Oct 11 '19 at 07:57
  • How do you know that $d(x,y)$ won't be equal to $\lVert x-y\rVert$, no matter how you define $\lVert\cdot\rVert$? I know that because my function $d$ isnot translation invariant, but you seem to be asserting it even without using that. – José Carlos Santos Oct 11 '19 at 08:05
  • Can there be a function that is not in the form of $\lVert x-y\rVert$, but still being equal to it? – Michael Munta Oct 11 '19 at 08:15
  • Is that a real question? It doesn't look like. But I will answer: no. – José Carlos Santos Oct 11 '19 at 08:22
  • Since my new question got locked it does in fact answer my question. At least I think so, I'm wondering if you can confirm. What I wanted to know is why is norm invalidated if a metric does not satisfy the two properties when in reality it needs to satisfy only the scaling property. $||x|| = d(x,0)$ is a perfectly valid norm as long as $d$ satisfies the scaling property, but not translation invariance. By the definition of norm the latter is a norm, right? You can check the comments of user blat in the closed question. – Michael Munta Mar 01 '20 at 21:31
  • Why should that be a norm? How do you deduce from that definition that $\lVert x+y\rVert\leqslant\lVert x\rVert+\lVert y\rVert$? – José Carlos Santos Mar 01 '20 at 22:13
  • Look at the answer here https://math.stackexchange.com/questions/2888192/norm-induced-by-distance-implies-translation-invariant-distance – Michael Munta Mar 02 '20 at 07:36
  • Done. What else do you want me to do? – José Carlos Santos Mar 02 '20 at 07:48
  • Well there is the deduction which you asked for in the above comment. So why would that not be a norm which satisfies all of the properties of a norm? – Michael Munta Mar 02 '20 at 13:24
  • Yes, there is that deduction for a specific distance. Not in the general case. – José Carlos Santos Mar 02 '20 at 13:36
  • So as I was saying, I only need confirmation that if I find a specific distance which satisfies scaling but NOT translation invariance I can do this $||x|| = d(x,0)$ and have a valid norm? – Michael Munta Mar 02 '20 at 13:43
  • And I don't confirm that assertion. – José Carlos Santos Mar 02 '20 at 13:46
  • Why? $////////$ – Michael Munta Mar 02 '20 at 14:21
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    Because I am not aware of any proof of the fact that if $d$ is a distance on a vector space which is satisfies the scaling property, then the map $v\mapsto d(v,0)$ is a norm. – José Carlos Santos Mar 02 '20 at 18:19
  • If I prove that that norm(which is defined in terms of the metric) satisfies the 3 properties of the norm in general then it is a norm. Translation invariance is not one of those 3 properties. How is this incorrect? – Michael Munta Mar 02 '20 at 19:03
  • Where is your proof? – José Carlos Santos Mar 02 '20 at 19:04
  • I am saying hypothetically. Translation invariance is not required to define the norm in terms of the metric if a viable metric is picked so the norm really is a norm. It is required though when you also want to know if that metric is induced from a norm. – Michael Munta Mar 02 '20 at 20:35
  • Just like the user in the linked answer did. He had a perfectly valid norm without translation invariance, didn't he? – Michael Munta Mar 02 '20 at 20:52
  • Yes. For the distance he was working with, it was indeed a norm. Are you trying to deduce from that fact that you will always get norm? – José Carlos Santos Mar 02 '20 at 22:28
  • Not at all, I was trying to explain that. I just wanted to know that translation invariance was not required in a specific case. So if I pick some concrete metric and I check that it satisfies all the properties of the norm then I have a norm. And it will still be a norm even if the initial metric doesn't have translation invariance. – Michael Munta Mar 03 '20 at 08:48
  • I understand that the user did not prove the general case, but he did prove that it is not required in every single case. So saying that there are norms which we can induce from non-translation invariant metrics is a true assertion. Right? – Michael Munta Mar 03 '20 at 09:12
  • Yes, that is correct. – José Carlos Santos Mar 03 '20 at 10:07
  • @JoseCarlosSantos: Is there a formula to recover the norm from a Normable metric? – MSIS Nov 14 '22 at 00:28
  • I suspect d(x,0) does recover the norm. I believe that's what op wanted. – MSIS Nov 14 '22 at 01:41
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    @MSIS Yes, if $d(x,y)=|x-y|$, then $|x|=d(x,0)$. – José Carlos Santos Nov 14 '22 at 07:21