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I have asked similar questions here, but I still don't understand the following:

If I have a metric e.g. $$d(x,y) = |x^3 -y^3|$$ why is translation invariance and homogeinity preventing me from simply defining the norm as $$||x|| = d(x,0)$$?

Michael Munta
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    For one thing, $\lVert x-y\rVert\ne d(x,y)$, so the (non-)norm you have defined induces, if any, another distance. –  Feb 29 '20 at 12:04
  • But why do I need to think about the converse induction when I am here inducing a norm from metric, not the other way around. I mean, so what if it doesn't satisfy the two properties? – Michael Munta Feb 29 '20 at 12:08
  • "so what if it doesn't satisfy the two properties? " -

    Then it's by definition not a norm.

     – blat Feb 29 '20 at 12:41
  • @blat what if it satisfies the scaling property, but not translation invariance? How would you check whether it is a norm if translation invariance is not a property of a norm? – Michael Munta Feb 29 '20 at 13:56
  • Not sure I understand the question. The linked answer above gives an argument that norms necessarily have these two properties. So if your construct doesn't satisfy one or both of these properties, then it can't be a norm. – blat Feb 29 '20 at 14:05
  • @blat Well if I setup the initial metric so its norm $||x|| = d(x,0)$ would satisfy the homogeinity property, but not the translation invariance property, how would you prove that that norm is not a norm? Translation invariance is not a property of norms in general it is just something that results from inducing a metric from norm. – Michael Munta Feb 29 '20 at 14:16
  • What exactly do you mean by translation invariance of the norm? Because $\lVert (x + w) - (y + w) \rVert = \lVert x - y \rVert$ trivially holds for any norm. – blat Feb 29 '20 at 14:33
  • Are you asking if the metric has to be translation invariant in order to define a norm? This is not the case as answered here. – blat Feb 29 '20 at 14:40
  • That is what I am asking. So a metric that is not translation invariant can be turned into a metric? – Michael Munta Mar 01 '20 at 11:35

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Let $a \in \mathbb{R}.$ Then by your definition $$ \lVert ax \rVert = d(ax, 0) = |a^3 x^3| = |a|^3 |x^3| = |a|^3 d(x, 0) = |a|^3 \lVert x \rVert, $$ but we need $\lVert a x \rVert = |a| \lVert x \rVert,$ which clearly doesn't hold for all $a$.

blat
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