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I've just started the study of functional analysis and I've a "philosophical" question. Kreyszig and other (at least is my impression), start (1) defining a metric space (for instance $\mathbb{R}^n$ with Euclidean metric), then (2) define a vector space on it and finally (3) define a norm on it , thus obtaining a Banach space.

$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$ Lets forget about completeness, my question is why they define a norm and deduce a metric from it (metric induced by the norm) instead of using the metric $d$ and defining $$ \norm{x}:=d(x,0), $$ where $0$ is the zero element of the vector space.

Thus why to define a norm and deduce a metric if we already have a metric?

EDIT Perhaps because one can prove generically that the induce metric is a metric; but for the induced norm from the metric one has to use the concrete formulation of the metric to prove the properties of the norm.

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In order that your definition yields a norm, you need $$ d(t \, x, 0) = t \, d(x,0) \qquad\forall t > 0, x \in X$$ and $$d(x-y,0) \le d(x,0) + d(y,0)\qquad\forall x,y \in X.$$ However, these properties are not satisfied by every metric. This means that not all metrics are generated by a norm. But, of course, every norm generates a metric.

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  • $\begingroup$ I don't believe the two conditions you give are sufficient to guarantee that $\ d\ $ is generated by a norm. Necessary and sufficient conditions for a metric $ d $ to be generated by a norm are: $$ d(t \, x, 0) = t \, d(x,0) \qquad\forall t > 0, x \in X$$ and $$ d(x,y)=d(x-y,0) \qquad\forall x, y \in X\ , $$ but I don't believe the conditions you give are sufficient to guarantee that the second of these will hold. $\endgroup$ – lonza leggiera Nov 28 '19 at 16:18

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