# Define a norm using a metric instead of the other way around

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.

$$\newcommand{\norm}{\left\lVert#1\right\rVert}$$ 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.

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.
• 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. – lonza leggiera Nov 28 '19 at 16:18