Can we derive a norm and an inner product from a metric?

Question

Given an inner product on a vector space, I can always define a norm and a metric (and a topology using that metric). Is the converse true? That is, given a metric on a vector space, can I define an inner product with it? And what about a metric space that is not a vector space?

Answer 1

Let $k$ denote the field of real numbers, or the field of complex numbers. Given a vector space $V$ over $k$, it is always possible define a metric on $V$ (say the discrete metric), or a norm on $V$ (by choosing a Hamel basis). However, if the norm you define has no relation to the metric you define, then there is no point in studying them simultaneously.

Thus, we should restrict our attention to metrics $d$ that are induced by a norm $\lVert\cdot\rVert$, in the sense that $d(x,y)=\lVert x-y\rVert$ for all $x,y\in V$. This way, the metric and norm “interact” with each other. A metric induced by a norm has two special properties:

• Translation invariance: $d(x+c,y+c)=d(x,y)$ for all $c,x,y\in V$.
• Absolute homogeneity: $d(\lambda x,\lambda y)=|\lambda|d(x,y)$ for all $x,y\in V$ and $\lambda\in k$.

Your question becomes: given a vector space $V$ over $k$ and a metric $d$ on $V$, must there be a norm on $V$ which induces $d$? The answer is negative. Consider, for instance, the vector space $\mathbb R^3$ (with the usual vector space operations). The discrete metric on $\mathbb R^3$ is not absolutely homogeneous, and hence is not induced by a norm. On the other hand, we do have a positive result: if a metric $d$ on $V$ is translation invariant and absolutely homogeneous, then it is induced by the norm $x\mapsto \lVert x\rVert=d(x,0)$.

A similar story occurs with inner product spaces. A norm induced by an inner product satisfies a special property, namely the parallelogram law $2\lVert x\rVert+2\lVert y\rVert=\lVert x+y\rVert^2-\lVert x-y\rVert^2$. Not every norm satisfies this, hence not every norm is induced by an inner product. But if a norm does satisfy the parallelogram law, then it is induced by an inner product (see here for a proof).