Answer to Question #297118 in Functional Analysis for Luqman

Question #297118

Example of metric space but not norm

1
Expert's answer
2022-02-14T16:35:23-0500

The simplest example of a metric space that is not a normed space is any metric space that is not a vector space. Indeed, the definition of the norm uses the possibility of adding and multiplying by a scalar, so if we consider a space that is not a vector space, it can't be a normed space :

  • "X=\\{0, 1 \\}," with the trivial distance "d(0,1)=1",
  • "X=\\mathbb{N}, \\; d(a,b):=|a-b|",
  • "X\\subseteq \\mathbb{R}^n" with "X" which is not a vector subspace of "\\mathbb{R}^n", with the distance induced by "\\mathbb{R}^n"
  • etc.

We can also consider some vector spaces with the distance that does not come from a norm :

  • "\\mathbb{R}" with "d(a,b) =1" iff "a\\neq b" and more in general
  • Any vector space with this distance (which is called the discrete distance)

But we could also consider some vector spaces with a metric that is natural but it can not be "extended" to a norm :

  • Let us consider the space of all functions from "\\mathbb{R}" to "\\mathbb{R}", this forms a real vector space. We will impose a following distance on this space : "d(f,g) = \\min(1, \\sup_{x\\in\\mathbb{R}}|f(x)-g(x)|)" with the convention that "\\min (1, +\\infty) = 1". This distance is well defined and satisfies the properties of a distance. Yet we can not extend it to a norm : let us consider a function "\\rho(f) = d(f, 0)" where here is the constant null function. If we could extend the distance to a norm, we know that "\\rho" would behave like a norm. Yet we see that it does not satisfy the multiplication property : "d(\\lambda f, 0)\\neq \\lambda d(f,0)" because the distance we defined can't exceed "1" by its definition, so for a scalar big enough this equality fails. In addition, we can't even remove the "\\min(1, ...)" from the definition, as otherwise the distance would diverge to infinity, so this distance can't be extended to a norm by any simple means.

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