Answer to Question #297118 in Functional Analysis for Luqman

Question

Example of metric space but not norm

Answer 1

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.