4

Can we provide an explicit example of a norm $\|\cdot\|$ on the vector space of continuous functions $C([0,1])$ such that there is no constant $M>0$ with the property $\|\cdot\| \le M\|\cdot\|_\infty$?

Some comments:

  • It is shown here that, without AC, any complete norm on $C([0,1])$ is necessarily equivalent to $\|\cdot\|_\infty$. So there is no explicit example of a complete norm $\|\cdot\|$ not dominated by $\|\cdot\|_\infty$.
  • A simple example of such a norm is $\|f\| := \|f\|_\infty + |\phi(f)|$ where $\phi$ is a discontinuous linear functional on $(C([0,1]),\|\cdot\|_\infty)$. But of course we cannot construct such a functional $\phi$ on a Banach space without AC.

I would assume therefore that the answer is no, and in that case I'm not looking for a rigorous proof or anything of the sort as I probably wouldn't be able to understand it anyway.

For context, I was inspired by this question which inquires whether the set $$E = \left\{f \in C([0,1]) : \int_0^1 f(t)\,dt = 0\right\}$$ is closed in $C([0,1])$. An answer pointed out that the OP initially did not specify the norm on $C([0,1])$ but then I realized I couldn't think of an example of a norm with respect to which $E$ wouldn't be closed. Of course it exists since $E$ is a kernel of a linear functional so it suffices to provide a norm which makes the functional discontinuous but this is probably impossible without AC.

mechanodroid
  • 46,490

1 Answers1

3

You won't be able to find such an explicit example, essentially by the same reasoning as in the answers to the first linked question.

For example, in the Solovay model (in which ZF + DC hold), every linear map from a Banach space into a normed space is bounded. In particular, the identity map from $(C[0,1], \|\cdot\|_\infty)$ to $(C[0,1], \|\cdot\|)$ is bounded so that in this model no such norm exists.

Rhys Steele
  • 19,671
  • 1
  • 19
  • 50
  • Thanks. I suppose it would be similarly impossible to explicitly state a norm on $C([0,1])$ with respect to which the linear functional $f \mapsto \int_0^1 f(t),dt$ is discontinuous? – mechanodroid Jan 01 '23 at 13:49
  • 1
    Yes, one way to see this is to use the property of the Solovay model I mentioned in the answer to see that in that model if a vector space $X$ has a Banach norm then all norms defined on that space are equivalent. In particular, since $C[0,1]$ admits a Banach norm, in the Solovay model; it admits only one norm topology so that continuity of linear functionals with respect to any norm topology is entirely characterised by the supremum norm. – Rhys Steele Jan 01 '23 at 14:08