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Let $(X,\Sigma,\mu)$ be a finite nonzero measure space, and let $L_0$ the set of (equivalence classes of) measurable real-valued functions on it.

This should be a classical question, but why $L_0$ cannot be normed?

Ps. Also a reference would be fine, since I imagine this is straighforward..

Edit, could we use this result (see here): A locally convex Hausdorff space is normable if and only if it has a bounded neighborhood of zero.

  • What do you mean by "can't be normed"? – nombre Oct 03 '16 at 13:49
  • The space of bounded, measurable functions can be equipped with the supremum norm (see here). If the space of (possibly unbounded) measurable functions cannot be equipped with a norm, then I suspect it has something to do with measurable functions which aren't bounded – measure_theory Oct 03 '16 at 14:00
  • @measure_theory, indeed I asked about $L_0$, and not $L_p$ with $p \in [1,\infty]$ (your example is precisely $L_\infty$..) –  Oct 03 '16 at 14:15
  • @nombre That you cannot define any function $|\cdot |: L_0 \to \mathbf{R}$ which is also a norm (hence, with the properties of nonnegativity, omogeneity, and subadditivity) –  Oct 03 '16 at 14:16
  • @Nduccio: Then it can, as can any real vector space $E$, since given a basis $\beta : I \longrightarrow E$ of $E$, $\sum \limits_{i \in I} x_i.\beta_i \mapsto \sum \limits_{i \in I} |x_i|$ is a norm, and $E$ admits a basis by virtue of the incomplete basis theorem. In the result you cite, the space already has a topology, or equivalently, you have some contraints on the resulting topology, which is different. – nombre Oct 03 '16 at 14:25
  • @nombre Maybe I am wrong, but the quoted result just says that the "non-normability" of $L_0$ would follow from the existence of a topology on $L_0$ itself making it a locally convex Hausdorff space with a bounded neighborhood of $0$..? –  Oct 03 '16 at 14:30
  • In this article, the space already has a topology (the product topology), and what they prove is that there can be no norm which induces the same topology. In your case, you didn't mention a topology for $L_0$ so the same can't be said. – nombre Oct 03 '16 at 14:36
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    Rephrasing nombre's concern: given a Hamel basis for your space (one exists by an argument which in general requires the axiom of choice), you can easily construct a norm for it. Thus as a vector space it is normable, as any real vector space is. It can only fail to be normable as a topological vector space, where you are already given some topology. – Ian Oct 03 '16 at 14:38

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