Let $E$ be a Banach space.

Consider the space of bounded functions $\mathcal{B}:=\{f:\Omega\to E:f\text{ bounded}\}$ equipped with the supremum norm $\|f\|_\infty:=\sup_{\omega\in\Omega}\|f(\omega)\|$.

Is it true that the simple functions $\mathcal{S}:=\{s:\Omega\to E:\# s\Omega<\infty\}$ are dense in $\mathcal{B}$?


Generally, no.

Consider $\Omega = \{ x\in E : \lVert x\rVert < 1\}$ and $f = \operatorname{id}$. Then $f$ belongs to the uniform closure of $\mathcal{S}$ if and only if $E$ is finite-dimensional.

Without any regularity assumptions on the functions in $\mathcal{B}$, if $E$ is infinite-dimensional, all that matters is the cardinality of $\Omega$, and then any infinite $\Omega$ can be mapped onto a dense subset of the unit ball of an infinite-dimensional separable subspace of $E$, and such an $f$ cannot lie in the uniform closure of $\mathcal{S}$ (since the unit ball of an infinite space is not totally bounded).

  • $\begingroup$ Identity function doesn't make sense for a general Banach space without multiplication but a constant function $\endgroup$ – C-Star-W-Star Jul 16 '14 at 17:35
  • $\begingroup$ The identity function ($x\mapsto x$) makes sense for every set. Did you perhaps confuse it with the constant $1$? $\endgroup$ – Daniel Fischer Jul 16 '14 at 17:38
  • $\begingroup$ Of course but not for $f:X\to Y$ $\endgroup$ – C-Star-W-Star Jul 16 '14 at 17:40
  • $\begingroup$ For an easy to write down example, I chose the domain to be a bounded subset of the codomain. As told in the next paragraph, if $E$ is infinite-dimensional, for every infinite $\Omega$ there are bounded functions $f\colon \Omega\to E$ that don't belong to the uniform closure of $\mathcal{S}$. $\endgroup$ – Daniel Fischer Jul 16 '14 at 17:43
  • $\begingroup$ Oh I'm sry my mistake ;) $\endgroup$ – C-Star-W-Star Jul 16 '14 at 17:58

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