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For a probability measure space $(\mathbb{R}, \mathscr{B}, \mu)$, the spaces $L_p^{+}(\mu)$, $1 \leq p < \infty$, are defined to be the collection of all $\mu$-equivalence classes of $\mu$-measurable functions $f: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ for which $\int_{\mathbb{R}_+} f^{p} \mathrm{d} \mu < \infty.$

If we define two subsets $E_c:= \{ f \in L^+_p(\mu) : \text{$f$ is $\mu$-equivalent to a real-valued non-negative function on $\mathbb{R}_+$ that is concave} \}$, $E_m:= \{ f \in L^+_p(\mu) : \text{$f$ is $\mu$-equivalent to a real-valued non-negative function on $\mathbb{R}_+$ that is monotone increasing} \}$,

are these two subsets $E_c$ and $E_m$ closed in $L^+_p$ space?

I thought that this subset $E$ of the $L_p$ space should be closed.

I'm curious to know how to show it rigorously. Could anyone help to show it please? Thanks in advance:)

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Since these are not functions but equivalence classes of functions, your definition doesn't quite make sense. Perhaps you mean

$f$ is $\mu$-equivalent to a function on $\mathbb R$ that is concave and non-decreasing

But this doesn't quite work either. Let $\mu$ be Lebesgue measure on $[0,1]$ and consider the functions $f_n(x) = \sqrt{x+1/n}$ on $[0,1]$. These are in your set, and converge in $L_p$ to $f(x) = \sqrt{x}$. But that is not in your set, because it is not $\mu$-equivalent to a concave function on $\mathbb R$. It is concave on $[0,1]$, but can't be extended to negative reals and remain concave.

You might want to restrict the concavity requirement to the support of $\mu$, but that is not necessarily a convex set, and I don't know how you want to define concavity for functions whose domain is not convex.

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  • $\begingroup$ Thank you for your answer. I'll think about it more carefully:) $\endgroup$ – Paradiesvogel Mar 17 '17 at 5:53
  • $\begingroup$ When we say a concave function, its domain should be a convex set. Otherwise, it doesn't make sense:) $\endgroup$ – Paradiesvogel Mar 28 '17 at 23:09

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