0
$\begingroup$

Let $\mu(\cdot)$ be a probability measure on the closed set $\mathbb{W} \subseteq \mathbb{R}^m$.

Consider $f: \mathbb{X} \times \mathbb{W} \rightarrow \mathbb{R}_{> 0}$ locally bounded, where $\mathbb{X} \subset \mathbb{R}^n$ is compact, such that for any $w \in \mathbb{W}$ the map $x \mapsto f(x,w)$ is continuous.

Define $\bar{f}:\mathbb{X} \rightarrow \mathbb{R}_{\geq 0}$ as $$\bar f(x):=\int_\mathbb{W} f(x,w) \mu(dw).$$

Prove (or find a counterexample) there exist $F_0 \in \mathbb{R}_{\geq 0}$ and $F:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ continuous, strictly increasing, with $F(0)=0$, such that:

$$ \bar f(x) \ \leq \ F_0 + F(|x|) \quad \forall x \in \mathbb{X}$$

(for some norm $|\cdot|$)

EDIT: $(x,w) \mapsto f(x,w)$ locally bounded.

  • 0
    If I understand correctly, $\bar{f}(x)$ need not even be finite. (For instance, take $\mathbb{W} = (0,1)$, $\mu$ Lebesgue measure, $\mathbb{X} = \{0\}$, and $f(0,w) = 1/w$.) Are there some more assumptions missing?2012-05-05
  • 0
    $\mu$ is a probability measure, so $\int_\mathbb{X} \mu(dx) = 1$. Also I forgot to write that $\mathbb{W}$ is closed. Of course, the question is relevant if and only if $f(x,\cdot)$ is integrable for finite $x$: I'll make this explicit.2012-05-05
  • 0
    $\mu$ is a measure on $\mathbb{W}$, not $\mathbb{X}$, correct? Note that my $\mu$ is a probability measure, and one can replace $(0,1)$ by $[0,1]$ with no essential changes, so making $\mathbb{W}$ closed does not help.2012-05-05
  • 0
    Requiring $f(x,\cdot)$ to be integrable doesn't help much because $\bar{f}$ can still be unbounded, while $F_0 + F(|x|)$ is necessarily bounded on the compact set $\mathbb{X}$. Find a sequence of functions $g_n$ on $(\mathbb{W},\mu)$ such that $g_n \to 0$ pointwise but $\int g_n\,d\mu \to \infty$ (make $g_n$ take large values on a set of small measure). Then take $\mathbb{X} = \{1, 1/2, 1/3, 1/4, \dots, 0\}$ and set $f(1/n, w) = g_n(w)$, $f(0,w)=0$.2012-05-05
  • 0
    I'm not clear about your comment. Since for any $x \in \mathbb{X}$ $f(x,\cdot)$ is integrable, we have $max_{x \in \mathbb{X}}\bar{f}(x) < \infty$. In your example, I don't see how $\int f(1/n,w) $ can go to $\infty$.2012-05-05
  • 0
    Take for example $\mathbb{W} = [0,1]$, $\mu$ Lebesgue measure, $g_n(w) = n^2 1_{(0,1/n)}(w)$. For each $n$, $g_n$ is integrable, but $\sup_n \int g_n = \infty$. Moreover, $g_n \to 0$ pointwise.2012-05-05
  • 0
    Maybe I see what do you mean. In your example you set $f(0,w) = 0$ while $f(1/n,w)$ is going to $\infty$. But this means that for fixed $w$ the map $x \mapsto f(x,w)$ is not continuous in $x=1/n$ because it jumps from $0$ to $n^2$. However, we had assumed continuity. Am I missing something?2012-05-05
  • 0
    I meant there is no continuity for $w=0$: you define $f(0,w) = 0 \ \forall w$, so we also have $f(0,0)=0$. Then we have $f(\epsilon,0) = 1/\epsilon^2$, i.e. $x \mapsto f(x,0)$ is not continuous in $x=0$. This fact holds true regardless of the function we use to approximate the "point mass" at $x=0$.2012-05-05
  • 0
    As I defined it, $f(x,0) = 0$ for every $x$. $f$ *is* continuous in $x$ (check it carefully) and you didn't ask for it to be continuous in $w$.2012-05-05
  • 0
    Ok, maybe I see: for $w>0$ we have $g_n(w)$ that we can made continuous in $n$ with an hal-Gaussian or something like that. Can adding the assumption of local boundedness of $(x,w) \mapsto f(x,w)$ help us?2012-05-05

1 Answers 1

1

Local boundedness will not help either.

Take $\mathbb{W} = \mathbb{N}$, which is a closed subset of $\mathbb{R}$ (all its points are isolated). Define $\mu$ by $\mu(\{n\}) = 2^{-n}$; this is a probability measure. Take $\mathbb{X} = \{1, \frac{1}{2}, \frac{1}{3}, \dots, 0\}$; this is a compact set. Note that all the points of $\mathbb{X}$ except 0 are isolated points.

For each integer $n$, let $f(1/n, n) = n 2^n$. Let $f(x,w) = 0$ for all other $x,w$. Now $f$ is (jointly!) continuous on $\mathbb{X} \times \mathbb{W}$. I want to emphasize that no smoothing or other modification to $f$ is needed to obtain this! $f$ is automatically continuous at points of the form $(1/n, m)$ since those points are isolated in $\mathbb{X} \times \mathbb{W}$. At points of the form $(0,m)$, we have $f(0,m)=0$. Now $U = \{(1/n, m) : n > m\} \cup \{(0,m)\}$ is an open neighborhood of $(0,m)$ in $\mathbb{X} \times \mathbb{W}$, and $f=0$ identically on $U$, so $f$ is continuous at $(0,m)$ also.

Being continuous, $f$ is automatically locally bounded, i.e. bounded on compact subsets of $\mathbb{X} \times \mathbb{W}$. Furthermore, $f(x, \cdot)$ is integrable for every $x$ (indeed, for every $x$, $f(x,\cdot)$ is a bounded function). We have $\bar{f}(1/n) = n$, and $\bar{f}(0) = 0$, so $\bar{f}$ is unbounded and the estimate you want cannot hold.

Fundamentally, the problem is that your conditions are ensuring that as $x_n \to x$, we have $f(x_n, \cdot) \to f(x, \cdot)$ pointwise, or if you require joint continuity, $f(x_n, \cdot) \to f(x, \cdot)$ locally uniformly. But what you need is something more like $f(x_n, \cdot) \to f(x,\cdot)$ in $L^1(\mathbb{W}, \mu)$ and that is a very different topology from the pointwise or compact-open topologies. To get it, you will need something like uniform integrability, and this will require some sort of global or uniform control.

  • 0
    Thanks. So maybe the result could be true if the family $\{f(x,\cdot)\}_{x\in\mathbb{X}}$ is Uniformly Integrable, right?2012-05-05
  • 1
    If the family is in fact uniformly integrable, then $\bar{f}$ will actually be bounded, and your estimate will hold with $F=0$.2012-05-05
  • 0
    What happens if we know that $x$ is "taken independently" to $w$? I mean in all the examples we had troubles because we considered $x = x(w)$...2012-05-07