Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu(\cdot)$ be a probability measure on $W \subseteq \mathbb{R}^m$, so that $\int_W \mu(dw) = 1$.

Consider a locally bounded function $f: X \times W \rightarrow \mathbb{R}_{\geq 0}$, with compact $X \subset \mathbb{R}^n$, such that $\forall w$ $f(\cdot,w)$ is continuous, $\forall x$ $f(x,\cdot)$ is integrable.

Find $f(\cdot)$ such that $$ \int_W \sup_{x \in X} f(x,w) \mu(dw) = \infty $$

share|improve this question
1  
Perhaps not stated properly? You do not assume $f(x,\cdot)$ is integrable for each $x$? –  GEdgar May 17 '12 at 0:32
    
I was thinking on $X = [0,1]$, $W = \mathbb{R}_{\geq 0}$, $f(x,w) = 1/x$ if $w \in [1/x,1/x+1]$, $0$ otherwise. But such $f(\cdot)$ is not locally bounded. –  Adam May 17 '12 at 0:32
    
Yes. According to GEdgar, I meant $f(x,\cdot)$ integrable. –  Adam May 17 '12 at 0:37

1 Answer 1

up vote 1 down vote accepted

Take $\mu(dw) = 1/w^2\ dw$ for $w \in (1,\infty)$, $X = [0,1]$, and $f(x,w) = x w^3 e^{-wx}$. Note that $\sup_{x \in [0,1]} f(x,w) = f(1/w,w) = w^2 e^{-1}$.

share|improve this answer
    
I guest that even if we have assumed that $\forall x \in X \ $ $\int_W f(x,w) \mu(dw) < \infty$, we can not find $\delta > 0$ such that $\sup_{x \in X} \int_W \sup_{\xi \in X \cap (\{x\}+\delta \mathbb{B})} f(\xi,w) \mu(dw) < \infty$... Right? –  Adam May 17 '12 at 0:57
1  
Assuming your $\mathbb B$ is the open unit ball of ${\mathbb R}^n$, that's correct: in my example take $x = 0$ and you get $\sup_{\xi \in [0, \delta)} f(\xi, w) = w^2 e^{-1}$ for $w > 1/\delta$. –  Robert Israel May 17 '12 at 1:34
    
To be precise, I meant $\mathbb{B}$ as the closed ball, but it should be the same thing: a $\delta$ does not exists. –  Adam May 17 '12 at 2:44
    
Sorry for asking again. It would be interesting to find a strictly positive $f:X \times W \rightarrow \mathbb{R}_{>0}$ such that, assuming that $\max_{x \in X} \int_W f(x,w) \mu(dw) < \infty $, does not exist $\delta>0$ such that $\max_{x \in X} \int_W \max_{\xi \in X \cap \{x\}+\delta \overline{\mathbb{B}}} f(\xi,w) \mu(dw) < \infty $. –  Adam May 17 '12 at 5:39
    
Can it be like $(x-a)(\text{something not integrale}) + \text{something integrable}$? So that for $x=a$ we get $\int_W (\cdot) < \infty$, but for $x=a+\delta$ we get $\infty$. Is this the idea behind an example like that? Thanks. –  Adam May 17 '12 at 5:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.