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Let $\mu$ be a probability measure on $X \subseteq \mathbb{R}$.

Consider a polynomial function $p_d: \mathbb{R} \rightarrow \mathbb{R}$ of degree $d \in \mathbb{Z}^+$.

I would like to know if the following is true.

$$ \int_X |p_d(x)| \mu(dx) < \infty \ \Leftrightarrow \ \int_X |x^d| \mu(dx) < \infty $$

In the case the result does work, I'm wondering how it could be extended to the multi-dimensional case $X \subseteq \mathbb{R}^m$, $m \in \mathbb{Z}_{> 1}$.

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Well, $p_d(a+bx)$ is just another polynomial of degree $d$, so you really only need to prove the case $a=0$, $b=1$. –  Thomas Andrews Mar 21 '12 at 21:23
    
Do you mean something like "a particular polynomial of degree $d$ is integrable iff all polynomials of degree $d$ are integrable"? –  Adam Mar 21 '12 at 21:27
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You really want to say $\int | \ldots | \ \mu(dx) < \infty$, not $\int \ldots \ \mu(dx) < \infty$, because you don't want to include cases where the integral diverges without diverging to $+\infty$. –  Robert Israel Mar 21 '12 at 21:33
    
No, I mean you've asked the following. For any triple $(a,b,p_d)$ with $a,b\in \mathbb R$, $b\neq 0$ and $p_d$ a polynomial of degree $d$, is it true that $\int p_d(a+bx)<+\infty$ iff $\int x^d<+\infty$. But I'm saying this is equivalent to asking it just about any triple $(0,1,p_d)$, since you can always convert the question about $(a,b,p_d)$ to a question about some $(0,1,q_d)$, where $q_d(x)=p_d(a+bx)$ is a polynomial of degree $d$. –  Thomas Andrews Mar 21 '12 at 21:42
    
Ok, I see. We can see it as $\int q_d(x) < \infty \Leftrightarrow \int x^d < \infty$. Thanks –  Adam Mar 21 '12 at 21:47

2 Answers 2

Hint: By Hölder's inequality, if $1 \le j \le d$, $$\int x^j \ \mu(dx) \le \left(\int 1^q \ \mu(dx)\right)^{1/q} \left(\int |x^d| \ \mu(dx)\right)^{1/p}$$ where $p = d/j$ and $1/p + 1/q = 1$.

The multidimensional case is trickier, because e.g. if $\mu$ is concentrated on $\{(x,y) \in {\mathbb R}^2: y = 0\}$, $\int |x^j y| \ d\mu$ will converge for all $j$ while $\int |x^j| \ d\mu$ might diverge for all $j \ge 1$.

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so $\int x^j \leq (\int |x|^d)^{j/d} \leq \int|x|^d$ ? –  Adam Mar 21 '12 at 21:34
    
but I don't see how it comes from: $\int |x| \leq (\int |x|^q)^{1/q} (\int |x|^p)^{1/p}$ with $1/q + 1/p = 1$ –  Adam Mar 21 '12 at 21:36
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Hölder says $\left|\int f g \right|\le \left(\int |f|^q\right)^{1/q} \left(\int |g|^p\right)^{1/p}$. Here you're taking $f = 1$ and $g = x^j$. –  Robert Israel Mar 21 '12 at 21:46
    
so I can prove that: $ |x|^d $ integrable $\Rightarrow$ $p_d(x)$ integrable. I need the other implication now –  Adam Mar 21 '12 at 22:01
    
I still don't see a rigorous proof. I need a proof instead of hints... –  Adam Mar 21 '12 at 22:28
up vote 1 down vote accepted

I'm wondering if the following argument could work.

By the Limit Comparison Test we have that

$$ \lim_{x \rightarrow \infty} \frac{ | \sum_{i=0}^d a_i x^i |}{ |x^d| } = a_d \in (0, \infty)$$

so $| \sum_{i=0}^d a_i x^i |$ is integrable iff $|x^d|$ is integrable.

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Given it works, what do you expect for the multi-dimensional case? –  draks ... Mar 22 '12 at 23:30
    
this is nice question! –  Adam Mar 23 '12 at 0:36

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