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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 $a=0$, $b=1$.2012-03-21
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    Do you mean something like "a particular polynomial of degree $d$ is integrable iff all polynomials of degree $d$ are integrable"?2012-03-21
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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 $+\infty$.2012-03-21
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    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 $d$.2012-03-21
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    Ok, I see. We can see it as $\int q_d(x) < \infty \Leftrightarrow \int x^d < \infty$. Thanks2012-03-21
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    Let's put $p_2=c_0x^2+c_1xy+c_2y^2+c_3x+c_4y+c_5$, so $m=2$. What would you expect at the RHS?2012-03-21
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    What do you mean? Is this the multi-dimensional case? First we have to solve the mono-dimensional case!2012-03-21
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    Can I use a sort of "limit comparison test"?2012-03-22
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    @Adam I thought you got that already settled. Sorry. Interesting question.2012-03-22

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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?2012-03-22
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    this is nice question!2012-03-23
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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 $\int x^j \leq (\int |x|^d)^{j/d} \leq \int|x|^d$ ?2012-03-21
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    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$2012-03-21
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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 $g = x^j$.2012-03-21
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    so I can prove that: $ |x|^d $ integrable $\Rightarrow$ $p_d(x)$ integrable. I need the other implication now2012-03-21
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    I still don't see a rigorous proof. I need a proof instead of hints...2012-03-21
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    Can I use a sort of "limit comparison test"?2012-03-22
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    If $x^d$ is not integrable, consider a subset $Y$ of either $(-\infty,0]$ or $[0,\infty)$ such that $\int_Y |x|^d \ \mu(dx) = M < \infty$. Hölder says $\int_Y |x|^j \ \mu(dx) \le M^{j/d}$ for $0 \le j < d$. Now if $p_d(x) = \sum_{j=0}^d a_j x^j$ we have $$\left|\int_Y p_d(x) \ \mu(dx)\right| \ge |a_d| \left|\int_Y x^d\ \mu(dx)\right| - \sum_{j=0}^{d-1} |a_j| \left|\int_Y x^j\ \mu(dx)\right| \ge |a_d| M - \sum_{j=0}^{d-1} |a_j| M^{j/d}$$ and the right side goes to $\infty$ as $M \to \infty$, so $p_d(x)$ is not integrable.2012-03-22
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    I don't see why if "$x^d$ is not integrable" then there should exist $Y$ (either $\mathbb{R}_0^-$ or $\mathbb{R}_0^+$ ) such that $\int_Y |x|^d < \infty$2012-03-22
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    If $\int |x|^d = \infty$, at least one of $\int_{(-\infty,0]} |x|^d$ and $\int_{[0,\infty)} |x|^d$ is $\infty$. Let's say it's the latter. Now $\int_{[0,\infty)} |x|^d = \lim_{N \to \infty} \int_{[0,N]} |x|^d$, and $\int_{[0,N]} |x|^d$ is finite.2012-03-22
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    I still don't see why $\int_{\mathbb{R}_0^+} |x|^d = M < \infty$: your set $Y$ is not compact! Do you mean "consider a compact subset $Y$ [...]"???2012-03-22
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    I didn't say $\int_{{\mathbb R}_0^+} |x|^d = M$. I said $\int_Y |x|^d = M$, where $Y$ is some subset of ${\mathbb R}_0^+$ or ${\mathbb R}_0^-$. I didn't specify whether it is compact, but you can certainly take it to be compact.2012-03-22
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    Well, $Y$ MUST be bounded so there is no need to take it not-compact as we are taking $\int_Y \cdot$2012-03-22