The integral of a polynomial with respect to a probability measure is bounded
-2
Let $P$ be a polynomial and $mu_n$ a sequence of probability measures. I'm having trouble in showing that the sequence of integrals
$$mu_n(P) := underset{mathbb{R}^N}{int} P ; dmu_n$$
is bounded. Can someone help me, please?
real-analysis measure-theory polynomials
asked Nov 28 at 9:44
g.pomegranate
1,062516
add a comment |
-2
Let $P$ be a polynomial and $mu_n$ a sequence of probability measures. I'm having trouble in showing that the sequence of integrals
$$mu_n(P) := underset{mathbb{R}^N}{int} P ; dmu_n$$
is bounded. Can someone help me, please?
real-analysis measure-theory polynomials
asked Nov 28 at 9:44
g.pomegranate
1,062516
$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18
add a comment |
-2
-2
-2
Let $P$ be a polynomial and $mu_n$ a sequence of probability measures. I'm having trouble in showing that the sequence of integrals
$$mu_n(P) := underset{mathbb{R}^N}{int} P ; dmu_n$$
is bounded. Can someone help me, please?
real-analysis measure-theory polynomials
asked Nov 28 at 9:44
g.pomegranate
1,062516
Let $P$ be a polynomial and $mu_n$ a sequence of probability measures. I'm having trouble in showing that the sequence of integrals
$$mu_n(P) := underset{mathbb{R}^N}{int} P ; dmu_n$$
is bounded. Can someone help me, please?
real-analysis measure-theory polynomials
real-analysis measure-theory polynomials
asked Nov 28 at 9:44
g.pomegranate
1,062516
asked Nov 28 at 9:44
g.pomegranate
1,062516
asked Nov 28 at 9:44
g.pomegranate
1,062516
asked Nov 28 at 9:44
g.pomegranate
1,062516
asked Nov 28 at 9:44
g.pomegranate
1,062516
1,062516
$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18
add a comment |
$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18
$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18
add a comment |
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$mu_n(P)$ may not even be finite for fixed $n$.
– Kavi Rama Murthy
Nov 28 at 9:45
Can you explain me why, please? I thought that if the measure is finite, then the integral is finite.
– g.pomegranate
Nov 28 at 9:46
If $mu_n$ puts mass $frac c {k^{2}}$ at $k$ for $k=1,2...$ what is the integral of $x^{2}$ w.r.t. $mu_n$?.
– Kavi Rama Murthy
Nov 28 at 9:52
A better example: consider the measure with density $frac 1 {x^{2}}$ for $x >1$. No polynomial (except a constant) is integrable w.r.t. this measure.
– Kavi Rama Murthy
Nov 28 at 10:09
@g.pomegranate If $mu$ is finite and $f$ is bounded then $int f,dmu$ is finite. Polynomials are not bounded (in general).
– David C. Ullrich
Nov 28 at 15:18