Mixing
Suppose $f$ is a measurable function st $|f(x)| leq c|x|^{-p} chi_{B(0,1)}(x)$for some $c > 0, p <n$....
Suppose $f: mathbb R^n to mathbb R$ is a measurable function such that $|f(x)| leq g(x)$, where $g(x) = c|x|^{-p} chi_{B(0,1)}(x)$for some $c > 0, p <n$. Prove $f$ is integrable.
This is from Bass exercise 11.21.
Let $tilde x=(x_1,...,x_{n-1}), x = (tilde x, x_n)$, and $f^{x_n}(tilde x) = f(x) : mathbb R^{n-1} to mathbb R$.
If $p leq 0$ then $g$ is integrable, so that $f$ also is, and we are done. So assume $p > 0$.
We are suggested to proceed by induction on $n$. The case for $n=1$ can be solved. Now assume the case holds for $n-1$, and let $epsilon > 0$ be so small that $p+epsilon$ is still smaller than $n$. Then $p-1 + epsilon < n-1$, and we can show that $|f^{x_n}(tilde x)| leq c|tilde x|^{-p+1-epsilon} chi_{tilde B(0,1)}(tilde x) |x|^{-1+epsilon}$. A person suggested to me that I should somehow use Dominated Convergence theorem and the fact that $|x|^{-1+epsilon}$ is integrable, but I am not sure how to proceed as such. Certainly, we may bound $g$ by $c|x|^p chi_{[0,1]^n}(x)$, but I am not sure what to do here. It seems like I have to use Tonell-Fubini somewhere...
I apologizw for mixing up the precise problem statement. The main problem is now correct; if there is any error in signs please construe as close you think it should be. Thank you.
Of course, this could be done in polar coordinates but I am specifically asking for one using indunction.
real-analysis measure-theory lebesgue-integral
asked Nov 20 '18 at 9:13
Cute Brownie
995416
add a comment |
Suppose $f: mathbb R^n to mathbb R$ is a measurable function such that $|f(x)| leq g(x)$, where $g(x) = c|x|^{-p} chi_{B(0,1)}(x)$for some $c > 0, p <n$. Prove $f$ is integrable.
This is from Bass exercise 11.21.
Let $tilde x=(x_1,...,x_{n-1}), x = (tilde x, x_n)$, and $f^{x_n}(tilde x) = f(x) : mathbb R^{n-1} to mathbb R$.
If $p leq 0$ then $g$ is integrable, so that $f$ also is, and we are done. So assume $p > 0$.
We are suggested to proceed by induction on $n$. The case for $n=1$ can be solved. Now assume the case holds for $n-1$, and let $epsilon > 0$ be so small that $p+epsilon$ is still smaller than $n$. Then $p-1 + epsilon < n-1$, and we can show that $|f^{x_n}(tilde x)| leq c|tilde x|^{-p+1-epsilon} chi_{tilde B(0,1)}(tilde x) |x|^{-1+epsilon}$. A person suggested to me that I should somehow use Dominated Convergence theorem and the fact that $|x|^{-1+epsilon}$ is integrable, but I am not sure how to proceed as such. Certainly, we may bound $g$ by $c|x|^p chi_{[0,1]^n}(x)$, but I am not sure what to do here. It seems like I have to use Tonell-Fubini somewhere...
I apologizw for mixing up the precise problem statement. The main problem is now correct; if there is any error in signs please construe as close you think it should be. Thank you.
Of course, this could be done in polar coordinates but I am specifically asking for one using indunction.
real-analysis measure-theory lebesgue-integral
asked Nov 20 '18 at 9:13
Cute Brownie
995416
1
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25
add a comment |
Suppose $f: mathbb R^n to mathbb R$ is a measurable function such that $|f(x)| leq g(x)$, where $g(x) = c|x|^{-p} chi_{B(0,1)}(x)$for some $c > 0, p <n$. Prove $f$ is integrable.
This is from Bass exercise 11.21.
Let $tilde x=(x_1,...,x_{n-1}), x = (tilde x, x_n)$, and $f^{x_n}(tilde x) = f(x) : mathbb R^{n-1} to mathbb R$.
If $p leq 0$ then $g$ is integrable, so that $f$ also is, and we are done. So assume $p > 0$.
We are suggested to proceed by induction on $n$. The case for $n=1$ can be solved. Now assume the case holds for $n-1$, and let $epsilon > 0$ be so small that $p+epsilon$ is still smaller than $n$. Then $p-1 + epsilon < n-1$, and we can show that $|f^{x_n}(tilde x)| leq c|tilde x|^{-p+1-epsilon} chi_{tilde B(0,1)}(tilde x) |x|^{-1+epsilon}$. A person suggested to me that I should somehow use Dominated Convergence theorem and the fact that $|x|^{-1+epsilon}$ is integrable, but I am not sure how to proceed as such. Certainly, we may bound $g$ by $c|x|^p chi_{[0,1]^n}(x)$, but I am not sure what to do here. It seems like I have to use Tonell-Fubini somewhere...
I apologizw for mixing up the precise problem statement. The main problem is now correct; if there is any error in signs please construe as close you think it should be. Thank you.
Of course, this could be done in polar coordinates but I am specifically asking for one using indunction.
real-analysis measure-theory lebesgue-integral
asked Nov 20 '18 at 9:13
Cute Brownie
995416
Suppose $f: mathbb R^n to mathbb R$ is a measurable function such that $|f(x)| leq g(x)$, where $g(x) = c|x|^{-p} chi_{B(0,1)}(x)$for some $c > 0, p <n$. Prove $f$ is integrable.
This is from Bass exercise 11.21.
Let $tilde x=(x_1,...,x_{n-1}), x = (tilde x, x_n)$, and $f^{x_n}(tilde x) = f(x) : mathbb R^{n-1} to mathbb R$.
If $p leq 0$ then $g$ is integrable, so that $f$ also is, and we are done. So assume $p > 0$.
We are suggested to proceed by induction on $n$. The case for $n=1$ can be solved. Now assume the case holds for $n-1$, and let $epsilon > 0$ be so small that $p+epsilon$ is still smaller than $n$. Then $p-1 + epsilon < n-1$, and we can show that $|f^{x_n}(tilde x)| leq c|tilde x|^{-p+1-epsilon} chi_{tilde B(0,1)}(tilde x) |x|^{-1+epsilon}$. A person suggested to me that I should somehow use Dominated Convergence theorem and the fact that $|x|^{-1+epsilon}$ is integrable, but I am not sure how to proceed as such. Certainly, we may bound $g$ by $c|x|^p chi_{[0,1]^n}(x)$, but I am not sure what to do here. It seems like I have to use Tonell-Fubini somewhere...
I apologizw for mixing up the precise problem statement. The main problem is now correct; if there is any error in signs please construe as close you think it should be. Thank you.
Of course, this could be done in polar coordinates but I am specifically asking for one using indunction.
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Nov 20 '18 at 9:13
Cute Brownie
995416
asked Nov 20 '18 at 9:13
Cute Brownie
995416
edited Nov 20 '18 at 10:34
asked Nov 20 '18 at 9:13
Cute Brownie
995416
asked Nov 20 '18 at 9:13
Cute Brownie
995416
asked Nov 20 '18 at 9:13
Cute Brownie
995416
995416
1
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25
add a comment |
1
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25
1
1
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25
add a comment |
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1
What makes $g$ integrable for $p leq 0$? For all $p>0$ $f$ is integrable and this hardly requires a proof (because it is a bounded measurable function with com pact support). I wonder if you have some mistakes in the statement.
– Kavi Rama Murthy
Nov 20 '18 at 9:35
The error was corrected: Just use the theorem on change of variables for $lambda^n$ by taking polar-coordinates. ($g$ depends only on the radius and then it is easy to see that we get the integrability in $x=0$ if and only if $n-1-p >-1$.)
– p4sch
Nov 20 '18 at 10:25