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Measure of set where functions are simultaneously bounded by their respective averages
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This pertains to an exercise in Maggi's "Sets of Finite Perimeter and Geometric Variational Problems"-- specifically Exercise 1.14:
If $u_k$, $1leq kleq N$ are summable on $mathbb R^n$ with respect to a finite measure $mu$, then $$ muleft(bigcap_k=1^Nleft u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kmathrmdmuright right)geqslant fracmu(mathbb R^n)2 $$
The result seems intuitive enough; the set of points where the $u_k$ are collectively not too much bigger than their averages takes up ''most'' of space. However, I've had a bit of trouble trying to prove this in full generality, and have only been able to do it in the case of one $u_k$ which is positive.
Indeed, let $A=left uleqslant frac2mu(mathbb R^n)int_mathbb R^n umathrmdmuright$. Then we compute that $$int_mathbb R^n ugeq int_mathbb R^nsetminus A u> int_mathbb R^nsetminus Afrac2mu(mathbb R^n)int_mathbb R^n u=2left(1-fracmu(A)mu(mathbb R^n)right)int_mathbb R^n u$$
If we assume to the contrary that $2mu(A)<mu(mathbb R^n)$, then we reach the desired contradiction.
Towards the general case (still considering only one function $u$), we can define the function $F(x):= frac2mu(mathbb R^n)int_mathbb R^n umathrmdmu-u(x)$, which is positive on $A$ (I wanted to reduce to the positive case because then I might be able to utilize monotonicity somewhere). I have been able to prove (under the assumption that $mu(A)<fracmu(mathbb R^n)2$) by integrating $f$ over $A$ that $$int_A u < int_mathbb R^n u < int_mathbb R^nsetminus A u$$ Clearly, this also implies that the integral over $A$ is strictly negative, and that the integral over the complement is strictly positive. I am having a hard time seeing any potential insight from this.
Unfortunately, this is about all I've drummed up so far, and any helpful nudges in the right direction (especially towards the general case) would be very greatly appreciated. I suspect that a full proof is going to involve a function somehow cooked up from all of the $u_k$, and that it will be fundamentally different from the proof of the positive case (since monotonicity played the crucial role). Thanks!
Edit: I've now been able to prove the result for two positive, summable functions as follows: Set $A_k=left u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kright$ and $A=cap A_k$.
In general, for $u_1,ldots, u_N$ positive and summable, we have for each $k$ that $$int_mathbb R^n u_k geq int_mathbb R^nsetminus A_k u_k>2Nleft(1-fracmu(A_k)mu(mathbb R^n)right)int_mathbb R^n u_k.$$ Assuming without loss of generality that $u_kneq 0$, we gather from this (by summing over the result for each $u_k$) that $$sum_k=1^N mu(A_k)>left(N-frac12right)mu(mathbb R^n).$$ Using now inclusion-exclusion in the case where $N=2$, we calculate that $$mu(A_1cap A_2)=mu(A_1)+mu(A_2)-mu(A_1cup A_2)>left(2-frac12right)mu(mathbb R^n)-mu(mathbb R^n)=frac12mu(mathbb R^n).$$
Unfortunately, the approach by inclusion-exclusion does not appear sharp enough to handle any other cases as far as I can tell, as I get trivial bounds from the approach.
real-analysis probability measure-theory geometric-measure-theory
asked 2 days ago
Hunter Stufflebeam
113
add a comment |Â
up vote
2
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This pertains to an exercise in Maggi's "Sets of Finite Perimeter and Geometric Variational Problems"-- specifically Exercise 1.14:
If $u_k$, $1leq kleq N$ are summable on $mathbb R^n$ with respect to a finite measure $mu$, then $$ muleft(bigcap_k=1^Nleft u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kmathrmdmuright right)geqslant fracmu(mathbb R^n)2 $$
The result seems intuitive enough; the set of points where the $u_k$ are collectively not too much bigger than their averages takes up ''most'' of space. However, I've had a bit of trouble trying to prove this in full generality, and have only been able to do it in the case of one $u_k$ which is positive.
Indeed, let $A=left uleqslant frac2mu(mathbb R^n)int_mathbb R^n umathrmdmuright$. Then we compute that $$int_mathbb R^n ugeq int_mathbb R^nsetminus A u> int_mathbb R^nsetminus Afrac2mu(mathbb R^n)int_mathbb R^n u=2left(1-fracmu(A)mu(mathbb R^n)right)int_mathbb R^n u$$
If we assume to the contrary that $2mu(A)<mu(mathbb R^n)$, then we reach the desired contradiction.
Towards the general case (still considering only one function $u$), we can define the function $F(x):= frac2mu(mathbb R^n)int_mathbb R^n umathrmdmu-u(x)$, which is positive on $A$ (I wanted to reduce to the positive case because then I might be able to utilize monotonicity somewhere). I have been able to prove (under the assumption that $mu(A)<fracmu(mathbb R^n)2$) by integrating $f$ over $A$ that $$int_A u < int_mathbb R^n u < int_mathbb R^nsetminus A u$$ Clearly, this also implies that the integral over $A$ is strictly negative, and that the integral over the complement is strictly positive. I am having a hard time seeing any potential insight from this.
Unfortunately, this is about all I've drummed up so far, and any helpful nudges in the right direction (especially towards the general case) would be very greatly appreciated. I suspect that a full proof is going to involve a function somehow cooked up from all of the $u_k$, and that it will be fundamentally different from the proof of the positive case (since monotonicity played the crucial role). Thanks!
Edit: I've now been able to prove the result for two positive, summable functions as follows: Set $A_k=left u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kright$ and $A=cap A_k$.
In general, for $u_1,ldots, u_N$ positive and summable, we have for each $k$ that $$int_mathbb R^n u_k geq int_mathbb R^nsetminus A_k u_k>2Nleft(1-fracmu(A_k)mu(mathbb R^n)right)int_mathbb R^n u_k.$$ Assuming without loss of generality that $u_kneq 0$, we gather from this (by summing over the result for each $u_k$) that $$sum_k=1^N mu(A_k)>left(N-frac12right)mu(mathbb R^n).$$ Using now inclusion-exclusion in the case where $N=2$, we calculate that $$mu(A_1cap A_2)=mu(A_1)+mu(A_2)-mu(A_1cup A_2)>left(2-frac12right)mu(mathbb R^n)-mu(mathbb R^n)=frac12mu(mathbb R^n).$$
Unfortunately, the approach by inclusion-exclusion does not appear sharp enough to handle any other cases as far as I can tell, as I get trivial bounds from the approach.
real-analysis probability measure-theory geometric-measure-theory
asked 2 days ago
Hunter Stufflebeam
113
As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This pertains to an exercise in Maggi's "Sets of Finite Perimeter and Geometric Variational Problems"-- specifically Exercise 1.14:
If $u_k$, $1leq kleq N$ are summable on $mathbb R^n$ with respect to a finite measure $mu$, then $$ muleft(bigcap_k=1^Nleft u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kmathrmdmuright right)geqslant fracmu(mathbb R^n)2 $$
The result seems intuitive enough; the set of points where the $u_k$ are collectively not too much bigger than their averages takes up ''most'' of space. However, I've had a bit of trouble trying to prove this in full generality, and have only been able to do it in the case of one $u_k$ which is positive.
Indeed, let $A=left uleqslant frac2mu(mathbb R^n)int_mathbb R^n umathrmdmuright$. Then we compute that $$int_mathbb R^n ugeq int_mathbb R^nsetminus A u> int_mathbb R^nsetminus Afrac2mu(mathbb R^n)int_mathbb R^n u=2left(1-fracmu(A)mu(mathbb R^n)right)int_mathbb R^n u$$
If we assume to the contrary that $2mu(A)<mu(mathbb R^n)$, then we reach the desired contradiction.
Towards the general case (still considering only one function $u$), we can define the function $F(x):= frac2mu(mathbb R^n)int_mathbb R^n umathrmdmu-u(x)$, which is positive on $A$ (I wanted to reduce to the positive case because then I might be able to utilize monotonicity somewhere). I have been able to prove (under the assumption that $mu(A)<fracmu(mathbb R^n)2$) by integrating $f$ over $A$ that $$int_A u < int_mathbb R^n u < int_mathbb R^nsetminus A u$$ Clearly, this also implies that the integral over $A$ is strictly negative, and that the integral over the complement is strictly positive. I am having a hard time seeing any potential insight from this.
Unfortunately, this is about all I've drummed up so far, and any helpful nudges in the right direction (especially towards the general case) would be very greatly appreciated. I suspect that a full proof is going to involve a function somehow cooked up from all of the $u_k$, and that it will be fundamentally different from the proof of the positive case (since monotonicity played the crucial role). Thanks!
Edit: I've now been able to prove the result for two positive, summable functions as follows: Set $A_k=left u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kright$ and $A=cap A_k$.
In general, for $u_1,ldots, u_N$ positive and summable, we have for each $k$ that $$int_mathbb R^n u_k geq int_mathbb R^nsetminus A_k u_k>2Nleft(1-fracmu(A_k)mu(mathbb R^n)right)int_mathbb R^n u_k.$$ Assuming without loss of generality that $u_kneq 0$, we gather from this (by summing over the result for each $u_k$) that $$sum_k=1^N mu(A_k)>left(N-frac12right)mu(mathbb R^n).$$ Using now inclusion-exclusion in the case where $N=2$, we calculate that $$mu(A_1cap A_2)=mu(A_1)+mu(A_2)-mu(A_1cup A_2)>left(2-frac12right)mu(mathbb R^n)-mu(mathbb R^n)=frac12mu(mathbb R^n).$$
Unfortunately, the approach by inclusion-exclusion does not appear sharp enough to handle any other cases as far as I can tell, as I get trivial bounds from the approach.
real-analysis probability measure-theory geometric-measure-theory
asked 2 days ago
Hunter Stufflebeam
113
This pertains to an exercise in Maggi's "Sets of Finite Perimeter and Geometric Variational Problems"-- specifically Exercise 1.14:
If $u_k$, $1leq kleq N$ are summable on $mathbb R^n$ with respect to a finite measure $mu$, then $$ muleft(bigcap_k=1^Nleft u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kmathrmdmuright right)geqslant fracmu(mathbb R^n)2 $$
The result seems intuitive enough; the set of points where the $u_k$ are collectively not too much bigger than their averages takes up ''most'' of space. However, I've had a bit of trouble trying to prove this in full generality, and have only been able to do it in the case of one $u_k$ which is positive.
Indeed, let $A=left uleqslant frac2mu(mathbb R^n)int_mathbb R^n umathrmdmuright$. Then we compute that $$int_mathbb R^n ugeq int_mathbb R^nsetminus A u> int_mathbb R^nsetminus Afrac2mu(mathbb R^n)int_mathbb R^n u=2left(1-fracmu(A)mu(mathbb R^n)right)int_mathbb R^n u$$
If we assume to the contrary that $2mu(A)<mu(mathbb R^n)$, then we reach the desired contradiction.
Towards the general case (still considering only one function $u$), we can define the function $F(x):= frac2mu(mathbb R^n)int_mathbb R^n umathrmdmu-u(x)$, which is positive on $A$ (I wanted to reduce to the positive case because then I might be able to utilize monotonicity somewhere). I have been able to prove (under the assumption that $mu(A)<fracmu(mathbb R^n)2$) by integrating $f$ over $A$ that $$int_A u < int_mathbb R^n u < int_mathbb R^nsetminus A u$$ Clearly, this also implies that the integral over $A$ is strictly negative, and that the integral over the complement is strictly positive. I am having a hard time seeing any potential insight from this.
Unfortunately, this is about all I've drummed up so far, and any helpful nudges in the right direction (especially towards the general case) would be very greatly appreciated. I suspect that a full proof is going to involve a function somehow cooked up from all of the $u_k$, and that it will be fundamentally different from the proof of the positive case (since monotonicity played the crucial role). Thanks!
Edit: I've now been able to prove the result for two positive, summable functions as follows: Set $A_k=left u_kleqslant frac2Nmu(mathbb R^n)int_mathbb R^n u_kright$ and $A=cap A_k$.
In general, for $u_1,ldots, u_N$ positive and summable, we have for each $k$ that $$int_mathbb R^n u_k geq int_mathbb R^nsetminus A_k u_k>2Nleft(1-fracmu(A_k)mu(mathbb R^n)right)int_mathbb R^n u_k.$$ Assuming without loss of generality that $u_kneq 0$, we gather from this (by summing over the result for each $u_k$) that $$sum_k=1^N mu(A_k)>left(N-frac12right)mu(mathbb R^n).$$ Using now inclusion-exclusion in the case where $N=2$, we calculate that $$mu(A_1cap A_2)=mu(A_1)+mu(A_2)-mu(A_1cup A_2)>left(2-frac12right)mu(mathbb R^n)-mu(mathbb R^n)=frac12mu(mathbb R^n).$$
Unfortunately, the approach by inclusion-exclusion does not appear sharp enough to handle any other cases as far as I can tell, as I get trivial bounds from the approach.
real-analysis probability measure-theory geometric-measure-theory
asked 2 days ago
Hunter Stufflebeam
113
edited yesterday
asked 2 days ago
Hunter Stufflebeam
113
asked 2 days ago
Hunter Stufflebeam
113
asked 2 days ago
Hunter Stufflebeam
113
113
As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago
add a comment |Â
As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago
As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago
add a comment |Â
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As stated, this is not true unless the $u_k$ are nonnegative. Indeed, consider the case where $mu$ is a Gaussian measure and where $N=n$, so that the coordinate functions $u_k =X_k$ are standard normally distributed and independent. Then the set you are interested in is $(-infty,0]^n$, which has measure $mu(BbbR^n)/2^n$. The problem is that when the $u_k$ are allowed to have zero integral, then it makes no difference whether you consider $2N/mu(BbbR^n)int u_k$, or $1/mu(BbbR^n)int u_k$.
– PhoemueX
2 days ago
@PhoemueX thanks so much! I believe I've got a functional proof for the case of a finite number of positive functions, which I'll post below.
– Hunter Stufflebeam
2 days ago
Ok I thought I had a functional proof but it has a hole... I'll keep trying, but are there any hints you'd be willing to share?
– Hunter Stufflebeam
2 days ago