A property of Darboux sumWhy is the upper riemann integral the infimum of all upper sums?Prove that for any $epsilon > 0, exists delta > 0,$ if $||P|| < delta $, then $|L(f,P) - I|<epsilon $ , and $|U(f,P) - I|<epsilon $Darboux Integrability epsilon-delta proofProving the converse of the Cauchy criterion for integrationIf a bounded function is integrable on each interval $[a,1]$, then it is integrable on $[0,1]$.Fitzpatrick's proof of Darboux sum comparison lemmaConverging of Riemann Sums with different partitionsWikipedia proof of “Darboux Integral implies Riemann Integral”Difference between Riemann-Stieltjes and Darboux-Stieltjes integralLower sum of partition and upper sum of another partitionIn what situations the sum of darboux sums can beHelp understanding the Darboux Integral definition
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A property of Darboux sum
Why is the upper riemann integral the infimum of all upper sums?Prove that for any $epsilon > 0, exists delta > 0,$ if $||P|| < delta $, then $|L(f,P) - I|<epsilon $ , and $|U(f,P) - I|<epsilon $Darboux Integrability epsilon-delta proofProving the converse of the Cauchy criterion for integrationIf a bounded function is integrable on each interval $[a,1]$, then it is integrable on $[0,1]$.Fitzpatrick's proof of Darboux sum comparison lemmaConverging of Riemann Sums with different partitionsWikipedia proof of “Darboux Integral implies Riemann Integral”Difference between Riemann-Stieltjes and Darboux-Stieltjes integralLower sum of partition and upper sum of another partitionIn what situations the sum of darboux sums can beHelp understanding the Darboux Integral definition
$begingroup$
I'm trying to show that:
$$overlineI:=inf _P S(f;P)=lim_lambda (P)to 0S(f;P)$$
where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=sum _i=1^nsup_Delta_if(x) cdot Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $lambda (P)$.
Facts already known and proved:
a) $s(f;P_1):=sum _i=1^n_1inf_Delta_if(x) cdot Delta x_i leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.
b) $0leq S(f;P)-S(f;widetildeP)leq omega (f;[a,b])cdot (Delta x_k_1+...+Delta x_k_m) $, where $widetildeP $ is a generic refinement of $P$, $ omega (f;[a,b]):=sup _x',x'' in [a,b]|f(x')-f(x'')| $ and $Delta x_k_1,...,Delta x_k_m $ are all the intervals of $P$ which contain points only in $widetildeP $.
My attempt to demonstrate the statement:
$ overlineI $ is well defined thanks to a). Being an $inf $, this implies that $forall epsilon >0$ there exists a partition $P_epsilon $ such that:
$$overlineI leq S(f;P_epsilon) leq overlineI + epsilon$$
to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_epsilon $ leads to $S(f;P) leq S(f;P_epsilon) $, but all I managed to get (using also b)) was this:
$$S(f;P) leq S(f;P_epsilon) +omega (f;[a,b])cdot mcdot lambda (P)$$
where $m$ is the number of points in $P_epsilon $ and $lambda (P) $ is the mesh of the new partition $P$.
I can't do better.
A little help, please?
Thanks in advance.
integration riemann-sum
asked Mar 17 at 13:02
NamelessNameless
5711
$endgroup$
add a comment |
$begingroup$
I'm trying to show that:
$$overlineI:=inf _P S(f;P)=lim_lambda (P)to 0S(f;P)$$
where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=sum _i=1^nsup_Delta_if(x) cdot Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $lambda (P)$.
Facts already known and proved:
a) $s(f;P_1):=sum _i=1^n_1inf_Delta_if(x) cdot Delta x_i leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.
b) $0leq S(f;P)-S(f;widetildeP)leq omega (f;[a,b])cdot (Delta x_k_1+...+Delta x_k_m) $, where $widetildeP $ is a generic refinement of $P$, $ omega (f;[a,b]):=sup _x',x'' in [a,b]|f(x')-f(x'')| $ and $Delta x_k_1,...,Delta x_k_m $ are all the intervals of $P$ which contain points only in $widetildeP $.
My attempt to demonstrate the statement:
$ overlineI $ is well defined thanks to a). Being an $inf $, this implies that $forall epsilon >0$ there exists a partition $P_epsilon $ such that:
$$overlineI leq S(f;P_epsilon) leq overlineI + epsilon$$
to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_epsilon $ leads to $S(f;P) leq S(f;P_epsilon) $, but all I managed to get (using also b)) was this:
$$S(f;P) leq S(f;P_epsilon) +omega (f;[a,b])cdot mcdot lambda (P)$$
where $m$ is the number of points in $P_epsilon $ and $lambda (P) $ is the mesh of the new partition $P$.
I can't do better.
A little help, please?
Thanks in advance.
integration riemann-sum
asked Mar 17 at 13:02
NamelessNameless
5711
$endgroup$
$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
$endgroup$
– RRL
Mar 17 at 13:12
$begingroup$
You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
$endgroup$
– RRL
Mar 17 at 13:34
$begingroup$
I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
$endgroup$
– Nameless
Mar 17 at 13:54
1
$begingroup$
See math.stackexchange.com/a/2047959/72031
$endgroup$
– Paramanand Singh
Mar 18 at 19:25
$begingroup$
Interesting, thank you @ParamanandSingh.
$endgroup$
– Nameless
Mar 19 at 20:51
add a comment |
$begingroup$
I'm trying to show that:
$$overlineI:=inf _P S(f;P)=lim_lambda (P)to 0S(f;P)$$
where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=sum _i=1^nsup_Delta_if(x) cdot Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $lambda (P)$.
Facts already known and proved:
a) $s(f;P_1):=sum _i=1^n_1inf_Delta_if(x) cdot Delta x_i leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.
b) $0leq S(f;P)-S(f;widetildeP)leq omega (f;[a,b])cdot (Delta x_k_1+...+Delta x_k_m) $, where $widetildeP $ is a generic refinement of $P$, $ omega (f;[a,b]):=sup _x',x'' in [a,b]|f(x')-f(x'')| $ and $Delta x_k_1,...,Delta x_k_m $ are all the intervals of $P$ which contain points only in $widetildeP $.
My attempt to demonstrate the statement:
$ overlineI $ is well defined thanks to a). Being an $inf $, this implies that $forall epsilon >0$ there exists a partition $P_epsilon $ such that:
$$overlineI leq S(f;P_epsilon) leq overlineI + epsilon$$
to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_epsilon $ leads to $S(f;P) leq S(f;P_epsilon) $, but all I managed to get (using also b)) was this:
$$S(f;P) leq S(f;P_epsilon) +omega (f;[a,b])cdot mcdot lambda (P)$$
where $m$ is the number of points in $P_epsilon $ and $lambda (P) $ is the mesh of the new partition $P$.
I can't do better.
A little help, please?
Thanks in advance.
integration riemann-sum
asked Mar 17 at 13:02
NamelessNameless
5711
$endgroup$
I'm trying to show that:
$$overlineI:=inf _P S(f;P)=lim_lambda (P)to 0S(f;P)$$
where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=sum _i=1^nsup_Delta_if(x) cdot Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $lambda (P)$.
Facts already known and proved:
a) $s(f;P_1):=sum _i=1^n_1inf_Delta_if(x) cdot Delta x_i leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.
b) $0leq S(f;P)-S(f;widetildeP)leq omega (f;[a,b])cdot (Delta x_k_1+...+Delta x_k_m) $, where $widetildeP $ is a generic refinement of $P$, $ omega (f;[a,b]):=sup _x',x'' in [a,b]|f(x')-f(x'')| $ and $Delta x_k_1,...,Delta x_k_m $ are all the intervals of $P$ which contain points only in $widetildeP $.
My attempt to demonstrate the statement:
$ overlineI $ is well defined thanks to a). Being an $inf $, this implies that $forall epsilon >0$ there exists a partition $P_epsilon $ such that:
$$overlineI leq S(f;P_epsilon) leq overlineI + epsilon$$
to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_epsilon $ leads to $S(f;P) leq S(f;P_epsilon) $, but all I managed to get (using also b)) was this:
$$S(f;P) leq S(f;P_epsilon) +omega (f;[a,b])cdot mcdot lambda (P)$$
where $m$ is the number of points in $P_epsilon $ and $lambda (P) $ is the mesh of the new partition $P$.
I can't do better.
A little help, please?
Thanks in advance.
integration riemann-sum
integration riemann-sum
asked Mar 17 at 13:02
NamelessNameless
5711
asked Mar 17 at 13:02
NamelessNameless
5711
asked Mar 17 at 13:02
NamelessNameless
5711
asked Mar 17 at 13:02
NamelessNameless
5711
asked Mar 17 at 13:02
NamelessNameless
5711
5711
$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
$endgroup$
– RRL
Mar 17 at 13:12
$begingroup$
You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
$endgroup$
– RRL
Mar 17 at 13:34
$begingroup$
I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
$endgroup$
– Nameless
Mar 17 at 13:54
1
$begingroup$
See math.stackexchange.com/a/2047959/72031
$endgroup$
– Paramanand Singh
Mar 18 at 19:25
$begingroup$
Interesting, thank you @ParamanandSingh.
$endgroup$
– Nameless
Mar 19 at 20:51
add a comment |
$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
$endgroup$
– RRL
Mar 17 at 13:12
$begingroup$
You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
$endgroup$
– RRL
Mar 17 at 13:34
$begingroup$
I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
$endgroup$
– Nameless
Mar 17 at 13:54
1
$begingroup$
See math.stackexchange.com/a/2047959/72031
$endgroup$
– Paramanand Singh
Mar 18 at 19:25
$begingroup$
Interesting, thank you @ParamanandSingh.
$endgroup$
– Nameless
Mar 19 at 20:51
$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
$endgroup$
– RRL
Mar 17 at 13:12
$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
$endgroup$
– RRL
Mar 17 at 13:12
$begingroup$
You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
$endgroup$
– RRL
Mar 17 at 13:34
$begingroup$
You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
$endgroup$
– RRL
Mar 17 at 13:34
$begingroup$
I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
$endgroup$
– Nameless
Mar 17 at 13:54
$begingroup$
I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
$endgroup$
– Nameless
Mar 17 at 13:54
1
1
$begingroup$
See math.stackexchange.com/a/2047959/72031
$endgroup$
– Paramanand Singh
Mar 18 at 19:25
$begingroup$
See math.stackexchange.com/a/2047959/72031
$endgroup$
– Paramanand Singh
Mar 18 at 19:25
$begingroup$
Interesting, thank you @ParamanandSingh.
$endgroup$
– Nameless
Mar 19 at 20:51
$begingroup$
Interesting, thank you @ParamanandSingh.
$endgroup$
– Nameless
Mar 19 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are on the right track. To help you finish, recall that it is to be shown for any $epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - barI| < epsilon$ or equivalently
$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$
when $lambda(P) < delta$.
As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that
$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$
We want to introduce any partition $P$, not necessarily a refinement of $P_epsilon$, and produce $delta$ such that (*) is satisfied if $lambda(P) < delta$. Next, we construct a partition $widetildeP$ that is a common refinement of $P$ and $P_epsilon$ by adding the $m$ interior points of $P_epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that
$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$
By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then
$$barI leqslant S(f;P) leqslant barI + epsilon$$
answered Mar 18 at 2:28
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
add a comment |
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$begingroup$
You are on the right track. To help you finish, recall that it is to be shown for any $epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - barI| < epsilon$ or equivalently
$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$
when $lambda(P) < delta$.
As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that
$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$
We want to introduce any partition $P$, not necessarily a refinement of $P_epsilon$, and produce $delta$ such that (*) is satisfied if $lambda(P) < delta$. Next, we construct a partition $widetildeP$ that is a common refinement of $P$ and $P_epsilon$ by adding the $m$ interior points of $P_epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that
$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$
By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then
$$barI leqslant S(f;P) leqslant barI + epsilon$$
answered Mar 18 at 2:28
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
add a comment |
$begingroup$
You are on the right track. To help you finish, recall that it is to be shown for any $epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - barI| < epsilon$ or equivalently
$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$
when $lambda(P) < delta$.
As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that
$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$
We want to introduce any partition $P$, not necessarily a refinement of $P_epsilon$, and produce $delta$ such that (*) is satisfied if $lambda(P) < delta$. Next, we construct a partition $widetildeP$ that is a common refinement of $P$ and $P_epsilon$ by adding the $m$ interior points of $P_epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that
$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$
By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then
$$barI leqslant S(f;P) leqslant barI + epsilon$$
answered Mar 18 at 2:28
RRLRRL
53k42573
$endgroup$
$begingroup$
Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
add a comment |
$begingroup$
You are on the right track. To help you finish, recall that it is to be shown for any $epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - barI| < epsilon$ or equivalently
$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$
when $lambda(P) < delta$.
As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that
$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$
We want to introduce any partition $P$, not necessarily a refinement of $P_epsilon$, and produce $delta$ such that (*) is satisfied if $lambda(P) < delta$. Next, we construct a partition $widetildeP$ that is a common refinement of $P$ and $P_epsilon$ by adding the $m$ interior points of $P_epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that
$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$
By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then
$$barI leqslant S(f;P) leqslant barI + epsilon$$
answered Mar 18 at 2:28
RRLRRL
53k42573
$endgroup$
You are on the right track. To help you finish, recall that it is to be shown for any $epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - barI| < epsilon$ or equivalently
$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$
when $lambda(P) < delta$.
As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that
$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$
We want to introduce any partition $P$, not necessarily a refinement of $P_epsilon$, and produce $delta$ such that (*) is satisfied if $lambda(P) < delta$. Next, we construct a partition $widetildeP$ that is a common refinement of $P$ and $P_epsilon$ by adding the $m$ interior points of $P_epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that
$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$
By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then
$$barI leqslant S(f;P) leqslant barI + epsilon$$
answered Mar 18 at 2:28
RRLRRL
53k42573
edited Mar 18 at 2:37
answered Mar 18 at 2:28
RRLRRL
53k42573
answered Mar 18 at 2:28
RRLRRL
53k42573
answered Mar 18 at 2:28
RRLRRL
53k42573
53k42573
$begingroup$
Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
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@Nameless: You're welcome.
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– RRL
Mar 19 at 22:09
add a comment |
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Thank you very much. (+1)
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– Nameless
Mar 19 at 20:50
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
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Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
$begingroup$
Thank you very much. (+1)
$endgroup$
– Nameless
Mar 19 at 20:50
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
$begingroup$
@Nameless: You're welcome.
$endgroup$
– RRL
Mar 19 at 22:09
add a comment |
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$begingroup$
See here for a proof when the integral exists. The part of the argument for the upper sums converging to the upper Darboux integral is all you need for this question.
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– RRL
Mar 17 at 13:12
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You will see you are almost there. Take the partition $P$ with sufficiently small mesh and form a common refinement with $P_epsilon$ to make the needed estimates.
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– RRL
Mar 17 at 13:34
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I have not understood the way I should follow. I already obtained the last inequality using a refinement of $P_epsilon$.
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– Nameless
Mar 17 at 13:54
1
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See math.stackexchange.com/a/2047959/72031
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– Paramanand Singh
Mar 18 at 19:25
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Interesting, thank you @ParamanandSingh.
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– Nameless
Mar 19 at 20:51