Prove about real function of bounded variation [closed]
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Let $X$ be a interval on $mathbbR$ and $gin C^1(X)$, show that, if $g$ is a function of bounded variation, then $forall gin C^1(X)$ we have:
$$int_X^|g'(x)|,dx,,=,,sup:(x_j)_jinmathbbN,,,,textmonotonic sequence of,,,X$$
Does anyone know how to prove it?
real-analysis integration bounded-variation
asked Aug 23 at 7:39
F.inc
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closed as off-topic by Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy Aug 24 at 13:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy
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up vote
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Let $X$ be a interval on $mathbbR$ and $gin C^1(X)$, show that, if $g$ is a function of bounded variation, then $forall gin C^1(X)$ we have:
$$int_X^|g'(x)|,dx,,=,,sup:(x_j)_jinmathbbN,,,,textmonotonic sequence of,,,X$$
Does anyone know how to prove it?
real-analysis integration bounded-variation
asked Aug 23 at 7:39
F.inc
3208
closed as off-topic by Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy Aug 24 at 13:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a interval on $mathbbR$ and $gin C^1(X)$, show that, if $g$ is a function of bounded variation, then $forall gin C^1(X)$ we have:
$$int_X^|g'(x)|,dx,,=,,sup:(x_j)_jinmathbbN,,,,textmonotonic sequence of,,,X$$
Does anyone know how to prove it?
real-analysis integration bounded-variation
asked Aug 23 at 7:39
F.inc
3208
Let $X$ be a interval on $mathbbR$ and $gin C^1(X)$, show that, if $g$ is a function of bounded variation, then $forall gin C^1(X)$ we have:
$$int_X^|g'(x)|,dx,,=,,sup:(x_j)_jinmathbbN,,,,textmonotonic sequence of,,,X$$
Does anyone know how to prove it?
real-analysis integration bounded-variation
asked Aug 23 at 7:39
F.inc
3208
asked Aug 23 at 7:39
F.inc
3208
asked Aug 23 at 7:39
F.inc
3208
asked Aug 23 at 7:39
F.inc
3208
3208
closed as off-topic by Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy Aug 24 at 13:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy
closed as off-topic by Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy Aug 24 at 13:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42
add a comment |Â
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42
add a comment |Â
1 Answer
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It is clear that RHS $leq$LHS. For the other way let $epsilon >0$ and choose $delta$ such that $|x-y| <delta$ implies $|g'(x)-g'(y)| <epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $x_i$ of $X$ with $|x_i+1-x_i|<delta$ for all $i$. Now write $sum |g(x_i+1)-g(x_i)|$ as $sum |x_i+1-x_i||g'(t_i)|$ for some $t_i$ between $x_i$ and $x_i+1$ Now $int_X |g'(t)|, dt=sum int_x_i^x_i+1 |g'(t)|, dt$. From this show that $|int_X |g'(t)|, dt-sum int_x_i^x_i+1 |g'(t_i)|, dt| < epsilon$. This gives $int_X |g'(t)|, dt< epsilon + sum |g(x_i+1)-g(x_i)|$. Conclude the proof from this.
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It is clear that RHS $leq$LHS. For the other way let $epsilon >0$ and choose $delta$ such that $|x-y| <delta$ implies $|g'(x)-g'(y)| <epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $x_i$ of $X$ with $|x_i+1-x_i|<delta$ for all $i$. Now write $sum |g(x_i+1)-g(x_i)|$ as $sum |x_i+1-x_i||g'(t_i)|$ for some $t_i$ between $x_i$ and $x_i+1$ Now $int_X |g'(t)|, dt=sum int_x_i^x_i+1 |g'(t)|, dt$. From this show that $|int_X |g'(t)|, dt-sum int_x_i^x_i+1 |g'(t_i)|, dt| < epsilon$. This gives $int_X |g'(t)|, dt< epsilon + sum |g(x_i+1)-g(x_i)|$. Conclude the proof from this.
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
add a comment |Â
up vote
0
down vote
It is clear that RHS $leq$LHS. For the other way let $epsilon >0$ and choose $delta$ such that $|x-y| <delta$ implies $|g'(x)-g'(y)| <epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $x_i$ of $X$ with $|x_i+1-x_i|<delta$ for all $i$. Now write $sum |g(x_i+1)-g(x_i)|$ as $sum |x_i+1-x_i||g'(t_i)|$ for some $t_i$ between $x_i$ and $x_i+1$ Now $int_X |g'(t)|, dt=sum int_x_i^x_i+1 |g'(t)|, dt$. From this show that $|int_X |g'(t)|, dt-sum int_x_i^x_i+1 |g'(t_i)|, dt| < epsilon$. This gives $int_X |g'(t)|, dt< epsilon + sum |g(x_i+1)-g(x_i)|$. Conclude the proof from this.
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is clear that RHS $leq$LHS. For the other way let $epsilon >0$ and choose $delta$ such that $|x-y| <delta$ implies $|g'(x)-g'(y)| <epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $x_i$ of $X$ with $|x_i+1-x_i|<delta$ for all $i$. Now write $sum |g(x_i+1)-g(x_i)|$ as $sum |x_i+1-x_i||g'(t_i)|$ for some $t_i$ between $x_i$ and $x_i+1$ Now $int_X |g'(t)|, dt=sum int_x_i^x_i+1 |g'(t)|, dt$. From this show that $|int_X |g'(t)|, dt-sum int_x_i^x_i+1 |g'(t_i)|, dt| < epsilon$. This gives $int_X |g'(t)|, dt< epsilon + sum |g(x_i+1)-g(x_i)|$. Conclude the proof from this.
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
It is clear that RHS $leq$LHS. For the other way let $epsilon >0$ and choose $delta$ such that $|x-y| <delta$ implies $|g'(x)-g'(y)| <epsilon $. [ Possible because $g'$ is uniformly continuous]. Consider a partition $x_i$ of $X$ with $|x_i+1-x_i|<delta$ for all $i$. Now write $sum |g(x_i+1)-g(x_i)|$ as $sum |x_i+1-x_i||g'(t_i)|$ for some $t_i$ between $x_i$ and $x_i+1$ Now $int_X |g'(t)|, dt=sum int_x_i^x_i+1 |g'(t)|, dt$. From this show that $|int_X |g'(t)|, dt-sum int_x_i^x_i+1 |g'(t_i)|, dt| < epsilon$. This gives $int_X |g'(t)|, dt< epsilon + sum |g(x_i+1)-g(x_i)|$. Conclude the proof from this.
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
answered Aug 23 at 8:08
Kavi Rama Murthy
23.7k31033
23.7k31033
add a comment |Â
add a comment |Â
Try using the mean value theorem to write any distance traveled $g(x) - g(y)$ as $g'(c)(x-y)$ for some $c in (x,y)$.
– 4-ier
Aug 23 at 7:42