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




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closed as off-topic by Did, Jendrik Stelzner, user91500, Theoretical Economist, amWhy Aug 24 at 13:49


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  • 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














up vote
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down 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




share|cite|improve this question












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
If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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












up vote
1
down vote

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1









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




share|cite|improve this question












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





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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
If this question can be reworded to fit the rules in the help center, please edit the question.




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
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 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




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










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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.






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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.






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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.






      share|cite|improve this answer






















        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.






        share|cite|improve this answer












        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.







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        answered Aug 23 at 8:08









        Kavi Rama Murthy

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