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If total variation function for f of bounded variation is absolutely continuous, then f is absolutely continuous
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Suppose that $f$ is a function of bounded variation on [0, 1], and let $V(x)$ be the total variation function for $f$, i.e., for any $x in [0, 1]$, $V(x)$ is the total variation of $f$ on the interval $[0, x]$. Prove that, if $V$ is absolutely continuous on [0, 1], then so is $f$.
My attempt:
First, I note that $sum_k = 1^n |V(b_k) - V(a_k)| = sum_k = 1^n |TV(f_[a_k, b_k])|$. Then I note that $|f(b_k) - f(a_k)|$ would be the variation for the crudest possible partition of $(a_k, b_k)$, hence $|f(b_k) - f(a_k)| leq TV(f_[a_k, b_k])$ for each $k$. Thus if we choose the $delta$ responding to the $epsilon$ challenge for the absolute continuity of function $V$, we have that if $sum_k = 1^n [b_k - a_k] < delta$ for a collection of open intervals $(a_k, b_k)_k = 1^n$ in [0, 1], then:
$sum_k = 1^n |f(b_k) - f(a_k)| leq sum_k = 1^n TV(f_[a_k, b_k]) = sum_k = 1^n |V(b_k) - V(a_k)| < epsilon$
real-analysis bounded-variation absolute-continuity
asked Jul 21 at 21:19
BMac
24517
 |Â
show 1 more comment
up vote
1
down vote
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Suppose that $f$ is a function of bounded variation on [0, 1], and let $V(x)$ be the total variation function for $f$, i.e., for any $x in [0, 1]$, $V(x)$ is the total variation of $f$ on the interval $[0, x]$. Prove that, if $V$ is absolutely continuous on [0, 1], then so is $f$.
My attempt:
First, I note that $sum_k = 1^n |V(b_k) - V(a_k)| = sum_k = 1^n |TV(f_[a_k, b_k])|$. Then I note that $|f(b_k) - f(a_k)|$ would be the variation for the crudest possible partition of $(a_k, b_k)$, hence $|f(b_k) - f(a_k)| leq TV(f_[a_k, b_k])$ for each $k$. Thus if we choose the $delta$ responding to the $epsilon$ challenge for the absolute continuity of function $V$, we have that if $sum_k = 1^n [b_k - a_k] < delta$ for a collection of open intervals $(a_k, b_k)_k = 1^n$ in [0, 1], then:
$sum_k = 1^n |f(b_k) - f(a_k)| leq sum_k = 1^n TV(f_[a_k, b_k]) = sum_k = 1^n |V(b_k) - V(a_k)| < epsilon$
real-analysis bounded-variation absolute-continuity
asked Jul 21 at 21:19
BMac
24517
1
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38
 |Â
show 1 more comment
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1
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up vote
1
down vote
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Suppose that $f$ is a function of bounded variation on [0, 1], and let $V(x)$ be the total variation function for $f$, i.e., for any $x in [0, 1]$, $V(x)$ is the total variation of $f$ on the interval $[0, x]$. Prove that, if $V$ is absolutely continuous on [0, 1], then so is $f$.
My attempt:
First, I note that $sum_k = 1^n |V(b_k) - V(a_k)| = sum_k = 1^n |TV(f_[a_k, b_k])|$. Then I note that $|f(b_k) - f(a_k)|$ would be the variation for the crudest possible partition of $(a_k, b_k)$, hence $|f(b_k) - f(a_k)| leq TV(f_[a_k, b_k])$ for each $k$. Thus if we choose the $delta$ responding to the $epsilon$ challenge for the absolute continuity of function $V$, we have that if $sum_k = 1^n [b_k - a_k] < delta$ for a collection of open intervals $(a_k, b_k)_k = 1^n$ in [0, 1], then:
$sum_k = 1^n |f(b_k) - f(a_k)| leq sum_k = 1^n TV(f_[a_k, b_k]) = sum_k = 1^n |V(b_k) - V(a_k)| < epsilon$
real-analysis bounded-variation absolute-continuity
asked Jul 21 at 21:19
BMac
24517
Suppose that $f$ is a function of bounded variation on [0, 1], and let $V(x)$ be the total variation function for $f$, i.e., for any $x in [0, 1]$, $V(x)$ is the total variation of $f$ on the interval $[0, x]$. Prove that, if $V$ is absolutely continuous on [0, 1], then so is $f$.
My attempt:
First, I note that $sum_k = 1^n |V(b_k) - V(a_k)| = sum_k = 1^n |TV(f_[a_k, b_k])|$. Then I note that $|f(b_k) - f(a_k)|$ would be the variation for the crudest possible partition of $(a_k, b_k)$, hence $|f(b_k) - f(a_k)| leq TV(f_[a_k, b_k])$ for each $k$. Thus if we choose the $delta$ responding to the $epsilon$ challenge for the absolute continuity of function $V$, we have that if $sum_k = 1^n [b_k - a_k] < delta$ for a collection of open intervals $(a_k, b_k)_k = 1^n$ in [0, 1], then:
$sum_k = 1^n |f(b_k) - f(a_k)| leq sum_k = 1^n TV(f_[a_k, b_k]) = sum_k = 1^n |V(b_k) - V(a_k)| < epsilon$
real-analysis bounded-variation absolute-continuity
asked Jul 21 at 21:19
BMac
24517
edited Jul 22 at 18:34
asked Jul 21 at 21:19
BMac
24517
asked Jul 21 at 21:19
BMac
24517
asked Jul 21 at 21:19
BMac
24517
24517
1
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38
 |Â
show 1 more comment
1
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38
1
1
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38
 |Â
show 1 more comment
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1
What does $TV(f,P_k)$ mean, exactly? The notation looks like it means something like the total variation of $f$ with respect to $P_k$, but that makes little sense.
– David C. Ullrich
Jul 22 at 0:23
Sorry, it's sloppy. I was trying to indicate the total variation of f taken over all possible partitions $P_k$ of the interval $(a_k, b_k)$. Fixed.
– BMac
Jul 22 at 18:31
Makes more sense now. Of course people are still wondering what your question is.
– David C. Ullrich
Jul 22 at 18:39
Just checking my work---with real analysis I'm still building self-confidence, because I tend to make blunders.
– BMac
Jul 22 at 19:32
Fine. You can get in big trouble for not following the Rules. One of the rules is that when you post a "question" you're supposed to ask a question. "Is the following proof correct?" is in fact an acceptable question...
– David C. Ullrich
Jul 22 at 22:38