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Let $f$ be a continuous monotone function. Show that $f$ must be absolutely continuous on [0,1]
$begingroup$
Let $f:[0,1]tomathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on (0,1) and $f'(x)$ is continuous on $(0,1)$. Show that $f$ must be absolutely continuous on $[0,1]$
and construct a counter example for the case without the "monotone" condition.
My attempt:
$f'(x)$ exists for all $x$ in $(0,1)$ anld continuous on (0,1), then $f'(x)$ is integrable on (0,1) and
$$f(x) = int_{0}^{x}f'(x) + f(0),forall x in [0,1]$$
Since $f$ is an indefinite integral on [0,1] hence $f$ is absolutely continuous.
I want a different approach with the direct proof using the definition and not using so many theorems.
for the counterexample I have:
$f(x) = xsin(1/x)$ and we must show that $f$ is not of bounded variation.
measure-theory lebesgue-integral monotone-functions absolute-continuity
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]tomathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on (0,1) and $f'(x)$ is continuous on $(0,1)$. Show that $f$ must be absolutely continuous on $[0,1]$
and construct a counter example for the case without the "monotone" condition.
My attempt:
$f'(x)$ exists for all $x$ in $(0,1)$ anld continuous on (0,1), then $f'(x)$ is integrable on (0,1) and
$$f(x) = int_{0}^{x}f'(x) + f(0),forall x in [0,1]$$
Since $f$ is an indefinite integral on [0,1] hence $f$ is absolutely continuous.
I want a different approach with the direct proof using the definition and not using so many theorems.
for the counterexample I have:
$f(x) = xsin(1/x)$ and we must show that $f$ is not of bounded variation.
measure-theory lebesgue-integral monotone-functions absolute-continuity
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]tomathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on (0,1) and $f'(x)$ is continuous on $(0,1)$. Show that $f$ must be absolutely continuous on $[0,1]$
and construct a counter example for the case without the "monotone" condition.
My attempt:
$f'(x)$ exists for all $x$ in $(0,1)$ anld continuous on (0,1), then $f'(x)$ is integrable on (0,1) and
$$f(x) = int_{0}^{x}f'(x) + f(0),forall x in [0,1]$$
Since $f$ is an indefinite integral on [0,1] hence $f$ is absolutely continuous.
I want a different approach with the direct proof using the definition and not using so many theorems.
for the counterexample I have:
$f(x) = xsin(1/x)$ and we must show that $f$ is not of bounded variation.
measure-theory lebesgue-integral monotone-functions absolute-continuity
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
$endgroup$
Let $f:[0,1]tomathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on (0,1) and $f'(x)$ is continuous on $(0,1)$. Show that $f$ must be absolutely continuous on $[0,1]$
and construct a counter example for the case without the "monotone" condition.
My attempt:
$f'(x)$ exists for all $x$ in $(0,1)$ anld continuous on (0,1), then $f'(x)$ is integrable on (0,1) and
$$f(x) = int_{0}^{x}f'(x) + f(0),forall x in [0,1]$$
Since $f$ is an indefinite integral on [0,1] hence $f$ is absolutely continuous.
I want a different approach with the direct proof using the definition and not using so many theorems.
for the counterexample I have:
$f(x) = xsin(1/x)$ and we must show that $f$ is not of bounded variation.
measure-theory lebesgue-integral monotone-functions absolute-continuity
measure-theory lebesgue-integral monotone-functions absolute-continuity
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
asked Jan 7 at 23:10
Richard ClareRichard Clare
1,066314
1,066314
add a comment |
add a comment |
1 Answer
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$begingroup$
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' geq 0$ on $(0,1)$. We can write $f(x)=f(frac 1 2)+int_{1/2} ^{x} f'(t), dt$ for $x<frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x to 1$ it follows that $int_{1/2}^{1} |f'(t)|, dt=int_{1/2}^{1} f'(t), dt <infty$. Similarly, $int_0^{1/2} |f'(t)|, dt <infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+int_0^{x} f'(t), dt$ for all $x in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(frac 1 {frac {(2n+1)pi} 2},frac 1 {npi})$ and apply definition of absolute continuity.
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' geq 0$ on $(0,1)$. We can write $f(x)=f(frac 1 2)+int_{1/2} ^{x} f'(t), dt$ for $x<frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x to 1$ it follows that $int_{1/2}^{1} |f'(t)|, dt=int_{1/2}^{1} f'(t), dt <infty$. Similarly, $int_0^{1/2} |f'(t)|, dt <infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+int_0^{x} f'(t), dt$ for all $x in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(frac 1 {frac {(2n+1)pi} 2},frac 1 {npi})$ and apply definition of absolute continuity.
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
add a comment |
$begingroup$
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' geq 0$ on $(0,1)$. We can write $f(x)=f(frac 1 2)+int_{1/2} ^{x} f'(t), dt$ for $x<frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x to 1$ it follows that $int_{1/2}^{1} |f'(t)|, dt=int_{1/2}^{1} f'(t), dt <infty$. Similarly, $int_0^{1/2} |f'(t)|, dt <infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+int_0^{x} f'(t), dt$ for all $x in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(frac 1 {frac {(2n+1)pi} 2},frac 1 {npi})$ and apply definition of absolute continuity.
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
add a comment |
$begingroup$
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' geq 0$ on $(0,1)$. We can write $f(x)=f(frac 1 2)+int_{1/2} ^{x} f'(t), dt$ for $x<frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x to 1$ it follows that $int_{1/2}^{1} |f'(t)|, dt=int_{1/2}^{1} f'(t), dt <infty$. Similarly, $int_0^{1/2} |f'(t)|, dt <infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+int_0^{x} f'(t), dt$ for all $x in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(frac 1 {frac {(2n+1)pi} 2},frac 1 {npi})$ and apply definition of absolute continuity.
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' geq 0$ on $(0,1)$. We can write $f(x)=f(frac 1 2)+int_{1/2} ^{x} f'(t), dt$ for $x<frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x to 1$ it follows that $int_{1/2}^{1} |f'(t)|, dt=int_{1/2}^{1} f'(t), dt <infty$. Similarly, $int_0^{1/2} |f'(t)|, dt <infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+int_0^{x} f'(t), dt$ for all $x in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(frac 1 {frac {(2n+1)pi} 2},frac 1 {npi})$ and apply definition of absolute continuity.
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
edited Jan 9 at 23:11
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:29
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
add a comment |
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
Why you choose 1/2 ? and why not applying that directly to [0,x] ?
$endgroup$
– Richard Clare
Jan 7 at 23:36
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
The equation $f(x) =int_0^{x} f'(t), dt$ cannot be justified without monotonicity. In fact we don't even know that this integral exists. @RichardClare
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:38
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
Hey, I think we should have that $f(x) + f(1/2) = int_{1/2}^{x}f'(t)dt$ ???
$endgroup$
– Richard Clare
Jan 9 at 21:31
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
$begingroup$
@RichardClare I have made some corrections in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 23:12
add a comment |
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