If $f$ is differentiable with $f'$ bounded, is $f$ absolutely continous? [closed]Existence of nonnegative bounded derivatives with a dense set of zerosDoes a nondecreasing, differentiable function have continuous derivative?Prove that $f$ is uniformly continousProve that a function is both differentiable and continous at a point $x_0$Is the space of almost everywhere differentiable function with bounded derivative embedded with uniform norm complete?When the function is continuous, bounded of variations, absolutely continuous?$ sum a_n$ converges absolutely if $sum a_nb_n $ converges absolutely for all bounded $b_n$Derivative of bounded functionSufficient condition for a function to be absolutely continuousConterexample for Product of Absolutely Continuous Functions is Absolutely Continous
Use of noexpand in the implementation of TextOrMath for eTeX
My singleton can be called multiple times
How to calculate the right interval for a timelapse on a boat
In the UK, is it possible to get a referendum by a court decision?
How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?
What's the meaning of "Sollensaussagen"?
Is this draw by repetition?
How seriously should I take size and weight limits of hand luggage?
Are British MPs missing the point, with these 'Indicative Votes'?
What is the fastest integer factorization to break RSA?
Finding the reason behind the value of the integral.
Why was Sir Cadogan fired?
Finitely generated matrix groups whose eigenvalues are all algebraic
Mathematica command that allows it to read my intentions
Were days ever written as ordinal numbers when writing day-month-year?
What do you call someone who asks many questions?
How exploitable/balanced is this homebrew spell: Spell Permanency?
How to coordinate airplane tickets?
Blending or harmonizing
How to install cross-compiler on Ubuntu 18.04?
Why didn't Boeing produce its own regional jet?
Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?
How to prevent "they're falling in love" trope
What does the same-ish mean?
If $f$ is differentiable with $f'$ bounded, is $f$ absolutely continous? [closed]
Existence of nonnegative bounded derivatives with a dense set of zerosDoes a nondecreasing, differentiable function have continuous derivative?Prove that $f$ is uniformly continousProve that a function is both differentiable and continous at a point $x_0$Is the space of almost everywhere differentiable function with bounded derivative embedded with uniform norm complete?When the function is continuous, bounded of variations, absolutely continuous?$ sum a_n$ converges absolutely if $sum a_nb_n $ converges absolutely for all bounded $b_n$Derivative of bounded functionSufficient condition for a function to be absolutely continuousConterexample for Product of Absolutely Continuous Functions is Absolutely Continous
$begingroup$
Let $f:[0,1]to mathbbR$ differentiable and $f'$ bounded. How can I prove $f$ is absolutely continous?
real-analysis
edited Mar 28 at 14:33
anomaly
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by YiFan, RRL, Adrian Keister, Umberto P., Strants Mar 28 at 20:25
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, RRL, Adrian Keister, Umberto P., Strants
add a comment |
$begingroup$
Let $f:[0,1]to mathbbR$ differentiable and $f'$ bounded. How can I prove $f$ is absolutely continous?
real-analysis
edited Mar 28 at 14:33
anomaly
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by YiFan, RRL, Adrian Keister, Umberto P., Strants Mar 28 at 20:25
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, RRL, Adrian Keister, Umberto P., Strants
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49
add a comment |
$begingroup$
Let $f:[0,1]to mathbbR$ differentiable and $f'$ bounded. How can I prove $f$ is absolutely continous?
real-analysis
edited Mar 28 at 14:33
anomaly
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $f:[0,1]to mathbbR$ differentiable and $f'$ bounded. How can I prove $f$ is absolutely continous?
real-analysis
real-analysis
edited Mar 28 at 14:33
anomaly
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 14:33
anomaly
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 14:33
anomaly
17.8k42666
edited Mar 28 at 14:33
anomaly
17.8k42666
edited Mar 28 at 14:33
anomaly
17.8k42666
17.8k42666
asked Mar 28 at 14:31
newbie11123newbie11123
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:31
newbie11123newbie11123
1
asked Mar 28 at 14:31
newbie11123newbie11123
1
1
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
newbie11123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by YiFan, RRL, Adrian Keister, Umberto P., Strants Mar 28 at 20:25
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, RRL, Adrian Keister, Umberto P., Strants
closed as off-topic by YiFan, RRL, Adrian Keister, Umberto P., Strants Mar 28 at 20:25
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YiFan, RRL, Adrian Keister, Umberto P., Strants
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49
add a comment |
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49
add a comment |
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Mar 28 at 14:33
$begingroup$
Start at the top: what does absolutely continuous mean?
$endgroup$
– Umberto P.
Mar 28 at 14:38
$begingroup$
To bound specific values of a function starting from a global bound of its derivative you can use the Mean value theorem. Assuming that $|f'(t)|leq M$ for all $tin[0,1]$, then $|f(y)-f(x)|=|f'(t)||y-x|$ for some $tin[0,1]$. Therefore, $|f(y)-f(x)|leq M|y-x|$. In absolute continuity you need to show that a sum like $sum_i|f(y_i)-f(x_i)|$ is small when a sum like $sum_i|y_i-x_i|$ is small. So, you only need to sum instances of the previous inequality to get this.
$endgroup$
– user647486
Mar 28 at 14:49